Applied Problems in Probability Theory

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E. Wentzel, L. Ovcharov

Applied Problems in

PROBABILITY THEORY

MIR PUBLISHERS MOSCOW

. Applied Problems In

ability

ory

E. C

BenT~enb,

JI A OB11apoa

llpnua~nLie aa~aqn

no Teopuo eepoaTuocTeu lfocxsa

Pa,nE o

D CBJtaJ.. •

Appli d Problems In • • eo r o a 11 E.Wentzel and L.Ovcharov

Tra nsl ate d from Russian by Irene Aleksanova

Mir Publishers Moscow •

F1rst pub],sh&d j986 R~vtsed trom the 1983 Russtan edltlon

©

ll3JJ.aTen~cTBO

«Panna u Cs.stabt 1983

© English translation,.

~br

Pubhsh ers, iQ86

•

AUTHORS' PREFACE

This book is based on many years of experience o~ teaching pr.obabilit y theory and its applicatio ns at higher educatiOnal estabhsh~ents. It contains many of the problems we omselves encounter ed m .our research and consultati ve work. The problems are related to a vanety of fields including electrical engineering, radio en~i.neering, dat~ transmission, computers , informatio n systems, reliability of techmcal devices, prev~ntive maintenan ce and repair, ac~uracy of apparatus , consumer service, transport, and the health service. . . . The text is divided into eleven chapters; each of winch begms w1th a short theoretica l introducti on which is followed by relevant formulas. The problems differ both in the fields of applicatio n and in difficulty. At the beginning of each chapter the reader will find comparati vely simple problems whose purpose is to help the reader grasp the fundamental concepts and acquire and consolidat e tlte experience of apply- · ing probabilis tic methods. Then follow more complicat ed applied problems, which can be solved only after the requisite theoreticallmowledg~ has been acquired and the necessary techniques mastered. We have avoided the standard typical problems which can be solved mechanica lly. Many problems may prove difficult for both beginners and experience d readers alike (the problems we believe most difficult are marked by an asterisk). In tbe interest of tbe reader most of the problems have both answers and detailed solutions, and they are given immediate ly after the problem rather than at the end of the book; we wrote the book for a laborious and thoughful reader who will first try to find his own answer. This structure is very convenien t and has justified itself in another book we have written, The Theory of Probabilit y (Nauka, Moscow, 1973), which bas been republishe d many times both at home and abroad (some problems in that edition have been repeated in this book). We believe that statement s and detailed solutions of nontrivial problems wh.ich dem?nstrat~ certain tl~e techniques of problem solving are part1cul~rly mterestmg . Our a1m 1s not just to solve a problem but to use the s1mplest and most general technique. Some problems have been given several different solutions. In many cases a method of solution used ha~ a general nature and can be applied in several fields. We have pmd special attention to numerical characteri stics which m~k.es it possible .to solve a number of problems with exception al simphclty. The applied problems using the theory of Markov stochastic processes has been given the greatest considerat ion. The brief theoretica l sections which open each chapter do not usually

o!

PreftJce

repeat 'vbat extst1ng textbooks on probabtltty theory present, but have been based on new methods Thus thxs book ts. In a certatn sense, 1ntermed1ate between a collec.. t1on of problems and a textbook on the theory of probabilities It should be useful to a w1de variety of readers such as students and lecturers at h1gber schools, engineers and research 'vorkers 'vho require a proba btlislic approach to their 'vork Note that the detatled solut1ons and the cons1deratton g1ven to problem solvtng techniques make this book :::u1 table for 1ndependent study 'Ve w1sh to express our gratitude to B V Gnedenh.ot Academictan of the Ukratntan Academy of Sctences, \vbo read the manuscript very attentively and made a number of valuable remarks t and 1lso to V S Pugacbev, Academtctan of the USSR Academy of St1ences, 'vhom we consulted frequently \vben we \\or ked on the book and whose methodtcal pr1nc1ples and notat1on we substantially followed

PREFACE TO THE ENGLISH EDITION

We wish to express our satisfaction at having the opportunity to bring our techniques of solving applied problems in probability theory to the notice of the English reader. We wrote this book so that it could be used both as a study aid in probability theory and as a collection of problems, of which are about 700. The theoretical part at the beginning of each chapter and the methodical instructions for solving the various problems make it possible to use the book as a study aid. The solutions of many of the problems in the book are important both from an educational point of view and because they can be used when investigating various applied engineering problems. The methodology and the notation in this book correspond, in the main, to those used in V. S. Pugachev's book [6] which was recently published in Great Britain. When the book was being prepared for translation into English,, a number of corrections and additions were made which improved the content of the book. During this preparatory work, Assistant Professor Danilov made an essential contribution, for which we want to express our gratitude. E. Wentzel,

~.

Ovcharov

CONTL.NTS

Authors'Preface 5 Preface to the Enghsh Ed1t1on 7 Chapter I FUnd 3l.aental Concepts o! Proba btlt t y Theory lat1on of Probahtlity 1n an Urn Aiodel 9

DU"ec t

Caleu-

Chapter 2

A1ge bra of E veu ts. Rules I or Ad d1ng and AI ul t1 plyn1g ProbablhtJes 29 Chapter 3 The Total Pro~ab1hty Formula and Bayes's Theorem

Chaptel' 4 Dl~rete

Random

Chapter 5

Continuous- and

Var~ahles

65

88

~ftxed Random

Variables 112

Chapter 6 8} stems ~f Random Variables (Rand om Vector.s)

t 44

Chapter 7 Numertcal Charac tenst1cs of Func t1on.s of Random Vatlables i 66

Chapter 8 D 1~ tr1hu t tons of • unc t 1ons of Rand om Variableli;.. The L1rm t Theo retns of Probabd1 ty Theory 217 Chapter 9 Random Functions

260

Chapter 10 Flov, s of Even~. ~farltov Stochas t lC Proces..~s Chapter 11 Queueing Theory 363 Appendices 420 B1hl1ogra phy 432

324

•

CHAPTER 1 •

Fundamental Concepts of Probability Theory. Direct Calculation of Probability in an Urn Model 1.0. The theory of probability is the mathematical study of r~nd_om phenomen~~ The concept of a random event {or simply event) is one o! the prmc1paf conc~pts m probability theory. An event is the outcome of an expenment {or a trwl). Six dot& appearing on the top face of a die, the failure of a dev~ce ~uring its service life, and a distortion in a message transmitted over a commumcatwn channel are all examples of events. Each event has an associated quantity which characterizes how likely its occurrence is; this is called the probability of the event. There are several approaches to the concept of probability. T~e "classi:a~" a:pproach is to calculate the number of favourable outcomes of a trwl and divide 1t by the total number of possible outcomes [see formula (1.0.6) below]. The f~equency or statistical approach is based on the concept of the frequency of an event m a long series of trials. The frequency of an event in a series of N trials is the ratio of the number of trials in which it occurs to the total number of trials. There are random events for which a stability of the frequencies is observed; with an increase in the number N of independent trials, the frequency of the event stabilizes, and tends to a certain constant quantity, which is called the probability of an event. The modern construction of probability theory is based on an axiomatic approach using the fundamental concepts of set theory. This approach to probability theory is known as the set-theoretical approach. Here are the fundamental concepts of set theory. A set is a collection of objects each of which is called an element of the set. A set of students who study at a given school, the set of natural numbers which do not exceed 100, a set of points on a plane lying within or on a circle with a unit radiu& and centre at the origin are all examples of sets. There are several ways of designating sets. It can be denoted either by one capital letter or by the enumeration of its elements given in braces or by indicating (also in braces) a rule which associates an element with a set. For example, the set M of natural numbers from 1 to 100 can be written J1!

=

{1, 2, ... , 100}

=

{i is an integer; 1 ~ i~ 100}.

The set C of points on a plane which lie within or on a circle with centre at the origin can be written in the form C = {x2 y 2 ~ R 2 }, where x and y are the Cartesiancoordinates of the point and R is the radius of the circle. Depending on ho'': ma~y elements i~ has, a set may be finite or infinite. The set M = {1, 2, ... , 100} 1s fimte and contams 100 elements (in a particular case a finiteset can. c~nsis~ of only one element). The set of all natural numbers N = {1, 2, ... , n, ... } Js mfimte; the set of even numbers N 2 ={2, 4, .. . , 2n, ... } is alo::o infinite An ~nfin~te set is said to be _countable if all of its terms can be enumerated (both of themfi~ute sets N ~nd 1V2 given above are countable). The set C of points within or on a c1rcle of radms R > 0 C = {x2 y 2 ~ R} (1.0.1)·

+

+

is infmite and uncountable (its elements cannot be enumerated). Two sets A and B coincide (or are equivalent) if they consist of the same elements (the coincidence of sets is expressed thus: A = B). For example the o::et of ro_ots of the equation x 2 - 5x 4 = 0 coincides with the set {1 4} '(and- also With the set {4, 1}). ' ' . The notation a E A means t~at an object a is an element of a set A; or, in other\\ords, a belongs to A. The notat10n a E A means that an object a is not an elem nt. of a set A. e

+

to 7

Applied Problems tn Probablltl'{l Thtoru I &

•

5 Est DIE E

An tm p ty set 1 s a set w1th no elem-ents It \ s des 'tg na ted 0 r. "tam. ple th-e set of p01nts on a plane whoso coordtnate.9 (x Y) satisfy the 1nequahty .x2y 1 ~ -1 1.! an empty set {z'! + yl ~ -t} = eJ All entpty sets are equlvalent A set B 1s said to be a subse-t of a set A tf all tho ele1nents oi B also belong to A. The notation IS B £A {or A 2 B) Example_, ~t, 2t t 100} ~ {I, 2, •., tOOO}t {t, 2 , 100} ~ (1 2 100} {x 2 + Y ::::;; 1} s:; {x' y'J G; 2} An empty set 1s a subset of any set A 0 s; A \Ve can use a geomettlcal1nterpretaban to depLct the. tncluston of sets the pomts on the plane are elements of the set (see F1g 1 0 1 where B IS a subs at of A).

+

+

Ftg t 0 2

The union (logical sum) of the set~ A and B Js a set C ~ A +B which consists oi all the elernents of A and all those of B (tncludtng those whu:.h belo,g to both A and B) In .short~ a unton of two ~ets Is a col!ectlon of elements belongtng to dl U:ast one of them Examples {t 2 100) +{50 51 200} =- {i 2 • 200}. i 2 tOO}+ {1 2 fOOO} = {1 2 iOOO} {1 2 tOO} + 0 = 1 2 tOO} The un1on o[ the two sets A and. B 1s sh')\Vn tn F1g 1 0 2. the 5 aded area IS A +B . We can suntlarly define a Union of any number of sets A1 +A 2 + +An= 1

n

~ A, 1s a set eonststtng of all the elements which belong to at least one of i=i

the sets A1,

, An \Ve can also cons1der a unlon of an Infinl te (countable)

number of sots

2J

00

Pi

+ {n.

Al=A 1 +A~+

+An+

•

Example

{1. 2}+{2. 3J+

;:;: {1~ 2 1 3~ ,. n } The lnttrs~tfon (log1cal product) of two sets A and B IS the set D = A B that coruilst!' of the elements whieh belong to both A and B Examples {I 2 t tOO) x {SO 51. 200} = {50~ 51 iOO) {1 2 tOO} {1 2, , 1000} =={t. 2~ ., 100}9 {tt 2 t iOO} 0 = 0 An Intersection of t'vo sets A and B lS shown In } 1g 1 0 3 We ea n s tm Ilar l y de fine the 1n ters~c t ton o [ any number of sets T b. e. set A 1 A 1 ~

{3t 4l+

n-t}+

41

41

n

An

An.

::=:;;

IJ A

i-t

1

consists of all the elefJlents '" h1ch belong to all the sets A 1 A 2

Slll1ul tan.eou.sl y

'The- d eftnl t 1on ean be extended. to an

1

nn n1t e

~.

{co un table)

00

number of sets

A1

IJA, 1s a set conQ.1st1ng of element.9 belonging to all the 5ets Au

t=l

An

s1multaneously

Two sets A and Bare .said to be dLSJalnt (non1ntersect1ng) 1f the1r 1ntersectton 1s an emptJ. set A B =::::; 0 1 e no element belongs to both A and B (F1g t 0 4). F tgure 1 0 5 1llu~ t tates several d l.SJO 10.t sets As tn the notation of an ord1nary mult1pltcation the SifPl 1s usually om1tted. It tS ~uffic1ent to have tht:; elementary knowledge of set theory in order to U.!:!e the set theoretical constructton of probabihty theory Assnm e that an ex per1 ment (tn al) is co nd ue ted whos a result 1s not known be Ioreband 1 e ts acc1dental Let us constder the set Q of all posszble outcomes of th&

Ch. 1. Funda menta l Conce pts of Proba bility Theor y

11

€Xper iment: each of its eleme nts OJ E Q (each outcom e) is known as an elemen tary event and the whole set Q as the space of elemen tary events or the sampl e sp~ce. Any ~ubset of the set Q is known as an event (or random event); and any event A IS a sub~et of the set Q, viz. A £ Q. Examp le: an experi ment involv es tossin g a die; the space of eleme ntary eve~ts Q = {1, 2, 3, 4, 5, 6}; an event A is a:r;t even score; A.= {2, 4, 6); A.£ Q. ~n partl: ular, we can consid er the event Q (smce every set IS a subset of 1tself); 1t IS sa1d to be a certain event (it must occur in every experi ment) . We can add the empty set J?f to the whole space Q of eleme ntary events ; this set is also a.n event and is said to:be an imposs ible event (it canno t occur as a result of an expen ment) . An examp le

A

A

8

B

B

ll Fig. 1.0.5

Fig. 1.0.4

Fig. 1.0.3

of a certain event: {a score not exceed ing 6 when a die is tossed }; an examp le of an impos sible event: {a score of 7 dots on a face of a die). Note that there are differe nt ways of defini ng an eleme ntary event in the same experi ment; say, in a rando m throw of a point on a plane, the positi on of the point can be defined both by a pair of Cartes ian coordi nates (x, y) and by a pair of polar coordi nates (p, bab1l1ty of the eve nt be fou nd as the ra tto of the num ber of ele me nta ry eve nts fa, our ahl e to A to the tot al number of ele me nta ry eve nts and 1f not ,vh y? Sol utio n. The e1e me nta ty e\e nts wh tch for m the set g (th e hea ds are clenotcd by the lett er h' and the tail s hy "t") are ru 1 = {h h}, (1) 2 =

=

=

Ch. 1. Fundamental Concepts of Probability Theory

17

{t t} ffis = {h, t, t}, ffi 4 = {t, h, h}; ffis = {h, t, h, h}, ffis = {t: h,' t, t} ... , the number of events is infinite but countable: Q = {ffi 1 , ffi 2 , (1) 3 , • • • }. The subset of elementary events favourable to A is A = {ffil, ffi2, ffis, ffi4}. • • • It is impossible to find here the probab1hty of the event A as the ratio of the number of elementary events favourable to A to the total number of elementary events since the elementary events w1 , w2 , w3 , • • • are not equipossible: each pair is less probable than its predecassor [see problem 2:18 to find P (A)]. 1.6. A polyhedron with /daces (k > 3) labelled 1, 2, ... , k is thrown onto a plane at random and it falls one or another face down. Construct the sample space for the experiment and isolate the subset corresponding to the event A = {the polyhedron falls the face not exceeding the number k/2 down}. Solution. The space Q consists of k elementary events: Q = { 1, 2, ... , k}, where the numbers correspond to the number of faces. The subset A consists of elementary eventsA = {1, 2, ... , [k/2]}, where [k/21 is an integral part of k/2. 1.7. Under the conditions of the preceding problem, the polyhedron is regular; the possible number of faces are k = 4 (a tetrahedron); k = 6 (a cube); k = 8 (an octahedron); k = 12 (a dodecahedron); k = 20 (an icosahedron). Find the probability of the event A for each polyhedron. Solution. For a regular (symmetric) polyhedron the appearance of each face is equi possible and, therefore, we can calculate P (A) by ·formula (1.0.6). Since the number of faces in each regular polyhedron is even, it follows that [k/2] = k/2 and so for each of them P (A) = 1/2. 1.8. There are a white and b black balls in an urn. A ball is drawn from the urn at random. Find the probability that the ball is white. Answer. a/(a + b). 1.9. There are a white and b black balls in an urn. A ball is drawn from the urn and put aside. The ball is white. Then one more ball is drawn. Find the probability that the ball is also white. Answer. (a - 1)/(a b - 1). 1.10. There are a white and b black balls in an urn. A ball is drawn from the urn and put aside without noticing its colour. Then one more ball is drawn. It is white. Find the probability that the first ball, which was put aside, is also white. Answer. (a - '1)/(a + b - 1). 1.11. An urn contains a white and b black balls. All the balls except for one ball are dra-wn from it. Find the probability that the last ball remaining in the urn is white. Answer. a/(a b). 1.12. An urn contains a white and b black balls. All the balls are drawn from it in succession. Find the probability that the second drawn ball is white. b). Answer. a!(a 1.13. There are a white and b black balls in an urn (a~ 2). Two balls . _ are dra\vn together. Find the probability that both balls are whit·-- -

+

+

+

2-05i 5

tS

A ppl1ed Problems in Pro bab!l~ t g I heary

Soluhon An event A outcow.es

n where

C/:'=

ml

=

=

{t\vo \Vhite balls} The total number of

C!+a = (a

(:~m)\

+ b)

(a

+ b - 1)1(1

2),

Is the number of combinations of J.. elements

taken m at a t1me The number of favourable outcomes

mA =

C! = a (a - 1)/(1

2)

The probab1ltty of e'\ent A P (A) = mAin = (a -

1)/[(a

+ b)

(a

+ b - 1)]

1 14 There are a v.h1tc and b black balls In an urn (a> 2 b # 3) F1ve balls are dta,vn together F1nd the probabiltty p that two of them are whtte and three are black Solubon The total number of cases («+b) {a.+b-1) ta+b-2) {«+b-3) (a+b-4) n = Ca+b === t 2345 :rt

TJ1c number of fa,ourable outcomes

ctc3mb0

P

=

a \tl-\) b \b-\) \b-~) 1 2

1 2 3

'

m tOa (a-t) b (b-1) (b-2) n- =-(a+ b} (a+ b- i) (a+b-2) (a+ b- 3} (a+ b-4}

1 15 A batch cons1~ts of k arttcles l of" luch are faulty A total of r arlrcles are taken from the batch for Inspection F1nd the probability p that exactl} s of the selected artJcles are faulty Ans\Ver p

= c;c~::.iJC~

1 t 6 A dte events

1s

A B

thro,vn once F1nd the probab•I1 t1es of the folio" Jng

(the score 1S even}, = {the score IS no less than 5} C = {the score ts no more than 5} =

P (A) = 1/2, P (B) =: 1/3, P (C) = 5/6 1 17. There are a whtte and b black balls 1n an urn (a> 2, b ~ 2) T"o ba[Ls are drawn at once \Vhtch of the foll{)Wlng events ts mnre ltkely Ans,~er

A

=

{the balls are of the same colour},

8

=

{the balls aTe dtHerent

1n

c~lout}1

Ch. 1. Fundamenta l Concepts of Probability Theory

19

Solution. a(a-1)+b (b-1) P(A)= c~~c~ = (a+b) (a+b-1) ' ca+b

c~c~

p (B)= c2 a+b

2ab (a+b) (a+b-1)

·

Comparing the numerator s of the fractions, we find that P (A)< P (B) i.e. (a -

for

a (a- 1)

+

+ b (b- 1)
P (B) for (a- b) 2 >a

2ab,

+ b, + b.

1.18. Abo): contains n enumerate d articles. All the articles are taken out one by one at random. Find the probabilit y that the numbers of the selected balls are successive: 1, 2, ... , n. Answer. 1/n! 1.19. The same box, as in the preceding problem, but each article is drawn and its number is written down, after which it is replaced and the articles are stirred. Find the probabilit y that a natural sequence of numbers will be written: 1, 2, ... , n. Answer. 1/nn. 1.20. Eighteen teams participat e in a basketbal l champion ship, out of which two groups, each consisting of 9 teams, are formed at random. Five of the teams are first class. Find the probabilit ies of the following events: A = {all the first class teams get into the same group}; B = {two first class teams get into one group and three into the other}.

Answer.

p (A)= 2qq3 _ 1

qs -

34 '

p (B)= C&Cl3+c~q 3 C~ 8

=!?

17 •

1.21. A certain Petrov buys a bingo ticket and marks 6 of the 49 numbers. Then the six winning numbers, out of the 49, are announced, Find the probabilit ies of the following events: Aa = A4 = A5 = As=

{he {he {he {he

guessed 3 out of 6 winning guessed 4 out of 6 winning guessed 5 out of 6 winning guessed all the winning

numbers}; numbers}; numbers}; numbers}.

. Soh~tion. The problem is equivalen t to drawing 6 balls from an urn m whtch there are 6 white balls (winning numbers) and 49 - 6 = 43 2*

20

Applied Problems tn Probablllty Theorf/

black balls (not 'vlnn1ng)

p (A,) = c~~J, ~ o 01765,

p

u

P (A5 ) = c~;:, ~ o 00001845,

"' are

(A~)= c~~i, ~ 0 000969, t

P (A 6 ) = c~,. ~ 0 715·10

7

labelled 0, 1, 2f 3, 4, 5, 6. 7 t 8 Two eards are drawn at random and put on a table 1n a successive order, and then the resulting number 1s read, say, 07 (seven), 14 (fourteen) and so on F1nd the probab1l1ty that the number ts even Solution M). Find the probability of an event A = {no more than one telegram will be sent over each channel}. Solution. The total number of cases is NM. The number of ways to choose Jlf channels out of N in order to send one telegram over each of them is Cl)i. The number of ways to choose one telegram out of M and to send it over the first of the channels is CAt = 111. The number of ways to choose the second telegram out of the remaining 111 - 1 is q 1_1 = Jlt! - 1, and so on. The total number of favourable cases mA

= M (M- t) ... 1

= M!.

P (A) = Clfi·M!JNM. 1.38*. A local post office is to send M talegrams and to distribute them at random over N communication channels. The channels are enumerated. Find the probability that exactly k1 telegrams will be sent over the 1st channel, k 2 telegrams over the 2nd channel, and so N

on, k N telegrams over the Nth channel, with

2j ki

= M.

i=l

Solution. The number of cases n = NM. Let us find the number of favourable cases m. The number of ways to choose k 1 telegrams out of M is C~}; the number of ways to choose k 2 telegrams out of theremaining M -k 1 . is C~f-n 1 , and so OIL The number of ways to choose kN telegrams out of

+ ... +

k

M- (k 1 kN_ 1) = kN is C;;~ = 1. These numbers must be multiplied together. m= _ -

-

c"tck2 111 111-h

1 • • •

c"N-1 .lii-(ki+ ... +kN_ )• 1 1

M! (M-kt)! [M-(kl+k2+ ... +kN-2)]! kl (M -kl)! k2l [Jif -(k1 +!.:2)]! . . . kN- 1 lkNl

Ml

.Ml

k1lk2! ... kNl-

N

IT kit i=1

.'

P (A)= min= M!/ (NM

IT ki!) .

i=1

v'1.39*. Under the conditions 'bf problem 1.37 find the probability that no telegrams will be sent over l 0 of N channels, one telegram will be sent over l1 channels, and so on; and all M telegrams will be sent over l;.r channels:

Zo

+ l1 + ... + Zu =

N; 0 • l 0

+ 1 • Z + ... + MZM = 1

Jlt!.

Solution. The total number of cases is n = NM. To find the number of favourable cases m, we must multiply the number of ways in which the channels can be chosen by the number of ways in which the tele-

24

Applit!d Problems zn Probabt llty Thtary

grams can be chosen The numbe r of ways to choose the channe ls ts M

l,1 l) =Nil

Nr/l 0 '1 1 r

0

lllJ

lt.:O

Let us find the numbe r of 'vays 1n 'vh1ch the telegra ms can be chosen They fall 1nto a numbe r of groups tha 1n1t1al group (0 telegrams} lS empty, the first group containS l1 telegra ms tn general kth group con tl!) The numbe r of ways to choose ta1.ns klh telegrams (k = t, 2 the groups of telegra ms

1s

.AJl

~11

(t 39 1)

~~~~~~--~~--~~---

(1 11)f (2l 2 ) I (3!1 ) I

(ftll ~t) 1-

.&1

II

(klk)l

l-1

Let us now find the numbe r Qf ways to choose the telegra ms from the kth group so that k telegra ms are sent over each channe l This num her of ways 1s (Al 1J1f (kfkr and the. numbe r of \vays tQ choose all the telegra ms ioT all the groups ts equal to the produc t of these numbers for differe nt A .AJ

II

ct 39 2)

(kl,.)r

Jd) l). { k-1

j\Iult1plying (1 39 1) and (1 39 2) .. we get the numbe r of ways tn ,vhtch the telegra ms can be chosen

.JI.f!

1\Iulttp lying th1s numbe r by the number of ways tn Vth1ch the channe l can be chosen 've fi.nd the numbe r of favour able cases m~

N1

Af

•

Ml

-M~--

II0 z,r II

'{;ma

h.-..a\

I\' log (1 - p)~log (1 - &), N~Iog (1 - &)!log (1 - p) .

(2.39)

s ist ns co l ne an ch n tio ica un mm co a er ov t sen ing be ge 2.40. A messa ted tor dis is n sig h eac on ssi mi ns tra the g rin Du ). ols mb (sy ns .sig of n of e sak the r Fo p. ity bil ba pro th wi ns) sig er oth the of tly en (in de pe nd th at ity bil ba pro the d Fin es. tim k ted ea rep is ge ssa me h eac ty, ili rel iab in ted tor dis be t no ll wi d tte mi ns tra ing be ges ssa me at least• one of the an y s1gn. be t no ll wi o-e ssa me te ara sep e on t tha ity bil ba pro So lut ion . The 11 of t ou e on st lea at t tha ity bil ba pro the ; p) (1 lo l ua eq is dis tor ted k messages wi ll not he dis tor ted is P (A)

=

1 - [1 -

(1 - p)n]k.

tppl ~a Problems in ProbabU ty Theory

2 4t Under the cond1t1ons of Problem 2 40 how many ttrnes must a message be repeated for the probab111 t} that at least one message \vxll not be dtstorted to be no less than !f? Answer By formula (2 39) we ha' e N ~ log (1 - !f)/log [1 (1 ~ p)"] 2 42 A Significant message 1s sent stmultaneously over n communi cat1on channels and ts repeated k times o'er each channel for the sake of rel•abiltty During one transml"'=Slon a message (Independently of the other mess1.ges) 1s dtstorted 'vtth probabtltty p Each communica t1on channel (1 n dependently of the other channels) 13 block.ed UJ>" 'v1 th no 1se 'vt th proba htl1 t y Q a blocked up channel cannot trane~m1t any me"isage.s F1nd the probabtltt:,. of the e\ ent A -

{a me-;;.sage 1s tr"tnsm•tted In a conect form at least once)

'Ve Introduce the e" ent B - {a message IS transm1lted over one communication channel w1th out d 1s tottton at least o nee) For the e'\er.t B to -occt\r flr~t the c.hannel must not be 4'bloc.ked 1.1.p With notse 1nd ~econd at least one message sent over 1t must not be distorted P (B) - (1 - Q) (1 - p~) Solut1on

The probahtl1ty of the event A 'vh1ch least o-... er one channel ts

P (A)

=

1

~

[1 -

P (B)]n

~

1-

IS

that the event B occurs at

(1 -

(1 -

Q) (1 -

pk)}n

2 43 An arr battle bet,veen tv.. o a1rcraft a fighter and a bomber IS go1ng on The fighter IS the first to fire It fires once at the bomber and hr1ngs 1t do\vn '' 1th prob a b 1l1 t) p 1 If the bomber 1s not brought clown 1t ftres once at the fighter and hr1ngs 1t do\vn wtth probabilJty p 2 If the fighter 1s not brought do,vn 1t fires at the bomber again and brtngs 1t do,vn \VIth probability p 3 F1nd the probabillttes of the folio'' 1ng out comes A = {the bomber

IS

brought down}

B - {the fighter IS brought do"n} C = {at least one of the aircraft IS brought down}

Ans"er P (A) - P1

+ (1

P (C) = P (A)

-t

- P1) (1 - P2) Ps P (B) - (1 - Pt) P2 P (B)

2 44 An a1r battle bet" een a bomber and t'vo attack1ng fighters 1s going on The bomber IS the first to fire 1t fires once at each f1gh t er and brJngs 1t down \\ 1th probabtllty p 1 If a fighter 1s not bro light do\vn 1t fires at the bomber lrrespectt\e of the desttny of the other fighter and

45

Ch. 2. Algebra of Events

brings it down with probabilit y p 2 • Find the probabilit ies of the following outcomes: A = {the bomber is brought down}; B = {both fighters are brought down}; C = {at least one fighter is brought down}; D = {at least one aircraft is brought down};

E = {exactly one fighter is brought down}; F = {exactly one aircraft is brought down}.

Solution. The probabilit y of one of the fighters bringing down. the bomber is (1 - p 1) p 2 ; the probabilit y that neither of theJ:?. brmgs down the bomber is [(1 - (1 - p 1 ) p 2 ]2, whence

P (A)= 1- [1-(1- Pt) pzJ2, (1 - p 1) 2,

P (C) = 1 P (E) = 2p 1 (1 - Pt).

P (B)= Pi,

P (D) = 1 -

(1 - Pt) 2 (1 - P2) 2,

+

+

Fa, F2 The eYent F can be represente d in the form F = F 1 where F 1 = {the bomber is brought down and both fighters are safe}; F 2 = {the first fighter is brought down and the second fighter and the bomber are safe}; Fa= {the second fighter is brought down and the first fighter and the bomber are safe}.

P (Ft) = (1 - Pt) 2 ![1 - (1 - P2) 2 ], P (F2) = P (Fa) = Pt (1 - Pt) (1 - P2), 2 2 1 2pt (1 - Pt) (1 - P2). pz) P (F) = (1 - Pt) ['1 - (1 -

+

2.45. The conditions and the questions are the same as in the preceding problem, with the only difference that the fighters attack only in pairs: if one of them is brought down, then the second one disengages. Answer. P(.4)=(1 -p 1)2[1-(1 -p 2) 2 ], P(B)=pi , P (C)= 1- (1- Pt) 2 , P (D)= 1- (1- p 1) 2 (1- p 2) 2 , P(E) = 2p 1 (1- P1), P (F)= (1- P1) 2 [1- (1- p 2) 2 ]

+ 2p

1

(1- p 1).

2.46. There are a white and b black balls in an urn. Two players draw one ball each in turn, replacing it each time and stirring the balls. The ~l~yer who is the first to draw a white ball wins the game. Find the proba~1hty 1\ of the fitst player (the one who begins the game) being the wmuer.

46

Ap p lted Problems n Pro bab l ty Theory

Solut1on The fir~t pla1 er can "1n e1ther 1n the first or 1n the th1rd selection (1n the latter case the first o t1mes blacl balls must be dr(nvn and the thtrd t1me a '\h1te ball must be dra\\n) ~nd ~o on

t''

!ft-

a a+b

+

(

b

a+b

)2

a a+b

+ +

(

b ) 2k a a+b .,a+b-;-

(lt 15 e\ 1dent that &1 > 1/2 for an) a and b) 2 4.7 There are t\vo 'vhtte and three black balls tn an urn T"'o play ers each dr'l'\ a ball In turn '\ 1thout replacing 1t The first player who 1S the first to dra'\ a "htte bnll ''Ins the game F1nd the probabtl1ty &1 tbat the first pla) er \?Ins the gan1e

So1ution

~+~.!. ~-a 5

5

4

3

5

2 48 An Instrument 1~ being tested Upon ea.ch tr1al the Instrument falls ' 1th prob1b1l1t} p -\fter the flrst fru]ure the Instrument IS re pa1red after the second fa1lure It ts considered to be unfit for operatton F1nd the probabth 1) th1.t the Instrument IS reJected el.actl) 1n the kth tr1al Solution For the g1\ en e' ent to occur It IS nece~~ar} first that the tus trumen t should fa 1l 1n the ;.., th tr1al the prob a b 111 t y of that e' ent being p In addttton 1t ts nece(O~ar:,. that In the preced1ng k __. 1 trials the tnst rumen t should fa 1l eA. actll once the prohabtli t y of that e' ent betng (k - 1) p (1 ~ PV~ 2 The required probahiiit) IS equal to (A - 1) p (1 - p)tt 2 2 49 111ss1les trre fired at a target The probahll1ty of each m1sstle h1tt1ng the target IS p the h1ts are 1ndependent of one another Each Dllc;;.slle "\Vb1ch reaches the target br1ngs Jt do,vn \vttb probability p 1 The mtsstles are fired until the target 1s brought do,vn or the mlSStle t:>eserve 1s exhausted The reserve con:nsts of n mt~stles (n > 2) Find the probabtl1ty that ~orne m1~s1les 'v1ll rema1n tn the reser' e Solut1on \Ve pass to the complementarJ e'ent A- {the \Vbole reser\ e ts e'.hausted) For A to occur the fi.rst n - 1 mi~Siles must not bring the target do,vn

-

P(A)-{1-pp1)n

1,

P(A)=1-(1-pp 1 )~ 1.

2 50 Under the conditions of the preccdtng problem find the proba b1l1ty that no le~s than t'\\o miSSiles rema1n In the reserve after the tar get 1s brought down Solution The complementary e\ent A = {less than two mtssiles remain 1n the reser' e} 1s equivalent to the first n- 2 mi~Siles not hr1ng 10g down the target

P (A) =:::: ( 1 - p p 1)n-2

P (A) == 1- (1- p p 1)n-2

. .

-.

-

Ch. 2. Algeb ra of Event s

-

47

2.51. Unde r the condi tions of probl em 2.49 find the proba bility that no more than two missi les will be fired. Solut ion. For no more than two missi les to be sent, _the targ~t must be broug ht down after the first two shots : the proba bility of th1s even t is equal to 1 - (1 - PP1Y·. . . . . 2.52. A radar unit track s k targe ts. Durm g a surve illanc e peno d the ith targe t may be lost with proba bility Pi (i = 1, 2, · · ·' k). Find the proba biliti es of the following event s ' A = {none of the targe ts is lost}; B = {no less than one targe t is lost}; C = {no more than one targe t is lost}. Answ er. k

P(A) =

11 (1 i=l

k

-Pi), P(B )=1 -

IT (1-P i), i=l

k

P(C) =

II 1=1

(1-p i)+p 1 (1-p 2)

•••

(1-p h)+( 1-Pt )P2( 1-ps )

... (1- P~t) + · .. + (1- Pt) (1- P2) · · • (1- P~t-t) P~t • The last proba bility can be writt en in the form k

P(C) =

II

k

(1-p i)

+~

"1 1=

"1 J=

k

1!_ip· 1

l1

(1-P i)•

"1 1=

2.53. An instru ment consi sting of k units is in use for a perio d t During that time, the first unit may fail with proba bility q 11 the second witl1 proba biiity q 2 and so on. A repai rman is summ oned to check the instru ment. He checks each unit and eithe r locate s and clears the fault , if one exists , with proba bility p or passes it as faultl ess with proba bility q = 1 - p. Find tho proba bility of the event A = {at least one unit remains faulty after the inspections}. Solut ion. The proba bility that the ith unit remai ns faulty is equal to the proba bility that it became faulty durin g the time t multi plied by the proba bility that the repai rman failed to locate the fault: qiq. The proba bility that this event will occur in at least one unit is k

P(A )=1- 11 (1-qi q). i=l

2.54. A new condi tion is added to those of Probl em 2.53: after time t a repai rman is not found with proba bility Q and the instru ment is used withot~t inspecti_on. Find the proba bility of the event A = {the instru ment 1s used With at least one faulty unit} . Answer.

m1

+ m:a + m,+ m + m13 + m~ 3 + m123 =

m T\\0 people arc. cho sen from the group at random \Vhat P~ tho prob1b1 h ty p that they can u~e one of the three languages to talk to each of l1er '\ 1 thout an Inter 12

preter? Solut1on. '\70 enumerate nil tho seven groupq of people and '" rtle 1n front of each group the language~ "h1ch all tho member-; of tho group

speaJ... denot1ng then1 h) the letter~ G, t\ and R (I) G (m 1 people), (1 I) A (m 2 people); (I II) I{ (m 3 people)~ (VI) G, A (m 12 people) (V} G, R (m 13 people), (yl) i\, 1\ {m 23 people), {VII) G t \ R (rnu: 3 people) For t \\ o pco pic to t nih. to each other~ they must belong to a pa1r groups \\ ntch h.a\ c a language In COnlmOn It lS Simpler lfl tlus ca~e to find the probab1ltty that t'' o people cannot talk to each other They must then belong to one of the folio\\ tng patr~ (I~ I I)~ (I .. II I)~ (I~ VI}, (II I I I) (I I~ V) (III IV) Th~ probab1l1 ties that one of the t" o sc Joctcd people belongs to one group nnd the other to another group aro

or

I II 2mtm~ p{ ) ~ m {m~t)

p(

1 IT

'

2.m1m3 I) == na (m -1)

p {II, HI)=

P (lilt IV)=

"'2;•:_a1) ' 2

p (T VI)=

2m!m2"3 m (m-1)

p (II V) ==

2m1mt1 n~ (m.-1)

,.

m,mu

m lm-1)

Adchng these probabJlttics and subtracting tho rec;ul ttng sum from. un1ty t '\\oe find the Iequ11ed probabtltty p p

=1-

2

m.\m.-t}

(m 1m 2

+ m m 3 + m~m 23 1

.J

m.1m 3 + m'2m 13

+

n1 3 m 1.,)

2 60. A factory manufactures a certa1n typo of art1cle Each artic!(t may hn ve a defect, the prob ab 111 t y of \V htc h IS p A manufactured articleIS checked successively by J.., Jnspectors, the Lth 1nspector detects a defect (tf any) '' tlh probabtllt) p 1 (t = 1, 2, , k) If a defect ts detected, the artl~le; 1s reJ~Cted F1nd tho probabi\ltleC) oi the follo'' tng o' ent.s

A

=

{an article 1S reJeCted} t

B = {an article IS reJected by the second 1nspec tor and not by the first} ,

C = {an a.rt1cle

lS

reJected by all h. tnspectors}

Ans\\er" ~

P(A)=p [1-

II (1-p,)l

t

i~t

P (B)= P (1- Pt) P:1~

k

P (C)==== P

[J i~t

Pl

'

. Ch. 2. Alg ebr a of Eve nts

51.

2.61. A fac tor y ma nu fac tur es a cer tai n typ e of art icl e. Ea ch .ar tic le pec ms one by d cke che is cle rti ~na P: y ilit bab pr? th wi ect def ma y hav e a tor who det ect s a defect w1th pro bab 1h ty p 1 • If he does no t det ect a defect he let s the art icl e pass. In add itio n, the ins pec tor ma y ma ke t en ev ich wh of ity bil ba pro the e, icl art s les flaw a ect rej and ke sta a mi is a. Fin d the pro bab ilit ies of the following eve nts : A = {an art icl e ic:; rej ect ed} ; B = {an art icl e is rej ect ed by mi sta ke} ; C = {an art icl e wi th a defect is passed to a lot of fin ish ed products}. Answer.

+

P (B) = ~1 - p) a,

(1 - p) a P (A )= PP1 P (C) = p (1 - P1).

t bu m, ble pro ing ced pre the in as e sam the are s ion dit con e Th 2. 2.6 each art icl e is checked by two inspectors. The pro bab ilit ies tha t the p and p are e icl art ive ect def a ect rej ll wi tor pec ins ond sec the 1 2 first and res pec tiv ely ; the pro bab ilit ies of rej ect ing a flawless art icl e by mi sta ke are a 1 and a 2 res pec tiv ely . If at lea st one ins pec tor rej ect s an art icl e, it goes to waste. Fin d the pro bab ilit ies of the sam e eve nts . Answer.

+

) (1 -a )], -a (1 [1 p) (1 )l P (1 ) p (1 [1 p )= 1 (A P 2 1 2 P (B )= (1 - p) l1 - (1 - at) (1 -a2)1, n. rf,.J,B~ p (C )= p (1 - Pt) (1 - P2).

r

of e lur fai a and 3), 2.6 g. (Fi its un n of ts sis con nt me tru ins 2.63. An y ma its un e Th . ole wh a as nt me tru ins the of e lur fai a to ds lea it un any

-

p

p

p

oe •

p

p

-~/ -=--·-~ ' ''- -- -- ... ., ~ . ~' , . v .~ n

-f iB3{ Fi~: ~:6~. ,___

'

~

p Fig . 2.64

•

per ree e-f lur fai of y ilit bab pro e (th y. ilit iab rel e Th y. ntl nde epe fail ind tru ins ole wh the of P y ilit iab rel the d Fin p. is it un h eac of ) nce ma for me nt. Wh at mu st be the rel iab ilit y p 1 of each un it to ens ure the rel iabil ity P 1 of the ins tru me nt? Re ma rk. Fro m now on ele me nts wi tho ut wh ich a sys tem can no t ich wh nts me ele ; ies ser in ted nec con ks lin as ed ent res rep are e operat rep eat each oth er are connected in par all el. Th e rel iab ilit y of eac h ele me nt is wr itte n in a res pec tiv e squ are . 71 r'"} )_ .~-\nswcr. P = p , p 1 = f!J\. ~.

\ _.:::;.

S' \

52

1pplted PrOblems In

Proba.b~htv

Theo ry

2 64 To 1ncreaso the re)Ia .btlt ty of an tnstr ume ntt 1t 1s repeated by another instrument of the same ktnd (F1g 2 64) the relia btlat y (the prob abtl1 ty of fa1lure free perf orma nce) of each Instrument lS p \Vhen the first Instrument fa1ls, the second Instrument lS Instantaneously S\\ ttehed on (the rel1abilit.y of the sw1tch1ng dev1ee 1s equal to un1ty) Ftnd the rel1 abtlt ty P of a system of two Instruments which repeat

each other Solut1on For the system to fall, the two Instruments mu~t fa1l sunulta.neously., the relta bittt y af the syst em P = 1 - (1 - p)~ 2 65 The sarn e prob lem, but the relta blltt y of the :;w1 tchtn g de' 1ce

Sw lS p 1 (F1g 2 65) Answ er The rel1ab1l1ty of the syst em P = t -- (t - p) (1 - p 1Pl 2 66 To Incre ase the relta b1l1 ty of an rnst rum ent 1t lS repe ated by (n- 1) rn.struments of the sam e kind '(F1g 2 66), the rcl1a btllt y of ~

p I

1-o

-p -

p ...........

p

f/

l

sw

• ••

1a,.._

J

p

.......

Fig 2.66

F•g 2 6-l

each Instr ume nt Js p Ftnl l the relia bilit y P of the syst em Hon man ) tnstr ume nts mus t be used to Incre ase the rel1a b1ltt y to the asstg ned 'alu e Pt? Ans l\er P = 1 - (1 ~ p)n n~ log (1 - P 1 )/Iog (1 - p) 2 67 T1le same prob lem but a devt ce \Vllh relia bilit y p 1 1S used to tnvi tch on ever y stan d by 1nst rum ent (F1g 2 67)

Ans "er P = 1- (1- P) (1- P1Pr"

1,

n;;:: log (t-P~)-log (t-p ) log (t- PtP)

+1

2 68* A syst em cons tsts of n untt s the rel1 abtlt ty of each of \Vhtcb 1s p r\ fatlu re of at leas t one untt lead s to the fa1lure of the \vhole syste m To 1ncrease the relia bilit y of the syst em 1ts o11eratto.n 1s repe ated , for wh1ch purp ose anot her n untt s are allo tted The s'vlt chtn g devi ces are com plete ly relia ble DPte rmJn e 1vh1eh of the fo1lolv!ng repe atin g tech ntqu es lS the more rel1.able (a) each un1t 1s repe ated (Ftg 2 S8a) (b) the who le syst em IS repe ated (Fig 2 68b) Solt1 hon The relra btlit y of the syst em repe ated by tech ntqu e (a) IS P.. = U ~ (t - P)~tn 1and that of th~ syst em repe ated by tech niqu e (b) 1 - (1 - pn) Let u.~ sho\v th(lt Pa > Pb for any n. > 1 and ts Pb

=

Ch. 2. Algeb ra of Event s


0) and the inequ ality assumes e orm

0

p

53

+

p

p p

-

•

-

p,

p -

-

p, -

p -

Pt

p

2-

p

p

p {a)

>n -1

p

p

•••

p

p

p

•••

p

•• • "-

>

- ••• p

...

= 1-

1-

'

(b) Fig. 2.68

Fig. 2.67

1 _ q)n or (1 + q)n + (1 - q)n> 2. Applyin~ the binom ial theor em, we note' that all the negat ive terms are elimm ated: (1+q )n+( 1-q) n=1 +nq +

+

n (n-1)

2

q2+ ..•

n(n-1 ) 2 1-nq + q 2

...

=2+ n(n- 1)q 2 + ... >2. and this proves the requi red inequ ality.

I I

'

I p2

Ps

/)I

'

Pz ._l

PJ

I

p, I

II

p4

Po

Pz

r-

I

m Fig. 2.69

Fig. 2.70

2.69. In a techn ical syste m not all units are repea ted but only some of them which are less reliab le. The reliab ilities of the units are show n in Fig. 2.69. Find the reliab ility P of the syste m. Answer. P = [1 - (1 - p 1 ) 2 1[1 - (1 - p 2 ) 3 ]p 3 p 4 [1 - (1- p ) 2 ]. 5 2. 70. An instru ment consists of three units . The first unit conta ins n 1 elements, the second conta ins n 2 eleme nts and the third conta ins n 3 ele-

5~

Apphed Problem3 in

Probabitl.ty Th.-eory

ments For the tnstrument to functton, un1 t I ts obviously necessar} the o other units 11 and 111 repeat each other (F1g 2 70) The re1Ia hlllty of the elements lS the samo and equal to p A failure of one e1e ment means a f atl uro of the 'vholo ttni t The elements far I tndependentl~ of one a no tber F 1nd the rella bill ty P of tho 1nstrument Solnfton The rel1abJI1ty oi untt I 1s PI = pnt, the reltabtltty of untt II lS p 1 1 = pn2 the rel1abtl1ty of untt Ill IS Pili = pnl, the reha b1l1ty of the repeaters (ll and Ill} 1S 1 - (1 - p"t.) (1 - p"a) the reliabJlity of the 1nstrument ts

t''

P~pnt(1-(1-pn•)(1-pha)]

2 7J There IS an electr1cal device whtch can fail (burn out) only at the moment when 1t 1s s\VItched on If the devtce bas been s'vttched on k - 1 ttmes and has no-t burned QUt, then the eond1t1onal probab1ltt~ of Its butntng out on the ktb S'\ 1tch1ng operation 1s Q4 ftnd the proba bll1t1es of the follo,v1ng events A = {tha de\ tee 'vJtbstands no less than n swttch1ng opera t1ons}, R = (the de, 1ce 'v1thstands no more than n switching opera trons}, C

=

{the devIce burns out e"t actll on the nth s" 1t chJng operation}

Solution The probabtltt} of the e"\ent Jl 1s ~qual to the prababtl1tl that the dev tee '' tll not burn out on the fus t n sw1 tchrng o perat 1ons n

p (;1} =

II (1-QA) h-1

To fi~d the probnb1l1ty of tha event B '' e pass to a compleroentarl e'en t B ~ {the de, 1ce w1ll wt ths tand more than n s'vl tchtng opera t1ons} For that event to occur~ 1 t 1s suffi.ctent that the devtce should not burn out on the first (n. 1) operatrons

+

P(B)~

n+ I

II

k~1

n+:S

(1-QA)t

P(B)=i-

IJ

~~~

(1-Q~t)

For the devtce to burn out exactly on the nth s\VItthlng operat1on 1t IS nece~sary that 1t ti:hould not burn out on the first (n -1} opetat1ons and should burn out on the nth operation n.-1

P (C)=Qn

n {i-QA}.

k~t

2 72 An Instrument consts ts of four un1ts t \\ o of them (I and I I) a.re obv1ousll neccs~.ary for the Instrument to fuuctton and the ot.hct t'\\ o (III and IV) repca t e1.ch other (FJg 2 72) The un1 ts can only fa11 \\hen the) are S\\o 1tchtd on On the kth S\Vttch1ng operation un1t 1 fatlc

Ch. 2. Algebra of Events

55

(independently of the other units) '':ith prob.abil~ty qr (k), unit II wi~h probability qn(k), unit III and umt IV fall w_I~h. the same probability qn 1 (k) = q1 v(k) = q (k). Find the probabilities of the events A, B C as in Problem 2.71. 'solution. The problem reduces to the preceding problem when t~e conditional probability Qk that a sound device will fail on the kth switching operation is found: Qh = 1 - [1 - qr (k)] [1 - qn(k)l X {1 _ (1 - q (k)) 2 ], and is substituted into the solution obtained. III 2. 73. An instrument consists of three I II units. When the instrument is switched c IJ! on, a fault occurs in the first unit with probability p 1 , in the second unit with probability p 2 , and in the third unit Fig. 2.72 with probability p 3 • Faults occur in the units independently of one another. Each of the three units is obviously necessary for the instrument to function. For a unit to fail completely at least two faults must occur. Find the probability of an event A = {the device withstands n switching operations}.

Solution. For the event A to occur, it is necessary that all the three units should function. The probability that the first unit withstands n switching operations is equal to the probability that no more than one p 1 )n fault (0 or 1) occurs when it is switched on n times: (1 np 1 (1 - p 1)n-l. The probability that all the three units withstand n switching operations is

+

+

3

P(A)=

IJ 1(1-pi)n+npi(1--pi)n-f]. i=1

2. 74. An instrument consists of three units, only one of which is necessary for the instrument to function, the other two repeat each other. Faults. only occur in the instrument when it is functioning and can occm m any of the components constituting the units with the same probability. The first unit consists of n 1 components, the second of n, components, t~e third of n 3 components (n 1 n2 n 3 = n). When a fault occurs m at least one component the unit fails . . Four faults arc knowT_l to have occurred in the instrument (in four ~tfforenl components). Fmd the probability that the:;;e faults prevent the mstrument from functioning . . Solution. An event A = {the instrument cannot function} is divided m two: A= A 1 A 2 , where

+ +

+

A 1 = {the first unit fails}; A 2 = {the lirst unit does not fail but the second and the third do}.

56

Appllrd Problem s in Probabtltty Theory

For the event A 1 to occur t t t ts necessa r) thn t at least one of the four faults be ID tho Drs\. Unlt n-n 1

-

P(A,) =1-P (At)= 1---n

n-n 1 -1 n-1

n-n 1 -2 n-2

n.-n 1 -3

n-3

•

To ftnd the probab ll1ty of the event A 2 tho probab tltty of the event A1 = {the first un 1t does not fo.tl} must be m u1tJ plJed by the probab1l· 1ty that the 6teco.nd and th1rd nn1ts f11l (\\ 1th rl ue regard for the fact that 1\l the four faults occur tn the second and th1rd Units) The last event may be 1n three varJllnt~ e1ther one fault occurs 1n the second un\l and the oth-er tl1ree \n the tb1rd un1t oy con'\ ersely; three faults occur 1n the second un1t and ono fault occurs 1n the th1rd un1t, or else tv.o faults occur In the second and tn the th 1rd un1t each The probo.b1h ty of the first vart 'l.nt 1s l

n,

n,

Cl

n~+ n 3 -1

n 3 +n 3

That of the second var1an t iS

C1 4 11 2

+n

n2 ~ 1

n1

n"

n~

3

+ na -1

n~

+n

3

-2

Hence p (tl2) - [ 1- p (At) 1{f!J I + !f z + ff 31

p (A} = p {.tit} = p {A2)

2 75 An Instrum ent \Vhtch must he 'cry rei table rs assemb led from/.. D h Before the as:,em b1y sto.rts eacl1 part I.S Inspect ed parts Dt D 2 thoroug h 1y "lnd 1 f 1 t pro' es to be of 'ery htgh qual1 t y IS tnc 1uded rn the tnstrum ent and lf not 1s re.pl ~ced h y a no.thet prrrt, ~ l\1ch tS also Inspect ed There Js a stock of spare parts of each type m t parts ot

type D ~ (1. === 1,

k

t

k), a total of m ~ ~ m, parts If there are 1-1

not enough spare parts, the assemb ly lS postpo ned The probab 1ltty that a part of typeD , proves to be of high quality IS equal to Pl and does not depend on the quality of the other parts F1nd the probab tllttes of the followi ng events

=

{the stock of spare parts ts suf&c.Lent for the assemb ly of the 1nstrument},. B = {all the spare parts have been used (tested end Included •n the Instrum ent or only tested) }, C ~ {ustng the a"\ a1lable stock of spare parts the fitter~ "dl assemble the Instrum ent and at least one spare part \vJll rema1n}

A

Ch. 2. Algebra of Events

Solution. The event A

k

II

=

51

A i, where A i = {at least one part

i=i

of type Di proves to be high quality}. k

P(A;)=1-(1-Pi ti,

The event B =

k

II

P(A)=

.IT [1-(1-PittJ. t=1

Bi, ,\rhere Bi = {all the p

+ {e+d), (.:--1) J (c+d-i}

3 45 There are three comrnunicat1on channels over whtch messages are sent at random (wrth equaJ probahtltty over each channel) The probab1l1ty of a message bc1ng dtstorted when 1t IS sent over the first channel zs p 1 over the .second channel, p 2 over the th1rd cbannelt p 3 A channellS selected and J... messages are sent over 1t, none of which are

Ch. 3. Total Probability Formula and Bayes's Theorem

85

+

1)th message, sent over distorted. Find the probabilit y that the (k the same channel, will not be distorted. Solution. The hypotheses are H 1 = {the messages were sent over the first channel}; H 2 = {the messages were sent over the second channel}; H 3 = {the messages were sent over the third channel}. P (A I H 1) = (1- P 1)k, P (A I H 2) = (1- P2)k,P (AI Hs) = (1- P 3)k

P(Hi lA)=

=

1/3(1-Pi)k 1!3[(1-Pr) k+(i-p2)k+ (1.-pa)kJ (1- Pr) 11

+

(1-Pi)k (1- Pz)k

+(1- P3)

(i=1, 2, 3).

k

The event A = {k messages are not distorted} and B= {the (k+1)th message is not distorted}. p (B I A)- (1-pr)k+1 +(1-p2)k+ l+(1- Pa)k+l • (1-p 1 )k+(1-p~)k+(1-p3 )k 3.46. There are m batches of articles with N 1 , N 2 , • • • , N m articles in each batch. The ith batch contains ni defective articles and N i - n 1 sound ones (i = 1, 2, ... , m). A batch is selected at random and k articles from it are checked. They all prove to be sound. Find the probability that the next l articles taken from the same batch will also be sound. Sotution. The hypotheses H1 , H 2 , • • • , Hm, where Hi = {the ith batch is selected} have equal a priori probabilit ies P (HJ = 1/m (i = 1, 2, ... , m). On the hypothesis Hi the conditiona l probabilit y of the observed event A = {all the k articles checked were sound} is

I Hi)=C;v.- n./C;v. (i=1, 2, ... ,

{3.46.1) By Bayes's formula the a posteriori probabilit ies of the hypotheses are P(A

I

I

I

m).

m

P (Ht I A) = P (A

I H1)/~

i=i

P (A I Hi)

(i = 1, 2, ... , m). (3.46.2)

The probabilit y of an event B = {the next l articles taken from the same hatch are sound} can be calculated from the total probabilit y formula using the a posteriori probabilit ies given in (3.46.2): m

P (B)= ~ P (H 1 lA) P (B I H 1A), where i=1

p (B I HiA) = N;-ni-k N 1-n;-k-1 • N1-k

Ni-k-1

. N1-n1-k- Z+1 (l = 1, 2, N 1-k-l+1 • · • l nvmark. Formulas (3.46.1) and (3.46.3) are valid .t:::= " ! . - n1 - k. If these conditions are not satisfied, 1 Ies WI 11 be zero.

· · ·' m).

(3.46.3)

only fork (ro) "\\here (I) f fJ The value of that iunchon depends on whlth ~lementary event m occurs as a result of an expertment \ Ve shall de not a ran. dom va r1 ahles by cap1 ta 1 1et t &s and no ora ndom v a rw. bles by !i mall Ietters The law of probability dtstnbuhon of a random vartahle Js the rule used to find the probahthty of the C\ent related to a random var1ahle for Instance the proha h1ht) that the variab!~ assumes a certa1n value or falls 1n a certa1n 1nter' al If a random var1able X has a g1ven d1str1button Ia" then It J:J sa1d to have such and .a uch a d1s tr1 but 1on The most genera I ror m of the d lS tr1 but I on la \\ Ls a dis trib u t 1on tunc t ton '' htch is the probabtllt) that a random vartahle X assumes a value ~maHer than a gL' en va1ue z 1 e

F (:t) - p {X


:

(the arith metic , or posit ive, value of the root is always impl ied).

(4.0.9)

BB

Ap plttd Probl~m' tn ProbabUttg Thtorv The kth momtnt a~out the orfgin of a randont variable X

1!

the kth PO""' ~r mean

value of that var1able

(4 0 fO)

For a dto10:crete random \ar1able X the first moment about the or1g1n can be cal culated by 'he formula. (4 0 tl) CCk IX]==~ Z'~PI t

\arl- :-"'• given that np = a = canst. We can use this u" ·-''- •t>r ke pro::nmat10ns when we have a large number of independent trials:~ ' ' ~-,,.. an event A occurs with small probability.

,,.

•

Applied Problt ms fn Proba blltty Thtary

The Po~.~~on distrtbubons can also be used In the ease of a number of point! fal ling 1n a g1\en region of space (one-d1menstonal two-.dtmen~Ional or three-thmen srona1) 1f the location of the po1nt! 1n space 1s random .nnd satisfies certa1n rest r1ettons The ()Oe d1menstonal variant 1S encountered when flows events are conszdered A flow .of ~"' en:U or t ra.[fi.e ts a sequence of homogeneous events occurr lng one after another at random moments in time (for more detatl see Chapter 10) The- averag-e number of events 1 occurring per untt time Is knovm as the 'nte"mflu of the no,, The quanttty 1 can he either constant or vartable :\ ::::;; )-. (t) A flo'\\ of events 1s satd to be without afterefftct.t tr the proba.b1hty of the numller of e' ents falling on an 1nterval of tame does not depend . .A hype rgeom etric distr ibuti on occurs when there is an urn whic h conta ins a whrte and b black ~ails and n balls are dra,¥11 from it. The rand om varia ble X is the numb er of whrt e balls draw n and its distr ibuti on is expre ssed by form ula (4.0.34). The mean of a rando m varia ble whic h has distr ibuti on (4.0.34) is M [X] = nal(a

+ b),

(4.0.35)

and its variance is (4.0.36)

"'l When form ula (4,0.34) is used, we must a:;sume that

c;;

=

o if r >

k.

92

-

-

Applted Probleml tn Probablllty Thtory

Ptoble1ns and Exerclses 4.1. Construct the drstributJon function of the lndtcator U of an e\ent prob~blltty lS

A whose

p

Solutton .

0 for x ~ 0, q for O 1t

F (x)= p {U C~qpn- 1 1 p > nq, '\hence p > nl(n 1) Let us consider the case v. ben 0 < m* < n, tn thts case t\\ o tnequah· tres must be stmultnneously sattsfied

+

c:'•pm*qn- m• C~ ... prrr• qn -m•

+

~ c~it+1 pm*+lqn-m•- t' ~

c:•- p"m* I

-1 qn- m*

+1

These two tn~qualtttes are eqntvalent to the following Inequalities (m*

+ 1) q~

(n- 1n*)

p.,

{n- m*

+

1)

p ~ m*q1:

whence m* must be an Integer satisfying the tnequal1t1es

(n

+

1) p - 1 ~ m* ~ (n

+

1) p

+

It can bo 'ertfied that thts condttton 1s also sattsfied tf p < 1/(n t) (m* = 0) or 1n another extreme ca~e If p > nl(n + 1) (m* == n) Since the rtght hand stde e'\cecds the left hand stde by un1ty, there lS only one Integer m * bet'' een them The only except ton ts tho case ,vhen (n + 1) p and (n 1 1) p - 1 are 1ntegers Then there are two very tmprobahle values (n + 1) p and (n + 1) p - 1 If np Is an Integer. then m* = np 4 10~~ Two marksmen fire at t l\ o targets Each of them fires one shot at h1s target tndcpendently of the other marksman The probabtltty of htttlng the t'lrrget 1s p 1 for the first marksman. and p 2 for the second T"o random 'artables are constderedt v1z X 1 the number of ttmes the first marksman htts hts target and X 2 the number of trmes the second marksman htts the target and thetr d1Herence Z = X 1 - X 2 Construct tbe ordered series of the random variable Z and find 1ts charactertstlCS mz and Var.z Solut1on The random variable Z has three posstble values -1,. 0

and +1 P{Z=- -1}=P{X,=0}P{X 2 = +1}=1JtP 2 t

P {Z =0} =P {X1 =0} P {X, =0}

+P {X

=

1} P {X s = 1} qtq'J + PrP2' P {Z ==-1} =P {X 1 = 1} P{X2 =0}=p 1q'l, 1 ::;;:__

'vhere q1 = 1 - p 1 , q2 = 1 - p 2 The ordered ser1es of the v ar1able Z has the form

z

-1 )

•q1P:~

0

j 91 qr+ P1P.z

I t l Ptfa

•

m" = (-1) fJtPt. + 0 (qtgl + P1P2J + iptp~ = - gtP:+ ip1q2 = p,.-. Pt

41

"'Ch. 4. Discrete Random Variables

95

We can find the variance in terms of the second moment about the origin [see (4.0.16)]: a 2 [ZJ = ( -1)2 • q1P2 02 ·(qtq2 P1P2) rz.. P1q2

+

+

+

+

= q1P2+ Ptq2=P1 P2-2P1P2 2 Varz=CX 2 [Z] -m~= Pi+ P2-2P1P2- (Pt- P2) = P1q1

+ P2q2.

4.11. Two shots are fired independently at a target. The probability of hitting the target on each shot is p. Tl1e following random variables are considered: X, which is the difference between the number of hits and the number of misses; Y, which is the sum of the number of hits and the number of misses. Construct an ordered series for each of the random variables X andY. Find their characteristics: mx, Var:\_, my~ Vary. Solution. The ordered series of the variable X has the form

X:

-2 1

o

1 2

I 2pq I

q2

p2

,

mx= -2q2 +2p2 =2 (p- q);

where

q= 1- p.

a2 [X]= 4 (q2

+ pz),

Varx= a 2 [X] -m~= 8pq. The variable Y is actually not random and has one value 2; its ordered series is

Y:

2

1-.

1

'

•

4.12. We have n lamps at our disposal, each of which may have a defect with probability p. A lamp is screwed into a holder and the current, is switched on. A defective lamp immediately burns out and is replaced by another one. A random variable X, the number of lamps that must be tried, is considered. Construct its ordered series and find its mean value mx. Solution. The ordered series of the variable X has the form

1]2131···1 i I ··· I n X: q I pq I p'!.q I ··· I pi-lq I ... I pn-1

'

where q=1-p.

(4.12.1}

4.13. A random variable X has a Poisson distribution with mean value. a = 3. Construct the frequency polygon and the distribution functwn of. the rand~m variable X. Find the probability that (1) therandom vanable X Will assume a value smaller than its mean· (2) the random variable X will assume a positive value. '

~6

Appl ed P oblem1 in Proba.bllltv Thtorv \ns~cr

(1) 0 423 (2) 0 960 4 14 ''hen a computer 1s operating 1t fatls (goes down) from tl.Dle to tune The faults can be considered to occur 1n an ele01entary flow The n\erage number of failures per day IS 1 5 F1nd the probahtiitles of the followlng events A - {the cotnputer does not go down for two days} B - {1t goes do,vn at least once dur1ng a day} C {1t goes do\\n no less than three t1me3 dur1ng a week}

=

Ans\\er

P (A) ~ 0 050

P (B) = 0 777 P (C) = 0 998 4 15 A shell landing and exploding at a certain pos1tion covers the target by a homogeneous Poisson field of spl1nters 'v1th 1ntens1ty l = 2 5 splinters per sq metre The area of the proJeCtion of the target onto the plane on \~ htch the field of spltnters falls ts S - 0 8 m 2 If a spltnter h1ts the target 1t destroys 1t completely and for certain F1nd the proba bility that the target '\VIII be destro) ed Ansl\er 0 865 4 16 The same problem but each spl1nter h1tt1ng the target may destroy 1t 'v1th probabtl1ty 0 6 Solution 'Ve corunder a f1eld of affec t1ng spltn ters With dens1ty "-* = 0 6 A. = 1 5 spltnters per sq metre rather than the assigned field --.....o--x- d ur1ng the 1nspect 1on has a 0 Po1sson dtstrtb ut1on \v1th para n1eter a Jf no faults are detect•

the n1a tn ten a nee lasts for an a" era.ge of t'' o hours If one or o faults are detected,. anothe r bali hour 1s spent el1m1nat1ng each defect 1 f mo-re than \W() faults Rre detecte d the car tales four hours on the average to be re-patred Ftnd the distrlb utton of the a' era go ttme T of matntenance and repa~r of the car and 1ts mean value .l'I {Tl Soluho n. ed

~

t'"'

6

T

J1iT) =e¢(2 +2.5a +f a'-)+G[t-e-~ ~

{1+a+ ~-}]

6-e-a (4 + 3 5a + 1 5a2).

4.22*- A group of antmal s 1s exam1n ed h} n 'et and each may he found to be stck \Vtth probah tltty p The e"inmt nation proced ure In\ ol ves a blood test on a sample blood taken from n an1mal s and mtxed The m1xtur e w1ll y1eld abnorm al test results 1f at least one of the n animals is s1ck A large numbe r N of animal s must be examtn ed T"o method s of examin ation are sugges ted (1) all N anlmal s are exam\n ed. 1n 'vh1ch ca~e .1V blood tests must he made, {2) groups of an1mals are examtn ed, first mrttng the blood of n ant mals If the test ytelds normal results , then all th~ an1mal s 1n the group are constde red healthy and the next n an1rnal s are- examtn ed If the results are abnonn al, each of then an1mals IS examin ed and then the next group o£ animal s lS examtn ed (n > 1). Ftnd '\Vhlch of the method s ts more advant ageous 1n the sense of the mtntm um average numbe r of tests F1nd n n* at 'vh1ch the m1n1mum average numbe r of tests 1s needed to examtne the animal s Solutio n .. A random variabl e Xnt wh1ch is the numbe r of tests needed

=

Ch. 4. Discrete Random Variables

99

for a group of n animals examined by the second method, has the following ordered series: q=i-p.

The average number of tests for a group of n animals examined by the second method is l\1 [Xnl = qn (n 1) (1 - qn) = n (1 - nqn).

+ +

+

When the first method is used, n tests are needed for n animals. For nqn < 1 the first method is, evidently, more. advantageous than the second, while for nqn > 1 the second method 1s preferable. Let us find the q for which the second method becomes more advantageous. What then is the optimal value n = n*? It follows from the inequality nqn > 1 that q > and hence that q > 0.694 since the minimum of 1f11 for an integer n is attained at n = 3. Let us assume that q > 0.694 and find the value n = n* for which the average number of tests per animal is a minimum: Rn = l\1 [Xnl/n = 1 - qn + 1/n.

11rn

n

We must find the least positive root of the equation dRnldn = - qn ln q - 1/n2 = 0, and having done so take the two closest integers, substitute them into the formula for Rn and then choose the optimal n*. By using the substitution -ln q = a, an = x, the equation -qn In q = 1/n2 can be reduced to an equation x2 e~-.: = a (a = - ln q < ln 3/3 = 0.366). For small values of a (and, hence, for small p = 1 - q) the last equation has a solution X ~ 11 a, whence n* ~ 1/V a. If the values of a are not small, then a direct comparison of the quantities R 2 , Ra and R 4 shows that R 3 is always smaller than R 2 and that R 3 < R 4 for 0.694 < q < 0.876; consequently, for 0.694 < q < 0.876 the optimal n* = 3. We can show that for q > 0.889 (p < 0.111) the formula n* ~ 1/lfp + 0.5 is a close approximation. 4.23. A number of trials are made to switch on an engine. Each trial may be successful (the engine is switched on), independently of the other trials, with probability p = 0.6. Each trial lasts for a time 't. Find the distribution of the total time T which will be needed to switch on the engine, its mean value and variance. Solution. The number of trials X is a variable having a geometric di~tri~ution beginning with unity (4.0.32) and T = XT has the distributwn

T:

•

I 2• I 3T I ... I

PI M [T] = 't~I [X]= 7:/p, i•

qp

l q p I··· 2

I ... lqm-lp t ••• ' m•

Var [T] = -r2 Var [X]= T2qfp2

(q= 1- p).

100

Applted Probfemr in Probability Theory

4 24 Under the condttxons of the proccd1ng problem the trrals are dependent and p 1 1s the probabdtt~ that the engJno can be s" 1tched on after , - 1 unsuccessful trtals, P1. be1ng a functton of lt 1 e p 1 = ftl (l) F1nd the dtstrtbut1on of the random vnrtable T = TX

Ans,,er

T

I

't 2T ~ 3"t f mT f --~--~----~--+-------~--~ m-1

I

n

qtPm

i-l

q1 = 1 - p, 4 25 \Vben a rel11.ble Instrument constst1ng of homogeneous parts 1:5 as.sembledt each part xs subJected to~ thorough 1nspectton as a result of'' brch 1t 1s etther found to be sound(\\ tth probab1ltty p) or l.S reJectec (w1th probability q = 1 - p) The classtficat1on of each part lS 1nde pendent The store of ptrts ts practically unlimited The selectton o: parts and thetr 1nspect1on go on unttl k h1gh qualtty parts are selected A random 'V'ar1able. X 1s the number of re-Jected parts F1nd the dtsttl button of the random variable Y P m = P {X ~ m} Solutton The possible" a lues of the random 'ar1able X are 0 1 m, 'Ve find their probablltttes p 6 == p {the r~rst k parts are sound} = pk ~here

P1

=

l' {one part out of the first "" ts reJected the (k + 1)th part IS sound} = Clq 1p" 1p =

Cl,.qlpl

•

•

~

•

6

•

•

Pm = P {m parts out of the first k -r m - 1 parts are reJected the (m k)th part IS sound} === Cli'+m-1qmpk (m = 1 2 ) 4 26 T'vo random 'artables X and Y mav be e1 ther 0 or 1 tndepend ently of each other Thetr ordered sertes are

+

X

y

Oil qx j Px

olt qy { PrJ

Construct the ordered series (1) of thetr sum Z = X + Y, (2) of thetr dtfference U = X ~ }~, (3) or their product V ~ XY Soluhon (1) The random vartabla Z has three possible values (l, 1 and 2 P {Z = 0} = P {A.=== 0, Y 0} = qxq11 P {Z = 1}=P {X= 1 Y=:O} +P {X =0, Y= 1}

=

= Pxqu + qxP1J1 P{Z=2}=P{X=1 Y=1}

z

I 2 qxqu J P-tqy + q~ Pv I P:xPu o

r

t

p~p11

101

Ch. 4. Discrete Random Variables '

'

By analogy we lind that

o

1 I 1-----T--------~~----1· -1 1

U:

q;xpy

I PxPy+qxqy

I

0

V:

qxqy+Pxqy +qxpy

Pxqy

1

•

I PxPy

• senes 4.27. A random variable X has an ordered

1 1 2 1 4

X:

o .5 1 o .2 1 o .3

·

Construct the ordered series of the random variable Y = 1/(3 -- X). Answer. ' - ~ 1 o.51 1 Y: • o.3 1 o.5 1 o.2

4.28. A random variable X has an ordered series -2 1 -1 1

X:

0.2

o

1 1

1 2

1

I 0.3 I 0.210.2 I 0.1

I"

Construct the ordered series of its square: Y = X 2 • Answer.

o

Y:

1

1

1

4

o.2 1 o.5 1 o.3

·

4.29. When a message is sent over a communic ation channel, noise hinders the decoding of the message. It may be impossible to decode the message with probabilit y p. The message is repeated until it is decoded. The duration of the transmissi on of a massage is 2 min. Find: (1) the mean value of the time T needed for the transmissi on of the message; (2) the probabilit y that the transmissi on of the message will take a time exceeding the time t 0 at our disposal. Solution. (1) The random variable X, the number of trials to send a message, has a geometric distributio n beginning with unity; T = 2X min. The ordered series T of the random variable is

T:

2 1 4 1 6 1 . . • 1 2m

q

(2) P {T > t 0} =

1 ...

I pq I pZq I ··· I pm-lq I ... · 2J pm-tq = q 2J 00

00

m=[fo/21+1

pm-1,

(4.29)

m=[fo/2]+1

where [t 0 /2] is the largest integer in t 0 /2. Summing up the geometric progression (4.29), we obtain

P{T>

t 0} =

q 1 -q

p[fo/2]+1

l

-p

= p[to/2].

,Appi ed

1.{)2

4 30 A

Pro~Iems

in Proba.bU ty Thtory

d1~crete random

vartabl e X has an ordered sertes

i :rR

X

A random 'arJabl e Z ts the mtntmu m value of t'vo numbe rs-the va1na of the random \ ar1ahlo Y and the number a 1 e Z = m1n (X a} ''here x1 s;; a~ rn. Ftnd the ordered ser1es of the random var1able Z Solut1o n By defin1 t1on z==:!f!

X

{Qr

Xa

The ordered sertes of the random vartable Z coincides '" tth that of \Vh l'C.h are smaller the random v ar ~able X for the ' al ues :r1 .x 2 than or equal to a P {Z - a} can be calculated as untty m1nus the sum of all the other prohab 1h ttes

P {Z =a) 4 3t

= 1-

~ p1

x,-~n

The ordered ser1es of a dtscret e random variabl e X 1S

X

t 1

a l s

1 1 1 9

o 1}02]0 a}o a)o

1

=

mtn {X 4} F1nd the ordered sertes of the random var1abl e Z Ans,ver Pro~eedtng from the solutto n of the preced tng problem \\e

ba\e

1

z

l

3 ' 4

Otl02 f07

If 32 T'vo random var1ables X and Y tndependently of each other assume a 'alue accordi ng to the follo\v1ng ordered series

X

o{t 12~ a 02loa Jo4Jo t

y

Constru ct tl e ordered sertes of the random \ art'1ble Z Solutio n

= mtn {.!, Y}

P {Z=O} = P{Y =0} ~0 2

P{Z= 1}=P{ X=1} +P{Y -=1 X>1}~01+07 05=0 65 P{Z 2}----. P{X-2 Y>2} =0 4 0 3-01 2 P{Z= 3}=P{ X 3 Y>J} -01 03~003

z

o It l2fa 0 20 ' 0 65 J 0 121 0 03

103

Oh. 4. Discrete Random Variables

-

4.33. Under the conditions of the preceding problem, find the ordered series of the random variable U = max {X, Y}. Answer. 1[2[3[4

I

.I

l \

I

I

U: 0.35 0.28 0.27 0.10

.

4.34. Ann-digit binary number is written in a computer cell and each sign of the number assumes the value 0 or 1 with equal probabilit y

independen tly of the other signs. The random variable X is the n~~~er of signs "1" in the notation of the binary number. Find the probah1ht1 es of the events {X = m}, {X> m}, {X< m}. Solution. The random variable X has a biriomi"al distributio n with parameters n, p = 1/2, P {X = m} = C~1 (112)n, n

P{X~m}=(1/2)n ~

l:=m

C'i:, P{Xm}.

4.35. A proper three-place decimal fraction X is considered , each sign of which may assume each of the values 0, 1, ... , 9 with equal probability independe ntly of the other signs. Construct the ordered series of the random variable X and find its mean value. Solution. The possible values of the random variable X are 0.000, 0.001, 0.002, ... , 0.999. The probabilit y of each of them is p = (0.1) 3 = 0.001. The ordered series of the random variable X has the form

X· o.ooo \0.001\0.00 2\ ... \ o.999

· o.001 1o.ooq o.ooq ... 1o.001 999

999

M[Xl=

·

XiPi=0.00 1 2J xi, 2J i=O i=O

where xi=i·0.00 1.

The numbers x, form an arithmetic progressio n consisting of 1000 terms with a common difference 0.001. Summing up the progressio n, we obtain

l\I [X]= +~· 0

999

·1000· 0.001 = 0.4995 ~ 0.5.

4.36. A random variable Y is a random proper binary fraction with n d~cimal places. Each sign, independe ntly of the other signs, may be e1ther 0 or 1 with probabilit y 112. Find the ordered series of the random variable Y and its mean value M [Y]. Solution. As in the preceding problem, all the values of the binary numbm· from 0.00 ... 0 to 0.11 ... 1 are equally probable and each of t!IC'm has a probabilit y 112'1 • The mean value of the random variable Y (m the decimal notation) M [YJ = 0.5-1/2n+ l. 4.3i. A message is sent over a communic ation channel in the binary code and consists of a sequence of symbols 0 and 1 alternating with an equal probability and independe ntly of each other ~a~r 0 0 1 1 1 0

. 1. 1' 1' 0 ' 0 ' 0 '0, 1, 0, 1, 1, 1, ....

'

"' '

'

'

'

'

'

Appltld Probltms fn Prob«bthty TheorY

i.04

A run of the s'lme symbo1s 1s constdered, say, 0 0, 0 or 1 J t 1 1 (or stmp\y 0 or 1. 11 the s-ymbo\~ a:re no\. T{rpeate:d) An1 one ~f th~e 1\\tl,. 'J. taken at random A random var1ablo X 1S the number of Slmhols 1n a run Ftnd tts ordered ser1es mean valoe and \ ar1ance Ftnd P {X# ~J Solution The random vartable X has a geomelrtc dtstrtbutton begin n1ng '\ 1th unlt}

X

1

l

2 {

o s J o s' 1

( m

l

J o sm

J

' CliO'

VarIX]== 0 5/0.51 = 2

P{X~k}= ~ 0 5m=0.5h-i m.~lt

4 38 Under the condtl1ons of the preced1ng problem P { 0"} ::::::: 0 7

P { r~) = 0 3 F1n.d the a\ erage number of s~ mbols !\{ lX1 tn a run of zero~ the 1' erage nun1ber of symbols .I\ I [Yl 1n a run of un•t•e~ and the over1Il a\ erage number of S} mhols ~~ [Zlln a run of S) wbols cho~en at random Flnd the var1anc~ Var {.A 1 Var {Y) Var IZ] So Iutton ~I (X I -- 1/0 7 - t0/7 'I [YJ == 1/0 3 _... 10/3 B} the formula for the complete expectat10n ''e find that '1 [Z] = 0 7 ~ +

0 3 ~ = 2 1 e the same as m the precedmg problem VarIX] d 0 3/0 72 = 0 612 Var [Y] = 0 710 32 = 7 778 To f1nd Var lZ1 \\e r~rst flnd a. 2 IZl \1 {Z2 1 by the formula for the complete e"t.pectatton [see (4 0 20)1 *) a 2 {Zl =0 7 et 2 lXJ + 0 3 a 2 iY} 'vhere a 2 L\ 1 = Var IX]+ ('1 IXJ)t~ 2 66 a 2 [Y] ~ Var fY] + (1\I [YJ)~ ~ 18 9 then a'" fZJ :::::: 7 53 Var {ZJ - a. 2 { Zl - (~1 lZ1) 2 ~ 3 53 1 e the last var\ance l.S larger than that ln the preced1ng problem The probability P {X~ l:t} can he found by the total probab1hty formula \VJth hypotheses ]/ 0 = {the frrst symbol IS 0 } H 1 = {the first symbol 1s 1 }

P (Ho) = 07, P (H1 ) - 03 P {X~! lH0 } = 0711 1 p {X~ a I HI} :== 0 3k f' then p {X~ k} = 0 7 0 7k + 0 3 0 3k-1 = 0 7ff. + 0 3A

i

4 39 A dev1ce cons1sts of m units of type I and n units of type II The reltablllty (a fallure free performance: for an ass1gned t1me 't) of each type I untt 1s p 1 and that of a type II untt IS p 2 The units fad lndependently of one another For the devtce to operate any two type I nn1ts and any two type 11 units must operate simultaneously lor time 1 Find the probab Il1 ty !P of the failure free opera t1on of the de' IC9

•>

B) the formula for the complete expectation 've seek the gecond moment ahout 1..h¢ or1g1n rather than the variance It~C~elf (~toce condztronal expectations 3.re d zff eren t i or d tlf eren t hypotheses)

· Ch. 4. Discrete Random Variables

105

Solution. The event A = {failure-free operation of the device} is • • a product of two events: ' A 1 = {no less than two out of m type I uriits operate without failure}; A 2 = {no less than two out of n type II units operate without failure}. The number X 1 of type I units operating without failure is a random variable distributed according to the binomial law with parameters m and p 1 ; the event A 1 corresponds to the random variable X 1 assuming a value no smaller than 2. Therefore

< 2} {X 1 = 1}

P (A 1 ) = P {X 1 ;:::: 2} = ·1 - P {X 1

= =

1 - P {X1 = 0} - P 1 - (qf.' + mqf_'- 1p 1 ) (qi = 1 - Pt).

Similarly, P (A 2 ) = '1 - (q~ + nq~- 1 p 2 ) (q 2 = 1 - p 2 ). The probability of a failure-free performance of the device

& =P(A)=P {A1)P(A2) = l1-(qf.'+mqT- 1Pt)1 I1- (q~+nq~- 1 P2)]. 4.40. An instrument is used (activated) several times before it fails (an instrument that has failed is not repaired). The probability that 1

0

P(k)

1

2

J Fig. 4.41

Fig. 4.40

having been used k times an instrument does not fail is P (k). The function P (7,·) is given (see Fig. 4.40). The instrument is known to have been activated n times. Find the probability Qm that it can be activated another m times and the mean value of the number X of future activations. Solution. Qm is the conditional probability that the instrument will not fail when activated the next m times provided that it has not failed the first n times. By multiplication rule for probabilities p (n m) = P (n) Qm, whence

+

00

Qm= P (n+m) p (n)

'

l\1 [X]= ~ mQmm=l

i06

Applied froble m1

111

Pro babiluy Th f!Crfl

4 4 t A 'vorh.er operates n machtne tools of the same type '"h1ch are located along a stratght llne v.1th 1nter~als t (F1~ 4 41) From t1me to t1me the tools stop {n 1th equal prob a b 1l1 t l and 1ndependen tly of one another) and must be adJusted Jfav1ng ad1uste(l one tool the "orker sta} s put un ttl another tool stops then he goes t c) that tool (If the same tool fails he sta:,. s put) A random vartable X 1s the dtstance the '\\Orker co\ ers bet'" een t" o adJustments Find the mean value of the random 'arJable A Sol ut1on '\ e apply the complete e"\ pee tat ton :formula 'v 1th hypothe SIS II t - {the '\ orker stays at the tth tool} (l :=:: 1, n) Accordn1g to the C()UUltlons of the problem all the hyp~theses are equally prob able P (H1 ) = P (H 2 ) ~ ~ P (lln) = 1/n Let us find the condtttonal e'tpect1t1on l\1 [X lll 1 for the lth hypothests Each tool -x}

=

•~

1 --- P {X< -x} ~ 1 - F~ (-:.t) To construct the graph of the funct1on F (-x) 1 Vte must cons1der the m1rror reflect\on of the. curve of F:r (x) a~ou.t the axts of ord1nates and subtract each ordtnate from un1ty (see tho dash l1ne ln F1g 5 4)4r 5 . . 5 .. Construct the d1.str1bUtlon funct1on F {:t) f-or a -random var1able which has a uniform distribUtion on theltnterval (at b).

Ch. 5. Continuous and lJ!i:ced Ran dom Variables

119

Sol uti on. :!:

F (x) =

) f (x) " -oo r

f(:c) = •

dx,

0 1/( b- a) 0

for for for

0 F( x) =

(x-a)l(b~a)

1

x< a, a< x< b, x> b, x~a,

for for for

a b

. , ). 5.5 . (Fig ht 'ng the to ing ord acc d ute trib dis is X le iab var m do ran e Th 5.6 . n tria ngl e law " in the int erv al (0, a) (see Fig . 5.6 ). (1) Wr ite the exp res sio f(x )

F(x)

1 -- -- -- -. ... -- -

o

a

0

b

Fig . 55.

a

:c

Fig . 5.6

cfor the pro bab ilit y den sity fun cti on f (:c); (2) find the dis trib uti on fun l fal l wil X le iab var m do ran the t tha y ilit bab pro the d fin (3) tion F (x); on the int erv al fro m a/2 to a; (4) ftnd the cha rac ter isti cs of the var iable X: mx, Va rx, ax, ~t 3 [X] . An sw er.

xE (O ,a) , o· (1) f(x )= { for x(f (O ,a) , j or more con cis ely f (x) = 2 (1- x/a )/a for x E (0, a), for x~O, 0 x (2 - xfa )/a for 0 < x~a, (2) F (x) = for x> a, 1 (3) P {X E (a/2, a)} = F (a )- F (a/2) = 1/4, 2 /18 , O"x=a/(3 V2}, by for mu la (4.0.15) a rx= Va (4) fnx =a /3, 2m i = a 3/135. !-La [XJ = a.a [X] - 3m xa 2 [X ] 2( 1- x/a )/a

for

+

the 5,7. A ran dom var iab le X has a Sim pso n dis trib uti on (ob eys "l.aw of an isosceles tria ngl e") on the int erv al fro m -a to a (see Ftg. 5.7a). (1) Fin d the exp res sio n for the pro bab ilit y don sity fun cti on-,

120

Applied Pro~lem• fn ProbtJbtlUg

Thtorv

(2) construct a graph of the d1strtbut1on function, (3) find mx, Varz ax . ~~ 3 I X] (4) find tl1e prob abd 1ty that tb e random vttria blo X '"Ill fa11 in the 1n terv a 1 (-a/2 a). Answer (1)

f{x)=

~{1-.;.)

for

--l-(1+.;)

for

0

for

D 0. fhe graphs of the density and the distributio n function arc given in Fig. 5.10a, b; (3) mx = 0; Varx = 2/'),_2. 5.11. A random variable R, which is the distance from a bullet holeto the centre of the target, has a Rayleigh distributio n, i.e. 1 (r) = Are-h'r' for r > 0 (Fig. 5.11). f(x}

1 ----f(r)

a f(J:}

D

0

(a)

(b)

Fig. 5.10

0

Fig. 5.11

Find (1) the factor A; (2) the mode QJ/t of the random variable R i.e. tho abscissa of the maximum of its probabilit y density function~. (3) mr and Varr; (4) the probabilit y that the distance from the bullet hole to the centre of the target will be less than the mode. ~nswcr. (1) A= 2h2 ; (2) QJ/t = 1/(h lf2); (3) mr =aft JJ/2 = 1fa/(2h); Varr=(4- a)/(4h2)=ak 2 (4-n)/2; (4) P{R 0 hence f (t) = F (t)= Ae i. t for t > 0 5 f9 Gt, en a Pot~son field of potnts on a plane w1th a constant dens1 ty ;., find the d1str1but1on and the numer1 cal charac:terlStlcs m, Var of the distanc e R from any po1nt tn the field to 1ts nearest netghour • Solutto n \Ve find the dlstrlbU tton funct1on ctrcle a g draWin by R e variabl tbe of (r) F •

• • • • •

• •

•

of ra d1us r about a po1n t 1n the fi.el d (F1g 5 f ~) For the dxstanr c R from the point to 1ts nearest netghb our to he .smalle r than r at least one othe!

point must fall \Vlthtn the ctrcle The proper ties of the Po1sson fields are such that the probab1l1t y o~ thts e. . . ant lS 1ndepende.nt of F•g 5 19 whethe r there IS a po1nt tn the centre of the field or not Therefore •

F (r)- 1-e... nr•).

(r > 0},

f (r)- 2JtA.re

(r > 0)

whence nl,-s

Thts distrib ution lS known as a Rayle1 gh d1~trthut1on (see Problem 5 11 0)

mr =

5r2tti.re-~r' dr- 2YA 1

0

OD

a2

[R] = Jr'2n,.re-nb• dr = {J

m:

1 nA ,

=-1/(n~) -1/(4A )- ~4- n)/(4nA ) Var, =a, [Rl5.20~ Po1nts are located at random 1n a three d1mens Jonal space The numbe r of potnts 1n a certa1n volume v of the space ts a random var1a.ble wh1ch has a Po1sso n dtstx1b ut1on wtth expectat1on a = ~u,

125

Ch. 5. Continuous and llfixed Random Variables

where I. is an average number of points per unit volume. We have to find the distributio n of the distance R from any point in the space to the nearest random point. Solution. The distributio n function F (r) is defined as the probabilit y that at least one point falls in a sphere of radius r: F (r) = P {R < r}= 1-e-i.v(r) , 4

v(r)=g~ra

where

is the volume of the sphere of raditt:; r. Hence

f(r) =

4nr 2 A.e

-~. ~ nr•

(r > 0).

3

5.21. The stars in a certain cluster form a three-dime nsional Poisson field of points with density I. (the average number of stars per unit volume). An arbitrary star is fixed and the nearest neighbour , the next (second) nearest neighbour , the third nearest neighbour , and so on, are considered. Find the distributio n of the distance Rn from the fixed star to its nth nearest neighbour . Answer. The distributio n function Fn (r) has the form n-1

Fn (r) = 1- ~ ~~ e-a,

4

h=O

••

the density function is

f 11 (r) =

(r > 0),

~r 3 /.

a= 3

where

dF n (r)

dr

=

an-1

(n-1)!

e-a4~1.r2

(

> r

0)

•

5.22. Trees in a forest grow at random places which form a Poisson field with density A. (the average number of trees per unit area). An arbitrary point 0 is chosen in the forest and the following rauuom variables are considered: R1 , the distance from 0 to the nearest tree; R 2 , the distance from 0 to the next (second nearest) tree; •

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

•

Rn, the distance from 0 to the nth nearest tree. Find the distributio n for each variable . . Solution. We found the distributio n function of the random variable m Problem 5.19 to be Fdr) = 1- e-n:r•A.

(r > 0).

Th_e. distributio n function F 2 (r) = P {R 2 < r} is equal to the probabihty that no less than two trees fall in a circle of radius r: F2 (r) = 1- e-nr•A._ ~r2A.e-nr•A. (r > 0). Reasoning by analogy, we obtain n-1

(r > 0),

(

126

Applle d Prablem.1 In Proba bllllV !'htor y

To get the denst ty ln. (r), we d•fter cnttat e Fr~ (r) w1th respe ct tor, n-l

n-1 n (r) dF ) ( fn r da

_tkJ - ( dr -

~ k

all-1

Jcl

LJ

c-a ....1. 0),

o- = 1, ar ( st whence

The second moment about the origin

2

1

) (2a•+2)

r (n + 1/2) =

= mx,

128

Apphed Problem! ltt Probabtlity Theory

'\Vltcnce (S

t :""-:- --- -

+ f) fl

------

2f" {

(

f )

B

1

•;l )

t

Some of the d 1s tr tbu tto ns of the form f, (x) have defin1 te names e g / 1 (z) Is lt.no,vn as a Rayleagb lhslrJh u linn. and f 2 (x), as a 'faxnell dtstr!b uhon I or a Raylei gh distrib ution (s = 1) \\ e ha\ e / 1 (x) = ax e-a:axt (x > 0)

a.=

~.x

.!=~

a= 2a.z=

l:"ir. ,

m~ [! -1 ].

Var,.,=

For a ]tlax" ell d1Slrtbut1on (s = 2) \ve have f 'J (x) = a..t 2e-a';.:•

a.= ne~nark

4~J.. 3

2

~ ( 3n Var~=m% 8 -

32

~· a=v-== 2 .n mx :t

r-' m~ 1 n

1)

All the dJstrlbUtlODS of the form

1. (z) = o.z'e

-a. .ax•

(.r > 0)

for ~ b !ven s have a ~nngle paramet er 1 e they depend only on one pd.rameter. whtch rna' bo e1tber the mean value or the variance ..

5 2fJ*. \[amen ts of a normal dt:;tr~butlon G1ven a random var1able X \Vh1ch has a normal dJ.StribUtlon With parameters m and a, fi.nd the ex pre~ s 1on for the q uant1 ty ex, [X], wh 1ch ts the sth momen t about the origln Solutio n Let us express the momen ts a 1 [XI .:=:= ?\f [X'] about tha or 1g 1n 1n terms ol the central momen ts f.'-, I X} = 1\.1 l (X - m)'} I

as=~~ [(X- m

~ c:Jlk [XJ m•-h + m)'] = ll-O

fl, IXJ

)oo

..

-'r~ = a l' 2n

odd

WI til

For central momen ts

(x- m)• e

-

s = 2n + 1 t

_1 (.x .... m )* u• 2

''

~-to [X]= 1.

e have

oo

J

dx =.. 1-....!.....- \ y~c a y 211: J

2oj II

'

dy ===== 0

-~

-oo

and With even s = 2n, by the formul as 1n the preced1 ng problem.,

l\e have

oo

Jl, [X]=

a

:2n

I y•e-

J

2n

(

oo

2o• :u• dy=

-~

r a

1

2_ a Y2tt

\

J

y•e-

0

2n.+t ~

For exampl e,

1-12 [XJ = az,.

a2 [~Y.J = a 3 [XJ

m2+a2 ,

= m3 +3a:m ,

y

o V2

)I

dy

Ch. 5. Contin uous and ll!ixed Rando m Variables

m4 + 6cr2 m2 + 3cr~, a 5 [X]= m5 + 10cr2 m3 + 5. 3cr!.m, a 6 [X]= m6 15cr2 m4 + 15. 3cr4m 2

a~ [X]=

ft~ [X]= 3cr4 , 6

fta [X]= 15cr

129

+

,

+ 15cr

6

•

5.27. A rando m varia ble X has a norm al distri butio n with param eters m and cr. Write an expre ssion for its distri butio n funct ion F (x) = P [X< x}. Write an expre ssion for the distri butio n funct ion · lJf (y) = P {Y < y} of the rando m variable Y =-X . Solut ion. In accor dance with the solution of Probl em 5.24 F (x) = 0, i.e. an exponential distribution which coincides with I (t) (see Fig. 5.35). Thus, the conditional distribution of the time 8 which remains till the occurrence of the event does not depend, for an exponential distribution of T, on the elapsed time. An exponential distribution is the only distribution which possesses this property. Recall that the time interval between two successive events in an elementary flow has exactly an exponential distribution: the time remaining till the occurrence of the next event does not depend on how long we have been waiting for it (this follows from the absence of aftereffects in an elementary flow). 5.36. The probability of a failure of a radio valve at the moment it is turned on depends on the voltage V in the circuit and is equal to q (V). The voltage T' is random and has a normal distribution with

134

Applied Probltms in Probablhly Theory

parameter s v0 and u 0 F1nd the total probabtltt y q of a fat lure of the ,al\e a the moment It 1s turned on. So1uttu-n. By the 1ntegrel total ptobah1lity formula (50 23) we ha'ie g=

l

J

q (v) j {v) dv =

Jf

-~

2._

2"t Ot

( J

(f'-'f:n}~

-

co

co

2a: dv.

fJ (v)e

-~

5 37 ~ random voltage V., \Vhtcb has a probabtltt y dens1ty I (v)t lS passed through a voltage ltmtter whrcb cuts off all voltages smaller than v1 and larger than v2 , lfl the first case r.a1s 1ng the voltage to v1 and 1n the second case lo\\ er1ng It to V2 F1nd the dtstrLbUtlon of the random ,., F(v) f(y) f(Y) variable V, the voltago whlch bas 1 .......... --~.,....-.-.... ....... .,..._--~-~ passed thro\1gh the l1m.1ter~ and / determtne ltS mean value and ,1-r"' 'ar1ance

-

......

Solution41

0

Vz

v=

v

V1

V

v2

l' < V1..., v1 < l' < v~,

for for for

V>v-z ,..,

Ftg 5 37

The random variable V lS a m1xed quan t1 t y and 1 ts two values t'1 and ti 2 ha' e nonzero probabtltt tes p 1 and p 2 For all the values between Vt and v~ the dts trtbu lion func t1on F (v) of the v ar 1a ble V 1s con t tn uous and

" F (v) = \ f (v) dv, ...

00

'l'

Pt =

-oo

JI (v) dv,

p2 =

J/(v) dv t,

-I)D

The graph of the funct1on F (v) 1s sho\vn 1n Ftg 5 37. tl

l\1 (VJ =

VtPt

+ VtP2 + Jvf (v) a~,•, 't1i

Var IV)= a~ tVJ- (l\l{V])2, t

r

a:, ll) = v1p, + v1p 2 + v''f (v) dv

"•

5~38. A message ol length l 1s be\ng bToadcast over a rad1o channel (F1g 5 38a) In order to eraze the message, a notse pulse tra1n of length b > l1s InJected 1nto the channel so that the centre 0 1 of the tratn coin· etdes \v1th the centre 0 of the message

Because of accidental errors, the centre of the pulse train proves to be d1splaced by X relat1ve to the centre of the message The random vartable X has a normal dtstrtbutJo n w1th parameter s m = 0 and o == l/2. F1nd the dtstr1butt an of the -random var labl~ U, the length of the

Ch. 5. Continuous and ll!ixed Random Variables

135

message which is eraze d, and the mean value mu, the varia nce V ar u, and the mean squar e devia tion au· Solut ion. The rando m varia ble U is mixe d; it assum es the value s 0 and l with nonze ro proba biliti es. On the inter val from 0 to l the distribut ion funct ionF( u) is conti nuou s. The noise train b does not touch F(t) 1 ---- -·-, ::-- --

X

}\\----;

l

frzuw

1'@,. Y)

-~

y

I X

I

(XY)

I

X

1J

$

fig G02

rig 601

(X1 X1 Xn> ts represented by a random point or a random v~ctor ln nn n d'men.ttanal spact 1'he jr)int pr~bability dulrtbuUon of two random vat1able-s (X, 'Y) (ot the pro~ .a.b1hty dlstr•button of a SJstem of two rand.om varJables) lS the probab1hty tha.t both lnequal1he~ X < ~andY

0.

(6.0.35}

For the indepen dent random variable s (X, Y, Z) the normal probabi lity dis· trihution in a three-di mension al space is 1 f (x, y, z) = (2n)3 / 2 cr xCJyOz 1 [ (x-mx) 2 oi, xexp { - 2

+

(6.0.36)

The probabi lity that a random point (X, Y, Z) will fall in a domain Eh bounded by an el~ipsoid of equal probabi lity density with semi-ax es a = kax, b = kay, c ka 2 IS (6.0.37)

=

Problenzs and Exercises 6.1. Two messages are being transm itted, each of which indepe ndently of the other, may be distort ed or not. The probab ility of an event 1 = {a message is distorted} for the first message is p 1 , for the second IS P2· We conside r a system of two random variabl es (X, Y) defined as follows:

6: { 6:

X= {

y =

if if if if

the first message is distorte d; the first message is not distorte d; the second message is distort ed; the second message is not distort ed

r2..

Conseque nt Iy

for

lx]

< r,

!or

(.x[

>r

Similarly 2

2

/2.(y)== 2Jfr -y h for IYIr 0 'The graph of the funct1on i1. (~).1s shown 1n F1q_. 6 6b For I.Y~e

have

1

f t (x Jy) == I t,(z(!I}II)

-

-

for

fzJ < V r2- y2~

for

lzi > Y r2-y2

I 1. A ~raph of the probability density function / 1 (x) is shown in Fig. 6.9b {Simpson's distribution). Similarly, 1-jxj ft(x)= 0

f2 (y) =

1-IYI 0

for for

IYI < 1, IYI > 1.

i 58

A p pl1td Probl~ mr in Pro babtlfty Thtory

Furthermore, for

I y I< 1, wo

have 1

for

lxl < 1- rvl,

for lzl > f -lui A graph of the probabtltty derunty function / 1 (z ] y) IS .shown 1n Fig. 6 ~~~ Slmtlarly, for r X r < 1' 've have IY[ < 1-lxJ,

for

for fYr>1-rxlThe random variables X and Y are dependent but not correlated 6 .. t0 The ]Otnt probab1ltty denstty of t~o random vo.rtables X and Y 1s gtven by the formula

I (:r:, Y) = 1 ~ll exp { -

k l(z-2) -1 2(x-2)(yJ.-3)+(!1+3) J} 2

2

1

Find the correlatton coefficient of the var1ables X and An~-w ~'f ~ T :-: v

Y.

-= G £>.

6.1 t A system of random v ar tables (X, Y) has a d1str1 but1on with a probnb1l1ty den~nty 1 (x, y) Exp~ess the probabil1t1es of the follow. . 1ng events tn terms of the probabtllty density f (x, y) (1) {X> Y} . (2) {X > I Y ,}, {3) { 1 X t > Y} and {4) {Y - X > 1}

!!

9 0

X

~)

(aJ

!J

X

X

(d)

(t) F1g 6 tf

Solut1on .. The domains D 1 , D 2 , D 3 , D 4 correspond1ng to the occur· :renee of even\s 1-4 are hatched 1n F1g 6 i1a d The -probablltttes or po1nts falling 1n these domains are 00

(1.) P{X>Y}~

S

%

1f(x. y}dxdy,

-ao -co

Ch. 6. Systems of Random Variables 00

(2) P{X> IYI}=

157

:X:

J~

f(x, y)dxdy,

0 -x 00

IX I

(3)P{IXI>Y}=) ) f(x, y)dxdy, -oo -oo

1~ 00

(4)P{Y-X>1}=

00

f(x, y)dxdy.

-oo x+i

6.12. A system of two random variables X and Y has a normal distribution with parameters mx = my = 0; cr x = cry = cr and r xy = 0. y

y

!/

0

X

(b)

(a)

(c)

Fig. 6.12

Find the probabilities of

A= {lYl 0 a n d - V!i--,. S j(z , 11 s} OJI t

-oo

Ch. 6. Systems of Random Variables y

:t

(5)

z

r .lr I (x, y, z) dx dy dz,

.lr

F (x, y, z) =

159.

J

-oo -oo -co :t

(6)

Ji\(x)=P( x, oo, oo)= ~

IX>

co

)

) f(x, y, z)dx dy dz,

-oo -oo -oo

(7)

:t

y

F 1, 2 (x, y)=F (x, y, oo) = )

)

-QJ

'

co

1f (x, y, z) dx dy dz.

-oo -oo

6.17. A shell is fired at a point airborne target. The point where th(?. shell explodes is normally distribute d with the centre of scattering at_ the target. The mean :square deviations Yn·

6.22. X is a discrete random variable with two values x 1 and x 2 (x2 > x~) which have probabilities p 1 and p 2 • A random variable Y is contmuous, its conditional distribution for X = x 1 being normal with me~n value x 1 and mean square deviation (xl, ·

(7 .0.14)

[q> (x11

(7 .0.15)

or

r

Vary =

(n)

•

J .. . ) -oo

•

-oo

In some cases, we do not even need to know the distr ibuti ons of the argu ment s hut only their chara cteris tics in order to get the chara cteri stics of funct ions. Here are the funda ment al theor ems on chara cteri stics . i. If c is a nonr ando m varia ble, then (7 .0.:16) M [c) = c, Var [c) = 0. 2. If c is a nonr ando m varia ble and X is a rando m varia ble, then Var [eX] = c2 Var [X] M [eX] = eM IX]

(7.0.17)

A pplltd Prohl em' In Pro fla!Jlllty Tl&tor v

t 68

3 The addttl on theorem for mean vaJoes Tl"' tn~an val~ c/ a 1u1n D( random uarlab ltr u ,qual to tA~ 1um of tbtlr m~an. values (7 0 18) AI!YJ, Y) ::!11: AI (X] ~~ [X

+

+

ell d-.

tn gene:a],. n

n

2J X 1) == lJ

ll (

{7 0 19)

Af CXtJ ..

·-·

i.al

4 The mean Yalua of a Ju1ear funct•on of eeveral random Yar1allles

where a 1 and lJ are nonrandom ecef.ficients Js equal to the

!aJlle

bnear functton ol

\heir mean valuea

tm 11 ~lr

1\

'ft

[L1 DtXt +b I== }]. tJ 1m~1 +b -.~J

(7 0 20)

j$1

where ms1 = ltf (X tl ibis .rule can be \\flU en w a. copcfee Iorm vu m:.nl, Xnlt ~ .L (m,.1,. m.r1 II [L (K1 X~

(1 0 21)

where L 1s a l1nea.r funct!On 5 ThQ mean value of t.he produe1.. af t-v.o randn m .,.arulbles X l\ttd Y ts ex prnsged /by the formula (7 0 22} Covxv• lU {XYJ ==At (XI II (Yl

+

wher-e Co"~v1! the eQ"lanance of the Tar!ables X and Y ~b1s formula t&n be telrntteu as [oJiows (7 0 23J Covz:!# = !! [Xl1 - m~m 11 cr• be-at1ng 1n mind that. l\1 [X Y] ~ a,lt •f X Y] as Co v~Y t = et:t1 1X Y] - m~m11 •

(1 0.2.4)

6 The mulhp hcaho n theDrem for mean values Tht mean z..al'll~ oJIAe pttJdud of two r.ouorrllllted random r-ariablez X Y ls equal to tht prculu.ct ~1 t1ult mto.n vo:!utl (1 0 25) ~f (XYJ ~ 1\I [X] Af (Y]

X n. are 1nd epen d ~.n t rand om \"v,r1ehl es then the mean value 7 ll X1 Xs of their product Is equal to the produ ct oi their mean values n.

l\t

f rJ

t=--1

n l'l

X 1] =

!tt 1X tl•

(7 0 26)

~[

S. The "Variance ~~the sum of t'\\o rando m vc;u·IalJles 1S ex}lressed 'by th& icnnula. (1 0 21) 2Covz11,. Var l¥1 Y] == Var !XJ VarIX 9 The var1ance of the sum. of sc•era l random v'nab les Is expressed by the lor"' mula n

Var [~

~=J

+

+

+

n.

XtJ== 2J

f:e:l

Var [X,]+ 2

2J Covzt:tl

i 1 this integ ral exists and is my ='AI ('A -1) for /.. ~ 1 it diverges, i.e. co

co

a 2 [Y] =

Jezxt..e-"x dx = ').. Je-0 for .\ -

1,. _ {-X

0

~

m11 = l\li

~

1rf (x} dx Jxf (x) d:r.

r ! = ~ 1:r 11 (z) dx = -

-~

-~

0

(IQ

-= \ x(f(x) +I ( -z) dxt 0 00

J j.:tf2J(x)dx-m;

Var11 =a 2 [Y]-m~= = a 2 fXJ

-m; = 'rarz + mi -m~

7 34 Ftnd the Jnean 'alue and varianc e of the tnodulu s of the random variabl e X" "h1ch ts norma lly dtstrtb uted \\ 1th parame ters m~ C1 Solution~~ It follo\\S from the preced1 ng problem that ~

m 11 = -

_ \ ze- 2cr• dx +

a lr 2n.

=

---

a l~'" 2"'t

J

iZ-m'' xe- 2°2 d:r

0

m)!a = t.., \\e get tl

s (tu+m) e -2 dt+ .r~'t J (ta+m.) e -T dt t•

0

4XI

m

--oo

1 { m

i2a 1 2r.t

f

-.oo

Chang1 ng the var1abl es (x -

m.==- v!n:

~

(~-m)

0

e -2

a

-~

)2

0'

....

+mlll ( m. ) 0

'

where ct> ts the error functva n Vary~ o 2

+m

2 -

m;

In parttcu lar, for rn = 0, 2

1

m11=V-./ :c:

cr~080cr '

2 2 1a2=1 V9r =a2- ~ n \ 11

)o2~036a1

Ch. 7. Cha ract erist ics of Fun ctio ns of Ran dom Vari able s

183

7.35 *. Ind epe nde nt ran dom vari able s X and Y hav e. pro bab ility densities / 1 (x) and f 2 (y). Fin d the mea n valu e and van anc e of the g modulus of thei r diff eren ce Z = I X - Y 1. 00

~~

mz=

lx- ylf dx )f 2 (y)d xdy .

1!

-oo

A stra ight line y = into two dom ains In dom ain I x > In dom ain II y > Hence

x divi des the x, y-p lane I and II (Fig . 7 .35). y, I x - y I = x - y.

Ix

x,

-

y

I=

x.

y -

Fig. 7.35

JJ(y- x)f dx )f

~ ~ (x- y)f i(x )f 2 (y) dxd y+

Tnz =

2

(y)d xdy

(II)

(!}

= .\' xfd x){

oo

oo

X

oo

J / (y) dy} dx- J Y/ 2

2

(Y){J

-oo

-oc

-oo

y

00

y

00

00

+ J Y/

2

fdx )dx }dy

(y){ .\' ft(. x)d x}d y- .\ xft (x) {J / 2 (y)d y}d x. x

-oc

-oc

-oo

We intr odu ce the dist ribu tion fun ctio ns X

F 1 (x) =

) / 1 (x) dx,

•

-oo

-co

The n we hav e 00

00

r

P

Tnz =

.\ x/1 (x) F 2 (x) dx -oc

-oo

y/2 (y) [1- Ft( y)] dy

-oc

oc

+)

j

00

Yf'2 (y) Fdy )dy -

Jxft (x) [1- F (x)]dx. 2

-oo

Co~nbining

the first inte gral wit h the fou rth and the seco nd wit h the thtr d, we obt ain 00

00

lllz =

~·

[2x f!(x )F 2 (x) -xf t(x )]d x+

-ex: oc

J [2yf

2

(y)F 1 (y) -yf (y)] dy

-co oc

=2 .\ xft( x)F2 (x) d:r -mx +2 .\' yf2 (y) Fd y)d y-m y. -oo

2

182

ApplH~d

Probltmt ln Ptobabillty Thtorf!

element Ax of the pto~e t1on results fron1 the proJeCtion of t'\\ o and onl\"' t"' o oppostte elements ot the contour L\l 1 and ~l 2 (see F1g 7 32), henee the a"era~ length of the contour pro jectto n 1s half the sum of the average lengths of the pl'O)CCtlons of the: e1ernentar) S~Cll.Ons al lOto '' hl.th the contour can. be dt' tded

Solution Stnce the contour

1\f{Xl

LS

con' e-.::,

~ac.h

=t ,3 2;1 =

~ .

7.. 33. There IS a random \ 8 r 1a ble X \VI t h probab1lttl- dens1 t) I (t) f1nd the mean 'alue and 'ar1ance of the random 'ar1able Y = l X [ Solution The nota t 10 n l,. = r X I means that II

i;g M I Y I= ~ I xI f(x) dx=- i xf(x)dx j rf(x) dx r = r-~ ~~~

mll=

~

tl

-~

-~

QO

0

oc

::::: ...~ x lf (x) +I ( -.x) dr 0

.ao

Jr

'arll =a: ll'J -m~ =

r~ f (x) dx- m;

.:t

-00

=a

2

{XJ- m~ = \"ar~+ mi-m~

7 34 FJ nd the mean ' al ue and '"ar1 a nee of the modulus of the random var1ab1e X, '\\ h1ch ts normally dt.StTihuted v.1th parameters mt o Solution It folio" s from the precedxng problem that 0 mil:;;;;:;:: -

a

f 1 1 2~

(%-m)

r \

.xe-

2o a.

d.r +

-~

(.:t.,.. ml•

o,

f

f

J

cr l/2"1

Cbang1ng the '\ar1ables (x- m)/o = tt

:XC-

2[

nab [X) _ - {a+b) 0). The device operates for a time T. Find (1) the mean value and variance of the number of units which will be replaced, and (2) the mean value of the total time T which will be taken by the repairs of the units that failed. Solution. (1) We designate as Xi the number of units of the ith type that fail during the time -r. This random variable has a Poisson disWe des= A,,-r. Its mean value mx.l = A. 1-r and variance Varx. tribution. .. .. l • Ignate as X the total number of units that failed during the time -r. We have

X=

xi' 21 i=1

n

n

n

mx =

2J i=1

mxi = T

At· 2J i=i

The variables X 1 being mutually independent, we have n

n

Varx = ~ Varx.1 = • 1 t=

T

~

• 1 t=

/... 1•

(2) W~ designate as Ti the total time needed to repair all the units o£ t-ype t that failed during the time -r. It is the sum of the times needed

(92

Appl~td

Problem1 tn Probahllltfl Tlaeorv

to repa1r all the units S1nce the number of un1t.s ts X h \Ve ha' e T 1 =Tiu+T,~

+

X

+ T~x,J = ~

T\l),

llz::l

f:

where lS a random vartable wh1ch has an exponential dlstrtbutton With parameter !-'-f .and the quantities Tlu, Ti2 ), are diSJOint Let us find the mean value of thn random var1able T l us1ng the 1n tegral formula for the complete expectation \Ve assume that the random vartab le Xi assumes a de fin 1te vnl ue m and then the mean \ al ue of the vartable T 1 m

M iTr I m} =

m

~ M t7'\kl1 == ~ ll~l

!, =

;,

11.:=1

\\ e multtply th1s cond1t1onal e"tpectation by the probabl11t-y Pm that the random variable X l assumes the value m sum the products and find the complete (absolute) ex: pee ta t1o n of the var table T ~

Ustng then the theorem for the addttton oE expectattans we obtatn

Note that the same result can be obtained by a less str1ct argument The average number of failures of the untts of the tth type durrng the t1me T ts At't' the average t~me taken to reparr one untt of that type JS 1/t-tt and the average t1me whtch 'vill be spent to repatr all the units of type i that fatl dur1ng the ttme -r IS A1't'I!J., The average t1me that 1'\

will he spent to repair all the umts of all types Is I

't

~ i-1

,_,

ill

7 52* The conditions of Problem 7 52 are changed so that each un1t that fails 1s sent to a repa1rsbop and the devtce IS stopped for the t1ma needed for the tepau~ When the dev1te does not operate other Units cannot fa1l Ftnd (1) the mean value of the number of stops of the ti.ttl..V.A r.b~l!\.TJ...~ \,lJR, t.,.\.WR. --r.. (L\ ~m YJ&O.'O. ... 1lu-e. 4 +.,~ 1r4fi'l•'-~ '0~ ~ tJ-tne 't' dur1ng \Vhlth the dev1ce w1ll be 1dle (It ts also the mean t1me spent on repairs) 1Solution (1) 'Ve designate the number of stops during the tune 't' as X and find tts mean value mx \Ve shall use the following not very strtct (but true) arguments to solve the problem \Ve represent the

,a

Ch. 7. Characteris tics of Functions of Random l'ariables

193

operating process of the device, infinite in time, as a sequence of "cycles" (Fig. 7.52), each of whic-h consists of a period of operation of the system (marked by a thick line) and a per~od of repair. The durat.ion of each cycle is the sum of two random vanables: Toper (the duratwn of operation of the device) and Trep (the time taken to repair it). The average

0

\

v

2nd cqcle

tst cqcle

1\

v

I

t

4th ~!fcle

3 rd C.lfcle

•'

Fig. 7.52

duration of the operation of the device mt oPer can be calculated as the mean time between two consecutiv e failures in the flow of failures of 1l

intensity 'A = ~ A.;; it is i=1

We seek the mean tirne of repairs mt rep . \Ve find it using the complete expectation formula on the hypothese s Hi ={a unit of type i is being repaired} (i = 1, 2, ... , n). The probabilit y of each hypothesis is proportion al to the parameter 1~~ i:

11

t.i/2J i=l

P (H;) =

·;.,i =

'AJA.

The conditiona l expectatio n of the time taken by the repair on this hypothesis is 1/~ti; hence 1l

I. i /.u' I.

-- ~ m 1rep-........! i=l

The average time of

fl.

1 ').

n

~ i=l

'),

.

I

~l i

•

cycle n

i. + ~- ~ :~: .

1=

I

n

~- (1 + ~ ~;~ ) . i=l

' z

Let us now represent the sequence of stops of the device as a sequence of random points on the t-axis partitione d by intervals equal to m 1 • The aYerage number of stops during the time T is equal to the aver~e~e b number of random points on the interval -c long; n

7n.1.=T/m,rep ='A-r:j (

1+ ~ ~;:). i=1

Applf td Proble m• fn. Probt: btltty Theory

f94

(2) Dur1 ng each cycle the devrce 'viii be 1dle (w1ll be repa1red)

"'

~ for t1me m,rep = ~ l'lr

,.., flJ

on the average; consequently, the average

f-1

idle t1me

7 ~53. A rando m varHt hle X Is norm ally dtstrt butcd '' tth character%~ tics m't' and a~ The rando m 'ar1a bles }~ and Z are related to A, as 2 3 F1nd the co' ar1an ces Co' .ru' Co\ :r:r and Cov 1 ~: X Y = X and Z = Soluh on To stmpl 1fv the calcu lation s \Ve pass to the standar(hzed var1a hles .r~nd u~e the fact that for a stand ard I zed norm al variab le X- nl '( all the cent-r al moment~ o{ odd ordet s are equal to tero X 0

=

Q

a;.

and ~I I X 2 1 -

0

3a~ (~ee Probl em 5 26)

r-.r (X"] =

S1nce

0

.,

Y ={X +m.x ) 2 -l\l [X2J= X 1 + 2Xm x+ m~- Varx -n1i Q

0

0

Q

= X2 +2Xm:.:-o~,

it follow s that 0

Q

0

0

0

Cov:cu = }.l {XY} = '1 [X (X?..+ 2Xmx -o~}] = 2aimx

Furth ermo re c

0

3 - 'J [XS] = mx) + z =(X

0

0

0

=

0

xa + 3X2mx + 3Xm~ + m~- (3mxo~ + M~) 0

xa + 3XZm~+ 3Xm i- 3m :co~

and there fore 0

~

0

C"

0

Covz: = !J! [XZJ = !J,J (X'J + 3m\.I\l [X3J + 3mi 1\1 [X 2 J -3m ~oit\I [X J ~ 3a~ + 3nt~o~ 0

1

and finall y c

2 + 2Xm x- o~) (X3 + 3 X2 nt~ + 3Xm~- 3n!xX2

parts that have been tee;; ted) F1nd the maxImum practically posstble number of parts whtch \Vtll need to be tested~ Xmlx Answer From the solut1on of the pre ced1ng problem we have

mx

= kp,

Var.:t

==:==::

Xmax ==

kql p 1 ,

kp

cr~

= Vkqfp

and

+ 31/kq/p.

1 61 According to a networl of management planntng the moment Y an operation begtns IS the ma"ttmum t1m.e two support Fig 7 61 1ng opera t 1ons X 1 and X 2 are f1n1shed Th~ random vartables X 1 and X 1 are mutually tndependen t and ha' e prob ab 1l1t y dens1 t1es f 1 (x1) and f 2 (x 2 ) Ftnd the mean value and var1ance of the :random var1ahle Y, as \vell as the

range of 1t.s pract 1call} pos.s 1bl e "al ues Solnllon for

The dnma1n I, '' h~re Y 1 are sho" n tn FJ~0' 7 61

> "\. 2 ,

X1 < X2 -

and the domrnn II, "\\here X 1

QC.,

m11 = M l1' 1= } xdtlxt) {

:C:l

j } (x 2

2)

dx2 } d.t:t


Xi

-co

~

ft(xi)la(xa)

-co

... In (xn) dx 1dx3

•••

dxn} dx 2 •

co

+ .. · +

xn

xn

1Xnfn (xn) { ~ (n-1) J fdxt)/2 (x2)

-oo

• .. fn-dXn-i) dx 1 dx 2

-oo

""

•.. dxn-i} dxn. In this case the (n - 1)-tuple integral decomposes into the product of n - 1 simple integrals, and, therefore,

i=l

-CX>

200

A r Dlt~d Pr ob le m ' in Pr ob ab rl ity Th eo ry

"' he re r l (x .) IS th e dt st rt bu tt on fu nc ti on of th e ra nd om 'a rl ab le XI (J = 1, 2,. , n) of the ar gu rn cn t z~ By analogy

" e find that

;\l{Y'-J =

_L, ) 1= ! -

fi

.r V ,( x1 )

E1 (x 1) dx ,;

l ~J~n

t)O

Var ll'"l = \l {Y2 ] -m;

1~J

If th e ra nd om va ru 1b le s X (J = 1, 2,. ., n) ha \C the snme d1strtb1 ut 1o n 'v 1t h den~Ht} f (x) an d dt st rl bu tt on fu nc ti on F (x ), th en

m 11 - n ~ .r f (x ){ f (xw-• dx . lXI

~~ [ Y 2 J an d \ ar

f} ] ca n be ca lc ul at ed b} an al og y oc

;\ I( Y 2 ]= n

J.r"/(x){F(.r)}- d.r 1

an d

-o o

V ar {Y J= [Y 2 J- m ;

7 ~63~ A de \ 1ce (d es tg ne d as a ma>..tmum 'o l tm et er ) reg1ste-rs the la rg er of t'" o vo lt ag es V an d lr T ho ra nd om 'a rJ ab le s l' an d V, 1 ar e m ut ua ll y In de pe nd en t an d ha2ve th e sa m e den~nty j (v) 1F1nd the m ea n \a lu c of th e vo lt m et er read1ng 1 e of th e ra nd om va rt ab le l' = m ax {V1 , V :} , 1f f (v) = J..e-tv (v > 0) IS an ex ponenl1=:1l dJstribUtton "1 th pa ra m et er A So lu ti on . In ac co rd a ne e '' 1t h th e so Iu tt on of th e pr ec ed tn g pr ob le m oc

J vf(v)F{v)dv=2 JAve-A"(1-e-:a.")dv

l\ I[ V J = 2

QO

-o o

0

00

- 2

J'J...ve-J.." d v - 2J.. ~ veco

0

2

0

=yIt ca n he se en th at et th er l l or v2

2'J.." c!v

t 2A

3

= 2). •

nl IV]

1s 1 5 ti m es as la rg e as th e m ea n va lu es of

1. 64 The mean t-alue an d the varlance of the sum of a random number y of random terms A ra nd om 'a rt ab le Z IS a su m Z = ~ X h "h er e the i- J ra nd om 'a r1 ab le s X 1 ar e In de pe nd en t an d ha vo tl1c sa m e dt st rt bu tl on \\ tt h m ea n va lu e mx an d vo.r1ance V nr :u th e nu m be r of te rm s Y I~ an In te gr al ra nd om va rt ab le '\h rc h do es no t de pe nd on th e te rm s X n has a me1n va lu e my an d a \a r1 an ce V ar y F tn d th e m ea n 'a lu e an d 'a rt an ce of th e ra nd om va ri ab le Z

Ch. 7. Characteristics of Functions of Random Variables

201

Solution. Assume that the discrete random _variable Y has an ordered • senes

Y:

1j2j ... ,k, ...

I P2 I ... I Pk I ... .

PI

We make a hypothesis {Y = k}. On this hypothesis h

l\1 [Z I Y = k] =

Lj M [Xd =

kmx.

i=1

By the complete expectation formula M [Z] = ~ kphmx = mx ~ kpk = mxM [Y] = m ...:my. k

k

Similarly, the conditional second moment about the origin of thevariable Z

M[Z21Y=lcJ=l\1[(~ 1 k

k

= i=i lj =

2] [ xi) =nl ~ k

1

J X1+2f]jxixj

a 2 (XI+ 2lj mxmx ;=: ka 2 [X]+ k (k-1) m~ i:my

k

+ mia

2

[Y]

= Varxmy +m~ (Vary+ m~) = Varxmv+m~Vary +m~m~,

Var [Z] =a2 [Z]-m~m~ = Varxmu+ m~Vary. Thus we have

+

Varz = Varx my m~ Vary. If the random \"ariable Y has a Poisson distribution with parameter a,. lnz =

mxmy,

then

+

Varz = a (Varx mi). 7.65. A message is sent over a communication channel in a binary code consisting of n symbols 0 or '1. Each is equally probable and independent. Find the mean value and variance of the number X of changes in symbols in the message as well as the maximum practically possiblenumber of changes. Solution. Let us consider the n - 1 changes from one symbol to the next to be n - 1 independent trials and consider, for each of them, a variable X i• which is the indicator of a clwnge of symbol. This value is equal to unity if a symbol is changed and to zero if there is no lnz

=

amx,

change.

n-1

"'' = --J ~ X i .. i\I[XJ = i=1

J\.

n-1

2J i=t

n-1

i\1 [X;]=

2J p= i=1

(n-1) p,

Appll'd Problem~ tn Probability Theory

202

"bere p IS the probabtltty that a symbol 1s changed at a g1ven (lth) Inter\ al Stmtlarly n-1

VariX)-~ Var(X 1)==(n-1)p(1.-p) i~l

1t 1s evident 1n thrs case that p = 1/2 and ?\II X) = (n- t)J2 Var [XJ = (n - 1)/4 By the three sigma rule Xma:.: = (n- i)/2 3 n-- 1/2 7 66 A group of four radar un1ts scans a reg1on of space In wh1ch there are three t arget.s S 1 , S 2 S 3 The region ts scanned for a t1me t During that t1me each untt 1ndependentl~ of tbe others. rtlay detect each target 'v1th probab1ltt, p 1 "h1ch depends on the ord1nal number of the target and transm1ts 1ts cooi'd,nates to the central control stat1on Ftnd the mean 'alue of the number of targets .A "bose coordinates w1ll be registered at the stat1on Solut1on \Ve des 1gna te as Xi the Indtca tor of the e\ en t A 1 ~ {the ltb target 1s detected}

+

,r

~:th target ts

1f the

detected

1f the tth target 1s not detected The mean value of the random variable A, target ts detected} -= 1- ( 1- p 1)"' a Stnce X= 2J X 1 t 1 t folio\\ s that

1S

(t = 1, 2, 3)

1\I [X,]= P {the ith

t.~1

3

3

t-1

t-1

~~[XI-= lJ [1- (1- Pt)'l = 3-}J (1- Pi)~t 7 67 The cond1t1ons of the prec.ed1ng problem are changed so that the proba hill t1es of detect 1ng a target by dt fferen t un1 ts d1Het the ]th untt detects the ~th target "1th probabllit~ PtJ (t = 1., 2, 3 1 = 1, 2 3 4) ,

4

(1- Ptj) lf 1=1 J--1

Ans\\er l\I [ \] = 3- 2J

7 68 A radar tnstallat1on scans a region of space 1n \VhLch there are four targets Dependtng on the ttme. of scann1ng the probabtltty of det ec t1 ng a target 1s a function p (t) and does not depend on the deteet1on of the other targets Ftnd (1) the mean value of the t1 me T 1 In \vbJch at teast on~ target'\ 1\\ be detected and {2) t'he mean value of \he trme T• 1n \\ h1ch all the four targets \Vtll be detected Sol u lion (1) The prob abtli ty that not a sIngle target \V Ill be detected 1n t 1me t JS [1 - p (t)J4. the probnblll t y that at least one target 'v1ll be detected 1n that trme lS 1 - (1 - p (t)]" Thts 1s s1mply a dlstrtbutlon funct1on F 1 (t) of a randum var1able T 1 As ,ve pro'\ed 1n Problem 5 28,

Ch. 7. Characteristics of Functions of Random Variables

203

for a nonnegative random variable T1 M[T1

00

00

0

0

]=) [1-Ft(t)]dt= J['1-p(t)]!'dt.

(2} The probability that all the four targets will be detected in time t is [p (t)l4. This is a distribution function F 4 (t) of the random variable T4 • Its mean value is

1[1-Fdt)]dt= ~ {1-[p(t)] }dt. 00

M[Td=

00

4

0

0

For example, if the probability p (t) is defined by the formula p (t) = 1 - e-at, then oc

J\1[T 1]=

,.

J

00

\'

1 [1-(1-e-al)]4dt-= j e-4at dt= rx, 4

0

0 00

M[Td=

~ [1-(1-e-cd) 4 ]

dt=

:;rx.

0

In this example the mean time of detecting all the four targets is €ight times as long as the mean time of detecting at least one target. 7.69. \Ve make n trials in each of which an event A may or may not 1) is equal to the probabilit y that each pair of carriages, between which this position is, has the same destinatio n. This may happen either if all the four carriages are bound for the same place (the Zth), or if the first pair is bound for the Zth place and the •second pair for the rth place. The probabilit y of the first variant is 1 2 and that of the second variant is ml(mz- ) (mz- ) (mz- 3) m (m-1) (m-2) (m-3) mz(mz-1)m r(mr-1) m(m- 1) (m- 2) (m- 3) .

1 . h h Tl1e to t a 1 pro ba b"l"t I I y t at t ere IS no uncouping at two non-adjace nt positions is k

qn.ad= m(m-i)( !- 2)(m- 3)

~

[ m,(mz-1 ) (m1 -2)(m1 -3)

1=1

+ ~ mz(m1-1)mr(m r-1)]. The covariance Covyiy i for two non-adjac ent positions is qn.ad - q 2 • With due regard for the number of adjacent (m - 2) and non-adjac ent 2) pairs of positions, we obtain (C~- 1 - m Var [X]= Var [Y] = (m-1) q (1- q) 2 (m-2) (gad- q2) + [(m- 2) (m- 3)- 2 (m-2)] (qn.ad -q 2), crx =Uy = lfVar [Y]. The range of the practically possible values of X is M [X] ± 3crx.

+

+

Remark. The calculations only have sense for m > 3. If in the formulas forqnd and qn.nd some of the terms are negative, then the correspondi ng terms areassumed to be zero.

"7.74."}._, A number of trials are made, in each of which an event A (success·) may occur or not occur. The probabilit y that the event A

occur~ at least once in the first m trials is defined by a nondecreas ingfu.nctiOn R (m) (Fig. 7.74)*>. Find the average number of trials which Wtll

be made before a success is achieved.

*) The function R (m) is defined only for an integer m hut to make the figure1 c earer, the points in Fig. 7.74 haYe been connected hy'lines,

'208

,.o bltli rru

,.1 , ptlt." t I

....

Pro bab ilttv Th~orp

In

lEU

Solution. Let us a!Jsume that the trtals are not termtnated alter a succeqs 1.s ach1c\ cd E~ch of them (e'tcept for the first) can bo "necessary,. tf a succes~ 1s not Jet ach1e" ed, and "supcrOuousu if a success is achieved Let us connect a random variable X 1 ~ tth each (itb) tr1al and assum(t 1 t to be un 1t ~ 1f the trial IS necessar} and zero tf 1t IS su perfl uou.s \\'e cons1der the ran(lom vartab\e Z, the number of tr1a1s whtch mm\ be made before a success IS achieved It 1s equal to the ~urn of all the random 'arutbles X h the frcst of \\ htch lS al\Va"}S equal to untty (the first tr1al 1s al,vays necessary}

x1

_____

R(m) 1 ---

OCI _....-~-

Z=X 1 +2} X 1 i~2

t

I I

I

J

'

j

1

2

The ordered ser1es of the random 1 ar1ablc A. 1 (t > 1) has the form

f

I l

0 R(t--i)

m

1

1

I t-R(t~t)

(rf a ~ueces~ 1s ach1eved 1n the pre .. crd1ng l - 1 trtals, the 1th tru1l IS superfl uoust If tt IS not, then the trtal Is necessary) The mean 'al ue of the random variable X 1 lS l\trX~]=O R(t-1).f 1

11-R (t-f)J=1--R(t-1) notatton for l\1 (X1 ) ~~I [1 I-

e can ovJdently use the ~same 1- R (0) (R (0) ~ 0) Hence 7 \\

00

ctO

\IIZJ=~ [1--R(l-1)J==: ~ [1-R(k)J 'i~1

~=0

7. 75* For the cond1t1ons of the preced1ng problem find the varH\nee O, for X=O, for X 0)? Solution. Proceeding from the solution of the preceding problem, we set Y = G-1 (X), where G-1 is the inverse of the required distribution function G (y) of the random variable Y. We have

g (r) = G' (r) = 2rlr6

y

G (y) =

Jt..e-'-v dy = 1- e-'·Y

(y > 0).

0

. Setting 1 - e-'-v = x and solving this expression for y, we get the m~·er~e function y = -(t..)- 1 In ('1 - x). Hence the required relationship 1s Y = -(t..t 1 ln (1 - X) (0 0). To what functional transformations must it be sub-

Appl ed Problemt fn ProbalHllty

230

Theory

Jccted to reduce 1t to a random vartable Y whtch 'vould have a Cauchy d1stt1hu\10n f z {y) = {n (1 + y1 )] 1 Solution

Fdx)=l-e b(x>O)

!

+

F!(y)=

!-[arctany+ ~]

J

Settmg [arctan y ~ = u and solvmg th1s equatton for g v. e find the tn verse fu nc t1on F, 1 ( u} y = F 1 1 (u) = tan (ttu - n/2) -cot '"1 u

=

From the precedu1g problem lVO get Y - F; (F1 (.X)) - --cot ~ (1 ~ e

)..X)

= cot ne

(X >0)

1X

8 31 T\vo pco ple agreed to meet at a certain place bet \veen 12 00 and 13 00 hours Each atrl,es at the meot1ng place Independently and y

!J

It-----

(A Y)

0

t

f t

Ftg 8 31

{a

f

z

0

,

z

(6)

Flg 8 32

wtth a constant probabl11ty donsLty 'lt any moment of the asstgned ttme Interval The first to arrPle ,vatts for the other person F1nd the prob abJll t} dlSttlbU tlOU funct lOU 0 f the \Val ttng t I m9 and the probabth ty that the first \Vlll \val't no less than half an hour Soluhon \Ve des1gnatc the arr1val tunes of the two people as T1

and T 2 and take 12 00 as our reference t1m~ Then each of the tndep~nd ~nt random vartables T1 and T2 Is dlslrtbuted \Vtth a constant density tn the tnterval (0 1} A random vartahle T 1S the watttng t1me T l T1

-

T,

=

J

Let us flnd the d1strtbut1on functiOn G (t) of th1s vartable \Ve tsolate on the p1ane t 10t 2 a domatn D (t) In 'vhxch J t 1 - t 1 I < t (the hatched doma1n 1n F1g 8 31) In th1s case the d1str1button functton G (t) ts equal to the area of tbts doma1n G (t) = 1 - {1 - t),. - t (2 - t) whence \Ve have g {t) = 2 (1 - t) for 0 < t < 1 P{T > 1/2) = 1 - G (t/2) = 0 25 8 32 A random pulnt (X Y} 1s un,formly dlsttlbut~d tn a square K With untt stdes (Fig 8 32a) Ftnd the d1str1button of the area S of the rectangle R With stdes X and Y Sotuhon \Ve 1solate on the z, y plane a domatn D (s) \Vtthln 'vh1ch xy < s (F1g 8 32b) In this case the dislrlbuttotl function IS equal to the

Ch. 8. Distributions of Functions of Random Variables

231

area of the domain D (s):

J Jdx d y =

G (s) = 1 -

1

1

1-

Jdx s

D(s)

~

dy = s (1 -1 n s).

sfx

Hence g (s) = G' (s) = -ln s for 0 < s < 1. 8.33. A random variable X has a normal distribution with parameters m = 0, cr. Find the distribution of the inverse variable Y =

1/X. Solution. y = cp (x) = 1/x; the inverse function is single-valued: x = 1/y. By the general rule 1

g (y) = a ]1"2n exp

(

1 ) 1 y2

2y2

-

or

1 - exp ( - 2 g (y) = 2 1 ay 1 2n

cr!y2) ·

For y = 0 the density function g"(y) has a discontinuity of the 2nd · kind (see Fig. 8. 33). It is interesting to note that the random variable Y does not have a mean value since the corresponding integral diverges. 8.34. A system of random variables (X, Y) has a joint density function I (x, y). Find the density function g (z) of their ratio Z = YIX.

g(f!)

o

!!

!f/X=Z

Y, Fig. 8.34

Fig. 8.33

Solution. We specify a certain value of z and construct, on the x, yplane, a domain D (z), where ylx < z (the hatched area in Fig. 8.34). The distribution function G (z) =

J1f (x,

0

00

00

zx

Jdx Jf (x, y) dy + Jdx Jf (x,

y) dx dy =

D {~)

-oo

zx

0

- oo

Differentiating with respect to z, we have 0

g (:::) = -

1

:t.j(x, zx) ax+

-oo

1 00

xf (x, zx) clx.

0

y) dy.

232

Applied Problem.r tn Probability Theory

lf the tandom var1ables X and Y are mutually 1nde.pendentt then ~

0

J

j

g (z) = -

:t/dz) / 2 (z~) dx+ zjt(x) / 2 (zx) d:r.

-~

0

8 35 Ftnd the distribution of the ratio Z = YIX of t~o Independent '\\~1m all~ u~s\t1.hut~d tand\)m v at\~b\~s X ann Y "1.\h thaiat.\~t\$\\t~ mx = mv = 0, Ox, f1u Solution. \Ve first cons1der a spec1al c.ase o~ = uu = 1 On the bas\S of the preeed1ng problem 0 ~

y(z)=-

r

1

f

J x . .2n e

-2 (~l+zo!:tl)

01'

=

:t

r

j

d.x+ J .x 2.-n. e

-~

t

I

-2 (~Z+z2x2) dx

0 :r2

QO

--2

le

u+tt)

t

z dx == n (t + 11)

(Cauchy's d1s tr1bu tton}.

0

In the general case the rat1o Z = XIY can be represented as Z = Yia 11 ba'e (Y 1 av)I{X 1 cr~)~ 'vhere the vartables X1 === Xfcr~ and Y1 a normal d1strihut1on 'v1th variance un1ty, therefore, Y

=

g (z) _

t

- n [1

In particular, tf a:Jt

+ (O'.x.:}~a;-•J

=a

11

g (~)=In ( 1

f

ax

a11

then

+ .zZ)J-l

8.3G. A candom po1:nt (X t Y) 1s untfor1llly Fig 8 36 distributed tn a circle K of radtus 1 F1nd the dlstr1button of the random var1able Z YIX Solution. In th1s case G (z) 1s the relative area of the domaJn D (z) (F1g 8 36)

=

G(z}=! (arctanz+ ~-)

=

I

G# (z) ~ (n (1 + z2 )}-1 (Cauchyfs dtstrihutton) \vhence g (z) 8.37 .. Erlang's dtstributton af order 2 Form a convolution of t\\O exponentral dtstrtbutlons wtth parameter A1 1 e find the dtstrtbutJOD of the sum of t\vo Independent random 'arJables X 1 and X 2 ~h1cb have pr oba btlJ t y denst ties

It (xl} = Ae-

1

~1

(.xl > 0),

I2 (.r2)-== Ae-ut

(x1 > 0)

Solution By the general formula (8 0 5) for the convolution of d1str1buttons '' e have Co

g (z) =

J fl (xt)fz (z-

Xt) d:tt.

Ch. 8. Distri bution s of Funct ions of Rando m Variables

23:5

Since the functions f1 and / 2 are zero for negat ive value s of the arguments, the integ ral assumes the form

If

~

%

g (z) =

f

1 (x 1) 2 (z- x 1)

dx1 = '),} ) e-1..x1e-Mz-x1> dx 1 = A,Zze-1..z (z > 0)~ 0

0

(8.37} The distri butio n with a proba bility densi ty (8.37) is know n as Erlang'sdistribution of order 2. It origin ates as follows. Assume that there is an elementary flow of event s with inten sity A. on the t-axis , and only every other point (event) is retain ed in this flow, the interm ediat e point s being deleted. Then the interv al betwe en adjac ent event s in the so rarefied flow has an Erlan g order 2 distri butio n. 8.38. Erlang's distribution of order n. Form a comp ositio n of n expon ential distri butio ns with param eter A., i.e. the distri butio n of the sum of n:

X A.___ __ _~nth point II

e; :y.=.! x

•

l 0

•

0

0

Xi 0

(a)

••

It;

v----/'-: :.1>---

Xn

"'"'

x+dx

(6) Fig. 8.38

i~dependent rando m varia bles X 1 , X 2 , • • • , Xn which have an expon ential distri butio n with param eter 'A. Solution. We could solve the probl em by conse cutiv ely finding the~on:olutions of two (see Probl em 8.37), three , etc. distri butio ns, but it IS Simpler to solve the probl em proce eding from the eleme ntary flow, retaining every nth point in it and deleti ng the interm ediat e point s n

{Fig. 8.38a). X = ~ X il where X f is a rando m varia ble which has an i=l

exponential distri butio n. We find the proba bility densi ty fn (x) of the ran.do.m varia ble X, first finding the eleme nt of proba bility fn (x) dx. Th1s 1s the proba bility that the rando m varia ble X falls in the eleme ndx). tary interv al (x, x For X to fall in that interv al, it is necessary that exact ly n _ 1 events fall in the interv al x and one event in the interv al dx (Fia. 8.38b) !he proba bility of this occurrence is In (x) dx = P 11 _ 1A. dx, where p · IS the proba bility that n - 1 event s fall in the interv al x. But the nu~: her of events of the elementar~' flo·w fallin g on the interv al x has a Pois-

+

23-t

Appl~ed Probltm.s ~n Probablltly ThefJry

~on dtstr•button '' tth parameler a =

In

fn 1M d \~\.t 1h'U\. \-0-r..

'Ax., and hence

(1.%')11•1 (x) dx = (n -t)l ~

(A.r)"-1 (n-t)l

(x) =

Ae-lx d.x,

e-u

""\t \~ \)t~lYnh \1\ t y d~-ns. \.t y

(8 38 t}

(X> 0}.

{& 3S 1) \~ k n(\w n a~ Er l 0}.

This express1on can be reduced to the tabulated functton R (m 1 a) {see Appendtx 2)

Gn (x) = i - R {n - 1, A.x) 8 39 Erlangts generalized d~str1bUt1on Form the convolutlon of two exponent1al dtstrJbuttons \Vlth dtfferent parameters / 1 (x1) = A1e-Atsl (x1 > 0), / 2 (z2) = A2e-At:.:s (:r2 > O) .

x,.

Solution \V'e designate X= XI + x2, ,vhere Xt and have the dtstributtons f 1.. (x t.l t f" (x,.) In accordance '\ Jth the general formula for a convolution of distrtbuttons OQ

g(x)

J

=

J fd:rt) fz (x-rl) d.t1

But 1n th1s case both dtslrlbUtions are nonzero only for a pos1ttve value -of the argument. and th1s means that J1 (x1) = 0 for x1 < 0 and / 2 (x - x 1 ) ==: 0 for x 1 > x For x > 0

Ch. 8. Distribution s of Functions of Random Variables

bution: g (x) = A.2xe-A:c

(x

>

235

0).

Remark. We can prove by induction that the distribution of the sum of n independent random variables X 1 , • • • , Xn, which have exponential distribution s with different parameters At• ... , An, i.e. Erlang's generalized distribution of order n has a probability density n n e -Ajx (x > 0). '),i ~ -n~--gn (x)=(-1)n -l

IT

IT u ) = P{ X >u t Y> u} ThlS 1s the pro ba btl tty tha t the ran do m po tnt (X Y) falls 1n the domain D (u) hat che d 1n F1g 8 41 It IS ev ide nt tha t 1- G (u) =1 - F (u, oo )F ( oo, u) + F (u,. u), wh enc e G (u) == F (u, oo) + F (eo . u) F (u. u) = F 1 (u) + Ft (u) - F (u, u) D1fierent1at•ng w1th respect to

Ch. 8. Distribution s of Functions of Random 'Variables

237

u, we have (see Problem 8.40) u

u

Jf(x, u)dx.

g(u)=/1 (u)+/2 (u)-) f(u, y)dy-

-co

-co

When the variables X and Y are mutually independe nt, we have g (u) = / 1 (u) [1 -

F2 (u)]

+ /2 (u) [1 -

F1 (u)] .

.If the random variables X and Y are mutually independe nt and have the same distributio n, then f 1 (x) = f 2 (x)= j (x) and g (u) = 2f (u) [1 - F (u)l. !! 8.42. The distribution of the maximal and the minimal of several random variables. Given n independent random variables X 11 X 2 , • • • , Xn distributed with probabilit y densities / 1 (x1), u density y probabilit the find } 2 {x 2), • • • , fn (xn), function of the maximal of them Z = 0 them of minimal the and max{X 1 , X 2 , • • • , Xn} U = min{X 1 , X 2 , • • • • • • , Xn}, i.e. of the ranFig. 8.41

0)

e the prohabtl•ty denstty of the distance from the nearest btt to the centre of scattering has the same form as that for each of them the only condttlon being that the parameter a 1s decreased ']13 t1me.s, 1 e replaced by o = cr/3 8 45 F1nd the dtstrtbut1on g tL (u) of the m1ntmal of t\\ o 1ndependent random 'ar1abl~s T1 and T 2 wh1ch ha\ e an exponential dtstrJhuttons 1

ft. (tt) = A. 1e-Al't

(t 1 > O)t

f 2 (t2) === A2e-lsit

(t 2 > 0)

Solution On the basis of the solutton of Problem 8 42

+/

Cu (u) =It (u) 11- F 2 (u)J 2 (u) [1- F 1 (u)J = llo-ltue-11U.+ Aae-)..~ue-l(u =(.AI+ A~) e-(A-t+"-I)U 1

-e the cl1str1but1on of the mtntmal of tv, o 1ndependent tandGm va.r1

abies "\\}nch have an cxponentu1l distributions IS also an exponential d1strihut1on 'vhose parameter ts equal to the sum of the parameters of the Initial distribUtions \Ve can arr1' e at th1s conclusion much easter 1f we use the concept of a flo'v o£ events Assume that there 1s an elementaty flo'v wxth 1nten s1ty A.1 on the t 1 a.xts and an elementary flow" 1th 1ntenstty At on the t 2 a.x1s \Ve br1ng these two flows together on tl1o t a:xts, 1 e we super-

Ch. 8. Distributions of Functions of Random Variables

23\Jl

impose them. It is easy to verify that the result of the superposition of two elementary flows is also elementary (the properties of stationariness, ordinariness and the absence of aftereffects are retained). The intensity of the combined flow is lv1 + A2 • The distribution of the distance from a given point to the nearest event of the flow is exponenA2 • tial with parameter lv1 The same is evidently true for the distribution of the minimal of any number of n independent random variables which have exponential distributions with parameters lv1 , lv 2 , • • • , An. It is exponential with

+

n

parameter

~ lv 1• i=1

8.46. From the conditions of the preceding problem find the distribution gz (z) of the maximal of the variables T 1 , T 2 • Solution.

gz (z) = ldz) F 2 (z) + 12 (z) F 1 (z) = .A1e-'-•z [1-e-'·2z] +A 2e-?.2z P- e-'-• 1 } = A,1e-'-•z A2e-'-2z _ (A, 1 A, 2) e- 0).

+

+

This is not an exponential distribution. For /, 1 =A,= A

gz (z) = 2/ve-'-z (1- e-z'-)

(z > 0) •

-

. 8.47*. A random variable X which has a probability density I (x) 1s subjected to n independent trials. The results of the trials are arranged in increasing order. A series of random variables Z1 , Z 2 , • •• , Z 10 • • • , Zn results. Consider the kth of them, Zk. Find its distribution function G11 (z) and probability density function gk (z). Solution. Gn (z) = P {Zn < z }. For the kth (in an increasing order} of the random variables Z1 , z~, ... , Z 111 ••• , Zn to be smaller than z ' it is necessary that at least k- of them be smaller than z: n

Gk (z) = Lj Pm, m=k wh~re Pm is the probability that exactly m of the values of the random v_a~Iab}e X in n trials are smaller than z. By the theorem on the repetitiOn of trials whence n

G,(z)=

where F (z) =

'

J1(x) dx.

Lj

m=k

C~[F(z)]m[1-F(z)]n-m,

.:240

Ap plled

Pra ble m.1 ln. Pro babtlitgo T lc.e.ory

The probabtltty denstty gA (z) can be found by differenttat1ng thts e~press1on and tak1ng 1nto account that

C"'

nm=

nf

(m-tH (n-m}l

~-1

=n n.-t' (m~ 10-6 X 104 -·-;:) )< 10-!! 8.12 X 10-4 -< 102

+

+

+

246

Applied Probl~ms 1n Ptobabilltu Thtory

tO a r< 106 = 22 3 I0- 2 rouble~~ 1 e the Ininnnum prtce of a t1clet ts about 23 J... o pee J. . s (2) JJ = (0 30 - U 2~3) A tOt = 77 x 10'3 roubles (3) Tho total '' tnn1ngs ~Y '' luch the organ1zers of the lottery must l DOO OQO

pal Is the sunt of the ''Innings of the partictpan t~ X=

~

i=i

X1,

'' h.erL X~, l~ the amount ''on h) the a.th partic1pant '' e assume that tht? p1rt1ctpants 1nark. thc1r numbers tndependentl~ .so that the 'artJ.bles .\. t (i == t ~ 2 , t 000 000) are 1ndependent '' e kno" from thP centrJl lun1t theorem that the sum of a sufflcientl} 1arge n u mher of 1n de pendent rando rn ' art a bles '' h IC h have the same dtstrlhUtlOJl has apprO'\.UUale}) a norma} distrJbUtlOn \\~e mUSt find out "hether the ntunber of terms n = 1 000 000 rs suffictent 1n this. case for the 'at1ahle A to be constd.ered normall~ dtslrtbuted \\ e find the me1n 'al ue nl"' and the mean square devtat1on ax of the r"tndom '~rtable ..\ Fot anl l = 1 2 , 1 000 000 '' e get m%i == 22 3 X 10 .. = 0 221 r.t 2 [ A , ] -== 2 25 10 -~ 8 12 ...L D 07 X 10~ 2 28 / 10-' - 2 3~ 10" v nrxt == 2 38 ' tOt ~ 0 22 2 ;:::::: 2 38 X to~ l{enc-e

+

nz - 106 .)..

(J 'I!

=

10 5 l/ 2 38 ;:::;: t 54

t 05 •

\Ve kno'' th 1t for the randont ''lrJahle A, '\ htch has a normal diS\r1bul1ont the Tang-e of pract u~all\ poss1ble '"alues 1s m'r ± 3a .:r In our Caii:e the lOl\ er bound of the po~~ 1ble ' al ues of the randotn \ ar1able X, pro'\ 1ded that 1 t " as norn1all) dIS tr tbu ted, \\ oul d be IJlx - 3a x = ~2 39 x 10~ The negatt' e 'alue of thts hound s1gn1fies that\\ e cannot consider the randon1 'artable A. to be normallJ. dtstrtbuted Since there cannot be negat1' e- '' 1Unkngs to

8 6! •. F1nd tho Jun1 t ltnl ~ a~ e '\ ,\here a IS a pos1 t1 ve Integer a-.oo

a

Solutwn.

L

:~ e-'

1s

m.~o

m

the probabthty that the rando1n ,anablt

rr=O

X '\ h1ch has a Po1'~·~on dl~tr\.button, ,, tll not e\.c-eed i.ts mean \ a1ue a But as the param~ter a tends to tnfint t}, the Po1 c:son dts trtbu t1on approache~ the norntal cltstrihutlon For the normal d1s .. trtbu t1on, the probtbt h t' that the randon1 'artable "1ll not e'1(ceed a

1ts mean

value 1s .112, and

~

am ~ ~ thrs means that l1m .LJ · , e- = 2 a:-oo 0 m m~

8 62. Indo pendent randoin \arJables X 1 , X 2 , .. have the same cx.ponent1al distrtbutton '' tth parameter 1- f (.r) = A.e-,._:r

Ch. 8. Distribution s of Functions of Random Variables

y

241

We consider the sum of a random number of these variables Z =

LJ Xi, where the random variable Y has a geometric distributio n i=!

}Jeginning with unity:

Pn = P{Y

=

(0

n} = pqn-l

< p
O),

k=O

i.e. the random variable Z also has an exponenti al distributio n, but with parameter 'Ap. Consequen tly

mz = 1/('Ap),

Varz = 1/('A2p 2 ).

8.63. The distributio n of the sum of a random, number of random terms. We consider the sum o( a random number of random terms Z = !I

~ X i• where X 1 ,

X2,

•••

is a sequence of independe nt random variables

i== 1

:'·hich have the same distributio n with density f (x); Y is a positive mteger-val ued variable, independe nt of them, which has a distributio n P{Y = n} = Pn (n = 1, 2, ... , N). Find the distributio n and the numerical characteri stics of the random variable Z. Solution. We suppose that the random variable Y assumes the value = n (n = 1, 2, ... , N) with probabilit y Pn- On this hypothesis n

~ Xi.

z

We designate the probabilit y density of the sum of n independ-

i==t

en~ random variables X 1 , X 2 , • • • , Xn, which have the same distribUhon, as j (x). \~7 e can find the densities f< 11 >(x) successive ly: we first 2 find f< > (x), i.e. the conYolutio n of two similar distributio ns f (x) and f (x), then} j< 3> (x), the conYolutio n of f< 2> (x) and f (x), and so on.

2..8

Applled Problems In ProbabllUy Theory

By the total probabtltty formula, the the random 'ar1able Z Is

probab1ltt~

dens1ty

q>

(z) of

N

fP (z) = }J

n-1

\~" e 10

fP") (z) P n

(8 63)

found \he numer1c.a 1 char ac ter1st1cs uf the random variable Z

Problem 7 64 ~~ fZ] =

m~myt

Var [ZI = Var~ m 11

+ nr! Var,,

"here m;ct m u, Var x, V .ar11 are mean 'al u~s and 'ar1a nees of the random 'artables X and Y 'Vhen certatn condtttons are fulfilled} '' e can a~sume, \vtth an accur acy sufficient for appltcat1ons that the random 'atiahle Z I.S normally d1strtbuted '~ tth the tnd1cated parameters l\J (Z]t Var [Z] Let us sho\v this \Vttbout v1olattng the general! ty of our reasoning, \Ve set mx = 0 for the sah.e of stmpllclt)' On the basts of the centrall1m1t theorem, ~e ca.n assume for terms ha' 1ng the same distrihutton (see Sec 8 0) that for n > m11 - 3cru > 20 the dtstribution dens1ty j(Ti) (z) tn formula (8 63) IS normal w1th parameters m 0.

for

(k!)2

k=O

We can express this density as follows:

V l.za)

cp (z) = 'A.e-J..z-aJ0 (2

for

z > 0,

where eo (x/2)~k

10 (x)= ~

(k!)2

k=O

is a modified cylindrica l Bessel function. Then, from Problem 8.66, we have M [Z] = mxmy = (a+ 1)/'A., Var [Z] = Varx my+ mi: Vary t)/1.2 • a//..2 = (2a 1)/1..2 = (a

+

+

+

8.65. Consider a system of random variables Xi (i related to a discrete random variable Y thus:

Xt=

1, 0,

=

1, 2, ..• , n)

if i~Y, if i>Y.

F.The distributio n function F (y) of the random Yariable Y is known. t~ the .di~tribution of each random variable Xi and the numerical c wract:nst1 cs of the system of random variables (X 1 , X 2 , • • • , X 11 ). Solution. The ordered series of the random variable Xi has the form o 1 1 Xi: P {Y < i} P {Y;;;;, i} ' (i='l, 2 ' ... ' n ) .

I

2.50

Appllt!d Problems tn Probability Theory

Srnce P{Y < ~J = F (z), the ordered series has the form

o r X t.

F (f)

It -

1 F (f)

t

"hence m:c 1 = 1- F (l),. Var.x 1 = F (L) [1- F (i) I Let us find the covartances of the random variables X f and XJt for \Vhtch purpo~e we determ1ne 11 [X ~X tl For 1 < 1 the product X1X1 can assume only two values VIZ. 1 tf X 1 = Jt and 0 tf X 1 = 0 Con sequently lt1tX 1X1)=mx1 =1-F(j) (i 1) F1nd the dtstrtbUtlon and the numerical charactertsttcs of the number X of .. ~fatlures'" 1n 'vbtch the e\ent A does not occur Solution~ \\ore find the probab1I1 ty that the random variable X assumes a \ alue k For that e'\ ent to occuT, 1t ts necessar) that the total number of trials should be equal to n -l- f... (J... outcomes are fallures and n out comes are successes) By the hypothesis,. the last trtal must be successful and 1n the precedtng n + k ---- i tr1als n- 1 successes and k fa1lures must be d1str1buted at random The probabtltty of th1s occurrence

P{X=k}=C!+Iit-tPntf,

IS

\Vhere q=1-p(k=O, 1, ~-)

The dtstribution obtatned Is a natural generalization of a geometriC d1stTibut1on, \\e shall -call1t a generalized geometrlc dtstrlbutl.on of order n. It lS a convolutron of n geornetrtc dtstrtbuttons 'vtth the same para n

meter p X = ~ X,, 'vhere each random var1able X, has a geometr1e s:::o;;z.f

d1str tbutJo n p { y. = h.} = pq~

(k

=

0' 1 t ~

ll

• )

41

Indeed, the total number of fatlures 1s the sum of (1} the number of fatlures t1ll the first occurrence of the event A., (2) the number of fa1Iures from the first to the second occurrence of the event A, and so on Hence \\e obta.ln the numer1cal charactei"tsttcs of the var1able X mx = nq!p, V ar;.: ::=:::- nq!p 2 8.,67. The hypothesiS Is the same as In Problem 8 66, except that the

random variable Y

the total number of trials (both successful and unsuccessful), made ttll the nth occurrence of the event A. F1nd the d1strtbut1on and the numerical characteristlcs of the random vaxtable Y. ts

251

Ch. 8. Distributions of Functions of Random Variables

Solution. Y = X + n, where X is the random variable in the preceding problem. Hence

p {Y = k}= p {X= 7,;-n} = c~:ypnqk-71 = C/::tpnqk-n

(k=n, n+1, ... ). The numerical characteristics of the variable Y are my= mx

+n

= nq/p

+ n =nip,

Varv = Varx = nqlp

2

•

8.68. There is a random variable Y which has an exponential distribution with parameter A.: f (y) = A.e-t..Y (y > 0). For a given value of the variable Y = y a random variable X has a Poisson distribution with parameter y:

P {X = k 1Y = y} = ~; e-Y

(k = 0, 1, 2, ... ) .

Find the marginal distribution of the random variable X. Solution. The total probability of the event X= k is 00

P {X= k} = \

J

00

_,_Y.,.-k e-YA,e-1.y

k!

dy = ....,..!.~ \ yke-< 1 +1-)y dy k!

0

0

-

J

f. kl (1+~)-(k+ilk! "' - (1

'},

+ /,)k+l

(k

+

=

0 1 2 '

'

)

' ....

+

If we introduce the designations A./(1 A.) = p, 1/(1 A.) = q = 1- p, we obtain P{X = k} = pqk (k = 0, 1, 2, ... ), i.e. the random Yariable X has a geometric distribution with parameter p = A./(1 A.).

+

8.69. A Geiger-Muller counter is mounted in a spaceship to find the number of particles falling in it during a random time interval T which ~las an exponential distribution with parameter ~t. Particles arrive 111 a Poisson flow with intensity/,; each particle is recorded by the counter with probability p. A random variable X is the number of recorded particles. Find its distribution and the characteristics mx and Varx. Solution. We assume that T = t and find the conditional probability that X = m (m = 0, 1, 2, ... ). P{X=mlt}= 0 ts equal to the sum of probab ilities that X and Y Will a~sume t" o 'a lues differin g by h. (l be 1ng greater than or equal to lfl')

The probab 1ltty that Z w1ll assume a nega\1 \ e value --k IS

For the random. var1ah le- U

m=O CIO

p {U = k} ~ ~ (ab)m (ah+bli} m:o=O

e- 0)

These ptohab1l1t1es can be '' r1tten b:y means. of a modtfie d eyl1ndr1ctd Bessel functto ns of the 1st ktnd 00

z ) "+2771 (k (2 1 " l h {x ) -J -- LJ m1 C!c+m)l - -k { X) ~

**) 0 ) ~. 2t 1t ==: t

m~o

.-.} lf the un1t of measure ments (the value ol the dlVJSlOn of the Instrument) IS small as compare d to the range of the poss1hle values of the random. var1ahle X' then mil~ m. Varv ~ Var~ ••} The taEies of cyhndri cal Bessel functiOns of the fst k1nd can he found 1n te[~rence hooks

Ch. 8. Distribution s of Functions of Random Variables

255

In this case

P{Z=k}= ln(2Vab)

(fr

12

(k=O, ±1, ±2, ... ),

e-

P {U = 0} = 1 0 (2 Vab) e- 0), f 2 (y) = ~e-IIY (y > 0). 00

Solution. g(z) =

11 (x) 12 (x-z) dx; 11 (x)

\ •

is nonzero for x> 0;

-oo

f2(x-z) is nonzero for x-z> 0.

(a) z> 0, :z:

Consequen ti y

g(z) =

+ ~tt l.f.te~:z: (/.. + ~tt Af.te-i.z (1..

1

1

The parameters of this distributio n are 1

1

').

!l

m.. = - - - = .

'·!l

'

for for

z > 0, z < 0.

256

1pplled Problems tn Probability Thtorg

The d1stribut1o n cur' e has the fornt shown 1n Fig 8 75a For l = JL -} e "' I (Fig 8 7ab) Tlus distrlbUttO n IS I~ no\\ n as a "e get g (z) L., pia ce d tstrJbu t 1on 8 76 A contolution of tuo dtstrlbut1ons of tu o nonnegative random tar1ables G1' en t'\ o nonnegat 1' e random 'artables .A and Y '\ tth pro h 1b1llt} densities / 1 (x) (x > 0) and f 2 {y) (y > 0) form the conloluhon of the tr d tstrtbu t1ons

;tp

g(z}

A/2

A.,u

z

0 (aJ

g(t)

z

0

(!J)

Fur 8 7;,

X _.. } \\ e find the probabtlit } denslt) g (z) Soluhon ~ssunte Z of the r1.ndom '1r1able Z B) the general formula (8 0 5) 100

g (z)-

Jfdx) !'I. (z-x) dx ~

TAh1ng 1nto account th'lt / 1 (x) x > z \\e obta1n

=

0 for x


O)

0

or

) crx~y ~ exp [ - ( 2~~ + 2~~- ~t + 2Z::~) ]au. ,.~ 11 2

g(z)= (

•

(8.77)

0

This integral can be expressed in terms of the error function 1). Solution. The total length T of m messages plus the intervals between them is a random variable which has a generalized Erlang's distribution of order 2m- 1 with parameters 1

A., A., ••• , A.; m times

p,, p,, •.. , fl· (m-1)

times

. The probability that no less than m messages will be sent during the hme t is none other than the distribution function of the random variable T: P{no less than m messages during the timet} = P{T < t} = G (t). The random variable T is the sum of two independent random variT 2 , where T1 has Erlang's distribution of order m n~les: T = T1 With parameter A.:

+

gt ( t) =

(/.t)m-1 (m-i)l

-], t

e ·

(t > 0);

258

Applitd. Probt!mt tn Probabtllty Theory

T 2 has Erlang s distribUtion of order m - 1 \vtth parameter f'. g (t} = (l.u)m.-t e-~e (m-2)J

!

ca~e A>~

Consequently, for the

(t > 0) ..

''e have

. {tl g ('t) g (t- T) d-e== e-J.t {nt~ (m.-1)! J t

o:)

g ( t) =

1

(~- ~)]ft1.-l

(A,'t)m.-l

r

2

2}~

e-J.!{f-1')

dt

0

0

m .... 2 Am.-t!J.m-5! ~ C' tm.-2-i ( -1)1 (m- t+()t (m -1)! (m-2)1 LJ rn- 2 (A- J.L)m+l

-

1;;;;;;;2 0

X [

1-

n -1+-\

~

H"--:r> tl" e-tl-1.1>1 Je-111

(t

> 0).

ll=O

For the case tJ. >A \Ve ha\ e I

g (t)

=j

C1

m-1

"'m-tpm-2

(t- 't) C2 (-r) d"t" =

(m- 1)1 (m.-2)J

~ C~-ltm-t-1 ( -1)' j:;;;;;;~J

0

(m

+ f) I

m-1+~

X (~-h)m+l+t

[ i

""

-

LJ

k=-0

For the caso A= fl we have g (t) =A

(A')tm-2 (Zm- 2)l e-At

(t

>OJ.

8 79 . A pendulum makes free continuous oscillations, the angle tp (F1g 8 79) varying '\\ 1th the time t accord 1ng to the harmonic law

oo, n/c = '). = const. ~n this case p -+ 0 but the average number of points falling on the mterval (a, b) is constant: l\1 [Y] = /.. (b - a) = const. We know (see Sec. 4.0) that in this case the limiting distributio n of the random variable Y is a Poisson distributio n with parameter d = ').. (b - a):

P{Y=k} =

dh kl

-d

e

(k=O, 1, 2, ... ).

T~le number of events in a stationary Poisson flow with intensity ').,, falling on the interval (a, b), has the same distributio n. Thus we can make the following conclusion : a stationary flow of eYer~ts with intensity ').. can be considered to be a limit case of a coll:c~wn of n independe nt random points on the interval (0, c), each of ''Inch has a uniform distributio n on that interval, provided that n - co, c ->- oo, but n/c = /.. = const. We shall need this model of an e1ementary (stationary Poisson) flow later on. 17•

CHAPTER 9

Random Functions

9 0 A functLon X (t)l! called a random function 1f its value 1s a random \arJab" for any argumant t B.xamples or random !unctlorn V (t} the supply \oltage! a computer dppend1ng on t1me t T (h) the ~ur temperature at a gtven polnt ; a g1 ven moment depend 1ng on the al h tude h abo\ e the ground Q (t) the numb1 of hmcs a computer fa.11s durtng the t1me from 0 to t The- concept of a random funct1on 1s a generahz1t1on of the concept of a randm var1able S1nce w·e can constdcr a random variable X to be a function of an eleme1 tary event (il (s~c Sec ' 0) X

{Xk (t)}

k=i

k=i

(i.e. the sum can he transformed term-by-term), and if (2)

£~0)

c£~0>

{eX (t)} =

{X (t)},

(i.e. a factor, which is independent of the parameter t with respect to which the transformation is carried out, can be put before the transformation sign). A transformation Lt is called a nonhomogeneous linear transformation if

Lt {X (t)} = Li 0 ) {X (t)}

+ qJ (t),

where QJ (t) is a nonrandom function. f If a random f11nction Y (t) is related to a random function X (t) by a linear transorrnation Y (t) = Lt {X (t) }, then its mean value my (t) results from mx (t) upon t he same linear transformation:

my (t) = Lt {mx (t)},

(9.0. 7)

and. to find the correlation f~nction Ry (t, t'), the function Rx (t, t') must he twice subJected to the correspondwg homogeneous linear transformation, once with respect to t and once with respect to t':

(9.0.8) .

IS

The cros~correlation function Rxy (t, t') of two random functions X (t) and Y (t) a functiOn . e

o

Rxy(t, t')=l\l[X(t) 1"(t')).

It follows from the definition of a crosscorrelation function that Rxy (t, t')

=

Ryx (t', t).

'

(9.0.9)

25 \

Ap pl ted Pr ob lem s ln Pr ob ab tll ty

Th e normed is a funct1on

croyscorr~latlon

Tx y

The ra nd om

The~ry

ju nc t ton of t\\ o ra nd om funct1ona X (t) and Y (t)

_ R xy (tt t ) _

(t ,. t ) -

Ox (t)

ou (t

)

Rxu (t~

-

t )

Y \ arx (t) Va r11 (t

)

(9 0 10)

•

funct1on~

X (t) .and Y (t) ar e un to "e la te d 1f R~JJ (l. t'} $ 0 If Z (t) = X {t) Y (t}, th en mz {t) = nlx (t) m9 (t) , R: (t t') = Rx (t,. t') R y (t'" t'') Rx y (t, t ) Rx11 (t, t') . If t\\ o ra nd om func t1ons X (t) an d Y (t) ar e unco rrelated, then

+

+

R~ (t~ t') =

If

+

+

R:c (t. t')

+

+ R t~ (t.

t')

(9 0 if)

n

Z (t) = ~ X k ( t) ~

(9 0 t2)

It- t

, Xn (t) ar e un co rre lat ed ra nd om fu nc ho ns , the n n

mz (t) = ~ mx (t)!o k=-1

k

R~ (t!o t ) =

n

2J

Rx (t, tt)

A= t

k

\Vhen tran.sformtng ra nd om fu nc tio ns 1t l! of te u co nv en ten t to 'M'tte them lD a co mp lex form A co mp lex ra nd om var1abl~ ts a ra nd om fu nc tto n of the form Z (t)

=

X {t)

+ tY (t).

(9 0 13)

wh er e X (t) an d Y {t} ar e re al ra nd om fu nc tto ns d t 15 a un1t unagiJJ.ary number. Th e m ea n value th e co rre lat iO n funct1on anan d var1anee of a complex random funct1on are defined th us m:r (t}

= mx (t) + im11 (t) ,

Rz (tt

= \f (X (t} X (f' )], 0

t )

0

{9 0 {4)

wh er e th e ba r ov er th e le tte rs de no tes a eo m pl ex co nJ ug ate qu an ttt y and Varz (t) R~ (t, t ) = AI {I (t) 111. (9 0 t5) Be fo re constder1ng co mp lex ra nd om va r1 ab les define va rta nc e as th e m ea n va lu e of th e .square an d fu nc tio ns it 1s necessary ta of th e modulus,. an d covartance a~ th e m ea n va lu e of th e pr od uc t of a centrerl ra nd om var1ance by the complex conJU· ga te of an ot he r ce nt re d var1able• * Th e canontcal decomposltton of a random va r1 ab le X {t) fs 1ts representaboD m th e form

=

X (t) = m~ (t)

X

m

+~

Vkq'!Jt.

(t)••>,

(9 0 16}

A=-:1

wh er e Y~r (k = :t, 2 , m) ar e ce nt re d un co rre la te d ra nd om varzable .s w1tb va ria nc es VarA {k = 1 2, t .m) an d 'Fk. (t) (.l = 1 t 2, •t m) ar e no nr an do m funct1ons Ra nd om var1ables Vk (k = t,. 2. , m) ar e kn ow n as co eff itlt nf8 an d fu nc tio ns Cf'tt (t) {k = 1 1 2. .. ., m) as co ordtnate fu nc tio

compositiOn

ns of a c.anon1cal de ·

•1 In \v ha t follows we sh al l1 nd tc at e th e co m pl ex na tu re of a ra nd om fu ne tto l !.very t1me, tf 1t lS uo t sp ec ifi ed , we Bhall co ns id a ra nd om fu nc tto n to he * •> In pa rtt cu la rt th e su m ca n be ex ten de d toer an in fin ite (c ou nt ab le) num er lf ter ms

rh

Ch: 9. Random 'Functions

.i'

265

If the random function X (t) admits of a canonical decomposition (9.0.16) in a real form, then the correlation function Rx (t, t') is expressed by the sum m

2j

Rx (t, t')=

Vark cpk (t) cpk (t'),

(9.0.17)

h=1

which is known as a canonical decomposition of a correlation function. If a random function X (t) admits of a canonical decomposition (9.0.16) in a complex form, then the canonical decomposition of the correlation function has the form m

2j

Rx (t, t') =

Vark (jlk (t) cpk (t'l,

(9.0.18)

k=i

where a bar over the letters denotes a complex conjugate quantity. The possibility of a canonical decomposition of a correlation function in the form (9.0.17) or (9.0.18) implies a representability of a random function X (t) in the canonical form (9.0.16), where the random variables Vk (k = 1, 2, ... , m) have variances Vark (k = 1, 2, ... , m). A linear transformation of a random function X (t) defined by the canonical decomposition (9.0.16) results in a random function Y (t) = Lt {X (t)} also in a canonical form, i.e. m

Y(t)=my(tl+

2j

VkWk (t),

(9.0.19)

k=i

where (9.0.20} That is, in a linear transformation of a random function, defined by a canonical decomposition, its mean value is subjected to the same linear transformation and the coordinak functions are subjected to the corresponding homogeneous linear transformation. . A stationary*> random function X (t) is a random function whose mean value Is constant, mx = const, and whose correlation function only depends on the difference between its arguments: Rx (t, t'\ = Rx (T), where T = t' - t. From t~e symmetry of the correlation function Rx (t, t') it follows that Rx (T) = Rx (-t), I.e. the correlation function of a stationary random function is an even function of the argument -r. The variance of a stationary random function is constant: Varx

=

Rx (t, t') = Rx (0)

=

canst.

(9.0.21)

The correlation function of a stationary random function possesses the property

I Rx (•) I ~

Varx.

(9.0.22)

The normed correlation function of a stationary random function is

Px ('t)

=

Rx (T)IVarx = Rx (T)!Rx (0).

(9.0.23)

The canonical decomposition of a stationary random function is 00

X(t)=mx+

'p·Y~ere.

:2}

(Ukcosrokt+Vksinrokt),

(9.0.24}

k=O

rk

(k = 0, 1, ... ) are qmtred uncorrelated random variables with nirw1se equal variances Yar [ Ukl = Var [ Vh] = V ar,.. Uh,

A Hepresentation (9.0.24) is known as the spectral decomposition of a function. spectral decomposition of a stationary random function is a~sociated with a series

*>

Stationary in a broad sense.

266

.A.pp lfed l'robltml In Proba btltty Thto ry

expatl.slon of IU correla. bon £unction· QQ

R :t ('t) =

2J Va r" cos IDJt 't 1i 0

(9 0 25)

;1;;;;11"

\Vhence 00

}2 \ ar);E.

Var %-=

~t~o

Set hog w0 = 0 "' e can rewTlte the spectral decomposition {9 0 24) of a .statzonary r"ndom !unct1on in a complex Iormt 1 e i ~ {t)

co

=m-1+

"'U

11~hct

L.i

i c.> Jt. I

~90

27)

J;~-1)11)

,.,. here

JJk=jl11-fV~ 2

'I'he- .rpectral d:nsUy ~f a s\ation3't'y Tandom funct1on X (t) 1s the l1mit of tM rauo oi the \ar1ance per a g1\ t'n 1ntprval of frequencies to the length nf the tnterval as the latter tends to zero The spPctral density Sj ((I.)) and the eorrelatzon lunct1on R:c (ri are related by Fourter s tra.nslormatlons n the real form tlus rfJat1on is ~

S" oo) = ~

JR.,

CIO

Rx {'t") =

{1') cos lil'f dt

JSac

({t)) CO.Hil1' d{,l

(9 0 28)

0

0

It follo\\ s from the last re latlon that

J

Oil!

\ arx= Rx (0)

=

Sx(c.J) dw

{9 0 29)

0

~;

In a tomplex [orm Fourier 5 transformahonst 'vbtcb relate the spectral df=n!ut.y (CJl.) and the correlation funchon Rx (T}, hale the form 00

j

CIO

Rx(t)e-""Tdt,

R~ {'f)=

j

s:(w)e'llYidw

(9 0 JO)

-oo

Appendtx 7 1s a table of relahons bet\\fen certa.1n correlatton functions and !pectral densl ttes Both S~ (oo) and S~ (C&l) are nonn~gattve real {unettons~ S~ (~) 1s an .even function defined. OU the interval (-oo +co) S:f. (ro) lS defined on the 10terval (0, +co), and on th.lS 1nwrval S~ {w} ~ {) 5S¥ (())) A normed sptctral dtnsfty 'x (m) [or sJ (w) tn a comples form] IS a spectral dell tlt} dlVlded b) the 'ar1ance of the random function X (t) .f,:t (fJl) = s X ((1))/Var~, s~ (~) = s; (~\Nar~ (9 0 3tl If a cros.scorrela tton funct1on R r. Y(t, tJ) of tVt o sta tJonary random functions X (t) and Y (t) JS a funchon of onll 't = t' - ~' then \\e say that they are station ar!J connected Tbe follo~.tnt .t~latlOWi t_hp.,n_ P:tJS.l bP~l\l.e..P.J:l.lhP.. cro!scorrelated function Rx!J (T) and the cross power spectral dtnstty S~v (w)t l\b1ch are defined by the Fou.r1e-r transform.at1on in a complex fol'Ifl: co

(!!)

t S *ru (oo} -~ 2:t

Jr

-OD

n- tW'f R .. J, (or) d.,... ~

.. . . ~

R" 11 { t") =

J

-~

e l ..,"~" S iv (Iii) dw

(9 0 32}

Ch. 9. Random Functions

267

If the normal stationary random functions X (t) and Y (t) are stationary connected, then the random function Z (t) = X (t) Y (t) is stationary with characteristics mz = m:-.·my

+ Rxy (0),

(9.0 .33)

00

S~ (w)=

S~ (w-v) Slj (v) dv+mxmy [S~y (w)Sux (w)]

) -oo

+i

00

S~y (w-v) Stx (v) dv+m~St (w)+m3S~ (w'.

(9.0.34)

-oo

In a special case, when Z (t) = X 2 (t), we have 7nz= m~+ Varx=m~+Rx (Ol,

(9.0 .35)

00

'

S~(w)=2 ~ S~(w-v)S~(v)dv+4mlS~(w).

(9.0.36)

-oo

White noise (or white noise in a wide sense) is a random function X (t), whose

two different (arbitrarily close) sections are uncorrelated and whose correlation function is proportiona l to the delta-functi on: (9.0.37) Rx (t, t') = G (t) o (t - t'). The variable G (t) is known as the intensity of white noise. The definitions and the properties of the delta-functi on are given in Appendix 6. Stationary white noise is white noise with constant intensity G (t) = G = const. The correlation function of stationary white noise has the form Rx (•) =

Go (•),

(9.0.38)

whence it follows that its spectral density is constant and equal to s~ (w) = G!2rr..

(9.0.39)

The variance of stationary white noise Varx = Go (0), i.e. is infinite. If a stationary random function X (t) arrives at the input of a stationary linear system L, then, some time later, the time being sufficient for the transition process to he damped out, the random function Y (t) at the output of the linear system will also he stationary. The spectral densities of the input and output functions are related as S*y (w)

=

S~ (w)

I cD (iw) 12,

(9.0.40)

where CD (iw) is the amplitude-f requency characterist ic of the linear system . .T~e stationary function X (f) is said to possess an ergodic property if its charac!m~tlcs [mx, Rx (-r)] can be defined as the correspondi ng t averages for one realIzatwn of sufficient length. The sufficient condition for ergodicity of a stationary ~ndom function (with respect to expectation ) is the tending of its correlation function to zero as -r-+ oo: lim Rx ('t) = 0. (9.0.41) "t-+00

F~r the random function X (t) to be ergodic with respect to variance Var it is sufficu.>nt that the random function Y (t) = X 2 (t) should possess a simila~.' (the same) property, i.e. lim Ry (T) = 0 as -r-+ oo.*l -t-oc

. .*>

For a random function to be ergodic with respect to a correlation function -r) possesses a simila; It IS necessary that the function Z (t, -r) = X (t) X (t property.

+

268

Ap pl ie d Pr ob ltm l ln Proboblllly Theory

If th e reahzaUon of a random functton Y (t) cros srs (u p" na rd s) a straJght lme "h1ch ts pa r 11leJ to th e t axJs an d Js ot the drstanee JS sa1d to ~ross the level a ("ee 1-Jg 9 0 5 1n '\\ hzeh a from 1t th en th e funct1on X (t) th e Je \ el cro~sr.nr;s ar e marked l:Jy crosses) ThP nu m be r of level cro'CI.~JDgs x(t) X d ur1ng th e tim e T ls a dJscretE- random 'A rla b] e 1f leove] cro~'O;rng~ ar e Infrequent, th en th~ var1a bl e ean be cons1d~red to ha Vf a Po1sson dt ~tr1 button Fo r a normal s ta tto na r) ra nd om ' an abl~ Y (t} th e av er ag e number of ero s "tngs of the level a pe r un Jt ttme 1s 0

t

1

Fi g 9 0 5 ~here

Aa- -- 2n: exp

{

-

(a -m t') J ) Cy

20'~

az

•

(9 0 42)

a11 ts the m ea n square de vt ah on of th e de nv at tv e of th e random functton y (t )=

~ y (t)

(9 0 43)

The average nu m be r of doV.ll?. ard cro~slng's of th e gl \e n le \e l 1s the ~arne

Probletns an d Exerclses 9 1 Cons1dcr1ng a no nr an do m funct1on of t1 m e cp (t) to be a special kt nd of r1 nd om funct1on ~\ (t) =

. The.random functio n X (t) is known as a Poisson process. Fmd the univar iate distrib ution of the Poisson process and its characteristics mx (t), Varx (t), Rx (t, t'), rx (t, t') as well as the charac teristic s of the stochas tic process Z (t) = X (t) - 'At. Solution. The distrib ution of the section of X (t) is a Poisson distribution with parame ter a = 'At and, therefo re, the probab ility that the random variabl e X (t) will assume the value m is express ed by the formula Prn = ('At) 111 (rn!)- 1 e-1.t (m, = 0, 1, 2, ... ). The mean value and varianc e of the random functio n X (t): mx (t) = Varx (t) = f..t. Let us find the correla tion functio n Rx (t, t'). Assume that t' > t ?nd consider the time interva l (0, t') (see Fig. 9.13b). We partitio n the lnterval into two portion s: from 0 to t and from t to t'. The numbe r of

•>

We assume that tho function X (t) is continuous on tho left.

18-0576

274

Applted Prnble1111 ln ProhtJbfllly

TA~ory

events O'er the" bole Interval {0, t') 1s equal to the sum of the numbers 1 of events on the Intervals (0, t) and (tt t )*) X (t') = X (t) Y (t - t) 'vbere Y (t' - t) 1s the number of events whtch occur In the Inter\al (t, t') Since the flow IS stationary,. the random function Y (l - t) has the same dJstributton as X (l) In a ddt t1on, accord1ng to the prop erttes of a Poisson flo,v of events, the random variables X (t) and Y (t - t) are uncorrelated \\ e have

+

41

= l\I 1x (t) x (t') 1= ni rX (t) (X (tl + Y (t' - t))l = 1\1 (X (t))'t] = Var:c: (t) == 'At. for t > t' "e get R x {t, t") = Af Thus Rx (t, t') =

R x (t, t')

S1mllarly min {At At } = A min {t, t' }, where min (t, t'} Is the smaller of the values t, t (If t = t "e can take e1ther t or t• as the m1n1mal value) Us1ng the notation of a untt funct1on 1 (x) (sea AppendiX 6) we can 'vrtte the correlatton function as

R :~: (t, t )

= At 1 (t' -

+ 'At' 1 (t -

t'). F1g 9 13c 1llustrates the surface R:.: (t, t') In the quadrant t > 0 and t > 0 the surface Rx (t, t') conststs of two planes whtch pass through the t and t' axes respectively and tntersect along the ltne 0-Var~tt the appltcates of ~bose po1nts are equal to the variance Var~ =At t)

The normed c.orrel at 1on funet1on

rx(t t')=

Rx(t t) YVarx (t) Var.1: (t }

=~/It(t V

t

-t)+ .. /I:I{t-t)

V

t

The surface r:r (t t ) IS sho\vn In Ftg 9 i3d A Potsson process IS a process w1th 1ndependent Increments .s1nce tts Increment at any tnterval 1s the number of events occurr1ng on that 1n terval, and for .an elementary flow the numbers of events falhng on nonover1apptng Inter\ als are 1nde pendent The process Z (t) = X (t) - At (see Fig 9 13e) results from a non homogeneous l1near transformation of the stochastic process X (t) Conseq~ently m: (t) == m:;r; (t) - 'At = O, Var: (t) = Varx (t) At, R: (t~ t ) = R:x (t, t ) == A mtn (t t 1 ) Since the corresponding homoge nco us ltnear transforma t1o n does not change the correla tton funct1on of the process X (t) 9 14 A stochastic process X (t) occurs as follows There IS a station... ary (elementary) Poisson flow of events w1th 1ntens1ty A on the taxiS The random functton X (t) alternately assumes values +1 and -1 Upon the occurrence of an event.,. 1t changes 1ts value JUmpwise from +1 to ~1. or Vlte 'ersa (F1g 9 1qa) Intttally the random funct1on X (t) may be e1ther +1 with probab1l1ty 1/2 or -1 'v1th the same prob

=

1

We neglect the po~.s1h1hty that an event m1ght occur exactly at the moment l s1nc'& the, pto ba h1l it y t b is oceurtenc e 1 s zero *)

oo

Ch. 9. Random Functions

275

ability. Find the characteri stics mx (t), Varx (t) and Rx (t, t') of the random function X (t). Solution. Tlie section of the random function X (t) has a distributio n represented by the ordered series

X (t)-

-11 +1 .

'

0.5, 0.5

Indeed, since the moments of sign changes are not connected with the value of the random function, there is no reason for either +1 or -1 0.5 = 0, Varx (t) = to be more probable. Hence mx (t) = -0.5

(-1)2/2

+ 12/2 = -,t

+

1.

Rx(t,t')

X(t)

To find the correlation function Rx (t, t'), we consider two arbitrary sections of the random function, X (t) and X (t') (t' > t), and find the mean value of their product: 0

0

Rx (t, t') = M [X (t) X (t')] = M [X (t) X (t')]. _The product X (t) X (t') is equal to -1 if an odd number of events (sign changes) occurred between the points t and t', and to +1 if an even number of sign changes (zero inclusive) occurred. The probabilit y t~at an even number of sign changes will occurr during the time "t' = t •! - t is 00

-

""' {f..'t)2m c-~-r- e-i.-r e'·-r+e-'·-r

Peyen - LJ (2m)!

2

-

'

m=i

sdim~larly, the probabilit y that an odd number of sign changes will occur unng the time

"t'

is

Podd

= e-i.-r

•

276

Applteri Probl~ms In Probabt ltty Thtory

Hence wber~

R"- (t, t') === ( + 1) Pe"rtn + (-1) Podd = e- 2 ~-r, Stmtla rly, for t' < t R:£ (t', t) = e-2~(-'l), 'vhere

-r = t'- t

T

= t'- t . .

Comb1n1ng these formul as, '\\ e get Rx (t, l') = R~ (-r);:;:: e-2~ l tl The graph of th1s funct1on IS shown 1n FJg 9 i4b The surface R~(t~ t )= e- ~A.~ t -t I IS sho,vn 1n F1g 9 14c The random functio n X (t) ts station ary Its spectra l dens1ty 00

1 S:c• ( (r) J = 2n.

rJ R

.x

(

> _ • T C

{J) T

d

T

t

=n

21

(2A.)'lL+ Cit'~

•

-oo

9 15 There 1s an elementar~ (statio nary Po1sson) flo\v of events wtth Intensity A on the t ax1s A stocha stic process X (t) occurs as fol lows "hen the ~;th event occurs tn X(t) X(t) the flov. (' = 1, 2, ), tt assumes J l a random value V 1 and retains that ' I X(t') 'alue ttl I the next event occurs lD I .. T f the flo" (F1g 9 15) At the 1D1hal 1 l - I ' "" I momen t X (0} = V 0 The random " ! l \. vartabl es Vo, VI' v2, , v., 0 t I T t t 1~ t are Indepe ndent and have the same -~1 I I distr1but1on w1th denstty

,

'I

whence it can be seen that the process is stationary . Its correlatio n function Rx ('t) = Var v e-h\'t\ does not depend on the form of distribution

0, t > 0. Let us co:r:s~der a random variable Y, which is the number of electrons arnvmg at the anode at time t. This variable has a Poisson distribution with

X(t) 1

;r

1 -

I

I I

0

T;

Tz

f

2

I I

I

I I

I

Ts Ts

t

~

T7

(a)

I-< Bz,., e, 0

Tt,

TJ

I

I I

•

J

l

I

t

!J .

el

t

(b) Fig. 9.19

parameter 'At. We represent the voltage X (t) as the sum of a random number of random terms: y

X (t) =

2J e-a(t-T1>1 (t-Ti)·

(9.19.1)

i=1

We have shown somewhat earlier (see Problem 8.80) that a Poisson flow of events on the interval (0, t) can be represented, with sufficient accuracy, as a collection of points on that interval, the coordinate of e~ch of which ei E (0, t) is uniformly distributed on that interval (see F1g. 9.19b) and does not depend on the coordinates of the other points. Consequently, expression (9.19.1) can be rewrilten in the form y

X (t) =

2J e-a(t-ei>,

(9.19.2)

i=1

wlwre.the random variables 0, are independent and uniformly distributed in the interval (0, t). We designate X 1 (t) = e-a(t-el) = e-ateaei, and then y

y

X (t) = 2j Xi (t) =e-at 2j i=1

eaei.

(9.19.3)

i=1

where X 1 (t} are independent similarly distributed random variables and _the random variable Y does not depend on the random variable~

280

Appll~d Problems in Probability

Theory ~olut1on

X 1 (t) e1ther In accordanc e with the

of Problem 7 64 v. c \\7 te

the e~pre~s1on~ for mx (t) nnd Var x (t)

m:1: (t)

=

m 11 (t)

mx 1 (t)

(9 19 4)

+

(919 5) Varu (t) m~ 1 (t) Varx (t) ~ mv (t) Varx 1 (t) S1nca tl•e random varHlble } has a Po1sson d1str1butJon \vlth para Let us find m.r 1 (t) meter At 1t fo11o,vs that m 0 (t} = Var u (t) =

"-t

r ' m~ ~ (t) = 1\ l (X 1 ( t)] = T J e 1

au

t-e

~) dx =

at

at

0

\.ext '\ e det erm tne the c:econ d moment :1 bout the or1g1n of the random var1able Y1 (t) t

l\1 rxr (t)] =

f\

(e

z

a(f

]2 dx =

i-e 2a;t

---=---2cxt

0

Con ~eqnen tl l

mx(t)-)

Varx (t) = J\t l Varx {t) +

1

-o a

o:f

m;i (t)}- At'llXF (t))-)

(9 19 6) f

-;a

2at

(9 19 1)

l\ ote that as t -... co the mean value and v arta.nce of the pr£K~S X (t) do not depend on t1me lun mx (l) = mx - A./a lim Varx (t) -Varx

= Al{2a)

' ...,.II)C

t-oo

To f1nd the distr1bUl1on of tho section of the stochastiC proce~s X (t) for m:.t - 'A.frz > 20 \\e :rearon as follow~ Cons1der1 ng a fintte but

suffictentl y large Interval (0 t) and assuming that a sufficientl y large number Y of electron emiSSions t1..he plD.ce on that 1nter' al "e '=ee that the process X (t) Isee formula (9 19 2)1 1s a sum of tndepende nt .stmllarly distribute d random var1ables ,,htch has an appro....:tmately

norma.l dlstrl.butlo n s1nce 1n thts Ca"-e the cond1t 1ons of tho centralltm it theorem are In fact ruifi11ed (sec Chapter 8) Consequen t!} the sectiOD of the stochastic proceco:s has a normal dtstrtbut 1on w1th characteriS ticS The process ,v1Il practically become n1:r = 'Ala and Var::t == A/(2a) sta t1onary In the t 1me "Tst = 3/a To f1nd tbe correlatio n functions let us cons1der t\V'O secttons ol the stochasttc process In questton at moments t and t (t > t) By vtrtue Gf the assumptio ns made \\e can assert that the- anode voltage 1. (t ) of the 'alve at the moment t 1s equal to the voltage X (t) at the moment t multiplied bl the e"tponent c o:o t) plus the voltage Y (t - t) "'\Vh tch results I rom the arr1 val of electrons at tbe anode 1n the t1 me 1nter' a1 (t t ) (9 19 8}

Ch. 9. Random Functio ns

28i

•

The stochas tic process es X (t) and Y (t' - t) are eviden tly indepe ndent since they are genera ted by electro ns arrivin g at the anode during different, nonove rlappin g time interva ls (0, t) and (t, t') respec tively. The same can be said of centred stochas tic process es X (t) andY (t' - t). Conseq uently, Rx (t, t') = M (X (t){X (t) e-a(t'- t> Y (t'- t)}] 0

0

0

0

= M [(X (t))Z] e-a{t'-t ) 0

Rx(t, t')=M [(X(t' )) 2 ]e-a(t- t')

+

0

0

for

t' > t,

for

t>t'.

Combining the last two express ions, we obtain Rx (t, t') = Varx (min (t, t')) (1- e2amin {t, t')] e-al t'-tf. Let us conside r the limitin g behavi our of the stochas tic process when t- oo, t' -+ oo, but the value of their differen ce 't = t' - tis finite. In this case A e-al-rl. R X ('t)=V ar X e-al-rl = 21X

Thus the stochas tic process X (t) we conside r in this problem is stationa ry and practic ally normal for t tending to infinity and 'Ala > 20. This problem is a more genera l case of Proble m 9.13. Indeed , as a -0, the anode voltage of the valve is a Poisson process since with the arrival of each new electro n the voltage only increas es by unity and does not decreas e with time. Conseq uently, the followi ng equalit ies must hold for anv• finite t and t': . . . 1_ 8 -2at 1 _ 8 -at =At, = hm Varx (t) = hm /, hm mx (t) =lim/ , 2IX a-o a-o IX ;:a-o a-o limRx( t, t')=lim A [1-e-2 amin( t,t'>]e -alt'-ti =Amin (t, t'). a-o 21X a-o The reader is invited to verify their validit y. 9.20. The functioning of a linear detector. Under the conditi ons of Problem 9.19, we assume that electro ns arrive at the anode in "bursts ", ~he m?men ts of arrival of the bursts formin g an elemen tary flow with mtens1ty A. The numbe r of electro ns in the ith burst is a random variabl ell'i which is indepe ndent of the numbe r of electro ns in the other bursts and h?s a distrib ution F (w) with charac teristic s mw, Varw. This problem IS equiva lent to that of defmin g the output voltage of a linear~elector, when positiY e pulses of the random variabl e (of the highes t \oltage) TVi arrive at its input at random momen ts defined by the Poiss~n flow,. and in the interva ls betwee n the pulses the voltage drops expone ntially (Fig. 9.20). Find the charact eristics of the process . Solutio n. 'Ve can represe nt this process by a formul a analogo us to.

(9.19.2):

y

v

.ll.

e-a(t-B .) (t)17 J ll1 i - L~ i=i

(9.20)

282

Appli~d Pro blem a tn

Pro bab thty The ory

\Vhere the ran dom \3t tab les Y, lV,, e, are mu tua lly Independent '\Ve des tgn ate Xt (t): :: lVle-~u-e1>, and the n i -e -o:t lt[ rx~ (t)] === mw at ' Co nse que ntly , m% (t)= Amw

As

t~oo,

1-e -cd t

tt

1_ 9 -2a t AICXl(t)]=(Varw+m~) at 2

1-e -2a t Var.:~:(t)=1{Varw+m:,) a 2

we hn\ e

l1m mx (t) = mx = Mn w t-.Ol l a.

1 A. (Va r +m ) , ltm Var~ (t) = Var~ = 2ua w , t . . . oo Rx {'t) = Va rx e-o: r""' Let us con s1d er the l1m 1ttn g beh avi our of the process X (t) \vhen the foll ow1 ng quant1t1es Increase 1nd efin ttel y the tnt ens tty of the Po1sson flow gen era t1n g the pulses X{t) (~ -+ oo), the var 1an ce of the am pl1 tud e of eac h pulse (Varw ~ oo) and the param ete r a (a; ... ~ oo) An tnfintte tncrease of the qua nti ty a I r s1gn1fies tha t the pulse voltage dro ps rap tdl y, 1 e 1n the l1m1tt 0 t as a )oa oo, the are a of the pul se ten ds to zer o. the speeds rig 9 20 at \Vhtch the quant1t1es ~ and Va r10 tnc rea se be1ng propor tto nal to the speed of Increase of the qua ntt ty a A = k a, Varw = 1 k 2a. \Ve obt aln (as t ~ oo) A ltm mx = ltm - mw = ktmw,

1 a ... oc

l1m

).. a Var to .. oo A. a,

hm

Varw . . . oo

1 tt-+o o

Va rx=

rx.

l1m

I.., a Var w-o o

Rx (-r) =

ltm

). 1

a Var w~OQ

~

(\'arw+m:C,) 2~

--+-)

OOt

k~, ae- a I 'I' I= kjk26 (T),

\vh ere rS (t) IS the del ta fun ctio n Thu s, In the l1m1t \Ve hav e 'vh tte not se, \Vhtch res ults fro m an tnfinltely fre que nt sertes of pulses 'vh tch hav e a fin1 te e'tp ect at1 on of the aro plttude of a pul se and an 1r:.fin1 te var 1an ce of tha t am plt tud e, as 'veil as an tnfi nitc s1m al dur atto n of the pul se Itse lf 9 21. The shot effect Let us con s1d er a sto cha stic pro ces s X {t) generate d by a Pot sso n pro ces s ltke the case tn Pro ble m 9 13 As the 1th eve nt of the Poi sso n flo\v occ urs at a mo me nt T h a non neg atP te vol tag e pul se J:Vi ar1ses 1n the ele ctri C ctr cut t, ,vhose ran dom val ue (am plit ude ) the n var1es acc ord tng to one and the sa me l "'v cp (rt) . '\vhere the var 1ab le

Ch. 9. Random Functio ns

283

11 is reckone d from the momen t Ti (Fig. 9.21) (

't} 1+P( T 2} - P{t > 2- -r} ~ 1 - (2- 't) - 't'--- 1 Pt ContinUing '\ tth our reasoning '' e get 1 - (l" - 2n) for 2n < "' < 2n + 1 p, ~ (T- 2n) ~ 1 for 2n + 1 < "T < 2n

+2

It can be seen that p 1 (and hence R~ (t t + "t) = (2p1 --- 1) as" ell depends onll on 't and ts 'lll even function of t' Consequently 1 g,v. f- J - 2T 1M 2n < 1' < 2n Rx (t" t + -rJ ~ R:x: (t) = 1 < T 1/ 2 th e cu r\ e R:r (-r) 1s re pe at ed pc rt od 1c al l) , at ta ln E ng local max1ma eq ua l to 1' (1 --- 1') at 1n te gr al va lu ed po 1n ts 9 31* '' e co ns id er a st at 1o na ry ra n d o m fu nc tt on X (t ), "b 1c h 1s a ~a\\ to ot h 'o l t 1.ge (F ig 9 3 ta ) T he re fe re nc e po tn t occuptes a random

m:

t

0

tlt-2

{~)

po st t 1on re i at 1' e to th e to ot h, Ju st as 1n P ro bl em 9 29 F tn d th e mean 'a lu e, . th e va rt an ce an d th e co rr el at to n function of th e ra nd om function X (t) Solut1on \\ e ca n e1Sily fi nd th e m e1 n va lu e na"t 1f " e ta ke 1nto ac~ount th n t X (t ) ha s a un tf or m dt st rt hU tl on on th e tn te rv al (0, t) fo r an} t H en ce mx = 1/ 2 T o .. find th e co rr el at to n fu nc ti on , ~ e ri g td l' co nn ec t th e sequence of te et h 'v 1t h th e t a"tls an d thro"\\ th e be gt nn tn g t of th e 1u te r\ al (tt t tt) at ra nd om on th e t ax ts (F ig 9 31 b) S tn ce th e te et h ar e pe rt od tc tt lS su ff ic ie nt to th ro '' th e po tn t t at ra nd om on th e first 1n te r' al (0 1) dt st rt bu tt ng 1t 'v 1t h co ns ta nt de ns it y A s ca n be se en fr om F1g 9 3l b. JD th at ca se X (t ) = t an d th e va lu e X (t + "t) 1s eq ua l to th e fr ac ti on al pa rt of th e nu m be r t t', 1 e X (t t- ~> = t + 1: E (t + t) , \Vhere E (t 't) 1s th e In te ge r pa rt of th e nu m be r (t 1:) If th e 1n te gr al pa rt of th e nu m be r -r 1s n (n ~ 't < n + 1) , th en n for t + -t < n 1,. E (t + 't )= n + i for t+ 1 '> n + 1 ,

+

+

+

+

+

Ch. 9. Random Functions

295

and, hence, for

t+1:-n X(t+'t)= t+'t-(n+1)

for

t:=; nx ( t t ) -+- t t n11 ( t t ) t' n xu ( t, t J) + t R :tv (t, t} = eo: (t +~ ) + t t ca ~ u t )t + a (t + t ) c- Ct f ' - ' t

+

l

9 34 Find the mean \ alue an{l the correla. tton function of the sum of t'" o uncorrelat cd random functtons X (t) and Y (t) 'v1th characte.ns tJCS

mx(t)=t~ Rx(t

t )=-tt', mll (t) = . . . . . t Ry(t 1 t') == tt' ea (t+t

)

Answerll mz (t) = m ~ (t) +my (t) ~ 0, R" (t, t') = R J: (t, t') ...t- Rr~ (tt t) = t t' r1 + eel( t+, >] 9 35 G1' en a complex randon1 funct1on z (t) = X (t) + L},. (t)t \Vhere t IS the 1mag1nary unlt X (t)i Y (t) are uncorrelat ed random funct1ons 'v1 th chn.racterl stles mx ( t) ':;:: t 2 t R:... (t , t') = e- a; 1 et - t >.a, R Y ( t t }=- e zo: J. H+t ) ~ my ( t) == 1 , find the charactert sttstics of the random functJon 7 (t) m: (t) Rl ( t, t') and Varz (t)

Ch. 9. Random Functions

297

Answer. mz(t)=t 2 +i, Rz(t, t')=e-a, 0 is p =

·' ) CJl (x) dx. The ordered series 0

of the random variable Y (t) has the form Y(t):

I +1 1-p I p -1

.

Hence my = 2p - 1, Vary = 1 - (2p - 1) 2 = 4p ('1 - p). Assume T = t' - t and t' > t. If no event occurred in the Poisson flow during the time ,; (and the probabilit y of this is e-'-'t), then the values of the random function Y (t) and Y (t') are equal to each otherand the conditiona l correlation function Ry (t, t') =Vary = 4p (1 - p). ~ow if at least one event occurred during the time 't', then Y (t) and ~ (t') are uncorrelat ed and the conditiona l correlation function R Y (t, t') 1s zero. Hence, for t' > t

Ry (t, t') = e-l.qp (1- p), and in a general case (for any t and t') Rv (t, t') = Rv ('t') = e-'- I 't 14p ( 1- p). 9.52. The random input signal X (t) considered in Problem 9.15 is transformed into a random output signal Y (t) by means of a relay

:304

Appl1ed Problems in Probabrllty Theory

w1th a de1d h1nd s1gn

lr(t)=

X ( t)

IX (t)l >

for

rx (t)[ < e,

for

0

B,

"here e 1s the tlca d band of the rela} F tnd the d 1s trtbu t 10 n of the sect ton of the random funct 1on Y (t) and 1 ts chnractcr1st 1c~ the mean v (l} ue .n nd the corre I at 10n function Solut1on The random \artablo Y (t) may assume one of the three 'a.I ues -1,. 0. t., for any t and has an orderod ser1cs

y (t)

-t Pt

1

o P2

'

1

'

+t p,

'

-e:

p 1 =P{..\(t)-oo

-T

T

JW

~ 2

2

--}f1+cos2(ro 1t-G)]dt=--}W2 •

-T

Let us fi.nd the spectral density of the random function X (t). We shall show that it is proportional to the delta-function: S x ( ro) = Varw6 (w - w1 )/2 (0 < ro < oo). Indeed, for such a spectral density the correlation function 00

Rx('t')=

JSx(ffi)COSffi't'dro 0 00

Varw COSffi1't' d ) ( Varw \ ffi= COS(J)'t' = j 6 (J)-(J)i 1 2 2 0

which coincides with the correlation function for X (t). And since the direct and inverse Fourier transformations define the spectral density and the correlation function one-to-one, the expression for S x ( ffi) written above yields the spectral density of the random function X (t). If we use the complex form of the Fourier transformations rather than a real one, we get the spectral density S~ (ffi) in the form s~ (ffi) = Varw [ 6 (ffi +COt)+ 6 (ffi -ffit)J/4

( - 00 0). 9.55. Show that the sum of elementary random functions of the form 00

X (t) = mx+ 2j W 1 cos (ffitt i=O

+ 8 1),

\vhere W 1 are centred random variables with variance Var 1 (i = 0. • 1, 2, ... ), 8 is a random variable uniformly distributed on the 1 ei ln~en·a~ (0, 2n) (i = 0, 1, 2, ... ) (all the random variables hemg Independent.), is the spectral decomposition (9.0.24) of the random function X (t). Solution. In accordance with the solution of the preceding problem

wi,

+

H'i cos (ffi 1t- 8 1) = U 1 cos ffi 1t V 1 sin ffi 1t, wh ere U1 and l'1 are uncorrelated random variables with zero mean Ya1ues. Consequently 00

X (t) =mx+ 2j (U 1 cos ro 1t+ Y 1 sin ffi 1 t), i=O

and this is what we wished to prove. 20•

308

Applltd Problems ln ProhaiJility Tk'!ory n

9 56 Cons1dcr n stochastic process Y (t) = ~ a 1X 1 {t) +b

where

i=d

X1 (t) aro stat1onary uncorrelatcd stochastic processes '' tth character 1sttcs m 1 R 1 ("t) and St (

0,

{2) k:c ( -'t) = R% (or).

(3)

J

llx (or) 1~ Rx (0),

CIC)

(4) S;(UJ) =

2~

JR= (r) e- ~1" d-r~O 1

for any

oo

-oo

Properties (1) and (2) are obviousll Let us ver1fy the others. (3) The funct1on R =-: ( -r) 1s even and, therefote, 1 t 1s sufficient to lnvesttgate 1ts behav1our for 't ~ 0.

R;,; {-r) = {-e-ta:-~n (

T+ 1)- f

e-Ctt+lllt (

i-_1) ,

Stnce Rx (0) = 1, this e"'{presston cannot e:tceed untty 10 absolute valuo For a,< ~ thts cond1t1on lS not fulfilled s1nco e- ca-~)1' 1ncreases tndeftnttely as 't ~ oo. If a ~, e get R:t (-r) 5:. 1, and for a > ~ we

= ''

1 Ch. 9. Ra nd om Functions ~31 ~ ~

~ ~ ~ ~ ·-------

have -R x (-r) ~

1. Th us pr op er ty (3) is fu lfi lle d on ly for a

~ ~-

00

(4) S~(ro)= in

00 ~d't= ~ i e~ Rx(-r) 2

R e { (i + f)

-o o 00

X )

e-(cx-Hiro)~ d-r

+ ( 1 - T)

00

j e-z+wz 2n~ = 2ex ex~~-~~~

n

·

[(ex-~)z+wz] [(ex+~)z+wz]

ve ha e ~w = a r Fo . r) be m nu ex pl m co a of for a~ ~ (Re is th e re al pa rt s~ (ro) = 6 (ro). 1 (c os h B-r X ~ 1 x e-c = ) (-r x R n tio nc fu e th Thus for a.,~ ~ n. tio nc fu n io at el rr co a of s tie er op pr e th l X sinh B I 't I) possesses al a, b. 61 9. g. Fi in n ow sh e ar p ,> a. r fo ) (ro S~ d an The graphs of Rx (-r)

+i

l

0

0

(h)

(a) Fi g. 9.61

se ho w (t) X n tio nc fu om nd ra ry na io at st 9.62. Show th at th er e is no - ' t11 -r1 ) an d ( al rv te in e m so in nt ta ns co is -r) ( x correlation fu nc tio n R equal to zero in its ex te rio r . nc fu om nd ra a is e er th at th . i.e , ry ra nt co e th e m . Solution. Le t us as su -r < l 1-r r fo 0 > b to l ua eq is n tio nc fu n io 1 tiOn X (t) for w hi ch th e co rr el at ity ns de al tr ec sp e th d fin to y tr l al sh e W • -r and to zero for• I 't 1 > 1 of the ra nd om fu nc tio n X (t): 'C

co

S x ( (J)) = ..!_ \ R ('t) cos (l)'t d-r = 2_ n

b cos

y

n

J0 -

(J)'t

d,; = !!... - sin 00 '1 n

w

•

0

e tiv ga ne is ro) ( x S n tio nc fu e th at th on It can be seen from th is ex pr es si al tr ec sp e th of s tie er op pr e th ts c. di ra nt co is th d for some values of ro, an

312

Appl1~d

Th~ory

Problem' tn ProbabtlUy

dens1ty and. consequently, there can he no correlatton functton of the ktnd descrtbed 9.63 Does the functiOn Ex ('t) = Var.r e-o: I" 1 ~ sm ~ I 't' I possess the propertieS of a correlation functton? Answer. No, 1t does not, Since the follo\ving t\vo cond1t1ons are not fulfilled Rx (0) > 0 and Rl'. (0) r R~ ("t) ] 9.64 A stationary random functton X (t) has the charactertsttcs m= and Rx (-r) Ftnd the crosscorrclation function Rl:l (t, t) of the random functton X (t) and of the random function Y \t) = 1 -X (t) Solution.

>

R~v (t, t ) = ~I [k (t) Y (t')l ~ ~~ [ - i (t) = -Rx (t, t') = -Rx ("t)

X (t')l

9.65. A random function X (t) has the charaetertstlCS Var:s: e-c; 1-r ~ F1nd 1ts spectral density Soluhon.

2~

J Rx ('t)

R,:, (-c)

=

,.

ext

SHffi) =

m;u

e-llllt

do= V~r., Re

I e-

(tJ.+foo)"r

dT t

0

-OD

where Re ts the real part .. 'Ve have OCI

rJ e-fet+

Ja.+t uo 00

t(J))t

d-r: =

0

e-la+iw)'l'

d (a+ lf:t1} T

0

(et+Ho)~=y

-

00

for 't=O, y=O

=

t a+tm

Je-u dy 0

J CIO

=

1 , Re a+ tro

e-ta+ial)'r

0

= Re

I

=

t

c:t+ ~(r)

1

d-r = Re - -a+ iro

a-loo -Re a-lw a- tw ct'+ w'

Re { a; a~ co~ - t

c:t'~ co' ) = a•~ ro•

•

Consequently, S~ ((r)) = Varx a n aa+ru~ • 9.66. F1nd the spectral density of the random function X (t) tf 1ts correlatton funct1on Answer.

Ch. 9. Random Functions

313:

+

e-if3o:)/2 and use the solutio n of Hint. Represent cos B• = (e 1!3' Problem 9.65. 9.67. Find the spectra l density of a station ary random functio n whosecorrelation functio n Solution. We have 00

~

1 S.., (ro) = -=-2n x

00

d't = _Varx • __.::::_ ~ e-(A.o:)Z-icoT d-r:. Var e- (',,:r )Z e- tcuo: 2n x

-oo

-oo

Using the familia r formul a -

00

Jf

e-AxZ± 2Bx-c dx

=

2

ACA-B-.,V/ An exp [ - -.....,-

J

(A>O )

-oo

and bearing in mind that fJ. = -1, we get ., / n S *'x(ro) = Varx 2n V 1..,2 exp

[

J-_ 4J...VarxVn exp [ -

2

ro -- 41..,2

2

ro 4J...2

J•

The graph of this functio n is like that of a normal distribution~

R (1:)= 2a sinW(C"

-r

2aw1 x a

s; (w)

I I

I

I

w, w

0

(6}

(a)

Fig. 9.68

9.68. The spectra l density of a station ary random functio n X (t) is chonstant on the interva l from -ro 1 to +w 1 and zero outside it, i.e~ as the form shown in Fig. 9. 68a: 1 ro 1) ( a for Iro I < rot _ x(ffi)- O f or Iro I> ffit -a1 1- rot . Find the correlation functio n Rx ('t) of the random functio n X (t). · Solution.

S 0, ~ > 0)

Solut 1on To solve the probl em \Ve shall we the appa ratus of gener ttl1zed funct 1ons, the rules for \Vhich are g1ven 1n Appe ndix 6

{a) R 11 (l') = -

tfl2 d•

e

ct

=a [ -ae

= ae

a 'tl

'tl

= _

t*l ( d

d

[ -ae-C lht d

dt

J:l ) + d~~:l 2

l'tl

d~

e-al20) and the condition 0 < mx- (1) ± 3 l Var~ {t) < n 1s satisfied n-e can approxunat ely assume that th& section of the random functlon X (t) 1s a normal random varlable "\\ 1th parameters mx (t) Yvar.-t(t} wh1ch "ere ohta1ned by solv1ng equations (10 0 24) and (tO 0 25) Formulas (J 0 0 24) and {10 0 25) rema 1n vahd as n-+ oo 1f the upper hm1t 1n the sums 1s replaced by oo

Proble1ns and Excrclses 10 1 T,,o elementar y flows are super1mpo$ed flow 1 w1tb 1n tens1ty A1 and flo~ II w1th tntens1ty A~ (F1g 10 1) Is flow III result1ng from the supe-rpos1t1on elementar y and )f 1t 1s. '\Vhat lS tts 1ntens1t)?

Ch. 10. Flows of Events. Markov Processes

....

333

•

Solution. Yes, it is since the properties of stationarity, ordinariness !· and absence of an aftereffect are conserved; the intensity of flow III j is equal to A-1 + '-2· :1 10.2. An elementary flow of events with intensity A. is thinned out ..: at random. Each event, independently of the other events, may be I

.-

•

I 0

I

I

>----.----- - - - - - - - t

I I I I

l I

I 1

I

I 1

l

--i-'~-t I I I

•

Fig. 10.1

conserved in the flow with probability p or removed with probability 1 - p (in what follows we shall call this operation the p-transformation). What is the flow resulting from the p-transformation of an elementary flow? Solution. The flow is elementary with intensity A.p. Indeed, under a p-transformation all the properties of an elementary flow (stationarity, ordinariness, absence of an after:ffect) are conserved and the intensity f(t) IS multiplied by p. il. - - - 10.3. The time interval T between events in an ordinary flow has a density l.e-i.(t-to) for t > t 0 , [ I (t) = l 0 for t < t 0 • t 0 (10.3)

The intervals between the events Fig. 10.3 are independent. (1) Construct a graph of the probability density j (t). (2) Is the flow elementary? (3) Is i~ a Palm flow? (4) What is its intensity 1:? (5) What is the variation coefficient v1 of the interval bel ween the events? So~utio~. (1) See Fig. 10.3. We call a distribution of this kind "exponential, displaced by t 0". (2) No, it is not, since distribution (10.3) is ~ot ~xponential. (3) Yes, it is, because of the ordinariness of the flow, Ie ,__Independence of the intervals and their similar distribution. 4 ( ) i. = 1/M [T], M [T] = 1/'A + t 0 , ~ = (1/'A t 0 )- 1 = 1./(1 /,t 0 ). (5) Var (T] = 1 _ 1 crt 1/'). _ 1 'j.2 ' at- 1. ' Vt = M [T) 1/l.+to - 1+1.t 0 •

+

+

334

Applied Problem1 fn Probab lfty Theory

tO 4 There 1s 1n eletnen tary flO\\ of events 'v1th tntenstty 1 on the t ax1s Another flow IS formed from 1t by tn~ert1ng another event at the m1dpo1nts of the 1nterval bet\\ cen e\ ery l\\ o adJacent e' ents (~ee Ftg 10 4 'vhere the Circles denote the pr1nctpal events and the crosc=es denote the IO~erted Ones) rtnd the diStribUtt on denstty f (t) of the 0

rlg 10 li

Inter\ al T bet" eon adJacent c' cnts Ill the ne'v flo'v Is the flo" elemen tary? Is 1t a Palm flow? ''hat IS the cooffictent of Yar1a lion of the 1nter val T bet" con the e\ onts~ Solution T - Ji./2 ''here X has an exponenti al distrlbutto n '\\llh parameter A / 1 (x) = A.e "'x (x > 0) Us1ng the solutton of Problem 8 1 and setttng a = 1/2 b = 0 In formula (8 1) \Ve obta1n f (t) = 2Ae- 2'- i (t > 0) {tO 4) Th1s ts an exponenti al dtstrtbut1 on "1th a coefficient of var1atton v, ~ 1 Neverthele(O:.s the new flo'\\ 1s not elementar y nor even a Palm flow \\ e shall first pro' e that 1t IS not a Palm flov. Though the In tervals bet" ecn the e\ ents ha' e the '=a me distributio n (10 4) thel are not Independe nt Let us constder t'\\O adJacent 1nter\als They may be Independ ent'' 1th probab!ltt y 1/2 or equal to one another '' 1th prob abiltty 1/2 and consequen tly dependent Thus the now flow oi events ls not a Palm flo" It 1s naturally not elementar y stnce an elementary flow ts a spectal case of a Palm flow Thus an exponenti al distributio n of an tn terval bet'\\ een events lS not a suffictent condttton for a f]o'' to be elementar y 10 5 The hypothesi s 1s the same as In Problem 10 4 except that the new flO\\ cons1sts only of cro~ses (the m1dpo1nts of the Inter' als) ~ Ansv,er the same questions as 1n the preceding problem Solutton The Intensity of the ne\v flo'v ,viii not ev1dently change as compared to the Intensity of the original flow and will rema1n equal to A The Interval bet~een t'\\ o neighbour ing crosses (F1g 10 5) Is T (Xi

+

X l+ 1)/2 1

(10 5)

where xi and xf+l are tVvo adJaCent Intervals lD the orrgtnal flow The v ar1abl es X 1 and X 1+1 h oth b a' e an ex ponentral dJstribu t1on w1 th parameter A and half theJr sum (10 5) \VIll ba\e a standardized Erlang dtstrihntio n of the 2nd order since the tnte:r' al T lS equal to the sum of two Independe nt exponenti ally distribute d random variables d1v1ded by2 two Thus the Interval T between two crosses has a density I (t) = 4"'- te - 2"-t ( t > 0)

Ch. 10. Flows of Events. Jlfarkov Processes

335-

ln the transformed flow all the neighbouring intervals are dependent since their components are the same random variables. However, thisdependence involves only neighbouring intervals. Flows of this kind are sometimes called flows with a \veak aftereffect. The coefficient of variation v1 for the random variable T is {seeformula (10.0. 7)] Vt = 1tV2 < 1. 10.6. The traffic on a road in the same direction is an elementary flow with intensity A.. A certain Petrov stands on the road and tries to stop· a car. Find the distribution of time T for which be will have to wait~ its mean value m 1 and mean square deviation CJt. Solution. Since an elementary flow bas no aftereffects, the "future''" does not depend on the "past", in particular on the time when the last car passed. The distribution of time T is the same as the distribution. of the time intervals between successive cars, i.e. exponential with parameter A.: f (t) = A.e- i.t (t > 0), hence m 1 = 1/A., Var 1 = 1//...2 ,. O). H ''htch IS none other than Erlang s F1g 10 10 dtstrtbut~on of the 2nd order [see for mula (10 0 3) for k = 2] 10 1 0*. There IS a Palm no,v of events ,v1th d ens1 t y f {t) of the Inter-val T bet\veen successtve events on the t axis A random pornt 1* '(an 'inspector') falls somewbore w1th1n the 1nterval T* (F1g 10 10) It d1v1des the 1nterval 1nto t'vo subintervals Q, from the nearest pre v1ous event to t*., and H from t* to the nearest succe~stve event Ftnd the dtstrl bu tions of the t \VO 1n t erv als

Ch. 10. Flows of Events. Markov Processes

337

Solution. Let the random variable T* assume a values: T* = s, and find the conditional distribution of the interval Q under this condition. We designate its density as fQ (t Is). Since the point t* is thrown on the t-axis at random (irrespective of the events of the flow), it evidently has a uniform distribution within the interval T* = s: fQ (t Is)

=

1/s

0

for

< t
t, we can write 00

00

/Q(t)=

f

J t

1 s f(s)ds= mt smt

IJ

t

1 f(t)dt= mt

[1-F(t)], ·

where F (t) is the distribution function of the interval T between the events in the Palm flow. Thus we have 1

('10.10.2)

fQ (t) = mt [1-F (t)]. It is evident that the time interval H = T* tribution:

Q has the same dis-

1 [1-F(t)]. fH(t)= mt

(10.10.3)

10.11. Using the results of the preceding problem, verify the solution of Problem 10.6, which we obtained from other considerations. Solution. We have f (t) = A,e-1-t, F (t) = 1 - e-7-.t (t > 0), mt 111.. By formula (10.10.3) fH (t) = '" [1 -

1

+ e-'·

1

]

= A.e-7-.t

(t

>

=

0),

i.e. the solution of Problem 10.6 is correct. 10.12. A passenger arrives at a bus stop irrespective of the timetable. The flow of buses is a Palm flow with intervals uniformlv distribute~ in the range from 5 to 10 min. Find (1) the distribution d~nsity of ll~e lnlerval during "·hich the passenger arrives; (2) the distribution dens~~~' of time H for which he will have to wait for a bus; (3) the average lime he must wait for a bus. Solution: We lHl.ve f (t) = 1/5 for t E (5, 10), mt = 7.5. II (1) B~: 1ormula (10.8) f* (t) = t/37.5 for t E (5, 10). The graph of lt' densrty f* (t) is shown in Fig. 10.12a. (2) F (t) =

0 (t- 5)/5

1

for t~5, for 5 < t~ 10, for t

>

10.

Appl!td Probl~ms

338

in Probablllty Thtorv

f*(t)

---t----.&;...,;~~-t

{1

10

;

0

5

10

t

{hJ

fa)

Fig 10 12

(3) The a' erage 'vatting t1me 5

mil= )

10

1t 5 dt

+ ~ t 1~.;-;t

dt

~611

m1n

1

0

10 13~ \Ve constder Erlang•s no,v of order h With a dtstrJbUtiOD density of the Inter\ al T bet" een events 1

.-.-_

•. .

0

___.A:------. v-....;; l t

'•

t

F•g 10 13

,. (~t}Jt.-l

t) = ( I"

(k- t)l

Find the

e

-:lt

dIS tr1bu tton

(t > 0)

(10 13 t)

funct1on F k ( t) of

thts Interval Sol ut1on \Ve caul d find the d 1s tri bu tion func lion ns1 ng the ordinary formula i

F,.(t)""' ) Ill (t) dt, 0

hut 1t

IS

s1m pler to find

1t

proceed tng from the d efin 1t 10n FA (t}

=

P{T < t} 1 '' e pass to the oppos1te event and find P { T > t) \\ e associate the or1g1n 0 '' 1th one of the events 10 Erlang's flo'' and lay off tv. o Intervals to the r1gh t of 1t T (the d 1stance to the ne"'\. t e'en t 1n Er langts flo,v) and t < T (Fig 10 13) For the Inequality T > t to be sattsfied, It ts necessary that less than h. events 1n the elementary flow w1 th 1nt ens 1t y 1 fa II on the 1n ter' al t {c1ther 0 or 1, , or k - 1) The probe btltty that m events fall on

Ch. 10. Flo ws of Ev ent s. Ma rko v Pro ces ses

339

the interval t is (A.t)m - M Pm - ml e .

By the pro ba bil ity ad dit ion rul e k-1

P {T > t} = ~ ('A~~ e-~t, m= O

whence k-1

Fk (t} = 1 - ~

e-~t = 1 - R (k -1 , /.t),

(J.t)m ml

(10 .13 .2)

rn= O

2). ix nd pe Ap e (se on cti fun ted ula tab a is a) , (m R 1 ere wh k er ord of w flo g's lan Er is es lur fai r ute mp co of w flo e 10.14*. Th the n he (w es lur fai the n ee tw be al erv int the of .1) .13 (10 ity ns de with the . ly} us eo tan tan ins n tio era op o int ck ba ht ug bro is it , wn computer goes do st fir the for its wa d an t* nt me mo m do ran a at s ive arr r" cto An "inspe he wi ll ich wh for H e tim the of ity ns de n tio bu tri dis the d Fin . ure fail have to wa it for a fai lur e an d its me an va lue mn. Solution. By for mu la (10.10.3) 1

fH (t) = mt [1 -F Jl{ t)] ,

e nc He .. k/1 = mt d an .2) .13 (10 la mu for by en giv is (t) Fh where k-1

k-1

m= O

m= O

_,. , _ 1 -v 'A ('At)m e-~t )m (/.t ""' A. = t) fH ( - k Ll [ml ml e k Ll

(t > 0). (10 .14 .1)

We rewrite for mu la (10.14.1) in the for m A. p.t)T-1 e-M (r- 1) !

(t > 0).

(10 .14 .2)

r=1

d" ixe "m a s ha H ble ria va m do ran the at th .2) .14 (10 m I~ ca~ be seen fro ord ers ; t en fer dif of ns tio bu tri dis g's lan Er k of ng sti nsi co on uti tnb ?Js bpro l ua eq th wi k , . .. 2, 1, ers ord of n tio bu tri dis lt .h?s Er lan g's nd fou be n ca ble ria va m do ran a ch su of lue va an me e Th . 1/k ;hl hty rom the co mp let e ex pe cta tio n for mu la k

mH

= Ivi [H ] =

~ ~

M [H I r],

(10 .14 .3)

r=1

~here ~~ lH l r]

is the co nd iti on al ex pe cta tio n of the ran do m va ria ble prov1ded tha t it ba s Er lan g's dis tri bu tio n of the rtb ord er.

340

lpplitd Probltms In ProhablUtrt Thtory

t rom the ftrst formula (10 0 4) '' e find that l\I [H 1r] k

mn ==

f

"'

k~ LJ r -

(k+f)' 2k~

=

k+ f. 2X: "

= ri'A, "hence (10 14 4)

10 15* A Palm flow of events \Vtth dtstr1bnt1on denstty I (t} for the 1nterval bet\veen the events 1s subJected to a p transformat1on (c:.ee Problem tO 2) A random varaable V ts the tnterval between the events 1n the transformed flow Ftnd the mean 'alue and vartance of the random var1able V Solut1on Tho random vartablo V ts the sum of a random number y

of independent random v1rtablcs (see Problem 8 63)

V = ~ T~ 1 1 ;!Ill;

\vhere Y IS a dtscroto random var1able 'vb1ch has a geometrtc dts-. tr1but1on P {Y == m} pqm-t (m= 1 2, ) q= 1- p and each of the random vartables T" has 1. d1str1bUt1on f (t) Then the succcsstvc Intervals between thee' ents tn tha p-translormed

=

flo'v are

where the random var1ables V1 V 1 are dtSJOlnt and the transformed Oo\V is a Palm flow In accordance ''1th Problem 8 63.,

\vhere

Jtf(t) ell, 00

m, =

0

J(t -m QC

Var 1 =

1)" f(t)

dt

0

Remark \Ve caD prove that a mult1ple p transformatton of a Palm Dow re.sult! in a flOW Which IS elose to an elementary now

10 16 Ftnd the dtstrlbulton of the Interval T bet\veen the events 1n a Palm flow 1f the random var1able T can be determtned from the y

express•on T = ~ T,

1

a

15

the sum of a random number of random

h--i

terms where the random vartables Tk are Independent and have an exponential d1strlbution 'v1th p1rameter 1 and the random var1able Y does not depend on them and has a geometriC dtstributton begtnntng WJth un1ty Pn = P {Y = n} = pgn 1 (0 p 1, n = 1 2 3 ) Solutaon As shown 1n Problem 8 62,. the random var1able T has an exponential dastrtbutton 'vtth parameter ~p and consequently the Palm flow be1ng cons1dered 1s an elementary flo\V wtth tntenstty ~p, 'vhtch results from a p transformation of an elementary fioW With 1n tensity 7v Thts confirms the correctness of the solution of Problem 10 2

<
QL-~-;¢ r' -r T'

-r

(c)

Fig. 10.17

to his destination. The distribution fT (t) is such that the random variable T cannot be smaller than -r -r' (Fig. 10.17b). Find the probability that the passenger will take a bus. Solution. We consider the opposite event A ={the passenger "\Vill not catch a bus}. This means that when the passenger arrives at the stop at a moment t* there are no buses at the stop, and no buses arrive ~uringthe period he waits. Each event, i.e. the arrival of a bus at the stop, lS f.ollowed by a period for the passenger to board, the width of the renod being T' -r (Fig. 10.17c). The .event A= {the passenger does not catch a bus} corresponds to t~e pomt t* falling outside the boarding period (Fig. 10.17d). The point t has a uniform distribution owr the whole length of the interval T*. !he pr?bability that it will fall on the interval T* - (-r -r'), which 15 0 \Itside the boarding period, is (by the integral total probability f ormula)

+

+

+

co

P(A)=

00

\

f*(t)dt=

~

1

mt

-r+•'

1 if (t) dt- '!;' 't+t'

"'

t

~

l+'t"

~t

\

r{e mt

f(t)dt

00

j

f (t) dt.

't+'t"

is the aYerage interYal between buses. le probnbility tlwt the passenger will catch a hus is P (A)

1- P (.4).

!~.18. An l OWin t

.

=

elementarY flow with intensity /.. is subJ·ected to thn fol• • " g ransformation. If the distance between adjacent events T 1

3'2

A pplitd Probltmr fn ProbabllHg Thtory

1s Cimaller than some perm ISStb le l1 m1t t 0 (a ' sa.fet Y tn tervar•), then tho e' ent 1s displaced by an 1nterval t 0 relat1 ve to the preced1ng event NO\\ 1f T f > t 0 • then tha event rem a 1ns put {F tg l 0 18) Is the transformed flo\Y formed by the potnts 9i, 9 2, •• ~~, on the t' -ax1s elemen tar:-,.? Is It a Palm no,v? Solution. The transformed flo'v 1s ne1tber elementary nor Palm's s1nce 1t has an aftereffect that can e"ttond arb1 trarll} far For example, 0

e,

'I

e~.~

I

I

l

I I I

l

I 0

82 9J

lo _............__

I

e'I

@i.i

911

•

It

I

I

I

to

l

lu

__...___~

e;

Oz

6, Ba 9s

85 BG

8~

9'5

8'5

t

to

I

fo

~ ..-""--. ~

e;

e'l

8.9

J

lr

e,a eft

Fig 10 18

the points 81 Oa e!h al 0 CfO\Vd the t' -a '\:IS and so e\ ery subsequent potnt on the t a "tts ts sh1fted by a t1me 1nterval \Vhtch depends both on when the events occurred and on the Inter' als bet,,een them 1n the past If t 0 ts much less than the a \erage d 1stance between the events 10 the orJgtnal flol.v 1 e t 0 < tli... then 've can neglect t1te s.ltereiiect in the transformed tlow ,

I

-

'

'I

I I I

T~

t j

-

,.~

Tg

....'

I

I

f

I

-.~.-·,..

•1 ;

t

T. A4

'I 'II J

l

l

l r, I r2 11jl

--·iF :

..

Tr

A

•I

I

• J

.~

1I

1 II ! ~H~~-.l l

Fig 10 19

10 19 T'" o Independent Palm flo\VS \VIth d1str1hutton denslttes It (t) and /r (t) for the Interval bet,veen the e\ ents are superimpo 01 ~ 0 110 P2 (3) ~ P2 (2) P22 +Po (2) P 0 ~ p 1 (2) Jl12 ~ 0 398

'\ e have Po (0)

+

+ +

+ Ps (3) = P3 (2) P~s +Po (2) P os + Pt (2) Pts + P2 (2) P:za ~0 482

Ch. 10. Flows of Events. Markov Processes

347

10.24. A computer is inspected at moments t 1 , t 2 and t 3 • The possible states of the computer are: (s 0) it is completel y sound; (s1 ) there are a negligible number of faults, which do not prevent the computer from functioning; (s 2) there are a considerab le number of faults, which make it possible for the computer to solve only a limited range of problems; and (s3 ) the computer goes down. ThB transition probabilit y matrix bas the form

Construct the directed graph of states. Find the probabilit ies of the states of the computer after one, two, and three inspection s, if at the beginning (t = 0) the computer was completel y sound. 0./i ,...... ~

So

O.'t

0.3

O.J 0.4

S't

~

1

s2

0.2

1

0. 7

I

liol.4 fs;l

83

3

(6) 0.2

Fig. 10.25

Fig 10.24

Solution. The graph of states is shown in Fig. 10.24. Po (1) Pt (1) P2 (1) Po (2) P2 (2) P3 (2) Pt (3) P2 (3) Pa (3)

(0) P 00 = 1 X 0.5 = 0.5, (0) P 01 = 1 X 0.3 = 0.3, (0) P 02 = 1 X 0.2 = 0.2, (1) P 00 = 0.25, p 1 (2) =Po (1) Pot+ P 1 (1) Pu = 0.27, =Po (1) P 02 p 1 (1) P12 P2 (1) P22 = 0.28, = P2 (1) P 23 + p 1 (1) P13 = 0.20, Po (3) =Po (2) P 00 = 0.125, = Po (2) P 01 p 1 (2) P11 = 0.183, =Po (2) P 02 + p 1 (2) P 12 + P2 t2) P2 2 = 0.242, = p 1 (2) P13 p 2 (2) P 23 p 3 (2) P 33 = 0.450.

=Po =Po =Po =Po

+

+

+

+

+

10.25. We consider the operation of a computer. The flow of failures ~malfll:nctions) of an operating computer is an elementar y flow with mtenstty A.. If the computer fails, the malfunctio n is immediate ly dt>te~te~ and the computer is repaired. The distributio n of the time of :ep~tr 1s exponentia l with parameter p.: q> (t) = 1-'-e-J.l.t (t > 0). At the 101 1Jal moment (t = 0) the computer is sound. Find (1) the probabilit y that at a moment t the computer functions; (2) the probabilit y that

•

348

tpplted Problems In Probablllty Theory

dur1ng the t1me (0, t) tho computer malfunctions ~t least once (3} the }Jmttang probnbtltties of the states of the cotnputer Solut 1on (1) 'fh e states of the sys tern (the computer) are- (s 0 ) 1t ts ~ound and functions, (s1) lt ls f~ul\.y and lS reptilie-d The mar\-.ed graph of states JS sho'\\ n In F1g 10 25a The Chapman holm ogoro\ equa t 1on (;1 for t be probabJltttes of stat~ Pa (t) 1nd p 1 (t) ha\ e the form

~~o =JlPI-l.Pot

d:/

=Ap0 -JlP1

(10251)

Etthlr of thP~e equ 'l.ttons can he deleted Since for any n1oment t Vt-e ha' e Po + p 1 - 1 SubstitUting p 1 = 1 - Po 1nto the first equat1on (10 2a I) "e obta 1n on~ d1 fferent1al equat1on "1th respect to Po dp 0 /dt = fl - (J.. + ~t) Po Sol\ 1ng th1s equ1t 10n lor the InJlial cond1tton Po (0) = 1, "c obtain Po (t) ::- __'!-

k+ ~\.

(1

+

1 e -()..+J.t) 1] \1

(

'

10 25 2)

\\hence

(10 2a 3)

-

(2) To ftnd t lH? prob"lbJilt~ p (t) that d ur1ng the t1n1e t the computer malfunctions 1.t leaqt once "e Introduce ne'' ctates for the s~ stem (sO:) the computer ne'er failed (s1 ) the compute-r malfuncttoned at lea~t

-

,....,

-

once The state s 1 Is the ab~orh•ng one (~ce F1g 10 25b) Sol' 1ng the Chapm1.n holmogorov equat1on dp 0 /dt

"""

= _, p 9

for the "" ,.., 1ntt1al cond1t1on Po (0) ~ 1 ''e get p 0 (t) :;;;;: t, ''hence p 1 (t) = 1 - Po (t) = 1 - e '-t Thus the probabtllly th1.t durtng the t1me t the computer malfunct1ons at ]cast once 1s 1 (t) = 1 - e 1n th1~ case the proba btltt y cuu Id La' e been calculated more qim pl} 1 e a.s the probabtl1ty that durtng the ttme t at least one e' ent (malfunctton) '" tll occur 1n 1.11 elemen tar} flo'' of malfunctiOns '' 1th Inten~tty 'A (3) As t -+- co "e get from cquattons (10 25 2) and (10 25 3) the l1mtt1ng probabllxt1es of states Po = ~t/(J._ + ~t) p 1 :::=::: J../('A ~t) wbtch could e \ en t uall l have been obtained d1rccti} from the graph of states ustng the btrth and death proce~s (v. e 1nv1te the reader to do th1s) 10 2G On the hypothesis of the precedtng problem the malfunctioning of the computer IS not lmmedtately not1ced but IS detected after an Interval of time ''htch has an exponent1al d 1.str1but1on w1th parameter v \Vrtte and solve the Chapman h.olmogorov equations for tho probabdt t1es of states Ftnd the !Imlttng probabilities of states (from the dtrected graph of states) Solut1on The states of the system are (s 0) the computer IS sound and operates (s1) the computer 1s faul ll but the ma]functton IS not detected, (sz) the computer 1S he1ng repa1red The graph of states ~~ shown tn Fig 10 26

e-'

~

-

p

'1

+

Ch. 10. Flows of Euents. llfarkou Processes

349

The Chapman-Kolmogorov equations for the probabilities of states

.are dpo dt

=

•

dpl

~Pz- "'Po•

dt

~

dp 2

=[\,Po- v P1'

dt

=

" v

Pt- ~tP2· ;;(10 · 26 · 1)

We shall solve this system using Laplace transforms. With due regard for the initial conditions for transforms :n:i of probabilities Ph equations (10.26.1) assume the form

s:n: 0

11n 2

=

s:n: 1 = 'An 0 SJt2 =

+ 1,

l.n 0

-

vrt 1 ,

-

~ll12.

VJtl -

Fig. 10.26

.Solving this system of algebraic equations, we get the following equations for the transforms: ~

'A.

~

••t= s+v ••o•

n;2-

v

s+ll

'V 'A.

n;-

n

1- (s+v)(s+~l)

o•

(s+ v) (s+ !!)

no= s(s2+s(f.t+v+t..>+vt..+vll+l.~t)' \i'e introduce the designations

a=!l+~+'·

~~~~----------2

+ 1/ (!!+:+1.)

-VA-V!-L-/.!1,

b = - ~t+~ +I. -l/(fl.+~+t.)z -v'A-v~ -All. Then the expressions for probabilities assume the form Po ( t) =

aeat_bebt a-b

Pdf)=l.

+ (v+ ~) 8 at_ 9 bt

a-b

eat_ebt [ 1 a-b +!-LV ab

+'All

1 p.'!.(t)=VA [ ab

[ 1 ab

+

+

+

beat-aebt ab(a-b)

beat-aebt ab(a-b)

J

J

beat-aebt ab (a-b)

J'

'

•

1o find the limiting probabilities, we can use either the transforms ()r the probabilities themselves: p 0 =lim p 0 (t) =lim sn 0 (s) t-eo

Pl =

/,~t

s ..... O

- 1-

= t.. ~t+~~~'+vtt•

,

-

'J.v (10 26 2) I.!J.+I.v+v~t ' P2Po Pt- lt..ll+lvv+''!l • . . \re can find the final probabilities of states directly from directed graph in Fig. 10.26: MP 2 = l.p 0 , l.p 0 = vpi, vpl = !1P 2 • We can delete Qne of these equations (say, the last one). Expressing p 2 in terms of Pr and P1 , i.e. p 2 = 1 - p 0 -PI, we get two equations

f.l (1 - Po - PI) = l.p 0 and l.po = vpp The solution of these equations yields the same results (10.26.2).

•

3:s0

tpptted Probltmr tn Prabablltty Theory

10 27 An electronic devu:e conslsls of t''o 1dent1cal repeat1ng untts For the dev1ce to operate It JS suffie1ent that at least one unit functtons \Vben one of the un1ts fa1ls, the device continues to funct1on normally due to the other unit The flow of fatlures of each unit is elementary

w1th parameter A \Vhen a un1t fa1Is, 1t ts Immediately repatred The t1me needed to repair the untt 1s exponent tal 'v1th parameter Jl lntt1ally (t = 0) both un1ts operate Ftnd the following characteriStlc.s of the operatton of the device (1) The probab1ltttes of sta.tes (as functions of ttme) (s 0 ) both untts are sound, (s1) one un1t 1s sound and the other ta be1ng repa1red, (s1) both untts are he1ng repaired (the device does not operate) 2.l

.......

So

-....

.

...

s,

,...,;

~

~

~

Sz

(h) ODes not

not

operate

Operates

cp~~ate

Operates

Does

,,-----..JA,.___--.. \

~------ow,

(c) Fig 10 27 lfllw

(2) The probability p (t) that d urtng t1me t the de' 1te ne'er falls (3) The l1m1 t 1n g proba b 111 ties of states of the de' tee (4) For the limiting (stationary) condtt1ons of the devtce the average rel a t1 ve t1 me for w h1ch the devtce 'viii operate (5) For the same llmtting cond1t1ons the a' erage time ~P of £allurefree operatton of the devtce (from the moment 1t 1S s'v1tcbed on after hetng repaired t1ll the next malfunctton) Solution The dtrected graph of states of the dev1ce ts shown lD Ftg 10 21a (2A. 1s put before the upper Ieft arrow s1nce o units operateand may fatl) for the same rea son 2~.t. Is put before the lower right arro\v stnce both un1ts are he1ng repaired) (1) The Chapman h.olmogorov equat1ons ha\e the form

t''

d:t

=

~tp 1 - 2Ap0

dft

1

-

2Ap 0 +211P:L-(A+~t) Pt• (10 27 1}

+ +

ln thts ca~e the following condttton must be fulfilled p~ p1 pj = 1 Solving this S}Stem of equat1ons for the Initial conditions Po (0) = 1,. Pt (0) -- P2 (0) = 0 ''e obtaJn erl I -e~t ae at :t.. 9 bt +(A...L.3n) Po( t)= '"',. --~q r «+b

.. [ t

+ 2f~. ;

beat_ aebl ] +~-~ eb 4b (a-b} '

351\

Ch. 10. Flows of Events. 'Markov Processes

where a= -(/•+f.l), a-b=A.+f-l, _ a2eat_b2ebt

Pt ( t)-

J.l(a-b)

+

b= -2 (l..+f-t), ab=2(l..+f-t)2,

(i..+SJ.l) (aeat-bebt) J.l(a-b)

+

2J.l (eat-ebt) a-b

2i..

+

()

Po t.

J.1

From the expressions obtained we get

p 2 (t) = 1 - Po (t} - p 1 (t}

(2) To find the probability

"' p,...., (t},

p 2 (0) = 0.

and

we make the state_s2 (the device

stopped operating) absorbing (s 2 ) (Fig. 10.27b). For the probabilities. of the states the Chapman-Kolmogorov equations are N

,...,

,_...,

-.,

AJ

-

,.._,

,....,

dpofdt = ftp 1 - 21..p 0 , dp 1 /dt = 21..p 0 - (!.. + J.L) p 1, dp 2 /dt ,...,= l..pl' Solving the first two equations under the initial conditions Po (0) = "" 1, P1 (0) = 0, 've obtain ,...,

Po(t)=

ae"t _ ~ 6 1lt

a-~

+(I..+J.L)

6 cd _

6 1lt

a-~

'

where -(Sf..+~~)+

y 1..2 + 6I..J.1+ J.1z '

2

(the quantities a: and ~ are negative for any positive values of 1.. and J.L)~ Furthermore, """

pt(t) =

1 J.1

a'J;

a/ +

+ 4}

Stnco Varx (4) - mx (4) and ~t = const ''e have Varx (t) = mx (t) on the ttme Inter,al t > 4 F1gure 10 35c sho" s a relationship bet" een mx (t) 1 e the average number of computers 1n ~erv1cc and t1me t The dash l1ne sho,\S a graph of the a\erage number of computers manufactured by a certatn ttme t "esus t1me t 10 36 \Ve no\\ cons1 der the pro cec:s of stor1 ng terms 1n a dtrcctory tn a data bank The proce sIS to 1nclude the terms 1n the dtrectory ,v] en

I sl7 r~o

t}..

l~

l !! ..... J

All

~ J";t /itA .. ~.. ~·{. .S.~n . ..._,tn

q-s"-1

Fag 10 36

they first appear For example the names of organtznttons for \\'htcb a factory has productJon orders are entered Into the data bank of the management 1nformatton S}Stem The entr1es for the names of the organisations Will be accumulated 1n the data banl. of the mandgement Information system 1n the order In "h1ch they nppear 1n the Input

records There are an a' erage of x terms In the dtrectory 1n every record fed 1nto the data banh. and the 1ntcns1 ty of feed1ng the .records Into the data banl IS A, (t) Consequently the 1ntens1ty of the flo'v of terms fed ~

Into the data bank IS A (t} = XA (t) \Ve nssume that tbts IS a Potsson flo\v The number of terms n IS fintte and nonrandom although 1t may not be kno\vn beforehand All the terms 1n the dtrectory may appear In a record 'v1th equal prob'lblltty 'vh1le naturally onl} those terms are fed 1nto the dtrectory that d1d not yet appear 1n the records F1nd the mean value and var1ance of the number of terms 1n the dtrectory Solution \Ve designate the number of terms In the dtrectory as X (t) The random process X (t) lS ev1dently a pure b1rth process w..tth ~ Jin1te number of states n 'vhose graph of states 1s sho,vn 1n Fig 10 36 To find the 1ntens1ty AJt (t) (k ~ 0 1 n - 1) we assume that the -pt.ol'!e~~ \S, 1.n. the ~ta.IJ,\ -tR B."1 dQ.n-n~..t.)..,Q.TJ. tn~ 1\IQhab.1..t~...t 'J of th.ts. ass u m"Q tton lS Pk (t) Provided that the ac:sumpt1on IS sat1sf1ed the 1ntens1ty of the flow of new (not yet 1n the d1rectory) terms 1s

Adt) ="- (t)

n:-k =A(t) ( 1- ! )

ss es ce ro P v o rk a M . ts n e v E f o Ch. '10. F lo w s •

361

fo rm e th e m u s s a ) 5 .2 .0 0 (1 d ) an 4 .2 .0 0 (1 s n o ti a u q e l a ti n The d if fe re n 1) .' 6 .3 0 (1 )' (t )A (t x m t) ( P k ( t) = 'A d m x( t) dt

=""

'A(t)

Ll

(1- kn )

lt= O

! )+ 2k'A (t) ( 1 - ~ )

n

d Va~; (t ) =

~

['A (t ) ( 1 -

lt = O

n

~

- 2 m x (t ) 'A (t )

(1-

= 'A ( t) - 'A (t)

mxn(t) _

)JpJ.(t) (1 0 .3 6 .2 )

2'}.. (t ) V a :x (t \.

hen w se a c le p im s e th r fo s Let us solve th e s e e q u a ti o n . 0 = ) (0 rx a V = ) (0 x s t, m n o c = n t, s n o c = A ' = 'A (t) A

mx

(t) = n ( 1 - e

--t

a ),

li m mx

(t ) =

n,

t- o o

A A t --t Varx(t)=n(1-e n )e n

(1 0 .3 6 .3 )

A

--t = m x ( t) e n ,

(1 0 .3 6 .4 )

li m V a rx (t ) = 0 . f...,.oo

n to g in d n te , y ll a ic n to o mon s e s a re c in ) (t x m n o ti c n - + oo t d n a 0 . Note th a t th e fu = t r fo ro e z ) is (t rx a V n o ti c n fu e th s a re e und fo e b n a c m the li m it , w h h ic h w tm f o e lu in v a a rt e c a r fo m u im x a m s it and a tt a in s 0 ). > (t 0 = t d )/ (t x r a V d n from the c o n d it io Hence

"

--t 0 .5 = e n m

;~

) tm-;::::;0.1n "' ·

:::; -;:: ) (t rx a V x a m e c n a ri a v m u im x a m e th tm For this v a lu e of im u m x a m e th d n a .5 0 = ) m (t 'x O , 5 .2 0 n = .7 -0 e 1l ( 1 - e-0.7) = ) m (t x n )h m (t 'x O n o ti a a ri e th in ry to c e ir value of th e c o e ff ic ie n t o f v d e th to rm s te f o w o fl e th f o , .. ') ty si n te , n s rm te f o If we know th e in r e b m u n l ta to th e d n a k n a b ta a d e th t a g me ti e g ra e v records a rr iv in a e th , y c ra u c c a ie n t ic ff u s a h it w , e i. in rm te e d then we c a n = rm -t -n e 1 . e i. , ry to c e ir e d th f o t n e c r e p 5 9 ll fi to trlll_ necessary O.9;), whence tf in -;:::::; 3ni'A. n a c e w ), e c ti c ra p in e c n u rr e c c o t n e u q e fr a is h ic h (w c tu a l a e th e in If n is u n k n o w n rm te e d e W s. w o fo ll s a n ty ti n a u q e th f o n te a m , . . . , m find the e s ti m , 1 1 m ry to c 2 e ir d e th in s rm d te te la u m u c c a e se th e f o th s e er b m u m s u s n a e 'V ). i+ -c 1 , • • • , -r 1 (-ci < -c , -c , -c t u n 3 c e c a f 2 o s 1 e ti ti n a at ea.cl~ m o m u q e g ra e e av th to l a u q e ly te a im x ro p p g in quantities to b e a lv o S . l) , . . . , 2 , 1 (i = -r i/ n ) -A e _ (1 n = i m ndo p s e rr o c e ~~lated te rm s: th r fo n , . . . , nn, 1 n s e lu a v l d n fi 1 e w , n r ~ fo m a te ti s e e us e q u• a ti o n . 'V ). -c , (m , • • • 1 1 Ill ), (m 2 , -c 2 ), -c , (m : s e lu a 1 v f 1 -" pans o 11 om th e fo rm u la fr

vn:

uvn:

362

Applr.ed Problem1 ln

Probabtlltu Thtory

10 3 7• r or the cond 1tlODS Of the proced 1ng problem find the ttme f.rnt needed to fill the dtrectory by 95 per cent and the probabtllty that after t" o years of accumulat1on the dtrectory '' lll contain less than 90 per cent of all the posstble terms 1f the total number of terms n = 100 000 a tot 11 of 100 000 records are fed 1nt o the data bank per year and each document conta1ns an average of 1 5 terms Solutaon \V' e find the 1n t ens1 t y of flo'\ of tetms per record f~d Into the data bank per annum A = 100 000 x 1 5 = 150 000 1/}ear \\ e can find the quant1 ty ttJn from the expression tttu = 3 n!i = 1 100 000/110 000 = 2 1 ears To find the probabtl1ty that after t" o years of a.ccu mula tt ng terms the dtrect ory '' tll con ta 1n no less than 90 per cent of 1ll terms, 1t 1s first of all necessar} to find mx- (2) and Var:r (2) [ ~ee formulas (10 36 3) and (1 0 o6 4)1 ml: (2) = 100 000 (1 ~ e i 5Y2) = 0 95 < 100 000 = 95 000 Varx (2) = D1 000 X 0 05 == 4710 crr (2) = 68 V \Jote that the max1mum var1ance Varx (tm) = 0 25n = 25 000, 0 (lm) === 158 At the moment t ::;;::;: 2 J e1.rs the number of terms tn the directory IS a random v1r1able A (2) \\ htch 1S appro>..Jmately normally dtstr1buted "1 th the characterlSttcs obta1nod above Therefore, P {X (2} > 0 Un} ~ 1 Since lnx - 3ax > 0 9n tO 38 \Ve constder 1 1nore general case for the operation of a data bank dtrectory The fi.rst comphcat1on as compared to the hypothesis of Problem 10 36 Is that the max11num number of terms n tn the dtrec tory IS not constant but 1. funct1on of t1me t n.. (t) (tn the case of a dl rectory of the names of org1.nn:at1on..:; th1s means that the total number of organ1sat1ons var1es '\ lth t•me 1ncreastng or decreas1ng) In ndd1t1on, at ~orne potnt In t1me a term fed tnto the dtrectory 1s deleted from lt because the term becomes obsolete It 1s assumed that tl1e flo'v of deletlon'5 1s ., Po1sson process \Vlth 1ntens1t} ~L (t), wh1ch 1~ the same for all the terms 1n the dtrectory In that case the 1ntens1ttes of the b1rtb and death flo\vs have the form ;t"

iwh ( t)

= i ( t) ( 1 -

k

n (t}

)

Jlk ( t) =

~ ( t) h.

(

1

10 38~ 1}

and equat1ons (10 0 24) and!{10 0 25) assume the form (the relat1ons of the function m~ (t), Varx.,(t) n (t),!l~ {t), 1.1 (t), Pk (t) and t1me t are o n11tted for the sake of brevt ty)

dm:.t/dt = A - mx f'A/n + ~t), dVaro:ldt = 1.-- m:t (Ain ~ f.L) - 2 (Ain + ~) Var.x• (10 38 2) li the qua.nt1\1es A., n, f! are constant {do not depend on t1me), then, as t ). oo, stattonary cond1t1ons are poss1ble for \Vhic.h dm~/dt ~ dVarxldt = 0, whence ~

-

m;a:=n(t

+JL:·r

1 ,

Var"=m"(t !--

~~

r

1 •

CHAPTER 11

Queueing Theory

ri v in g ar ) ds an em (d s er m o st cu e rv ned to se ig es d e on is em g e, st sy an ch ng ei ex eu e n qu o h p le te 11.0. A a e ar s em eu ei n g sy st u q of s le p am x E ei n g . eu ts u en Q . m o em m st sy r te u p at random m co e iv ct an in te ra , 's er ss re rd ai h a , ce fi of gin ok sy st em s. g n ei eu u q a garage, a bo in ce la p g in k ta s processe c ti as ch e o er st h e T th ). h el it nn w ha s al (c de er ry rv se theo a ed ll ds) is ca an em (d s er m o st cu s. es em rv st se sy h g ic n h ei eu u q ) el n A facility w an ch ilt u rv er (m se ilt u m d an ) el nn sha sy -c ne er v (o er r -s le g n si a f o are single-serve le p am is an ex w o d in w e th t a an m a e n o of h le it p w am ex an is s w A booking-office o d in w en a t th e m al er v se h it w ce fi of gin ok tem, while a bo d delay an ) ls sa fu re multi-server system. h it w s m te ys (s s em estion syst ng co n ee w et b e at ti se rv er s n e re th fe l if d al so en h w We al em st sy n o ti es g t a co n a g in v ri ar er m o st ce ed cu ro A p s. er h em rt st fu y an in queueing sy rt a p g in k it h o u t ta w ts ar ep d d an e ic rv b u sy e se ar d s se er fu rv re se e th l al ~re busy is en h w g in v m er ar ri o st cu a , em st sy g n come ei be eu u to q er ay rv el se e th r fo mgs. In a d ts ai w d q u eu e an e th s in jo t u b em . If st ed sy it e m th li n e u av r o ed it m does not le li be er th e m ay ei eu u q e th in m s ce d ed la n p u o of b er b be m n ca e eu u q free. The nu A . em st es ti o n sy g n co a to in s rn d tu an em e) st eu sy qu ay e th of th g n m = 0, a d el le r o ze it (t h e si in s er m o st cu of er b h it m u w n s e th em st of y "s as n w o n ~oth in terms k e ar is k in d th of s em st y (s e m ti g n ei eu u q m ter~s of th e e. cu st o i. , ne li ip Impatient customers"). sc di e eu qu e th of s in te rm ed id iv d b su so al e r so m e ar o s , em er st rd o sy m o d Delay n ra a in r o al v er of ar ri rd o e th in er th ei p ri o ri ty ed A . rv e) se ic be rv se ty ri o mers may ri p (a s er th o e before ic rv se in ta b o to le ab be ay m s cust?mer s. to n ro p f o s k n ra r o s, n io at d sernce may h av e se v er al g ra of a q u eu ei n g sy st em ca n be p er fo rm ed in a si m p le st en ta ry em el e ar e at st The an al y ti c in v es ti g at io n to e at st om fr ri n g it er sf an tr ts en ev of ts in s en ow fl ev e e th th l n ee w et b s way ~vhen al al rv te in e t th e ti m a th s n ea m is h T si ty ). n 's te n in so e is o th P to al u eq (stahonary er et am ar p w it h a n o ti u ib tr is d al ti th en o n b o p t a ex th s an n ea m n o ti p the flow have m su as is sy st em th g n ei eu u q a r o F . rm te ow fl e g th in e nd us e po 'V . 's n so is o of the corres P y ar n are st at io s es oc pr e ic rv se e th cone d on an s y b es oc er th o an r te af the arrival pr e on ed s ar e serv er m o st cu g in v ri ar me e ti th e n ic he rv w se s e th if ly n o s;n·zce proces 's n so is o io n ar y P at st is s es oc pr is h . T n . o er ti u rv ib se tr sy is d al ti en n o tfnuously bu p ex an w h ic h h as le b ia ar v m o d n 0 ra e: m a ti is e r ic se rv T se e ag er av a customer e th of in v er se an is n o ti u ib tr is d is th f 2. 1-L ast is The P~ameter s es oc pr e ic rv se e h "t g in d of sa y ea st In · rl se [T M s we = ow er ll rs fo re at h he w In ". al ti en ~~ = lf ts er , w n o p ex ti m e is e ic rv se e h "t y sa n io n ca at e st w e " ar 's s on se es oc pr e th t;onary Poiss l al h ic h em in w st sy g n ei eu u q y er d ea l ev , ly y n it v ai re m b l r al sh e w r te s !all call, fo ap ch is th st em . In sy g n ei eu u q ry ta en em el an s n' ary1Poisso 1 ei n g sy st em s. ng ei eu u q a in s es oc pr " 1 I elementary q u eu e th en th , 's P o is so n y ar n io at st e ar ts en e. ev m ti of s s w ?u u flo n ti n co If all the ~l an es at st dis~rete .b it ~v s es oc pr c ti as ess ch oc o pr st ts v th ko r fo ts ts ex e at st ~;htem is a_ Mar ry a n m 1 t st a tw h a . d le il lf fu e ar th e s n of io s it d ic n st ri co te ac ar ch er . en certam th o e th r st at es n o e th of s ie it il ab b ro p e th er th IU which n ei it ie s il ab b ro p e process depend on ti m e. th g in nd fi e d u cl in w it h ls ea d ry eo th g n ei eu th e u q n e ee w th et s b m ip sh n o ~ proble ti Th la f re a g in sh d es ta b li an em st sy 0 g n ei eu disu q e a th , of 1.. s s te es R oc pr al v Yanous st ri ar e th of in te n si ty e th n, s er rv se f o of er b cy m en u ci n fi ef he (t e th of s ic st P~ram.eters ri te ac ar ch ) an d th~ on so d an e, m ti e ic rv se e th ~hlbt~ho~ of er: 1d ns co n ca e w . le p am ex r e o F m ti it n u er p em st e -e rn ce sy st em . sy g n ei eu u q ed b y a rv se A s er m o st cu f o er b m u n e or l~~ averag e; ic rv se r fo em st sy e th of ty ci l e absolute capa

••

361}

l ppll cd P ra ble ms 1n Proba. bllUy Th tory

the probab1hty of serv1ng an arr1v1ng customer Q or the rtlative capadty o! the .s:,. stern for serY tee Q = AlA tho probab1hty of a refusal Prl!f 1 e the probab1hty that an arrtving custome-r w1ll not be served Pr~J = 1 - Q the average number or customers present 1n a system (betng .ser;;ed or 1n the

-

-

queue) :

the average number of customers 1n the queue r the average v.n.iting llme of a customer (tho average t1me a customer JS present 1n the- s~stem) t" tho average queue1ng time of a customer (the avera.ge ttme a customer spends

-

1n the queue) tq _ the a\erage number of bu:sy ser' ers k In a general case all these character1sttcs depend on ttme But many serv1ce Slstems operate under the snme eondillons !or a sufficJently long bme and there fore a st t ua tton c] o.se to ~ta tJ onn ry b established \Vt thout speetfy1ng 1t every tl.Dle \\ e sha 11 every'' here ea IcuI ate l1 m 1t1 ng pro bah 1ltttes of the states and the h ml t1ng charactertstlcs of the efficJeocy of a qucue1ng system w1th respect to 1ts lun1t1Dg steady state operation For any open •} queue1n g system opera t 1ng under I1m 1t tn g s t a t1 o nary cond 1hons th.e n 1o erage ~\o.aJ tJng tJme Dl a customer ~~ Js expressed In terms oJ t.he .avPrage num. her of customers tn the system by Ldtle s formula

-

(11 0 t)

"'here A Js the

1nten~nty

of the arrtval proees!l

A stmJlar formula (also kno'Wll as L1ttle s formula) relates the average queue1cg time of a cu~tomer tct and the average number r of customers 1n the queue tq

(ii 0 2)

r )..

L1ttle s forn1ulas are 'ery useful s1nce they mal.e 1t pos~nble to calculate one of the t" o character1.~ttcs of the efficH~nc} (the- a' erage queueing tlme for a customer and the average number of customers) wlthout nece~Q'arlly calculahng the other It should be emphasized that formulas (1 1 0 1) and (ii 0 2) are vahd for any Dptn queueing systtm (s1ngle 5erver multi server and for any kind of arr1val and serv1ce process) the only requ1rement he1ng that the arr1val and serv1ce processes mu~t be stah.onary Another very Important e'tpress10n for an open queue1ng system IS a formula v.h1ch expresses the average number of busy :servers/... ln terms of the absolute capac1ty of the S) Qtem for service A

-

J.

= Alfl

(1 t 0 3)

\v here J..t = 1/i;er 1s the 1n tens 1t y of the Berv1 ce process 'Ianl problems Ill queuetng theory conccrn1ng elementary queue1ng systeiil.S can be so]ved using a h1rth and death cha1n (see Chapter 10) If the drrected graph of states of a queueing system can be represented as tn F1g 1 t 0 t then the hmttlng probah1htles for tho states can be e"tpressed h) formulas (tO 0 23) 1 e Po

{t+ i..o + lo~t + flt

~1 r.Li

+ h~At J.ldt1

h~ llli

t

+

1

AoA1 Jlt Jl2.

~n

1 }-

1

J.l )1

A queue1ng system ts sa1d to he open lf the 1ntens1ty of the arnval process does not depend on the state of the system 1tse1f., *)

Ch. 11. The Queueing .Theory

365

(11.0.4) When deriving a formula for the average number of customers (in a queue or in a system), we often use the technique for differentiating a series. Thus if z< 1, then 00

00

00

~

kxh=x

d d xh=x dx dx

~

~ 1Ye get p 0 as a function of t:

Po(t)=

hence

f..t!l [1 +

Pt (t) = 1- Po {t) =

When t --+- oo, we get limiting probabilities, Yiz.

Po

=

~t/(1,

+ !1),

P1 = /J(/.

+ ~t).

(11.1.3)

370

Applied Problem1 !n ProbalJflity Theory

\\ e can find them much more east] y by c:.ol v 1ng aIgebr a 1c Iinear equa t lOns for the 11m tlJng probah1l 1t 1es of state~

Po + P1 = 1 'Vc can re't\ rt te formulas (11 1 3) tn a more conctse form 1f we lntroduce a deslgn 1tlOD r = Alfl

A.po

=

llP1

Pe = 1/(1 + p) P1 == p/(1 + p) The char"'lctertsttcs of the efficiency of the queue1ng system a:re A=?l.pa=

t~p

Q=

i~p

-

Pnr=Pt= t+.p ' (f f 1 4)

p

k=1-po = t+p j

1 2 G1' en , t e) epl one 11n n "h1 ch 1s a

~1ng Ie

server congest 10n sys

tom calls arr1' e at t c; tnput ln a stnttonary Poisson proce s The trafiie 1ntens1ty 1s "'- - 0 4 calls per m1nute The a' erage duration of a con versat1on t5 er = 1 mtn and has an exponential distribution F1nd the l1mlt1ng prob1btl1t1es of the states of the system p 0 and p 1 as '\\ell as A Q P ret and k Compare the capac1ty of the system for servu~e and Its rated capac.1ty If each con' crsnlton lasts exactly three minutes and the calls arrt ve regularly one after another Without a break Solution l ~ 0 4 ll - 1fls~r = 1/3 p = All-l = 1 2 By formulas (111. 3) Pfl =::: 1/2 2 ~0 455 p 1 ~0 545 Q ~0 455 A= '-.Q ~ 0 182 k = p 1 ~ 0 54.5 Thus on the a" erf)ge tl o telephone ltne wtll serve 0 455 calls 1 e 0 182 calls per mInute The r 1ted ca pac.I t y of the server would be (1f the calls arr1ved and '\ere ser\ ed regularly) A rate = iftaer = 1(3 ~ 0 333 callslmtn and that 1s almost t'' 1ce as large as the actual ca pactty A 11 3 \\ e aro given a single serv-er congest1on system \Vtth customers aJ"rlvJng Jn a qtatJonary Po1sson process w1th 1ntens1ty A. Tbe .se.rvJce t~tr

..t .

®

D

it

~

=~

•=•

'-

t,~ 'toT

r

®= ~

-'t2

•

tser ~.

-v

@~

. . "ts

A

•

lj_

,

t

Fig t 1.3

t1me ts nonrandom and lS exactly tser = 111-1 F1nd the relattve and absolute capac1ties of the S) stem for service 1n the l1m1t1ng stationary state Solutton Let us constder a stattonary Potsson arr1val procecos \VJth tntens1ty k on the t ax1s (F1g 11 3) \Ve shall denote all the customers \vho are served by ctrcles Assume that a customer "ho arrtved at the moment t 1 1s he1ng served Then all the custome_rs who arrtve after htrn during the t1me t ser wtll not be served The next

Ch. 11. The Queueing Theory

to be served

371

be the CUStomer who arrives at ar moment t 2 such that t2 - t1 > tser· Let us consider the interval T between the end of the first customer service period and the moment t 2 the next customer to be served arrives. Since there are no aftereffects in an elementary flow, the distribution of the interval T is the same as that of the interarrival time in the general case, i.e. exponential ·with parameter 'A. The average length of the interval T is m t = 1/"A. Thus nonrandom busy periods of the server (tser = 1/f.t long) and random idle periods (average length 1/lv) will alternate on the t-axis A fraction of all the customers '\Vill

f.. f..+

1/!-1

1/fl+ 1/f.. -

f1

'

will fall on the first intervals and a fraction f.t/(lv

+ f.t)

=

11(1

+ p),

where

p

= lv/f.t

will fall on the second intervals. The latter quantity is the relative capacity of the queueing system for service

Q = 1/(1

+ p),

(H.3.1)

whence

A = lvQ = !vi( 1

+ p).

(1'1.3.2)

Note that formulas (11.3.1) and (11.3.2) coincide with (11.1.4), which co~re.sponds to the exponential distribution of the service time. And th1s IS natural since Erlang's formulas remain valid for any distribution of the service time with the mean value equal to 1ht. 11.4. Using formula (11.0.5), prove that for an elementary singleserver system with an unbounded queue the average number of customers being served iR = p/(1 - p), where p = lv(f-1-, and the average number of customers in the queue is = p2 /(1 - p). (k ~lution. By formulas (11.0.12) p 0 = 1 - p, Pk = pk (1 - p)

z

r

- 1' 2, ... ).

We designate the actual (random) number of customers in the system as Z: 00

Z=M{Z]=

2J

00

kpk=

k=O

2J

00

kpk(1-p)=(1-p)

k=1

2J

kpk.

k=!

By formula (11.0.5) for p < 1 00

"" kpkp LJ - (1-p)2 . k=1

~h_:nce Z= p/(1 -

p). The average number of customers in the queue IS" • . "~Inns the average number of busy servers k = Alf-L = 'J../~t = p, I.e. r [p/(1 _ p)] _ p = p2/(1 _ p).

=

2~·

372

AppJJed Problems In ProbabJIJty Theory

1f 5 A rat I\\ ay

~o.hunt1ng J ard 1s

a sLng1e-ser' er queuetng system \Vlth an unbounded queue nt "htch tratns ntrt' e 1n 'l stationary Po1sson process The traffic 1ntens1ty JS .A ~ 2 tra1ns per hour The .serv1ce t1me (shunting) of a tt-a1n at the lard has an c'tponent1al distribution v, 1th mean value = 20 mtn ftnd tho lim1t1ng probahtllties of states of the S) stem, the average number .z of tra 1ns at the yard, the average number r of trruns In the q11eue the a, crage "a1t1ng t1me tw of a tratn 1nd the average queuetng ttme fq of a tratn Solution ).. = 2 traJn.s/h ~er === 1/3 h ~L = 3 tratns/h, p = 'JJp. = 2/3 D} formulas (11 0 12) "e have Po = 1 - 2/3 1/3 p 1 ~ (2/3) (t /3) = 2'9 p 2 = (2/3)~ (1/3) = 4127, , Pk = (2/3)~ (1/3) etc By formulas (11 0 13) and (11 0 14) = p/(1 - p) = 2 tralnSf = 4/3 tratns Tw = 1 h and fq == 2/3 h 11 6 The h} pothe~ts of tho preccdtng problem ts compllcated by the condttton that nQ more than three tratns can be present simultaneously In the ~hunting ~ ,rd (Includtng the tra1n being served) If a train arr1ves at a moment 'vh~n there f{fe three trains 'l.t the yard, tt has to JOln the queue out~ 1de t 1u} 5 1rd on a s1d e tcacl. The sta. tton has to pay a fine of a roubles p~r hour 1f a tra1n st 1.ys on the stde track F1nd the a' erage ftne per day the stat1on has to p1) for tratns staying on the stde track Solution '' e calculate the a' eragc number of tra1ns Zs!de on the

t;er

-

-

=

z

r

stde track

-

co

Zstde

= i P~t.

+ 2ps +

'OD

co

~ kp = p ~ 11

~~~

-

h~i

B, LJttlo s formu1a

00

(.'0

= 2.J 1.. p,_ - ~ kp" Po = Po }] ;{p11 , h~'-

ll=-l

h=i

OG

:p ,l = p ~

d p"~p d

dp

A~l

-

~p~t.

dp LJ

11=6

the average t1me n trazn stays on the su}e tr.ack

1 18/~ = 1 18/2 = 0 59 h On the n'erage 24 A = 48 tra1ns a.rr1 ve at the st a tJ on C\ ery 24 hours The a' erage datly fine JS 48 X 0 59 a~ 28 3a 11 7 Us1ng the btrth and death q,cheme, calculate dtrectly from the directed graph of st ,_ te.s the l1m1 t1ng pro ba b 1l1 t1es of states for a SliD pie two ser' er queueing system (n = 2) wtth three places 1n the queue (m = 3) for k ;;:: 0 6, 11 = 0 2 and p = 'AfJ.t = 3 Ftnd the character lStlCS of thiS queUeing system l e Z, tw, tq Wlthout USlDg formulas (11 0 26) proceedlng rather from the ltmtttng probabtlrttes, and com pare the results 'vtth those obtatned from (11 0 26) tsJd@ ~

r

Ch. 11. Th e Q zw ue in g Th eo ry

373

is em st sy ng ei eu qu e th of es at st Solution. The di re ct ed gr ap h of e th t ge e w p, = 1-1 A/ n io at gn si de e shown in Fig. 11.7. In tr od uc in g th ..;. .

So

~

Sz

St

S~t

SJ

Ss

2p

2p

2p.

~

Fi g. 11.7

n: ai ch h at de d an th ir b a g in us following,

Po= { 1 Pt =

+p +

+ + 23 + 2

p2 2

0.074, ~ ~ 40 58 p4 =

p2 =

10 .1 5 ~ . 40 58

p5 } -1

p~

p3 22

0.250,

p5 =

(40.58)- ~ 0.02::>,

:0 ~5 8 ~ 0.165, 5

p3 =

15.18 ~ _ 40 58

•

0.375,

0.250 + 5 X 0.375~3.67, , 11 6. ~ .6 0 z/ = tw , 79 1. ~ 5 37 0.

X 0 .1 6 5 + 4 X

r= 1 X 0 .1 6 5 + 2 X 0 .2 5 0 + 3 X

I

=

0.111, ~ ~ 6: 4 8

;= 1 X 0 .0 7 4 + 2 X 0 .1 1 1 + 3 I

-I

-

1

tq = rJo.G ~ 2.98.

) I

1. > x or 1 < x y an r fo id al v 11.8. The fo rm ul a fo r- r (H .0 .2 8) is . 0) 0/ e lu va ed in rm te de un an ng di el yi For x = 1 it is no lo ng er va li d, .l

So

~

s,

,1,

.r---._;.;1.

eoe :1 t:eo•:n;ti Sk

Sz

Sn

~oee np

kJL

Fi g. 11.8

e th of s ie it il ab ob pr e th ve ri de n ai ch Directly from th e b ir th an d de at h ac ar ch cy en ci fi ef e th nd fi d an se ca is states p 0 , p 1 , • • • , Pn +m for th . tw , tq k, • er Pr , Q , A e. i. , teristics of th e qu eu ei ng sy st em ha s em st sy ng ei eu qu e th of es at st of h th Solution. The di re ct ed gr ap th ir b e th r fo as ul rm fo l ra ne ge e th e form sh o\ m in F ig . 1'1.8. U si ng in ta ob e w p, = -1 !J1 g in at gn si de d an and death scheme

r, z,

l

' •

011+1 n· nl

pn+m } -1 ' -'-n=m-.n-:-1 • • • -L

+ 1! + 2! + · · · + nl + · + , )z (P r r n r p r t { = + 11 + 2! + •.. + n! ~ n + n

Po= { t ,I

pn

P

P2 2

I

For x = p in = 1 •

.'

P o= { 1

+

p 1!

p2

+ 2!

+ .. .

pn T n!

+

.. •

-1 ]} )m (r • + n

mpn } nl

1

'

( 11.8.1)

-- -- -

-

~--

874

Appllet!

Prcbl~mJ ln.

Prt:Jbabllitv Theorp

L

(11 8 3)

z:=:r+k, Tw=zi'A, tq=r!Jv. t 1 9. A petrol station has two petrol pumps (n = 2), but 1ts forecourt

can only hold four \Va1t1ng cnrs (m = 4) The arr1val of cars at the stat1on 1s statJon.nry Poisso n's WJth 1ntens1 ty :h.= 1 earlmi n The service t1me of a car 1s expone nttal "1th mean values tser = 2 m1n F1nd the ltmlt· of the states of tho petrol stat1on and 1ts characteristics 1ng probab tlltlcs - ........... 1 A , Q ! re rt k z, r, t vn t q Solut1o n A, = 1 11. = 1/2 = 0 5, p = 2, x = pin = 1 From formulas (11 8 1) (11 8 3) \\C find that 1 21 2' 1 = 13 ' 2l 4 Po = { + 2 + 21 ~

}-1

+

2

Pt = P2 ~ P3 = p., = Ps = Pa = 13 ' Pret = 2/13t

-

r=

23 21

Ptef = 11/i3, A .= lQ = 11/13 ~ 0 85 cars/m1n, J: = A/1• = 22/13 ::::1- j 69 pumps, Q = 1-

4 (4+ i} 2

1

13

~154

cars,

.z=r+A.~323cars

server conges tton s}stem at a rate of ;., = 4 custom ers per hour The a\erag e service t1me tser = 0 8 h per custom er The Income fron1 every custom er served c = 4 roubles The upkeep of every server 1s 2 rouble s/h Dectde whethe r It Is advan tageou s to 1ncrease the number of servers to three Solut1on By Erlang 's formul as (11 0 6)

t 1 10 Custom ers arrl\ e

Pa={1+32+

3

rt

2

Q = 1 - p2

'lt a two

1

=-"Qg tbe.m .bJ" /?F}NW!i)

p; =

{

1

+

3 3 2+ .}/

Q'

+ T' } 3 3

-t

~ o 0677.

p; ~ 5 48 x o 0677::::::0 371,

P3 :;:::::; 0,629, A I = 4Q' ~ 2 52, D' =A' c ~ 10 08 rouble s/h

~ 1 -

Ch. 11. The Queueing .Theory

375

The increase in the income is D' - D = 2.88 roubles/h, while the increase in the cost of the upkeep is 2 roubles/h. It can be seen that the transition from n = 2 to n = 3 is economically advantageous. 11.11. We consider. an elementary que'lleing system with a practically unlimited number of channels (n-+ oo ). The demands arrive with intensity 'A while the intensity of the service process (per channel) is 1-L· Find the limiting probabilities of states of the system and the average number of busy channels k. Solution. This que.ueing system is neither a congestion nor a delay system, but it can be regarded as the limiting case for a system with refusals for n-+ oo. Erlang's formulas (11.0.6) yield co

"\1

pk } -1

"'

pk

-

=e-P, where p=!l' Pn= kl e P=P(k, p)

Po= { L.J k! k=O

•

(see Appendix 1). For an infinite number of channels A = "A and k = t.f~t = p. 11.12. We consider a single-channel congestion system to which demands arrive in a stationary Poisson process with intensity "A. The service time is exponential with parameter 1-L = 1/~er• The operating channel can fail from time to time (refuse), the failure process being

(a)

-+-1----{@>--~~,.--{@ 0

1

t

)( X

@>--4X~XE--~)-E-(.....,@-@~)(~-

t

X -service process p @

-refusal process v {b)

Fig. 11.12

sta tionary Poisson's with intensity v. The reconditioning (repair) of a c1lannel begins immediately after it fails, the time of repair T r being exponential with parameter I' = 1/I;. The demand which was being ser~ed at the moment of failure departs from the system unserved.

Fhd the limiting probabilities of the following states of the system:

~b' t e channel is idle, s1 , the channel is busy and in working order, s 2 A e channel is being repaired and the characteristics of the system are and Q. . Solu~ion. The directed graph of states of the queueing system is gi\'en In Fig. 11.12a. The algebraic equations for the limiting proba-

Applfed Probltm• in Probabllzty Theory

376

btltttes of states are

f1P1 + 'YP2t (~t + v) P1 = i..po, ~P1 = 'VP2' (11 12 1} they are summed '' 1th the norm 'l}tztng cond1tton (1112 2) Po Pt + P2 = 1. \Ve el'pre~ tho prob'lhJlJtJes p 1 and p~ from (11 12 1) In terms of p11

APo

=

+

AV

~

A

Pt = ~~+v Po' P2 = y P1 = 'Y (~"+v) Po Subst1tut1ng p 1 and p 2 1nto (11 12 2), '' e get Po = { 1 + )./(~L + \) + Av/[1' ((t v)l}-1 (11 12 3} To frnd the rclat1 vo capac1 ty for servtce Q, the probabtltty PD that the demand 'v1ll be occepted for service must he mult1pl1ed by the cou· diltonal probab1l1ty p that the demand ,vbtch lS accepted for serv1ce \VIll actually he served (the channel 'v1ll not fa1l dur1ng the servtce ttme} \Ve ~hall use the tntegrnJ total probabtltty fo-rmula to find the condttlonal probab111ty \\ e ad' a nee a hypothcs1s that the serv1ce t1me of a demand has fallen on the 1nterv1l from t to t ;-- dt, the probabd1ty of th1s hypothesis he1ng approxiniately j (t) dt. 'vhere f (t) Is the dts tr1button denstty of the servtce time,. 1 e J (t) ~ ~Lew-J.l.t (t > 0) The cond1 t1onal proba btlt t y that the ch nnnel v. 1ll not fa 1l dur1ng the ttme t

+

1s

e

'Y t

hence 0::.

[.'1(]1

p

= )

!1P

~le-VI dt = )

J!e-(P+'Y)I

dt = ~~'V

,

0

0

Thts condlt1onai probahlllty can be found much more eas1Iy 1t 15 equal to the probabtl Lt y that once 1t hns begun, the .serv1ce procedure '' 1ll be completed before the channel fa1ls. 'Ve super1mpose on theta 'tiS (F.Jg 11 12b) the two processes, v1z the service process w1tb Jnten~Jty 11 (denoted by crosses) and the refusal procc~s \VIth 1ntens1ty v (denoted by circles) \\ e f1x a potnt t on the t a~ 1S and find the probab1lt ty that the first cross follo\v1ng the fr'\ed potnt w1ll arrtve earlter than a ctrcle It ts ev1dently equal to the rat1o of the 1ntens1ty of the of croc;:ses to the total 1ntens1ty of the floVts of crosses and Circles, ~L/(Jl v)

no"

Thus

,... } Q--PoP ,. ---- ( J-1+ v

I (1 + p.+A + v

h'V

1' UL+v)

) ..-

+

p.

ll+v+A (i+"'IY)

f

A=AQ. (11124) J 1.1341 The cond.1t1ons of Problem 1 f 12 are repeated w.1th tbe only ddlerence that the channel may fatl dur1ng an 1dle period too (\vith tntens1ty -v' < v) Solution The d1rected graph of states of the queue1ng system shown In F1g 11 13 From the equations

(h

+ \'') Po =

f!Pt

+ YP1' Pa

(!-1- +rv) P l

+ P1 + Pt

=

'Ap ru ~P2 =

= 1

"VPt

+ v' Po•

IS

Ch. 11. Th e Q ue ue in g Th eo ry

37 7

we find the li m it in g p ro b ab il it ie s

i + tJl+'),. v + !vv+Jl!l+Vv'+vv' } -1 ' { = Po lvv

'). P t= Jl + v .; P o •

P z=

Q = p o Jl ~ v ,

+ JlV' + vv ' ~t+v

A.Jl

A = 'A Q = Po Jl + v

Po•

.

em st sy g n ei eu u q el n an h -c le g n si 11.14. We co ns id er an el em en ta ry qu eu e, m = 2. T h e op er at in ge th in es ac pl of er b m u n ed it m li with a -

Sao foE

ft

fl

ft J)

}'

p

'}'

j)

J_ ,.

Stt

}'

Szt

+

+

+ +

+

' '

J)

•

m en t o m e th at ed rv se g in be d an em d channel m ay fa il (refuse). T h e if d an , es ac pl d ie p u cc o n u e ar e er of failure joins th e w ai ti n g li n e if th . T he in te n si ty of th e ar ri v al ed rv se n u ts ar ep d it es ac pl no e ar there ses oc pr re u il fa e th of t a th ~~, is es s process is A., th at of th e se rv ic e pr oc y. is s es oc pr r) ai ep (r g in n io it d n co Ef the channel is v, an d th a t of th eemre, find th ei r li m it in g p ro b ab il it ie sst sy e th of es at st e th e at er um •n = 1• "' d an 5 0 = 'V 1 = lL 2 = A ' r r • ' ' rand determine A ' k ' ' ' tW ' tq fo Sol ution. The st at es of th e sy st em are: r; de or g in rk o w in is el n an ch e 8 oo -t he sy st em is id le , th u eu e; q no is e er th , er rd o g in rk o w in 5 o -t h e ch an ne l is b u sy an d 1 is d an em d e on ; d re ai p re g n ei b 5 u -t h e ch an ne l h as fa il ed an d is Waiting to be se rv ed ; gin be is d an em d e on r; de or g in s2o-the ch an ne l is b u sy an d in w o rkbe se rv ed ; to ' nO ti ai w is e on er th o an d an ed rv se ear s d an em d o tw ; d re ai p re g n ei b . s21-the ch an ne l h as fa il ed an d b is In the queue; in e ar s d an em d o tw r; de or g in o rk th Sso-the ch an ne l is b u sy an d in w se rv ed . g n ei b is d an em d e on d an e eu e qu he T 4. .1 11 . ig F in n w o sh is Th~ graph of th e st at es of thbeabsyilstitem e ar es at st of s ie ro p g in equallons for th e li m it P1o• v) 'A + (~ = oo l"P Pu Y + ~lP1o = l•Poo, ~P2o o• P2 v) A ' L (~ = o P1 A 1 P Y 2 ~lPao + Pao• ) ' \ ' + l (~ = o P2 V o• Ps ) ·v + L (~ = AP2o 1• P2 Y = o Pa V o P2 V pn 'A w P "P1o = (1' + 'A) = '1. o Pa 1 P2 o P2 u P 0 P1 Poo

rz

I

•

Fi g. 11.14

Fig. 11.13

I

S3o

Szo

SIO

+

+

+

+ +

+ +

Applied Problems tn Prohabllltv Thtorv

878

Sol vt ng the~e equa t tons at ~ p 00

p 30

= 3/61 ~ 0 049 = 56/183 ~ 0 306

=

2 11 - 1 v

= 0 5 and y ==== 1

p 10 = G/61 ~ 0 098 p 11 = 1/Gt ~ 0 016

P20 ;::;:

we obta1n

14/61 :::::; 0 230

p 3 t = 55/183 ~ 0 301

Hence

z = 1 + P t) + 2 (p + P + 3p = 383/183 ~ 2 09 r = 1 (p + p + 2 {p + p ~ 89/61 ~ 1 46 k = 1 (pte + Pto + Pao) - 116/183 ~ 0 63 (p1o

20

1

20

21 )

11)

21

30 )

3o

The abc;tolute capac1t' A of a queue1ng system 'v1th nonrefus1ng channels can be found by mult1ply1ng kby J.t In our case the producttv 1ty of one channel (the number of demands actually served per unrt time) can be found by m ul t 1pl) tng k by the pro bab1l1 ty ~-tf (I-t v) that the servtce procedure'\\ 1ll be completed t e A - kt-t ~t/(~ v) k)l21(J-L + -v) ~ 0 42 1 t 15 There are three cha1rs (n 3) 1n an outstation dental surgery and three cbatrs tn the ' tttng room (m - 3) Pattents a:rrtve 10 a statronary Potsson proce~s 1.t a rate of A - 12 patients per hour The servtce ttmP (for one pat1ent) s exponential "1th a. mean value ta~r = 20 mtn If all the three chntrs n the 'va1ttng room are occupted an arr1 ving p'lt 1ent does not JD1n tl1o queue F1nd the average numbPr ol pattents that the denttsts serve per hour the average fractton of the arrlvtng pat1ents that are actuall} served the average numb~r of occu pted cha1rs In the \\a t1ng room the average \Va 1t1ng time tw of a pat1ent (Including both the dental surgery and 'va1t1ng room) and the a' erage "'a1t1ng t1me g1 ven that the pat1ent 'vtll be served Solution p 12/3 ~ 4 x = p/3 4/3 n = 3 m = 3 From for mulas (11 0 26) (11 0 30) 've find that

-

+

+

-

41

+ s-+

>'}

1 413 -< 1"Jo 0121s~o 012 P0 ={1+4+ 2 6 3X6 t-4/3 ~ ,...., Pt = 4 X 0 01218 :::::r 0 049 p 2 - 8 X 0 01218 ~ 0 097

Ps =

4

4 "

32 0 a X 01218 ~ 0 130

Ps+2- {- P3+1 ::::::

0 231

P3 +s

1

4' 3 p 3 +1 =as X 0 01218 ~ 0 17

=-} P +z ~ 0 307 3

The averago fract1on of pattents betng served 1s Q = 1 - Rret = i - Ps+s ~ 1 - 0 307 - 0 693 The average number of patients ser,ed by the dent1sts per hour lS A AQ ~ 12 X 0 683 ~ 8 32 By formula (11 0 27) the average number of busy servers (denttsts) k = 4 (1 - Pa+a) ~ 2 78

-

Ch. 11. T he Q ue ue in g T he or y

379

e u e u q e th in ts n e ti a p f o r e b m u By formula (11.0.28) th e a v e ra g e n 4 3 ""' "' 1 56 ) /3 (4 3 + ) /3 (4 X -4 1 8 - = 44X0.0121 2 (1 -4 /3 )

3x6

r

;"o ../

'

•

-=2 h. 6 3 . 0 ~ . A / z = w f , h 3 1 . 0 tq r/11.~

z=r+k~4.34,

in jo t o n o d ts n e ti a p e m so se u a c be ll a sm re a tq d n a 7w of es lu va e e Th m ti g in it a w e g ra e v a lf a n n d it io o c e h T . d e rv se n u rt a p e d d an the queue e th d n a , h 2 .5 0 ~ Q ! w t = fw is ed, rv se is e h t a th d e id v ro p t, n ie at p of a = tq ) n io it d n o c e m sa e th r nde (u e m ti g in e u e u q e ag er av l na ~onditio rte tq!Q ~ 0.19 h. e d in n a ld ie y ) 8 .2 .0 1 (1 d n a 6) .2 .0 1 (1 s la u rm fo 1 = x r o F 6. .1 11 and rm fo te a in rm te e d in e th f o e v a lu e th d in F . /0 0 rm fo e th of y ac in m = 1. x r fo d li a v re a h ic h w s la u 'OI'l'ite form le ru s l' a it sp o 'H L y B n. io ut Sol ~

~

.

hm Y. -+ 1

f

1 l = o P

1 -x m 1-x

=

-m x m -1 -x

pn n!

p

+ +···+ 11

Ph=

~~

P n+ i = P n+ 2 =

Po

+

=m, pn+lm } - 1

'

n ·n l

(11.16.2)

(1~k~n),

•· · =

P n+ m =

pn nl

(11.16.1)

Po•

(11.16.3)

m• n+ P h it w g in d n e d n a n P h it g w in n in g e b s, ie it il b a b ro p e th l al i.e. are equal. . e m sa e th in a m re k d n a f ), 8 .2 T~e ~ormulas fo r A , Q, P re .0 1 (1 la u rm fo in rm fo te a in rm te e d in e th f o e lu a v e th g dm Fm we get .4) 6 .1 1 (1 . il + (m m po +l pn = x m + m x m + l - m (m + 1 )

lim 1 -( m + il

11~1

l

(1 -x )z

r

2

'

2 n ·n l

e. m sa e th in a m re ) 0 .3 .0 1 (1 d n e th Formulas (11.0.29) a g in d n fi t u o h it w ) .3 6 .1 (11 ).1 6 .1 1 (1 s la u rm fo e iv er d ld u e. m e . We co h sc th a e d d n a th ir b e th g n si tu u b rm fo te a in rm te e d in e th of e 'alu , T , k t• re P , Q , A s ic st ri te c ra a ch cy en ci fi ef e th te la u lc a C ) (1 . 7 _11.1 in s ce la p e re th h it w m e st sy r e " e rv -s le g n si ry ta n e m le e n a r fo tw , ~, tq r = 1/f.L = ;e 't; d n a r u o h r ~e s r e ~ o t us ber m u n 6he queu~ (m = 3) g iv e n '). , = a4ra ccte e th n e h w e g n a h c l 1l w s ic st ri £5. (2) F m d o u t h o w th e se c h . 4 = m to d se a 0 re c in is e u e u q ) 6 .1 places in th e .0 1 (1 )2 .1 .0 1 (1 s la u rm fo y B 2. = /f.L A. = p , 2 = L f. . n ~ Q f Solutio 'A = A , 4 8 .4 0 ~ Q , 31 6/ = ·1 p , 1 /3 1 = o P e 4 v a h e w 3 = or m ~ z , rs e m o st u c 9 .1 2 :: ::: r:: , 8 6 .9 0 ~ pQ = k r, u o h er p s er m o st cu 1.93 3.1G customers, ~ ~ 0.55 h a n d fw :::::::0.79 h .

'

380

Appll~d

Problems fn Probabilltv

Th~ljry

(2) For m = 4 '' e have Po = 1 63 ~ 0 0158 p 5 = 32/63 ~ 0 507 Q ::::;: 0 493 ,t ~ 1 96 customers per hour ~ 3 11 customers ~ 4 09 customers tq ~ 0 78 h and tw ~ 1 02 h Thus an 1ncreac.e In the number m of places from three to four leads to a neg l1gible 1ncre'lso 10 the nbsol u te (and rela t1' e) ca pac1 ty of the system for servtce It doe$ tncrea~e some~hat the average number of customers 1n the queue '\nd 1n the S}stem and the correc;pondtng a'erage t1mes Thts ts natural since some of the customers who \\ ould he refu~ed 1n the first var1ant JOin the queue tn the second 11 18 Ilo\v 'vtll the cfftctenc)' cbar acterts tics of the system 1n the prccedtng problem change tf A nnd p. rematn the ~arne, m = 3 but the number of ~ervers 1ncrea.se to n ~ 2? Solution x 1 from formulas (11 16 1) and (11 16 2) \Ve have Po - 1/11 P 1 Ps = 2/11 Q - 1 - 2/11 ~ 0 818 A ~ 3 27 customers/h 12/tt ~ 1 09 customers f = Al~t ~ 1 64 i""+h. ~2 73 cu~ton1crs tq ~ 0 27 h nnd tw ~ 0 68 h 11 19 The queueing system 1s a ra1l \vay book1ng of flee "1th one

r

z

=

r-

r

'vtndo\v (n t) and an unbounded queue The man 1n the 'v1ndo\\ sells t1ch.ets to potnts 1 nnd B On the a\ erage three passengers e'\ery 20 minutes arrr\ c to bu:-;. '"\ ttckct to po1nt A and t\\ o p1s~engers every 20 mtnutes arr1' e to buy a ttcket to B The arrt\ al process can be con stdered to bo stationary Potsson s On the a' erage three passengers are served 1n 10 mtn The "erv1ce t1me IS e"" ~ 2.316 min.

Adding together the average numbers of customers in each queuet ~·e obtain the total average number of customers in the queues, i.e.

r=~+rz+Ts~ 15.5. _B~~ analogy we find the average number of customers in the system '== zl 2 + 3 ~ 21. 5. The average queueing time per customer

+z z

tq = t~' +t~> +t~':;:::;; 20.6 min. The average waiting time per customer tw=t#'+tW+tW~28.6 min. {1) The service can be improved by decreasing the waiting time of ~~usto~er in the first phase, which is the weakest phase in the system. he easiest \\'ay to do this is to increase the number of assistants, i.e. e number of servers in the first phase. For instance, a simple unit ~c~asc i~ the number of assistants (i.e. a transition from n1 = 4 to 1 - 5) Ywlds an essential gain in time. Indeed, for n 1 = 5 we obtain

384

Applted Pr The servtce tnt~ of a train T,e 1 IS dtstrtbuted o'er an Interval frc 1 0 to 1 b \vtth denstly rp (t), shown 1n F1g 11 : Us1ng formulas (11 0 34) (t 1 0 37), a ppro"trmf · 2 the efficJenc} cltaractcrtStlc..~ of the lard, 1 e t ~ ..... a' erage n nm her : of tra tns at the yard and 1n t ... -!"f' queue r the nver,go "a1 t1ng t1me lw of a tr1 · --..,;...~""""""'".:;...;;,..;.....__ t and the a' eragc queue 1ng t 1m e lq of a tratn " n ' Solution 1- or the dlstrtbution f.P (t) we h1~ t,er = t/3 and p - i.-l~t = 0 8 < 1 For a 10 F1g II 25 order Lrlang proc~ss vl = (itVk)~ = 0 it vJ ( f + ..!:.._ + p

n (n-1) p:~

1

n

p"~

+

~ ~

+ +

p" nl

+

pn ~ ) -1 nl 1-x

J-t t-x

+

+~n!

p 1.

+ _nn + P"' 1-x X

x

)

-I ~ X.n ( 1 _ X)

•) Note that A. here 1s the 1ntens1tl of Erlang '- tlo'v and not that of the i ~bich "as thnmed out tfJ get the Erlang no"

Ch. 11. The Queueing Theory ~

the other hand _!_J.. n Pn < ( 1 p ' p2

+

+ ···+

nn-l pn

X )-1 __

+ 1-x

-

385

xn(1-x)n xn (n -1) 1 ·

+

1ce [xn (1- x) n]/[xn (n- 1) + 1] < Pn < xn (1 - x), the inlily indicated in the problem is satisfied too. Note that the last 1ality can be used to approximate all the characteristics of the eing system in question . .27. A railway booking-office has two windows in each of which 'ls are sold to two destinations, Leningrad and Kiev. The arrival esses of the passengers who want to buy tickets to Leningrad and ~hm the same intensity A- 0 = 0.45 pass/min. The average service ~per passenger (selling a ticket to him) tser = 2 min. rationalization proposal has been made, i.e. that in order to dese the queues (in the interests of the passengers), the two bookinges should specialize, one to sell tickets only to Leningrad and the !r only to Kiev. If we consider, as the first approximation, all arrival processes to be stationary Poisson's verify whether the posal is reasonable. ~lntion. (1) Let us calculate the characteristics of the queue for "o-server queueing system (the existing variant). The intensity of arrival process').,= 2A. 0 = 0.9 pass/min, It= 1/~er = 0.5 pass/min, ~AI~t == 1.8 and x = p/2 = 0.9 < 1, thus limiting probabilities l\. From formula (11.0.21) we have Po={1+1.8+ 1.S

2

2

1 1 11- 09 . J

~0.0525,

1from formula (11.0.23) -_ 1.8 3 ·0.0525 r- 2. 2 . _ ~7.7 passengers, 0 01

7. 7 . tq= _ ~8.56 mm. 09

t~l In the second (suggested) variant we have two single-server queue-

·t systems

with p = A, 0 f~t = 0.45/0.5 = 0.9 < 1. _r;ormula (11.0.13), the average length of the gueue at a window -p·/(l- p) = 0.92/0.1 = 8.1 pass, and thus the total length of rue at the two windows = 16.2 pass. ··~\q~eueing_time of a passenger, according to (11.0.14), tq ~ ri'A_ = in th- 18. ~lll, and this is almost twice as long a queuemg t1me •JJnclns· e BXIS hng Vanan • t , v1z. · 9 . 3 mm. · . . .t 1 . JOn: the "rationalizati-on" proposal must be reJected smce ~ llcaUv d· · · d t · f !elfi · • lmJmshes the efficiency of the system. The re uc Ion -:e1 ~~~ncy of the system caused by going from a two-server sy~tem \a~ ng v~ri~nt) to two single-server systems (the pro~os~d varrant) 'i1nts eth division of the booking-office into two spec1ahzed offices 11.28* e men at the windows to repeat each othe: · ". . , -' "'er. An elementary multi-server queueing system wLth LmpatLent ,.,._ s and an unbounded queue. \Ve have an elementary

2r

°

'. ' ~

•

Applud Probl~mt tn Pral!alll!Ug Taeorfl_

386

n ~er\ er queueing system, the 1nten~:aty of the arr1' al process ts A, and the 1ntens1ty of the s~rvJce process tSJJl,. ;.· 1/~tr The queuetng tlme of a customer lS J1m1ted by a certain random l1me T 'tv. b1ch has an expo n en t 1a1 d 1st r1 but ton 'Vl th parameter 'V ( e \ ery customer 1n the queue rna) depart an(l the departure process bas an 1ntens1ty v) \Vr1te the formulns for the ltmit1ng probabtlttles of states, find the rclat1 "e capaCJtj of the system for c:ervi.ce Q. tbe average length of the queue the 3\eragc qucUelng time fq of a CUStomert the average num her o£ customer~ 1n the system and the average "atttng ttme tw per

z

rt

customer

Solut•on As bcfoie, we enumerate the states of the

dance ~ tth the number of customers --~

1n accor--

tbe system The d1reeted graph

Ill

l

l

.l

~)stem

l

J

[s~):P-~ ).;1 S2 ~ • ••: ~ sk J;: •~ ·{ Sn .J ~~~ rt:-.._: :[sn rj ..=• .. 2p Jp. kp np np, J

Sr

4

..

~

r1)l ,.~

sho,vn 1n Ftg 11 28 Us1ng the general huth and death formulas and Introducing the designations p = A/~L ancl ~ v/p, of states

JS

=

·ne obtaln p

Po~ { 1 + 11

P'

+ 2J +

+ -

p

Pt-- 11 Po ' Pn+~ ~ p"

-

nl

P'"

(n+ ~)(n+2~)

(n+r~) Po (r;:o.t),

(H 28 2)

Tb.e first formula (11 28 1) Includes an 1nfin1te sum, whtch 1s not a geometric progre~~Ion but whose terms decrease faster than the terms of d geomctr1c progre~ston ''e can prove that the error due to the t1 uncat1on of the sum to the first r - 1 term~ JS smaller than

~ {PIP )r

nl

e

PI~

rl

\\ e ac.~ume that \ve

ha' e calculated the probnbtltttes p 0 , Pt'

' P'n + r' (l.w ,v ~ .ll"\J'\V $lh.r-.V Mv 5.\.s . .f~rul t,1~ ~:luv·..ar teriStics of the queue1ng system the relative capac1ty for ser\ 1ce Q, the average number of customers 1n the queue rand others \Ve shaH f1rst f1nd Q All the customers '' tll be served except for those who will depart ahead of ttme Let us calculate the average number of customers '' ho depart un~erverl p~r unit tJmo The 1ntens1ty of the departure ,- A-.,

Ch. 11. Th e Q ue ue in g .Theory

387

y it ns te in e ag er av l ta to e th d an process per cu st om er in th e qu eu e is v en ce H is e eu qu e th in s er om st cu e th of the departure pr oc es s pe r al l se rv ic e r fo em st sy e th of ty ci pa ca te lu so the ab .2 ) 28 1. (1 , vr ... ') A = and the re la ti ve ca pa ci ty fo r se rv ic e ) .3 28 1. (1 ... o = Alf... = 1 - -WJf. di nd fi d ul co e w ch hi w ow kn l al of t Thus, to find Q w e m us t fi rs r n+ rp . .. + 2 n+ 2P t n+ 1P = r a rectly from th e fo rm ul r be m nu te ni fi in an ns ai nt co it at th is But the dr aw ba ck of th is fo rm ul a e ag er av e th r fo on si es pr ex e th g of terms. T hi s ca n be av oi de d b y us in 28 .2 ) 1. (1 g in k ta , or lfL A = k e. i. , A of s number of bu sy se rv er s k in te rm into account, .4 ) 28 1. (1 . ~r p = L )lf vr k = (f .. .-

vr.

r,

+

From (11.28.4) we ge t

-r = 0) ,

=

of

9.8168-1 9.0842 7.6190 5.6653

9.9326-1 9.5957 8.7535 7.3497

3. 7116 2.1487 1.1067 5.1134- 2 2.1363

5.5951 3.8404 2.3782 1.3337 6.8094-2

R(m,

a) as fo llo w s: P (m , a) =

=

Contln,ued m

i

a~l

a~a~2

f

f

li ~98-1 8 3082-6 1 3646

9 10 11

.QCII3

Q~5

f

i025

8 t32z-s

3 1828

2 923.'1-~ 7 t387-f. i 6149 3 4019-'$

2 8399 0 1523-'

1 3695 5 4531•3

2 7372

7 6328- 1

2 0189 6 9799~

t 9932

2 2625

4 8926- 4 i 1328

6 OOJS- 6

1.

i2 t3

f

tJ.~i::a;f.

1'15 17

t 9S69 5 4163-'

t8

1 4017

16

m

l

o: ..... 6

(J;c=,

t

9 9752-l

9

9 82G5 9 3800

9 9270 9 7036

84880

9 t823

4 5 6 1 8 9 10 ff

1 1.494

8

S M32

3 937()

6 9929 5 5029

2 5002 t 5276

4 0129 2 70[)1

8

1

0 l

2 3

3924-~

5 3350

12

8 8275-:S

~3

3 6285

27000 f 281 I

14

1 4004

iS

5 09iQ- 4

16 t7 18

1 7488 1 7597

19 20 21

I

22 23 24

25 26 27

5 1802-1 t 4551

68663 5 470i 4 0745

2 8333-l 1 1 6 3

5 7172-1 2 4036 9 5St8-,l 3 6178 t 2935

5 8917-1 a~e

9 0~31 8 0376

69~0-l 8521- 2

2 0092

I

9 9G9g 9 8825 9 5762

~1~\

9

Q5:1

I

4 1 4 1.

,

9 995(}-l

990~-l

4 26:!1

m

I

ai!;I;;I,B

(

8411 1192 3797-1 4181

a ... 9

I

Cl

!!!a

10

9 99S9- 1 9 9377 9 9317

9 !)995-1 9 9950 9 9123

9 7877

9 8966

9 ~~~ 8 8\31 1 9322 6 7610 5 4~35 4 t259 l 2 9t01 t 9o99 i 2\23 7 33'51-2

9 70.15 9 3291

8 6938 7 7978 6 6718 5 4207•1

4

109~

3 0322 2 054-1 f 355~

1 7257 8 231Q-3

4 1~56 2 2035

8 3-LSS

3 7180 i 5943 6 5037...,

t ttOB

2 70-i2 t 4278 7 1865-1

o.==S

4 874.0

s 3J!Jli_, 2 4251:

'

t1~9

!

nt:a!O

a 3407

1 0360 4 3925-l 1 7495

3 45l3 1 5'38.3 6 996.5

1 f385

6 agzg-s

3 7255... 8 2 J722

2 4519 8 6531-&

2 9574 1 2012

4402-1 4495 5263-e

2 5291 9 3968-6

3543

2 9414

I

I

4 £919-a 1 7690 6 4229-8 2 2535

A pp en di ce s

423

e or m no r cu oc l il w A t en ev an at th ity Ex:am:Ple•. We must find the probabil than twtee If a = 7. We have 4. 96 02 0. = 1 6 03 97 0. 1 = 36 70 9. R (2, 7) = 1 - R - (2, 7) = 1 e th of nt ne po ex e th en th , nt ne po ex has no e bl ta e th in r be m nu a If 1. . rk ma ~e 19) 3 (3 R e pl am ex r Fo nt ne po ex , • ber in the column is the · previous num i.2067 X 10-3, by ed at im ox pr ap be n ca a) , (m R 2. For a > 20 the p ro b ab il it y R (m,

a)~

( m + ~~-a ) + 0 .5 ,

1

. 5) x di en pp (A on ti nc fu r ro er e th where O T=O, 't

(-r) {, (-c) d't' = 1Jl (0) if the functi on 1Jl (t) is contin uous at the point 0);

o O+e ) 1Jl ('t) 0 ('t) d-r = 11Jl ('t') 0 ('t') dT = 0-e

o

+1Jl

(0)

if the functi on 1Jl (T) is contin uous at the point T = O. These defini tions yield the following proper ties which hold for any real any odd functi on cp ('t): (1) 1T l =T sign T, (2) T= I T I sign T, (3) cp (T) = cp (I T I) sign T, (4)