Stochastic Calculus for Finance I: The Binomial Asset Pricing Model (Instructor Solution Manual, Solutions) [1 ed.] 0387401008, 9780387401003

Citation preview

Steven E. Shreve

stochastic Calculus for Finance | Instructor’s Manual: Solutions to Exercises December

16, 2004

opringer Berlin

Heidelberg

Hong Kong London Milan Paris Tokyo

New York

Preface

This document contains solutions to all the exercises appearing in Stochastic Calculus for Finance I: The Binomial Asset Pricing Model, Springer, 2003. Steven E, Shreve December 2004

Pittsburgh, Pennsylvania USA

Contents

1

2

The

Binomial

1.7

Solutions

Probability 2.9

3

State 3.7

4

4.9 5

6

Theory

Prices Solutions

Walk

Solutions

00... 0 0c

cece

0,0.

ccc

................ cere

ccc cece

2.0. 00...

cee eee cee.

to Exercises

beep.

cec ceee cece e eee.

Securities

.....................-000...

0.0... 0... cece

cece

cece cbc e eee.

....000 0 0. oc eec eee eee. to Exercises

Solutions to Exercises...

0.0.0...

Assets

ccc ccc

]

cee cee cee.

cee

to Exercises

Interest-Rate-Dependent 6.9

Pricing Model

on Coin Toss Space....................

to Exercises...

Derivative

Solutions

Random 5.8

to Exercises...

Solutions

American

No-Arbitrage

1

13 13 35 35 51 51 69

ce ec eee eee ee eee.

69

....... 0c. ..... .c cece ca ee.

89

0... cc ccc cece

cece

cece eee p eee.

89

The Binomial No-Arbitrage Pricing Model

1.7

Solutions

to Exercises

Exercise 1.1. Assume in the one-period binomial market of Section 1.1 that. both H and T have positive probability of occurring. Show that condition (1.1.2) precludes arbitrage. In other words, show that if Xg = 0 and X,

=

49S)

+ (1 + r\{Xo

~

AgSo),

then we cannot have AX strictly positive with positive probability unless X, is strictly negative with positive probability as well, and this is the case regardiess of the choice of the number do. Solution

When

X,

= 0, we have

X (Hf)

=

AoS\(H)

=

u—({l+r)

AgSo,

X1 (Ty = AoSy(T) = (d— (1 +7)) Ap So.

According to (1.1.2), u~-{1+7) is positive and d—(1+7) is negative. Therefore, the only way X,(H) can be positive is for Ap to be positive, and the only way X,(T) can be positive is for Ap to be negative, Either one of these portfolio values is positive and the other negative, so both a positive and a negative portfolio value at time one have a positive probability of occurring, or else both X;(#) and X,(T) are zero. Exercise 1.2. Suppose in the situation of Example 1.1.1 that the option sells for 1.20 at time zero. Consider an agent who begins with wealth X, = 0 and at time zero buys Ap shares of stock and Jy options. The numbers dp and Io can be either positive or negative or vero. This leaves the agent witb a cash position of -44» ~ 1.20Iy. If this is positive, it is invested in the money market: if it is negative, it represents money borrowed from the money

2

1 The Binomial No-Arbitrage

Pricing Model

market. At time one, the value of the agent’s portfolio of stock, option and: rooney market is

. X, = oS: + Io(Sy -5)7 - 1 (4Ao + 1.20I).

Assume that both 4 and T have positive if there is a positive probability that X, probability that X, is negative. In other when the time-zero price of the option is

probability of occurring. Show that is positive, then there is a positive words, one cannot find an arbitrage 1,20,

Solution. Considering the cases of a head and of a tai] on the first toss, and utilizing the numbers given in Example 1.1.1, we can write: = 84g

o

+ 329 - ~ (449 + 1.201), oe

Ai(H)

X(T) = 24,+0-Io- 7 (4do + 1.205) Adding these, we get. X (A) + X41 (T) = 10Ap + 829 — 110A — 3m

= O,

or, equivalently,

X1(H)} = -X,(T). In other words, either X,(H) and X,(T) are both zero, or they have opposite sigos. Taking into account that both p > 0 aud q > 0, we conclude that if there jis a positive probability that 1 is positive, then there is a positive probability that X, is negative. Exercise want

1.3. ]n the one-period

to determine

binomial model of Section

the price at time

zero

of the derivative

1.1, suppose we security

Vj

=

94,

i.e., the derivative security pays off the stock price. (This can he regarded as a European cal] with strike price K = 0). What is the time-zero price Vp given by the risk-neutral pricing formula (1.1.10)? have

M(H) +

tf

Vo

l

{|

We

L+r

tf

Solution.

l+r

1

(7)

ipSi(H) + GS, (T)) [puso

+ Gd So}

up + dg = So

I+r

_

~

u((l1+r)—d)

8

= So,

+d(u—(14-7))

(14r)(u—a)

1.7

Exercise 1.4. In the proof of Theorem pothesis that X nti (Wy we ..

Wal)

Solutions

to Exercises

3

1.2.2, show under the induction hy-

= Vat (Wiwe.. Wy J'),

Solution. As in the proof of Theorem 1.2.2, use the notation X,(w,w2...W,) = Xn,

Xn+}

(ww

tae wn)

Xnoi(T)

=

(1

=

+

Xnar(T),

Xn

+

=(1l+r)V, +

Bn

and

(d

-

similarly

(1

+

for

Vn

and

Vai:

We

have

r))

(Vi4i(H)- Var (T)) (d ~ (1 +7)) u-d

= (l4+7)Va — BVaai (A) + Dna (T) =

DVn+

(7)

+

WVnsi(T)

~~ PVn+

(H)

+

PV ns

(T)

= Visi (H).

Exercise

1.5. In Example

1.2.4, we considered an agent who sold the look-

back option for Vo = 1.376 and bought 4p = 0.1733 shares of stock at time zero. At time one if the stock goes up she has a portfolio valued at ¥;(H) = 2.24. Assume that she now takes a position of 4,(H) = see a in the stock. Show that at time two, if the stock goes up again, she will have a portfolio valued at V2(H A’) = 3.20, whereas if the stock goes down, her portfolio will be worth V2(HT)} = 2.40. Finally, under the assumption that the stock goes up in the first period and down in the second period, assume the agent. takes a position of 4;(HT) = Speer in the stock. Show that at time three, 1f the stock goes up in the third period, she will have a portfolio valued at V3(A 7TH) = 0, whereas if the stock goes down, her portfolio will be worth V3(A7T) = 6. In other words, she bas hedged her short position in the option. Solution.

We

bave

_

(HH)- V5(HT)

_ 3.20 - 2.40 |

V(HTH)-V3(HTT)

0-6

S,(HTH)-Sy(HTT)~

8-3

OUD = SHA) SET) A.( HT)=

~ 1e—4 = 90887 7"

Assuming the the first toss results in A, at time 1 the agent’s portfolio value is X\(H) = 2.24 = V\(H). She rebalances the portfolio by taking a position 4,(A) = 0.0667 in stock, which means she has a cash position

X1(H) — O;(4)Sy(T) = 2.24 - (0.0667)(8) = 1.7064. Then at time 2, the portfolio values in the two possible cases are

4

1 The Binomial No-Arbitrage Pricing Model

il

X2(HH) = (0.0667)(16) + (1.25)(1.7064) = 3.20 = V2(HA),

X(HT) = (0.0667)(4) + (1.25)(1.7064) = 2.40 = V2(HT), as it they should be. Finally, we consider further the case of H on the first toss and T on the second. At time 2 the agent rebalances her portfolio, taking a position Ay(HT) = —1 in the stock, which results in a cash position of

Xo(HT) - O3(HT)Sz(HT) = 2.40 -- (-1)(4) = 6.40. At time 3, the portfolio values in the two possible cases are

X3(HTH) = (~1)(8) + (1.25)(6.40) = 0 = V3(HTH), X3(HTT) = (—1)(2) + (1.25)(6.40) = 6 = Va(HTT). Exercise 1.6 (Hedging a long position — one period.). Consider a bank that has a long position in the European call written on the stock price in Figure 1.1.2. The call expires at time one and has strike price K = 5, In section

1.1, we

determined

the time-zero

price

of this call

to be

Vo

=

1.20.

At time zero, the bank owns this option, which ties up capital Vo = 1.20, The bank wants to earn the interest rate 25% on this capital until] time one, i.e., without investing any more money, and regardiess of bow tbe coin tossing turos out, the bank

wants

to have 5

7 1.20 = 1.50 at time one, after collecting the payoff from the option Specify

bow

the bank’s

trader should

invest

in the stock

(if any) at time one. and

money

market

to accomplish this. Solution.

The

trader

should

use the opposite

of the replicating

portfolio

strategy worked out in Example 1.1.1. In particular, she should short 4 share of stock, which generates $2 income. She should invest this in the money market, At time one, if the stock goes up in value, the bank has an option worth $3, has $(2 2) = $2.50 in the money market, and must pay $4 to cover the short position jn the stock. This leaves the bank with $1.50, as desired, On the other hand, if the stock goes down in value, then at time one the bank has an option worth $0, still has $2.50 in the money market, and must pay $1 to cover the short position in stock. Again, the bank has $1.50, as desired, Exercise 1.7 (Hedging a long position — multiple periods.). Consider a bank that has a long position in the lookback option of Example 1.2.4. The bank intends to hold this option until expiration and receive the payoff V3. At time zero, the bank has capital Vp = 1.376 tied up in the option, and wants

1.7 Solutions to Exercises

to earn the interest rate of 25% on this capital until time three, (i.e., without investing any more money, and regardless of how the coin tossing turns out, the bank wants to have 5\3 G)

‘1.376 = 2.6875

at time three, after collecting the payoff from

the lookback

option at time

three). Specify how the bank's trader should invest jn the stock and the money

market to accomplish this.

Solution. The positions in the stock the trader should take are the negatives of those given by (1.2.17). Using the option values computed jn Example 1.2.4, this leads to the formulas

OWHH)= — ao

= SAT)

*~

a

= 0.533539,

=

-—

= 0.333333,

ast) = BATH WIT) 9-8, vir Segre « “Sopa =?

A(TH) =

ee

Kirn

4,(H) =

58 ae

nee

A(T) = Ao =

Se i

os aoS

= —0.0666667.

= aS ee =

ae

= 0.466667, = —0.173333,

The bank starts with initial capital Xo = 0 and holds the above positions in the stock, Gnancing this by investing in or borrowing from the money market. At each time n, where n = 0,1, 2,3, the value of the portfolio obtained by this

trading, plus the value of the lookback option, should be (£)” - 1.376. Then at time n = 3, the bank will have achieved the desired goal.

To see how this works, we begin at time zero. The bank takes a position Oo = —0.173333 in the stock. Because this is a short position, it generates income

Xq ~ 459

= 0.173333 > 4 = 0.693332,

which

is invested

in the money

market. By time one, this grows to (1 +7r)(Xq

— AoSo)

= 1.25 - 0.693332 = 0.866665.

Therefove, the value of the portfolio at time one if there is a H ow the first. toss is

X\(H) = O05) (H)+(1+7r)(X9— OoSp) = -0.17333 8 + 0.866665 3 = —0.52.

5

6

I The Binomial No-Arbitrage Pricing Model

If there is a T on the first toss, the vaiue of the portfolio at time one is X(T)

= 42S) (T) + (14+ r)(Xg

One can check

— AoSo) = -0.173333

: 2 + 0.866665

= 0.52.

that

X\(A)+V,(8)

= ~0.52+2.24=1.72

= (>)

- 1,376,

and X(T)

+ V,(T)

= 0.52 +1.20

= 1.72 =

@

- 1.376.

At time one, we rebalance the portfolio, taking either position 4;(H) = —0.066667 or 4:(T) = 0.466667 in the stock, depending on the outcome of the first coin toss. We then toss the coin again and compute the portfolio values at time two. These are

[!

it

X2(AT)

—(),066667 - 16 + 1.25(—-0.52 + 0.066667 - 8) —1.05,

if

X7(HH) = 4,(4)S2(H A) + (147) (Xi(#) - 4:(4)51(4))

A, (H)S2(HT) + (1+ 7)(X1 (A) - 41( 4); (8))

i

—0,066667 - 4 + 1.25{—0.52 + 0.066667 - 8)

X3(TH)

—0.25,

A(T) So(TH) + (147) (X(T)

-— A:(T)S,(T))

= 0,466667 - 4 4+ 1.25(0.52 — 0.466667 - 2) = 1,35,

Ay(T)Sa(TT) + (1 + 1)(Xi(T) - O1(T)S)(T))

i

X2(TT)

0.466667 - 1+

1.25(0.52 — 0.466667 - 2)

= —0,05.

In each case we Lave X2 + V2 = (3)° * 1.376 = 2,15. Indeed, XxAA)

+ V2(AA)

= -1.05 4 3.20 = 2.15,

Xo(HT) + V2(AT) = —0.25 + 2.40 = 2.15, X7(TH)+V(TH)

=

X2(TT) + V2(TT

\=

1.354+0.80

= 2.15,

—0.05 + 2,20

= 2.15.

_ Finally, we consisler the eight possibilities at time three:

1.7

X3(HEH)} = = = X3(HHT) = =

X3(HTH) = = = X3(HTT) =

Solutions

to Exercises

42(HH)S3(HHH) + (1 +1)(X2(HH)- 4o(HH)S3(HB)) 0.333333 - 32 + 1,25(-1,05 — 0,333333- 16) 2.6875, 4,(HH)S3(HHT) + (1+7)(Xo(HH) ~ 42(HH)S2(HH)) 0.333333 - 8 + 1,.25(—1.05 — 0,333333 - 16) ~5,3125, 42(HT)S3(HTH) + (14+1r)(X2(HT) — 43(HT)Sy(HT)) 1-8+1.25(-0.25 — 1:4) 2.6875, A2,(HT)53(HTT) + (1 +17)(X2(HT) - 4o(HT)Sy(HT)) 1-2+1.25(-0,25 - 1-4) ~3.3125,

X3(TAT

—

A3(THH) = 42(TH)S3(THH) + (1 +1)(X2(TH) -- O2(TH)S2(TH)) = 0.333333 8 + 1.25(1.35 ~ 0.333333 - 4) = 2.6875,

= Ao(TH)S3(THT) + (1 +7)(X2(TH) - 42(TH)S2(TH)) = 0.333333 - 2 + 1.25(1.35 — 0.333333 - 4)

= 0.6875,

——

A3(ITH) = = = X3(TTT = = =

O2(TT)S3(TTH) + 1-24 1.25(-0,.05 0.6875, O2(TT)S3(TTT) + 1-0.50 + 1.25(-—0.05 -—0.8125.

| (1 +7) (X2(TT) - 42(TT)S2(TT)) 1-1) (1 +1) (X2(TT) ~ 42(TT)S2(TT)) - 1-1)

Recall that we wish the hedge to work so that ¥,;+V4 = (s)° 1,376 = 2,6875,

regardless of how the coin tossing turus out. We now verify that the hedge performs as desired: Y3(HAA)+V3(HHH)=

2.6875 +0

= 2.6875,

X3(HAT) + V3(HHT)= -5.3125 + 8 = 2.6875,

X3(HTH) + V3(HTH)= 2.6875+0

= 2.6875.

X3(HTT) + V3(HTT)= —3,3125 +6 = 2.6875,

7

8

1 The Binomial No-Arbitrage Pricing Model

X3(VHA)+Vs(THH) = X3(THT)+V3(THT) =

2.687540 0.6875+2

= 2.6875, 1, the “up factor” upn(wjw2... wp), the “down factor” d,(wiw2...w,), and the interest rate r,(ww.. .W,y) are allowed to depend on n and on the first n coin tosses wiw2...W,. The initial up factor ug, the initial down

factor dg, and the initial jnterest rate To are oot.

random. More specifically, tbe stock price at time one is given by _

ug So

if wy

= H,

S1(w1) = {aes ifw, =T,

10

1 The Binomial No-Arbitrage Pricing Model

‘and,

for n > 1, the stock price at time n+ 1 Un (WW

Sn 1 (WW, g -..WaWng1) = (eae

is given by we Wn)

if Wna+]

= H,

io Wy Sy ) (WyWe...Wy)

1 Wr) Sn (Wy We

if way

= T.

One dollar invested in or borrowed from the money market at time zero grows to an Investment or debt of 1.+ rp at time one, and, for n > 1, one dollar invested in or borrowed from the money market at time n grows to an investment or debt of + ry(wyw2...Wn) at time n +1, We assume that for each n and for all wyw2...W,, the no-arbitrage condition 0 < dy,(wyw.. . Wy) holds. We

also assume

ES;

= 6, =

9,

ES3

=1.5-

ES.

=

13.50.

Exercise 2.3. Show that a convex function of a martingale is a submartin-

gale. In other words, let Mo,My,...Mn , be a martingale and let » be a convex function. Show that (Mo), (My),...,y(My) is a submartingale. Solution Let an arbitrary n with 0 p(E, Mn+1).

Combining these two, we get En

¢(Mn+1)

2

p(M,),

aod since 7 is arbitrary, this implies that the sequence of random (Mo), o(My),...,e@(My) is a submartingale.

variables

Exercise 2.4. Toss a coin repeatedly. Assume the probability of head on each toss is 5, as is the probability of tail. Let X; = 1 if the jth toss results in a head and A; = —1 if the jth toss results in a tail. Consider the stochastic process Mo, My, M>,... defined by My = 0 and MM,

=3°X;,

n>

l.

j=l

This is called a symmetric

random

walk; with

each bead, it steps up one, and

with each tail, it steps down one. (i)

Using the properties martingale.

of Theorem

2.3.2, show

that

My, M),M]o,...

is a

(ii) Let o be a positive constant and, for n > 0, define TL

S,

"

= e7Mn

2

ev +e?

Show that So, 5),$3,... is a martingale. Note that even though the symmetric random walk M,, has no tendency to grow, the “geometric sym-

metric random walk” e?™=

does have a tendency to grow, This is the

result. of putting a martingale into the (convex) exponential function (see

Exercise 2.3). In order to again have a martingale, we must “discount” the

geometric symmetric random walk, using the term ——4-; as the discount rate. This term is strictly less than one unless o = 0. Solution

(i) Clearly, M,, depends only on the first n coin tosses (the process is adapted), and Xn+: is independent of the first n tosses, so E,M, = M, and E,Xn-z) = 0 for all n. We have

Bn Matyi = En(Mn + Xnai) = ExnMn t+ Ey Xana =

M,+EXy4)

=M,+0=M,.

2.9

Solutions to Exercises

17

(ii) We have l

En, Sng) = Ey

ntl

lent

ooo

|

ev +e?

I =F,

, eo Mn pt Xana

n+]

(

e7 + en? oe

_—

9 =o!

eo Mn

n+] E.

[e°*n+s

|

ey +e?

(Taking out what is known.) —

Sn

2

RetXnai

ev + e~?

(Independence.) =

2

S)

Exercise 2.5. Let Mg, My, Mz,... 2.4, and define Jo = 0 and

(Fe

+

1 _,\_=

5°

S,.

be the symmetric random walk of Exercise

—_—

T™

In

1,

(25)

=

M;(Mj41

- M;),

n=1,2,....

l

n

jJ=0

(i) Show that I, = =

gn

Mi - -. 2

(ii) Let n be an arbitrary nonnegative integer, and let f(i) be an arbitrary function of a variable z. In terms of n and /, define another function g(t) satisfying Lf (Tn41)]

En

=

g(In).

Note that although the function g(J,,) on the right-hand side‘of this equa-

tion may depend on n, the only random variable that may appear in its argument is /,; the random variable M@, may not appear. You will need to use the formula in part (i). The conclusion of part (ii) is that the process is a Markov

To. J), Jq,...

process.

Solution

(i} We have n=

w=

> =O I=

l

n~]

(Mya

or M;)°

=

SM}, 3=0

n=-1

n=

- 25° g=0

MyaiM,

+ S” j=0

Mi.

18

2 Probability Theory on Coin Toss Space

We make the change of index k = 7 +1 in the first sum on the right-hand side and

use the fact that Mp

= 0 to write

n

n=]

m=]

n= SOME -25 0 MjaiMy + > — M} k=0

=O) |

ra

=Mi+)

P=0 n~]

n-j

j=

j=0

Mi-250 MjiiM; +)Mj}

= n=]

=

n—~]

M242

1

Faas

Therefore, J, = $M? - 3, (ii) We note that M, and J, depend only on the coin tosses 1 through n, and Mr+1 ~ M, depends only on the (n+ 1)-st coin toss. According to the Independence Lemma 2.5.3 and the fact that J,4, = Jn+M,(Mns:—M,), we have Ey lf (In+1)]

= EB, lf (Jn + Mn( Mas

— M,.))]

=A(I,, M,,),

where h{i,m)

= Ef (+ +m(Mnsy

- M,,))

= -

flit m) + = f(i-m),

Ifm > 0, then f(t +m) = f(i+|m]) and f(i~m) = f(i—|m|). On the other hand, ifm < 0, then f(i+m) = f(i—|m|) and f(i—m) = f(i+|ml), In either case,

Alim) = SF (E+ lal) + 5 fle= (ml). From part (i) we bave M? = n+2/,

and hence |M,| = /n+ 2/,. There-

fore,

Bn [f(Inea)] = AUIns Ma) = 5f(In + |Mpl) + 54 (In — Mal) =

=f(In

4

V/ Th +

We have thus managed to write

an)

+

=f (In

-

J/nt+

ln).

2.9

Solutions

to Exercises

En ( f(Tn41)| = 9(J.),

19

(2.9.1)

where

g(t) = “fit Vn ¥ a)

+ ofli- Vasa),

The only random variable appearing on the right-hand side of (2.9.1) is Jn, which shows that the process Jp, J1,J2,... has the Markov property.

Exercise 2.6 (Discrete-time stochastic integral), Suppose Mo, Mi,...,

Mwy is a martingale, and let Ao, 4;,.,., On, be an adapted process. Define the discrete-time stochastic integral (sometimes called a martingale transform) lo, /1,.-.,J~ by setting Jp = 0 and i In

=

S>

As(My4y

— M;),

n=

1,...,N.

3=0

Show that Jo,J,,...,Jn isa martingale.

Solution. Because Jn4) = Jy + 4n(Mya, — My) and Jn, dy and M,, depend on only the first n coin tosses, we may “take out what is known” to write

Ey [Inti] = En (In + On(Masr — Mn)] = In + On (En[Masi] - Ma). However, E, (Mnai] = M,, and we conclude that Ex(J,41] = Jy, which is the

martingale property.

Exercise 2.7. In a binomial model, give an example of a stochastic process which is a martingale but is not Markov.

a t

Solution Let Mo, M,M2z,M3 be the symmetric random walk of Exercise 2.4 with three tosses. For any adapted process Ao, 4, Ao, the discrete-time stochastic jntegral

j

(t o

In = NS ;(Mjay — Mj), n= 1,2,3, Is a martingale; see Exercise 2.6. As in that exercise, we set Jy = 0. We can choose the integrand process Ap, A), 42 so that the discrete-time stochastic integral is not a martingale by allowing A, to be path dependent. For example, suppose we take Go = 4\(A#) = 4;(T) = 1, but 4.(H HX) = A(HT) = 1, 42(TH) = 42(TT) = 0, so that Az depends on the first toss rather than the second. Then

K(A)=M(A)=1,

(HH) =2,

A(T) =M(T)=-1,

1h(HT)=1)(TH)=0,

(TT) =~—2

20

2 Probability Theory on Coin Toss Space

and in(HAA)

ee

= ),(HH)+

AHH)

=

241

=3,

= Ip(HH)~ 4,(HH) = 2-1

=1,

Now take f(z) = x* and compute

Eo[f(Is) (HA) =

$:38%+3-1?

=5,

Ex(f(s)(HT)=

F-1?+4.1?

9 =1,

Ex(f(Ja)/(TH) =

$0? +4-0?

=0,

Ea(f(Js)(TT) = 4. (~2)? +4. (2)? = 4 If there were a function g(z) such that

g(Io(wywe)) = Ex[f(J3)}(wywe) for all wywe, then we would

have

9(0) = 9(I2(HT)) = E,[f(Zs) (HT) = 1 9(0) = g((n(TH)) = EB [f(Is)\(TH) = 0 which is impossible. Therefore, the discrete-time stochastic integral Jo, Jy, J2, Js

is not a Markov process. Exercise 2.8. Consider an N-perjod binomial model. (i)

Let Mo, My,..., My and Mj, Mj,..., My be martingales under the riskneutral measure P, Show that if Mn = M,, (for every possible outcome of the sequence of coin tosses), then, for each n between 0 and N, we have

M, = M,, (for every possible outcome of the sequence of coin tosses),

2.9

Solutions

to Exercises

21

(ii) Let Vy be the payoff at time N of some derivative security. This is a random variable that can depend on all 2’ coin tosses. Define recursively Vn~w1,VN—2,-.+, VY by the algorithm (1.2.16) of Chapter 1. Show that Vo,

Vy VN=1 VN Tero *(l4r)¥-P (14 r)¥

is a martingale under P.

(iii) Using the risk-neutral pricing formula (2.4.11) of this chapter, define ~

v= Ea | Show

V, sd (l+7r)4

ae

that

Y,

Vu-1

VN

Ler OO (LET NAY (Dp ry is@ martingale. (iv) Conclude that V, = Vj for every n (i.e., the algorithm (1.2.16) of Theorem 1.2.2 of Chapter 1 gives the same derivative security prices as the riskneutral pricing formula (2.4.11) of Chapter 2). Solution.

(i) We are given that M, = Mj,. For n between 0 and N — 1, this equality and the martingale property imply

M, = E,(Mwy] = En[Mu] = Mp. (ii) For n between 0 and N — 1, we compute the following conditional expectation: ~ En

V,, ae

(W1W2...Wy) = 5 cori (wiwa

Wnt)

(l+r)n*)

g Vntilwiwie

wn)

(l+r)7*

_ Val wz... Wy) ~ (l+r)"

where the second equality follows from (1.2.16). This is the martingal property for Ta (iii) The martingale property for aay follows from the iterated conditioning property (iii) of Theorem 2,3.2. According to this property, for n between Qandn-1l,

22

2 Probability Theory on Coin Toss Space

7

Vit

= ire Sar

7

]

_

Vn

“= t ryan [ee lo

o=

“ae

En Fait

1 (l+r)"

fe"

maT |

Vy

a=

Vy res

|

Ve

(L+r)™ (iv) Since the processes in (ji) and (iii) are martingales under the risk-neutral] probability measure and they agree at the final time NV, they must agree

at all earlier times because of (1).

S3(HH) = 12 51(7)

a

1

= 8

m1(H) =}

:oN

Sa(HT) = 8

So = 4

mo

4

S2(TH) = 8 NN

Si(T) = 2 r1(T)

=i

Sx(TT) =

Fig.

2.8.1.

A stochastic volatility, random

Exercise 2,9 (Stochastic volatility, random

interest rate model.

interest rate).

Consider

a two-period stochastic volatility, random interest rate model of the type described in Exercise 1.9 of Chapter 1. The stock prices and interest rates are shown in Figure 2.8.1. (i)

Determine risk-neutral probabilities

P(HH), P(HT), P(TH), P(TT), such that the time-zero value of an option that pays off V2 at time two js given by the risk-neutral] pticing formula =

M% =E 7 nea

=i

|

2.9

Solutions

to Exercises

23

(ii) Let V. = ($2 ~—7)*. Compute Vo, Vi(H), and Yj(T). (tii) Suppose an agent sells tLe option in (ii) for Vp at time zero. Compute the position 4, she should take in the stock at time zero so that at time one, regardless of whether the first coin toss results in head or tail, the value of her portfolio is Vj. (iv) Suppose in (iii) that the first coin toss results in head. What position 41(F) should the agent now take in the stock to be sure that, regardless of whether the second coin toss results in head or tail, the value of her

portfolio at time two will be (S, — 7)*? Solution.

(i) For the first toss, the up factor is uo = 2 and the down factor is dy = 7 Therefore, the risk-neutral probability of an H on the first toss is L+79 — do l+i-4 yo=eoo FO * ug — do ~ 2-5

1 ~

9

and the risk-neutral probability of T on the first toss is .

Ua

~l=To

—_—

2~1-~}

1

—

uj -do

2-4

~

9!

If the Grst toss results in H, then the up factor for the second

toss is

it follows that the risk-neutral probability of getting an H on the second toss, giveo that the first toss is an H, is

;

_ltn(H)-d(H) _ l+qg-1 1

Pili)

=

u)(H) — d\(8)

7

g—]

~

9?

and the risk-neutral probability of J on the second toss, given that the first toss is an H, is

-

n=

_u(H)-1-n(H)

—R)-ale)

3-1-7!

Pod

If the first toss results in T, then the up factor for the second toss is

24

2 Probability Theory on Coin Toss Space

m= and the down

\=

So(TH)

ST

factor for the second d(H)

=

=

8

a

=

toss is

So(TT)

2 =—- =,

It follows that the risk-neutral probability of getting an AH on the second toss, given that the first toss is a 7’, is

-

_ltn(T)-d(T)

1 a= _14+5-

AT)= a(t)

_i

and the risk-neutral probability of T on the second toss, given that the first toss is a TJ, is

5 (r) = uD

De

a ton(T)

4-1-3

a(t) -d(T)

5

4-1

6

The risk-neutral probabilities are

P(HH) = popi(H) = 4-3 =}, P(AT) = foui(H) = 5-5 =F, PITH) = @A(T)

= 5:2 =

P(TT) = GoH(T)

=5°2=

%-

(ii) We compute

| Vi(A) = Te rtHy Pee)

+ §(H)V2(HT)]

a8! yp imtal ie nt iL aa-neabenn' = 2.40,

Vi (T)

=

1

— T$r(P)

i = 31;

15 7 V. IPi(L)Vo(TH) + A(T)VA(TT)]

(8 ~ee 7)

+ ee 5 (2

ee

= 0.111111,

Yo = ==

——

]

-

-

(PoViC A) + GOV (T)]

4/1 [5240+

= 1.00444,

1 50.2111]

|

2.9

* We

can confirm

this price by computing

according

“pricing formula in part (i) of the exercise: _ “= 8

25

to ihe’ risk-neutral

V2 iy! Vo(AT)

Dp

Vi(H#)

_

PF) * Gata

“ Teryientey

Gtr\ieny

_(2-7)" 1 thy 4d) 4° (8 —7)*

PED 3

+ Teper Ty PED 1

(@-7)* tba 42) 1

5

Va(TT)

P

VAT H)

_

Solutions to Exercises

4

(2-7)*

5

Q+)0Q4+4) Ro aehasy

72

= 0.80 + 0.16+ 0.044444 0 = 1.00444. (iii) Formula (1.2.17) still applies and yields Bo

_V(A)-V(T) — 2.40~-0111111

— S(H)- S(T) ~

= 0.381481,

8 =2

(iv) Again we use formula (1.2.17), this time obtaining Vi(HA)~V2(RT)

— (12-7)+ - (8-7)*

aH) = Sx(HH)- $,(HT) ~

12-8

_

=h

Exercise 2.10 (Dividend-paying stock). We consider a binomial asset pricing model as in Chapter 1, except that, after each movement in the stock price, a dividend is paid and the stock price is reduced accordingly. To describe this in equations, we define Yes

(Wh

Wnts)

= 4

if w ’

4 1

=T —™

.

Note that Y,4, depends only on the (n+1)st coin toss. In the binomial model of Chapter 1, Y¥.415, was the stock price at time n+1. In the dividend-paying mode] considered here, we have a random variable An41(w,...WpWnet), taking values in (0,1), and the dividend paid at time n+1 is Ana) Yna15n. After the dividend is paid, the stock price at time n +1 is Ont

=

(1

—” An+1)Yna15n-

An agent who begins with initial capital X, and at each time n takes a position of 4, shares of stock, where A,, depends only on the first n coin tosses, has a portfolio value governed by the wealth equation (see (2.4.6))

26

2 Probability Theory on Coin Toss Space Kn

(i)

=

An

=

OnYn+

Ont)

+

(1

+

Sn

+

(1 + r)(Xn

r)(Xy

—

A,Sn)

+

OnAnsei

ne

~ 4,5).

on

(4.3.2)

Show that the discounted wealth process is a martingale under the riskneutral measure (i.e., Theorem 2.4.5 still holds for the wealth process (2.8.2). As usual, the risk-neutral measure is stil] defined by the equations

_ l¢r—-d

| w-ie-r

Peed?

a

(ji) Show that the risk-neutral pricing formula still applies 2.4.7 holds for the dividend-paying model).

(i.e, Theorem

(iii) Show that the discounted stock price is not a martingale under the riskneutral measure (i.e, Theorem 2.4.4 no longer boids). However, if Ang,

is a constant a € (0,1), regardless of the value of n and the outcome of the coin tossing w,...Wn4), risk-neutral measure.

then Toasty

is a martingale

under

the

Solution.

(i) We use linearity of conditional expectations, and

the fact that Y,4, 7

depends

An+}

En

|

=

[4nYna1Sn

= Ea

A,Sn

xX,

| + En Tr oe

=


C]

(c,.X

ys

X (Dr...

xPlwnci

Wye

» aya) n+]

Danas

= Ona1,..-

(G1

YY

$op

(Dy ..

WN

1

Gy)

+

(Oy

1

WyWn4y

...@y))

By)

=Gn lw, =],-..,u,

= Dy}

...DnWagi-..Gy)

datees uN

xP{wn4t

= Ongi,.-.,Wn

= Onlu,

= cy En(X](Q1...0n)+ coE,[Y](@1...Tn) (ii) Taking out tosses, then

C2 Y

what

is known.

If X

= Dy,...,Wn

depends only

= Gn}

on the first n coin

90

6 Interest-Rate-Dependent E,(XY]( Gy .

=

Wn)

x, Wr bli

Assets

Wn)Y (By ...DnDng1.. Dy)

Gly xP{watt

= Wri,

WN

onl

=

Wy...

Wr

=

Wn}

Wrgli--

XPlonat = Ong.Wn = Onley =T1,...,n = Dp} (iii) Iterated conditioning. If0

an

= Gn tise Wm = Wms Wma

Ley

=

= Omtis.:-,Wn

Beem)

{0

= Ome1,...,WN = ON

nie

En[X](D...Om

Waeloe

Vn

Mens

DatirerWm

Seay

Wm

= Wm Pen

S Base

Oe)

Wm4is-

(Blonny o Sstss--suy xPlwns

= Wntly-++

Wm

where we have used the definition

© Byle = Om |W

= Bi,---,tim = Bm) = Wy,..-

Wy

= Oy},

6.9

Solutions

to Exercises

91

En ([X]}(@...On) =

sO Waly

xin.

mt 1...

Oy)

WN

x P{W

= Omtiy WN = ON|Wy

=D),...,Um = Dm}

in the last step. Using the fact that Plwm4y

=Wmdi, WN

x Plunsy

= Wat,

= Plwnat

= Gn iW) = Wj,...,Wm

= Om}

++) Wry = Dm |W} “2 W1,...,Wn

= Wn}

= Watts Wm

lu) = Dy... Wn = Tn}

= Wm, Wma)

= Wm4yi,--.WN ,

= Wy

= Plun+1 =Wnti,.WN : = Wy lw) = y,...,Wn = Dn},

we may write the last term in the above formula for En [Em[X]] (G1... Gn) as

E, [E.(X]] (@ ...0,) M4

YX

(Gh... OnTm41 Dy)

Dt tictlen Weng dieenN x Plwra

=

>»,

= Ona... WN = Only

= O1,...,Wn

= On}

= @],...,0n

= Wn}

X(@...0N)

Wrgdacens WN

x Plwns

= Wnt, + WN = UNi[W)

= E,(X1(@,...0,). (iv)

Conditional Jensen's inequality. the proof of part (v) in Appendix A.

This follows from part (i) just like

Exercise 6.2. Verify that the discounted value of the static hedging portfelio constructed in the proof of Theorem 6.3.2 is a martingale under P.

Solution The static hedging portfolio in Theorem 6.3.2 is, at time n, to short zero coupon bonds maturing at time m and to hold one share of the Eo asset with price S,,, The value of this portfolio at time k, where n < k < m, 1S

Sn

Xn =Sk~

Bam,

K=nnti,...,m

nym

Forn