Introduction to Ratemaking and Loss Reserving for Property and Casualty Insurance [Solutions Manual, 2 ed.] 1566983959

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Solutions Manual for

_ Introduction to Ratemaking and Loss Reserving for Property and Casualty Insurance Second Edition by Robert L. Brown, Ph.D., FSA, FCIA, ACAS Leon R. Gottlieb, FCAS, MAAA

ACTEX Publications, Inc. Winsted, Connecticut

Solutions Manual for

Introduction to Ratemaking and Loss Reserving for Property and Casualty Insurance Second Edition by Robert L. Brown, Ph.D., FSA, FCIA, ACAS Leon R. Gottlieb, FCAS, MAAA

ACTEX Publications, Inc. Winsted, Connecticut

Copyright © 1993, 2001 by ACTEX Publications, Inc.

All rights reserved.

No portion of this book

may be reproduced in any form or by any

means without the prior written permission of the copyright owner. Requests for permission should be addressed to: ACTEX Publications Inc. P.O. Box 974

Winsted, CT 06098

TABLE OF CONTENTS

Manufactured in the United States of America

1098765432

ISBN: 1-56698-395-9

CHAPTER 1 1.1

(a) The law of large numbers states that as the number of observations ingreases, the difference between the observed relative frequency of an event and the true underlying probability tends to zero.

(b) The risk to the insurance company is not equal to the sum of the individual risks (variance of total outcome) transferred to it.

es

u(x) =

u(x) = u(x) =

k-logz

— ka! —ka~?

Since u’(z) > 0 and u”(z) < 0, this decision maker is risk averse. 1.3.

To reflect the risk attribute, we use utility value rather than monetary value.

EUV(A) = 1(.6) + .5(.1) + 0(.3) = .65 EUV(B) = .9(.5) + .8(.3) + .2(.2) =.73 A risk avoider would choose Proposal B.

1.4 (a) EMV(X) = 50,000(.35) — 20,000(.65) = 4500 EMV(Y) = 5,000(.55) — 5,000(.45) = 500 Both choose X based on expected monetary value.

Chapter I

2

(b) A:

EUV(X) = 1.00(.35) + .30(.65) = .545 EUV(Y) = .55(.55) +. .45(.45) = .505

Businessman A chooses X based on expected utility value.

B:

EUV(X) = 1.00(.35) + .55(.65) = .70750 EUV(Y) = .77(.55) + .709(.45) = .74255

Businessman B chooses Y based on expected utility value.

1.5

(a) WP) = VP — 1000 = (P—1000)!/2 ul(P) = 4(P—1000)"' u"(P) = — 4(P-1000)-? Management is risk averse since u’/(P) > 0 but u”(P) < 0. (b) (i) EMV(A) = 3000(.10) + 3500(.20) + --- + 5000(.10) = 4000 EMV(B) = 2000(.10) + 3000(.25) + --- + 6000(.10) = 4000 The EMV is the same for both proposals, so management would be indifferent on this basis. (ii) First we find the utility value of each profit amount.

Proposal A

Proposal B

Profit

Utility Probability

Profit

3000 3500 4000 4500 5000

44.72 10 250,00-— 520 54.77 40 59.16.20 63.24 ~—.10

2000 3000 4000 5000 6000

Utility Probability

31.62 44.72 ATT 63.24 7T0at

10 “ad indO = 10

EUV(A) = 44.72(.10) + 50.00(.20) + --- + 63.24(.10) = 54.536 EUV(B) = 31.62(.10) + 44.72(.25) + --- + 70.71(.10) = 53.654 Management chooses A based on expected utility value.

NOTE: We can also see that Proposal B has the larger variance.

Chapter 1

1.6

3

(a) EMV (no insurance) = 10,000(p) + 30,000(1—p) EMV (insurance) = 20,000(p) + 25,000(1—p) Equating and solving for p we have

10,000p + 30,000(1—p) = 20,000(p) + 25,000(1—p), which solves for p = fs =e 7 (b) The utility values of the various profit amounts are as follows. Freeze No Freeze No Insurance 71 158 Insurance 123 14] EUV (no insurance) = 71(p) + 158(1—p) EUV (insurance) = 123(—p) + 141(1—p) Equating and solving for p we have

Tip+ 1580—p) = which solves for p =

1.7.

hee

“9

123p+ 141(1—p),

is = .2464.

ee

S0000

With insurance your utility position is (2 rpo00=)

ane :

Without insurance your expected utility position is

[7 82)’ (sob) # - (19) (rxtoo)(sozboa)($0.000-2)'*10"°*

= (75) (rohc0) (sohoa) [(50,000)!9 — (20,000)!.9]. Equate the two expected utility positions and solve for G.

(50,000—G)?

(75) (; An5) [(50,000)'-9—(20,000)!9} 12,258.46

Then we have G = 50,000 — (12,258.46)!9/9 = 15,109.54.

Chapter I

With no wager your wealth is 20,000 and your expected utility

position is | — exp(- (06-000) = .1812692. If you wager an amount w, you end up with either 30,000 — w (if you win) or 20,000 — w (if you lose). The expected utility posie ath f 30,000—w re © 20 O06") Equating and solving for w we have

1812692 == 1 — een 5(.7408182)

exp(ee 1¢b00)

= 4(8187307) exp( rotp00 } Ee

Ww .8187308 = (.779774) - exp( roteo 00 |

exp( 19-400) = 1.049958 w = 4,875.05 W

19

t=

With no wager the wealth is 3000, with expected utility value

10,000(3000) — (3000)? = 21,000,000. With the wager the wealth is either 5000 (if she wins). or 3000— w (if she loses), with expected utility value

.30[10,000(5000) — (5000)*] + .70[10,000(3000—w) — (3000—w)?]. Equating and solving for w we have

21,000,000 = 7,500,000 + .7[21,000,000 — 4000w — w], which solves for

nee

1.10

if 2800 +

—— 2 +4,800,000(.7) _ 390.46.

The criteria to review (see Section 1.4) include the following: Economically feasible (OK) Economic value is calculable (common; should be) Loss must be definite (OK)

Loss must be accidental (should be if no profit) Exposures in risk class homogeneous (OK) ae a Units spatially and temporally independent (OK)

Yes, the insurance purchase is appropriate.

Chapter 1

1.11

5

(a) This is really the same as Exercise 1.10; the risk is insurable. (b) The net single premium is the expected value of the present value of the insurance payment. If Z denotes the random present value of payment, then

NSP = E[Z] 8000(1.10)- 1(408) + 5000(1. 10)-?( 4308) + 0 1924.44

(c) Var(Z)= E[Z?] — (E(Z1)°

E[Z?] = (8000(1.10)"° (48 §) + [s000(1. 10)-2/° (4) = 11,215,081 Var(Z) = 11,215,081 — 3,703,469 = 7,511,612

1.12

1.13

Gambling:

Creates risk where none exists or needed to. Takes dollars of high marginal utility. If you win you get dollars of lower marginal utility (if risk averse).

Insurance:

Transfers risk through pooling techniques. Takes dollars of low marginal utility and protects dollars of high marginal utility. In total, society has higher total utility with insurance than without.

p= o

—

o =

Vo? = 2000/2/3 = 1632.99

u =ptol6 = 4000 +1932-29 — 4972.17 The gross premium is 4500. Since this exceeds the expected utility loss, no insurance will be purchased.

Chapter 1

1.14

(a) The gross premium is 1.10% of the expected loss, so we have

GP = 1.10- E(L) = 1.10[10,000(.15) + 20,000(.04) + 50,000(.01)] = 3080 Utility with insurance = U(525,000 — G) = 13.16527 Utility without insurance = .80u(525,000) + .15u(515,000)

+ .04u(505,000) + .01u(475,000) = 13.16571441 .. Do not buy insurance. (b) This time the company's gross premium is

GP = 1.10-E(L) 1.10[20,000(.04)(.50) + 50,000(.01).50)] = 715 Mr.

Smith's

expected

utility

position,

with

insurance,

is

(G=715)

80u(525,000 — G) +.15u(525,000 — 10,000 — G) + .04u(525,000 — 10,000 — G) + .01u(525,000 — 25,000 — G) = 13.16564312. His expected utility position without insurance is 13.165714, from

part (a). .. Do not buy insurance Risk:

A measure of variation in economic outcomes

e.g.:

risk of a monetary loss if house burns down

Peril: A cause of risk e.g.: fire, wind, theft, illness

Hazard: A contributing factor to the peril €.g.: poor wiring, location, moral hazard

CHAPTER 2 2:1

All-risks or comprehensive covers everything except what is specifically excluded. Specified perils only covers the named perils. All-risks cover will exclude several perils, such as nuclear radio-activity, war, wear and tear, so it is not absolutely all-risks.

22

A loss is covered by a policy. only if a covered peril is the proximate cause of a covered consequence (both are needed). A covered peril is the proximate cause if it is the cause that initiates an unbroken sequence of events leading to a covered conse-

quence. 2.3

Reasons for policy limits: 1. Clarifies obligation of insurer. 2. Provides an upper bound on risk to insurer, decreasing the probability of insurer insolvency. Also decreases the premi-

um. Makes sure that policyholder cannot profit from a loss. he had Allows the policyholder to choose the most appropriate

coverage at an appropriate price. 2.4

(a) Objectives of the coinsurance clause: 1. It encourages insurance to value. 2. It creates premium equity between insureds. 3. The overall rate level can be smaller but still adequate.

(b) Disadvantages of the coinsurance clause: 1. Not well understood by policyholder.

2. A policyholder who buys less than full coverage is only 3. 4.

penalized if there is a claim, since he or she can pay a lower premium and get away with it. The 80% coinsurance clause discriminates against those who carry higher levels of insurance. Because of the misunderstanding of the coinsurance

clause, some meaning.

costly disputes arise over

its use and

Chapter 2

5. With high rates of inflation in real estate, a homeowner may unwittingly fall below the coinsurance percentage requirement.

6. A coinsurance percentage of less than 100% implies a recommendation to policyholders to buy less than full coverage.

2

Advantages:

1. Gets rid of small claims and their expenses. 2. All losses are reduced by amount of deductible, so premium is lower. 3. 4.

Provides an economic incentive for the policyholder to prevent a claim. Policyholders can optimize the use of their limited premium

dollars by using deductibles to save money where the utility value of the coverage is not as great. Disadvantages: 1. Insured may be disappointed by being put at risk. 2. Can lead to misunderstandings and bad public relations. 3. Makes the marketing of the coverage more difficult. 4. The insured may just inflate the claim to recover the deductible which, in turn, penalizes the honest policyholder because of the resulting higher premium.

2.6

(a) Salvage: Once the insurer has paid the policyholder full compensation for damaged property, it assumes ownership of the property and can sell it for its salvage value. This decreases the premium for the coverage. (b) Subrogation: The insurer, having indemnified the policyholder, acquires the legal rights of the policyholder to sue the

party at fault and recover costs. This will lower some premiums (e.g., homeowners dwelling coverage or auto collision), and raise the corresponding premium on the liability cover.

Chapter 2 20

(a) Contributory negligence: It used to be that if a worker contributed in any way to the injury or sickness, then the worker could not seek compensation. (b) Fellow-servant: If a fellow worker contributed in any way to the worker’s injury or sickness, then the employer was not at

fault and the worker could not seek compensation. (c) Assumption-of-risk: The ability of the worker to sue was often restricted if the worker had advance knowledge of the

inherent dangers of the job.

2.8

Objectives of workers compensation: 1.

Broad coverage of workers for occupational injury and disease. Substantial protection against loss of income. Sufficient medical care and rehab services. Encouragement of safety (through lower premiums). ar An efficient and effective administrative system. ieee

25

Workers compensation benefits: 1. Medical care benefits (normally unlimited). 2. Disability income benefits (after waiting period). 3.- Death benefits including a burial allowance plus cash-income payments to eligible survivors.

4. Rehab services and benefits.

2.10

Find X such that

400,000

@

x (0% of 700000 )= 320,000. The equation solves for X = 512,000.

Chapter 2

10

Find X such that

120,000

XVof 200,000: ween

ee

—

SS

The equation solves for X = 80%.

212

120,000

i The payment would be pe 70% of 200,000

175,000 == 150,000.

But the payment will be limited to the policy limit of 120,000.

213 Tose

250

750

1000

Deductible

250

2%

0

The deductible is X = 2(0) + 4(250) = 83.33, so the payment is 750 — 83.33 = 666.67.

(a) Pay (.80)(12,000) = 9600, which is within the policy limit. (b) Pay (12,000 — 1,000) = 11,000; but the payment is limited to the policy limit of 10,000.

(c) Cie

5000

12,000

15,000

Deductible

5000

Xx

0

The deductible is X = (75) 6000) a (5) — 1,500; the payment is 12,000 — 1,500 = 10,500.

Chapter 2

245

11

Let L denote the loss. If L d, the claim payment is LD — d. d

10

E{L] =4a OdL + Me

eat

= th}pe at| i.ty [50-104 + 4d| Since this must equal 2, we find d from = 4 [50—10d+ 5e|, which solves for d = 3.68.

2.10

250 Deductible: 1

E(L] = =p

250

5000

if odL+

(L — 250)dL

1 [42 172 ~ 2505 a e sobs "| = 2256.25 500 Deductible: 1

500

5000

E[L] = x99 / odn+ | (L — 500)dL l

li?

a soon WN | - soo

se

|| eeallesin

The expected loss payment will be reduced by

2256.25 — 2025.00 = 231.25.

CHAPTER 3 2

This could result from one or more of the following, which is not intended to be a complete list. 1. Trend factor greater than zero.

2. A change in the permissible loss ratio, and hence in the 3.

3.2

expense ratio. A change in the externalities (e.g., speed limit), where the effect is not yet felt.

One-half of six-month written premium will be fully earned, and the rest will be half earned. For one-year policies, written premium will be half earned by year-end. Thus we have

Unearned premium

= +]4(24,000,000) " (120,000,000)] = 66,000,000

But Dollars earned = 144,000,000

be

— 66,000,000 = 78,000,000

Given Gross Premium: 500 Old Loss Cost: .67(500) = 335 New Loss Cost: 335 New Commission: 50 All Other Expenses: .21 Then the new average gross premium will be a3 + 50 = 474.05. (Note error on text page 95: should be 15%, not 5%.)

The general expenses percentage

Chapter 3

3.4

13

The current gross premium expense table.

is 1000, leading to the following

Old Basis Loss Cost Commission General Expenses Taxes Profit

640 20% 80 3% 50

New Basis 640 12% 80 3% 50

The loss cost, general expenses, and profit amount remain the same, but they now represent 85% of the gross premium. There-

fore the new gross premium is 640

3

-~ 30 _ 905.88.

The claim is not true. The trend in loss cost (i.e., trend factor) is from average date of incurral in past historical period to average date of incurral in future exposure period. Inflation in the loss development factor is to cover the period from average date of incurral in future exposure period to average

date of settlement.

3.6

(a) Z=

fe fob 0?

Ce WwW

aa

(Z) ae aS & (4) =

4 (EK)? =-}(EK)3/? from which

we find Q + K = S$ —Q,sothatQ = 5(S—K). (b) For Q < E < S, the function follows the tangent line with

6% Kk : slope (Q+K)(S—Q) since (Q+K) == (S—Q) lope —;>—va vr = (Q+K)” (SQ). Using a point-slope formula with Z = 1 at FE= S, we have

eo as.

K

o>

Maer:

S-B ~ (Q+K )\(S-O) & (On k \e7 where Q = Ss —K). Substituting for Q we have

Chapter 3

15

leo, Oe

fae

MK

(OLR

Then

z — 1_ S-E)GK) (S+K) _ 8°4+2KS+ K*-4KS+4KE (S+KY

BS hs

TAKE

(S+K)? _ (S-K/+4KE 3.8

(S+K)*

The mid-point of policy year 1998 is January 1, 1999, the midpoint of policy year 1999 is January 1, 2000, and the experience

period mid-point is November 1, 2002.

Thus the 1998 loss cost

will be projected 3 years, 10 months (3 years) and the 1999 loss cost will be projected 2 years,

10 months

a

years).

The

weighted projected loss cost is

(.40) (200) e619)23/9) + (,60)(217)e(1907/6) — 290.22.

3.9

For Company A, the gross premium is EC , where LC denotes the loss cost.

For Company

\=1/2 B, the gross premium

is pel

Ree since

expenses other than profit and contingencies represent 28.5% and profit and contingencies represents 5%, a total of 32.5%. Equating and solving for 2 we have

Peme eich: Had

Lo

_- Lodi)?

OO

a

675 PL

?

(292) ete 7544.

Chapter 3

1998 1999:

Jhae ee 2000

April 1 2001

3 at 1.08P and g at P, producing a factor of

4(1.05) + $(1.134) = 1.1235. Then the 1991 earned premium at current rates is 2927(1.1235) = 3288.48. 2000:

4 at P and 4 at 1.08P, producing a factor of $(1.134) + (1.05) = 1.0605. Then the 1992 earned premium at current rates is 3301(1.0605) = 3500.71.

2001:

“ at 1.134P and a at 1.08P, producing a factor of

$5(1.000) + $3.(1.05) = 1.0359375. Then the 1993 earned premium at current rates is 3563(1.0359) = 3691.

The mid-point of the 2001 accident year is July 1, 2001, and the mid-point of the future exposure period is August 1, 2003. This is a distance of 2 years and 1 month, so t = 245 years.

State A has a uniform distribution of renewals.

Assuming a

uniform distribution of claims, the average accident incurral date

is January 1, 2002. State B has all policies renewing on January 1, 2001. Assuming a uniform distribution of claims, the average accident incurral date is July 1, 2001.

Chapter 3

3.13

Li

Using Class 1 as the base class, the new Class 2 differential is

1.10(-280) = 1.276.

3.14

The student has used the loss cost method. We should also try the loss ratio method with Class A as the base class. Then, from Equation (3.12), (New Differential); = (Existing Differential);

- ie :

producing the following results. Class

Loss Ratio Indicated Differential

A B Cc D E F G

1.000 1.092 1.122 1.176 1.404 1.428 1.560

The loss cost method is subject to distortions if there exist heterogeneous distributions of cross classifications (e.g., territory versus class). The loss ratio method automatically adjusts for such heterogeneity, so the loss ratio method is recommended.

18 3.15

Chapter 3 The weighted average class differential is 1.25 (constant for all territories). The present base rates are as follows: Territory

| Present Base Rate

1 2 3

Old Differential

152.00 130.40 96.00

1.0000000 8578947 .6315789

Next we find the new territorial differentials using the loss ratio method. Territory

1 Pe 3

Old Differential

Loss Ratio

1.0000000 8578947 6315789

.720 .620 .770

New Differential

1.0000000 .7387426 .6754385

The overall rate change is +7%, or a factor of 1.07.

The old

average differential is .8836841, and the new average differential

is .8567104. The new base rate for Territory 1 is 152 x 1.07 x earn

= 167.76,

and the new base rate for Territory 3 is 167.76(.67543 ry Ed SE

3.16

First we find the existing base rate for each territory. Territory

1 Bs; 3

Average Class Differential

Base Rate

Existing Differential

1.043 1.03 1.0114286

215.65496 193.54839 177.9661

1.000000 897491 8252353

Next we find the new territorial differentials using the loss ratio method. Territory

Existing Differential

1 Z 3

1.000000* 897491 8252353

Loss Ratio New Differential

.700 .660 .720

1.000000 8462058 8488135

“One could easily argue that Territory 3 should be the base rate, but there would be no difference in the final answer.

Chapter 3

19

The overall rate change is + 3%, or a factor of 1.03. There is heterogeneity in the class distribution, so the balanceback factor must be determined. Old Average Differential:

{[2000(1.00)+150(1.10) +600(.90) +100(1.25)+150(2.00)] + 897491 [800(1.00)+150(1.10)+--- + 50(2.00)]

+ .8252353[2250(1.00) + 200(1.10)+---+50(.200)}} i.8000 = .9303055 New Average Differential:

{[2000(1.00) + 150(1.10) +--- + 150(2.00)] + .8462058[800(1.00) + 150(1.10)+---+50(2.00)] + .8488135[2250(1.00) +200(1.10) +--- +50(2.00)]} ‘i8000 = .9308023 Then the new base rate for Territory 1, Class A is

215.65496 x 1.03 x Reriiee = 222.01, and the new rate for Territory 3, Class B is 222.01 x 1.10 x .8488135 = 207.29.

ele

We will use the 1999 policy year data. The mid-point of the experience period is January 1, 2000. The price is to be set July

1, 2001 for one-year policies, so the mid-point of the exposure period is July 1, 2002. Therefore the trend factor is 1.30. To determine loss development following incurred losses data.

factors,

we

first

note

Incurred Losses as of Report Number Policy Year 1996 1997 1998

1999

0 2 800,000 900,000

1,000,000

1 660,000 880,000 990,000

Z 693,000 924,000

3 693,000

the

Chapter 3

20

This leads to the following loss development factors:

1/0

1.10

2/1

oo/2

1.05

1.00

Using the loss development factors we next find Expected Losses (Trended and Developed) = 1,000,000(1.10)(1.05)(1.30) = 1,501,500. At this point we can proceed using either the loss ratio or the loss cost approach. Loss Ratio Method: Earned Premium at Current Rates:

10,000(80) + 8000(150) = 2,000,000 Expected Loss Ratio: man Permissible Loss Ratio:

.700

Indicated Rate Change:

ae

= .75075

= 1.0725, or +7.25%

To develop the territorial differential change, we set Territory A

differential at 1.000. Territory A

Existing Differential

Territory B

1.000

130 — 1.875

Earned Premium at Current Rates

800,000

1,200,000

Incurred Losses

520,000

Loss Ratio Indicated Differential Credibility Differential

.650 1.000 1.000

480,000

.400 1.1538462 12259615

Old Average Differential: 1.38 New Average Differential: 1.1004273 Territory A Average Rate: 80 x 1.0725 x jess Territory B Average Rate:

= 108.29

108.29 x 1.2259615 = 132.76

Chapter 3

Pa

Loss Cost Method: Expected Losses:

1,501,500

Loss Cost per Unit Exposure: Average Gross Premium:

Lee

= 83.416

Seen = 119.16

Territorial Differential Change: Territory

Existing Differential

LossCost per Unit

A B

1.000 1.875

52 60

New Average Differential:

Indicated Differential

Credibility Differential

1.000 1.1538462

1.000 1.2259615

1.1004273

Then the Territory A Average Rate is pte

= 108.29, and

the Territory B Average Rate is 108.29(1.2259615) = 132.76.

Loss Cost Method:

Trend Factor: (1.08)! Loss Development Factor: (1.10)(1.05) = 1.155 Time Factor: July 1, 2000 to February 1, 2003, which is 2 years and 7 months, or 2.583 years Expected Dollars of Losses:

|

1,570,000(1.08)*583(1.155) = 2,212,210 New Premium per Exposure Limit: aa Average Gross Premium:

=

122.90

feet = 153.62569

Differentials (adjusting for heterogeneity in cross distribution)

Territory

I

Loss Cost

1,100,00 a elt To00NTy+ 400NT.D) = 74:324324

Differential

ST eA ne

1.000

470,000 2000(1) + 2000(1.2) ~~ 106.81818

1.4371902

Eee =

Chapter 3

Class

Loss Cost 900,000 peOe ee 10,000(1)+2000(1.1)

Differential 1.000

~~ 73.77049

6 70,000 hh ee a = 10806452 FOO)odSALEEDOONTA)

1.4648746

Ee

Average Differential:

10,000(1) + 4000(1.465) + 2000(1.437) + 2000(1.465 x 1.437) 18,000 = 1.2746936 Base Rate for Territory 1, Class 1: Peek

= 120.5197

Base Rate for Territory 2, Class 2: 120.5197 x 1.437 x 1.465 = 253.72 Loss Ratio Method:

Trend Factor: (1.08)! Loss Development Factor: (1.10)(1.05) = 1.155 Time Factor: July 1, 2000 to February 1, 2003, which is 2 years and 7 months, or 2.583 years Expected Dollars of Losses:

1,570,000(1.08)*>83(1.155) = 2,212,210 Expected Loss Ratio:

2,212,210 res, T00(10,000)+120(4000)+1 10(2000)+132(2000) ~ !: Indicated Rate Change: 112638 — 14079747 New Base Rate (unadjusted): New Differentials: ; : Territory Loss Ratio

| 7

1,100,000

140.80 fe 8): Differential

{0,0007100) + 40007120)

eS bel nt

ee

ee

= .74324324 470,000 ANE ae el = .9710743

«1-00

New Differential

1.0000000

Chapter 3

23

2 Territory

F Class Loss Ratio

My

Old Differential

et 10,000(100) + 2000(1

New Differential

1100 sae 120000000 :

10)

= .7377049

670,000 4000(120) + 2000(132) = 9005376

1.20

1.4648746

Old Average Differential:

10, 000(1) + 4000(1.2) + 2000(1.1) + 2000(1.32) _-, 18,000 = 1.091 3

New Average Differential: 10,000(1) + 4000(1 .4648746) + 2000(1.4371902) + 2000(1 .4648746 x 1.4371902) 18,000

= 1.2746936 New Base Rate (Balanced-Back):

1091) 140.79747 x yyaq6936 =2 120.5197

Then the new base rate for Territory 2, Class 2 is

120.5197 x 1.437 x 1.465 = 253.72. NOTE: Earned Exposures:

Territory |

Territory 2

Class 1

10,000

2,000

Class 2

4,000

2,000

Old Rates: Class |

Class 2 New Rates:

Class 1 Class 2

Territory 1 100

Territory 2 110

120

132

Territory 1

Territory 2

120.52 176.55

173221 253.73

Old Premium Income: 1,964,000 New Premium Income: 2,765,280 Ratio: 1.4079837, as required (within round-off error)

24 3.19

Chapter 3 (a) Calendar Year 1998 Occurrence #1 1,000 Occurrence #2 0 Occurrence #3 0 1,000 (b) Calendar Year 2000 Occurrence #1 1,000 Occurrence #2 5,000 Occurrence #3 5,000 11,000 (c) Accident Year 1999 (as of 12/31/2000) Occurrence #1 0

3.20

Occurrence #2

5,000

Occurrence #3

5,000 10,000

Only two events contribute to the answer: Claim #2: 200 (Paid in 1991) 4200 (12/31/92 Reserve)

400

300 (Paid in 1992)+

300 700

Claim #3: 0 (12/31/92 Reserve) = Total

Chapter 3

321

25

Policy year 1997 mid-point is December 31, 1997 (or January 1, 1998), and policy year 1998 mid-point is December 31, 1998 (or January 1, 1999). The mid-point of the effective period is December 31, 2001 (or January 1 2002). Then the policy year 1997 expected loss ratio is

1,000,000(1.05)* _ =:6077531, 7,000,000 and the policy year 1998 expected loss ratio is

2,000,000(1.05)? _ 3,000,000

——.

TAS:

Then the weighted loss ratio is

.30(.6077531) + .70(.77175) =.7225509, and the indicated change is

Weighted Loss Ratio _ ‘Permissible Loss Ratio

.7225509 ¥ 1.20425 16, .600

a change of +20.4%.

Spas

Incurred Losses

Earned Premiums £ ~

Paid Losses + (1992 Reserve — 1991 Reserve) Written Premium + (1991 Unearned — 1992 Unearned)

_ 90,000 + (140,000 — 160,000) ~ 100,000 + (50,000 — 40,000)

_ ~

63 °

26

3.23

Chapter 3

The new commission will be a constant 50, with effective date

July 1, 2001. Assuming one-year policies, the mid-point of the exposure period is July 1, 2002, and the mid-point of the experience period is January 1, 2000. The following values are then developed.

Trend Factor: (1.09)°* = 1.2404128 Dollars of Losses (Trended and Developed):

5,200,788

Loss Cost per Unit Exposure:

170.55678

Permissible Loss Ratio: Average Gross Premium:

oe 30,493

1 — (.05+.03+.05) = .87 [Qo 0r8 = 196.04228

Base Rate before Commission: Base Rate with Commission:

ee

= 156.83382

156.83 + 50 = 206.83

CHAPTER 4 Exercises in this chapter were done on a pocket calculator. Students are encouraged to use Excel, but the answers will be slightly different than the ones presented here due to rounding.

Note:

4.1

The gross IBNR is the pure IBNR, plus the development in known

claims, plus the files that are closed but may reopen. 4.2

Salvage: The insurance company receives any salvage value of, for example, a car for which full payment has been made to the

insured. Subrogation: The insurance company acquires the policyholder's right to sue a third party and has this right to recover monies. Both salvage and subrogation will reduce factors, and may make them less than one.

loss development

4.3

It would be better to use incurred loss development. The increase in the retention limit will be reflected immediately in the incurred data, but will not be reflected for quite some time in the paid data.

4.4

The loss reserve would be developed in the following two steps:

4.5

1.

Expected Ultimate Losses = (Actual Earned Premiums) - (Expected Loss Ratio)

2.

Loss Reserve = (Expected Ultimate Losses) — (Paid-to-Date)

For all three models the case reserve is

(419.7—231.3) + (465.5—301.7) + (456.7—349.8) + (442.2—380.2) = 521.1, The cumulative incurred minus the paid-to-date.

From the text

data we first find the loss-development factors (for each model),

and then the total loss reserves (see tables on page 26). Finally, the gross IBNR is the excess of the total loss reserve over the case reserve.

Chapter 4

28

Accident

Ratio of Successive Development Years

Year 1993 1994 1995 1996 1997 1998

1/0 1.250 1.250 iPa hae 94 1.240 1.246

2/1 1.100 1.085 1.093 1.100 1.107

3/2 1.082 1.088 1.094 1.100

4/3 1.008 1.016 1.024

Average:

1.240

1.097

1.091

1.016

4-Year: Mean:

1.236 1.240

1.096 1.098

1.091 1.092

1.016 1.017

Accident

Year 1996 1997 1998 1999

Development Table (Average)

1

520.6

2

3

510.7 5711

498.3 5502 623.1

Recerve

4 449.4 506.5 566.3 633.3

for Year 69.2 156.7 264.6 402.0

Total Loss Reserve

Accident Year 1996 1997 1998 1999

Accident Year 1996 1997 1998 1999

Development Table (4-Year Average) 1

518.6

A

510.3 568.5

3

520.4

514.0 5713

Reserve

4 _ for 449.4 4983 506.5 556.8 565.9 620.3 630.5 Total Loss Reserve

Development Table (Mean) 1 2 a 498.7 558.1 623.9

892.5

Year 69.2 156.7 264.2 399.2 889.3

Reserve

4 _ for Year 4498 69.6 507.3 157.5 567.6 265.9 634.6 403.3

Total Loss Reserve 896.2

Calculations carried to more decimal places than those shown.

Chapter 4

4.6

29

Given the data in the text, we can construct the following tables: Accident Year 1993 1994 1995 1996 1997 1998 Average: Dav earn Mean:

Ratio of Successive Development Years 2/1 3/2 4/3 5/4 1.502 1.075 1.077 1.020 1.003 1.308 1.074 1.077 1.054 1.011 1/0

1.366 1.580 1.596 1.812 1.527 eho32: 1.519

1.155 1.285 1.146

1.117 1.128

1.044

1.147 1.100 ‘1.039 1.147, — b18800-eid.039 1.153 1.104 1.042

Development Table (Average)

(a) Accident

2

Year 1994

3

4

1995

1996 1997 1998 1999

(b) Accident Year

5982

5881 6214 6832 6862-7545

6477 6113 7102 ~—7843

5

1994

1995

1998 1999

6002

6214 6885

5881 6832 7570

6477 6113. FA0Z 7869

1.007 1.007 1.008

1.000 1.000 1.000 Reserve

6 for Year 5199 6766 6521 6155 7150 7896 3979

6766 = 6521 6155 7150 7896 Total Loss Reserve

Development Table (5-year Average) Z 4 5 3

1996 1997

6/5 1.000

6766 = 6521 6155 ~—-F150 7922

Reserve

6 for Year 5199 0 6766 46 6521 290 6155 806 1733 7150 4005 IIe:

Total Loss Reserve

Development Table (Mean) 4 2 3

(C) Accident Year

1994 1995 1996 1997 1998 1999

5949

6247 6861

5906 6898 1515

6495 6156 7190 7896

2 6772 6545 6204 7246 7957

6853

6879

Reserve

6 for Year 5199 0 52 6772 314 6545 6204 855 7246 1829 4040 7957

Total Loss Reserve 7089 Calculations carried to more decimal places than those shown.

Chapter 4

30 4.7

Using the data from Exercise 4.6, we can develop the following payment patterns, first undiscounted and then discounted.

(a)

Accident

ear

1994 1995 1996 1997 1998 1999

1

Average Factor Model 2 3 4 5

2065

46 246 44 232,841 270 48 298 53

Accident Year

1994 1995 1996 1997 1998 1999

532 797 618 880683. l

2

on ooooo

3

4

=

14> S585 STi

44 238 40 210 935 228. G38 2359939

Reserve

6 for Year 0

FIQ. 795

1996

44 278 759 Oo°oO 1594 o.oo 0 3642

Total Loss Reserve

(b)

6317

5-Year Average Factor Model Accident Year

1994 1995 1996 1997 1998 1999

l

3

4

5 46

2085

Accident Year 0

1994 1995 1996 1997 1998 1999

Z

a3 797 —«618 883685 2

2016

770 798

246 232 270 299

44 4l

48 53

3

5

514 =558 578

44 40 35 38 39

238 210 228 236

oy; oooco

Reserve for Year

0 0 0 0 0 0

Total Loss Reserve

0 44 278 759 1594 3667 6342

Chapter 4

31

(C) Accident Year 1994 1995 1996 1997 1998 1999

Mean Factor Model 0

Accident Year 0 1994 1995 1996 1997 1998 1999

1

Z

3

4

830 912

557m 651 7-715)

264 250°. 292 321

1

2

3

4

1964

802 824

538 588 604

2032

5

6 0 52210 SO 0O AT 55 0 6! 0

Reserve 6 for Year 0 0 50) «6(—O 50 2555 «450 300 226 40 0 804 247 43 O 1680 253 45 0 3690 Total Loss Reserve 6524 5

Entries have been rounded to the nearest dollar

Chapter 4

32

4.8 (a) Given the data in the text, we can construct the following tables:

Cumulative Payments (in thousands) Year l Z 3 = 5 1994 192 443 596 741 839 839 1995 205 485 680 830 932 1996 230 575 805 1017 1997 5 288 698 973 1998 398 961 1999 530

Accident

ecient

Loss Development Factors

Year 1994 1995 1996 1997 1998

1/0 2.307 2300. 2.500 2.424 2.415

2/1 1.345 .0h402 1.400 1.394

3/2 1.243 122% 1.263

4/3 = 1.132 Tie

5/4 1.000

Average:

2.402

1.385

1.242

= 1.128

Mean:

2.408

1.388

1.244

1.127

1.000 1.000

Accident Payments by Development Year (Average) ] Year 2 3 4 5 1995 932.0 1996 1146.7 1146.7 1997 1208.9 1363.1 1363.1 1998 1331.3 1654.1 1861.1 1865.1 1999 1273.2 1763.8 2191.4 2471.0 1471.0 Total Loss Reserve

Recarve for Year 0.0 129.7 390.1 904.1 1941.0 3364.9

Accident

Raterve for Year

Year

1995 1996 1997 1998 1999

Payments by Development Year (Mean) ]

1276

2

3

4

1146 1210 1364 1333 1658 1869 1771 2202 2483 Total Loss Reserve

129 391 908 1954 3382

Chapter 4

33

(b) From the data in part (a), we can construct the following tables: Non-cumulative Payments by Development Year (Average) Accident Year 1 2 3 4 1995 1996 129.7 1997 235.9 154.2 1998 370.32 322000." 21170

1999

743.2

490.6

427.6

279.6

Discounted Payments by Development Year (Average) Accident Year

1995 1996 1997 1998 1999

l

2

2

4

124.23 225.95 135.50 354.68 283.57 170.10 711.86 «431.11 344.72 206.80 Total Discounted Loss Reserve

Reserve for Year

124.23 361.45 808.35 1694.49 2988.52

Non-cumulative Payments by Development Year (Mean) Accident Year

1996 1997 1998 1999

1

746

2

3

4

372 495

237 325 431

129 154 211 280

Discounted Payments by Development Year (Mean) Accident Reserve Year

1996 1997 1998 1999

l

715

Z

So. 435

=

fH) 285 348

4

for Year

124 135 170

207

Total Loss Reserve

124 362 812 1705 3004

Calculations were carried to more decimals than those shown.

Chapter 4

34 4.9

First we find the expected number of claims in each accident year using a normal chain ladder approach. Accident

Year 1994

Loss Development Factors

1/0 1.125

3/2 1.042

4/3 1.010

1995. 35925501.0554 1.042, 1996. .1.12522,1.035541,042 TORT ed 1258 1.055 1998; 1.125 1999 —

L010

Average 1.125

2/1 1.055

1.055

1.042

Ultimate

5/4 1.000

Number 6203

6372 6505 6512 7523 8250

1.010

1.000

Now the payment pattern of Exercise 4.8 can be presented in the form of payments per claim incurred (either cumulative or noncumulative). For example, one common.approach used in industry is to divide each entry of the cumulative payment triangle by the estimated ultimate number of claims by accident year. This produces the following average severity triangle. Accident Year

1994 1995 1996 1997 1998 1999

Payment per Claim Incurred at Development Year t 0

30.95 32.17 35.36 44.24 52.91 64.26

1

2

3

4

5

71.42 76.11 88.41 107.22 127.76

96.08 106.72 123.77 149.46

119.46 130.26 156.37

135.26 146.26

135.26

From this point we can develop this triangle of payments per claim incurred so as to ascertain an estimate of the ultimate payment per claim incurred (PPCI) for each accident year. Then for each accident year

E [Dollars of Ultimate Claims] = E[PPCI]- E [Ultimate Number of Claims] . Working with the severity triangle often proves to be more manageable and accurate than working with total claims dollars. Using the average loss-development factors we obtain the following payment per claim incurred development factors.

Chapter 4

35

Accident Year

1994 1995 1996 1997 1998 Average:

Payment per Claim Incurred Development Factors 1/0 a) 3/2 4/3 5/4 2.303 1.346* 1.243 1 132 1.000 2.363 1.402 L22iee 1, 123 2.497 1.400 1.263 2.425 1.394

2.414 2.400

1.386

12431. 128

1.000

*1.346 = 96.1/71.4 Accident Year

1996 1997 1998 1999

Future Cumulative Payments per Claim 4 l 2 3 5 176.3 176.3 209.3 185.6 209.3 L7TON 21S:8" 248.0 248.0 213.6 265.2 299.2 2992 154.1

*219.8 = 127.7 x 1.386 x 1.242

Accident Paid-to- Ultimate Year

Date

Paid

1994 1995 1996 1997 1998 1999

135.3 146.3 156.3 149.4 127.7 64.2

135.3 146.3 176.3 209.3 248.0 299.2

Liability

per Claim

No. of

Claims

0.0 6203 0.0 6372 20.0 6505 59.9 6512 120.3 7523 235.0 8250 Total Loss Reserve

Reserve for Year

0.0 0.0 130.1 390.1 905.0 1938.8 3364.0

Chapter 4

36 4.10

(a) Accident

Cumulative Proportion Closed

Year

0

l

y

3

4

1995 1996 1997 1998 1999

.4000 .4000 Jot .3800 4000

.7000 .6583 |g Orse -7000

8500 8333 8500

.9300 .9500

= 1.0000

Most recent

.4000

.7000

8500

9500

=:1.0000

* 3571 = 3 (b) Accidenf{ umulative Closed on Most Recent Speedsp timated

Year 1995 1996 1997 1998 1999

0 400 480 560 600 600

1 700 840 980* 1050

2 850 1020 1190

3 950 1140

4 1000

Ultimate 1000 1200 1400 1500 1500

*980 = 1400 x .07 Accident Year

1995 1996 1997 1998 1999

Cumulative Dollar Losses on Most Recent Speeds 0 1 2 3 4

2,000 2,600 2,666 3,284 3,800

6,000 12ms= 9,243 ~—-:10,800

9,000 1Ee 14,400

11,441 15,600

14,000

*7273 = eae x 6840 = Most Recent x Actual Cumulative

Chapter 4

37

(c) @) Accident

Year 1995 1996 1997 1998 Average

Unadjusted Loss Development Factors 1/0 2/1 3/2 4/3 3.0000 1.5000 1.2444 1.2500 2.6308 1.5965 1.4286 3.7647 1.6071* 3.4615 3.2143 1.5679 1.3365 1.2500

*1.6071 = 14,400 8,960

using the given cumulative paid losses

Development Using Unadjusted Loss Development Factors Accident

Year 1996 1997 1998

1999

Z

3

4

19,500 15,600 24,058 14,400 28,289 10,800 31,993 3,800

3,900 9,658 17,489 28,193

Total Loss Reserve

59,240

19,246* 16,933 22,631 12,214 19,150 25,594

*19,246 = 14,400 x 1.3365 UL a a Sa

Paid- Reserve to-Date for Year

Snare” 28. Sas

Accident L°SS Development Factors on Most Recent Speeds Year 1/0 2/1 3/2 4/3 1.5000 1.2712 1.2237 1995 3.0000 1.5314 1.4006 1996 Pathe 5379 3.4670 1997 3.2887 1998 1.5298 1.3359 V2231 Average 3.1383 Development Using Average LDFs on Most Recent Speeds Accident Year

1996 1997 1998

1999

2

3

4

Paid- Reserve to-Date for Year

19,090 15,600 23,540 14,400 27,009 10,800 29,824 3,800

3,490 9,140 16,209 26,024

Total Loss Reserve

54,863

19,237 16,522 22,072* 11,926 18,244 24,372

=22,072-= 16,522 x-1.3359- =10,800 x-1.5298 x 1.3359

38

4.11

Chapter 4

Claims to be Paid in the Future

_=

a1; Ultimate Total ( . Afee ) == 650 x 1000 x (1ae or)

=7122 81

4.12

(a) Expected Loss Ratio Method

Accident

Earned

Year

Premium

1996 1997 1998 1999

25,000 29,750 33,000 38,000

— .70(Earned

Paid-to-

Premium)

Date

17,500 20,825 23,100 26,600 Total Loss

17,000 20,475* 21,750 17,475 Reserve

Reserve

for Year 500 350 1,350 9,125 11,325

*20,475 = 12,050 + 6,025 + 2,400 (b) Chain Ladder Method Accident Year

0

1996 1997 1998 1999

10,000 12,050 14,500 17,475

Newident

Cumulative Claims Paid ] Z

15,000 18,075 21,750

1/0

2/1

3/2

1996 1997 1998

1.500 1.500 1.500

1.133 L533

1.000

Accident

1997 1998 1999

17,000

Loss Development Factors

Year

Year

3

17,000 20,475

Chain Ladder Development 0

l

2

24,642 26,212 29,699

Darts)

Recercs

to-Date

for Year

20,475 21,750 17,475

0 2,892 12,224

Total Loss Reserve

15,116

3

20,475 24,642 29,699

Chapter 4

39

(c) Bornhuetter-Ferguson Method Accident Year

.70(Earned Premium)

.70(Earned Premium) x (1 — 1/fur)

Futt

1997 1998 1999

20,825 23,100 26,600

1.000 1.133 1.500 x 1.133 Total Loss Reserve

0 Derae 10, 948 13, 660

Months

Incremental Percentage Paid

Discount Factor

Product

12 24 36 48 60 72

20 5% 15 15 15 10

= (1.08)-'/2 (1.08)-3/2 (1.08)~5/2 (1.08)~7/2 (1.08)~9/2

24.0563 13.3646 12.3746 11.4580 7.0728

4.13

3

68.3263 23 = 45-420 Then the discounted loss reserve is

16,000,000 x “oe = 10,932,208. 4.14

The

mean

loss-development

factors

are

1.51874,

1.15325,

1.10421, 1.04233, 1.00771, and 1.00000. Then from the original data we find, in turn, the cumulative loss payments (left-hand table on page 40), the estimated loss reserves (right-hand table on page 40), the allocated loss reserves (left-hand table on page 41), and the discounted allocated loss reserves (right-hand table on page 41).

Also shown on page 41 are the factors used to discount the allocated loss reserves.

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V| Chapter 4 41

42 4.15

Chapter 4 If future cash flows are negative (e.g., because of salvage or subrogation) then discounted loss reserves could exceed undiscounted loss reserve.

CHAPTER 5

5.1

c andd

a2

b, c ande

5.3

$50,000 2 Z= 1900,0 +850, 00000 = 459 Experience Loss Ratio = Langage

425,000 _ = 850.000 = 0 Premium = Manual Premium ‘ | ( oS Experience Loss es Expected Loss Ratio

— 500,000 x |(459 y 32) 5.(1—459)| — 500,000 x [.353 + 541] = 500,000 x [.894] = $447,000

ney

2)

44

5.4

Chapter 5

ere es Limit

ay

Losses

Losses up to

Limit

Losses Greater

Limit

Than Limit

Total

12,100 (1—.879) x 100,000

21,117

100,000

9,017 (107 + 3,305+5,605)

1,000,000

48,062 16,000 (107+3,305+5,605 (1—.984) x 1,000,000 +7,395+8,450+ 18,200)

a) ILF = b)

59,062

577777 = 2.80

:

_

(Total from Above)?

Risk charge =

T,000,000

Limit. 100,000 1,000,000

;

.

Risk Charge 446 (=21,1177 x 10-5) 3,488 ( = 59,062? x 10-°)

_ 59,062+3,488 — 62,550 _

Risk Adjusted ICF =

117+ 446

= 21.563 = 2.90.

59,062

Chapter 5

5.5

45

Total Losses for Policy Limits of $100,000 and higher

Limited to $100,000 = 27,000 $50,000 = 20,000 Ratio = 1.350 Total Losses for Policy Limits of $500,000 and higher Limited to $500,000 = 24,000 $50,000 = 11,000 Ratio = 2.182 Limit 50,000 100,000 800,000

Pure Loss Rate Fixed Cost 240 30 324 ( = 240 x 1.350) 30 524 ( = 240 x 2.182) 30

Variable 30 39 62

Total 300 393 616

Note: Variable Expenses are 10% since for base class: VariableExpense

Premium

_ 30 _

800

wees

or a = 5 of pure loss & fixed expense

ILF 1.310 2.053

Chapter 5

46 5.6

Stabilize: Excess of loss, stop-loss, catastrophe Leverage: Quota-share

One of the major purposes of reinsurance is to limit the losses for the ceding insurer due to either high severity (excess of loss), frequency (catastrophe) or either (stop-loss). Since the

reinsurer is taking on higher risk and must make a return on its capital, the reinsurance premium will be greater (often appreciably so) than the present value of the expected losses. The insurer is willing to forego some limiting its potential losses.

profits in return for

On the other hand, proportional reinsurance is a sharing of the primary insurer’s book of business with the ceding company. The reinsurer often shares the policies on the identical basis as the ceding insurer, with premium and losses ceded in the same proportion. This reduces the net premium of the ceding insurer and thus reduces its leverage ratio (premium to surplus). In fact, if the ceding comission is high enough, the ceding insurer’s surplus might even increase.

Dake

Since the last policy under the treaty will incept on December 31, 2000 and the maximum policy term is 12 months, a loss can occur on December 31, 2001.

5.8.

Losses above $100,000 and $5,000,000 can have different loss development patterns and trends. Also an attachment point of $5,000,000 implies that the covered policies have even higher

limits.

Companies or individuals that buy high limits usually

have different risk profiles than those that buy lower limits,

such as would be included with lower attaching treaties.

* ~

2 >

NWU VOLV

4

Symi