NOVEMBER 2004 COURSE 3 SOLUTIONS

Question #1 Key: D

[ ] ( ) 0.06 0.08 0.050 0

0.070

120

1 100 520 7 7

t t t tt t x

t

E Z b v p x t dt e e e dt

e

µ∞ ∞ − −

∞−

= + =

⎛ ⎞ ⎡ ⎤= − =⎜ ⎟ ⎣ ⎦⎝ ⎠

∫ ∫

( ) ( )22 0.12 0.16 0.05 0.09

0 0 0

0.090

1 120 20

1 100 520 9 9

t t t t tt t x

t

E Z b v p x t dt e e e dt e dt

e

µ∞ ∞ ∞− − −

∞−

⎡ ⎤ = + = =⎣ ⎦

⎛ ⎞ ⎡ ⎤= =⎜ ⎟ ⎣ ⎦⎝ ⎠

∫ ∫ ∫

[ ]25 5 0.04535

9 7Var Z ⎛ ⎞= − =⎜ ⎟

⎝ ⎠

Question #2 Key: C Let ns = nonsmoker and s = smoker

k = qx kns+b g px k

ns+b g qx k

s+b g ( )s

x kp + 0 .05 0.95 0.10 0.90 1 .10 0.90 0.20 0.80 2 .15 0.85 0.30 0.70

( ) ( ) ( ) ( )1 2

1:2ns ns ns ns

x x xxA v q v p q += +

( ) 21 10.05 0.95 0.10

1.02 1.02× 0.1403=

( ) ( ) ( ) ( )1 2

1:2s s s s

x x xxA v q v p q ++

( )( )2

1 10.10 0.90 0.201.02 1.02

+ × 0.2710=

1

:2xA = weighted average = (0.75)(0.1403) + (0.25)(0.2710) = 0.1730

Question #3 Key: A

( ) 0.03xP A µ= =

2 0.030.202 2 0.03

0.06

xA µδ µ δδ

= = =+ +

⇒ =

( ) ( )( )

( )( )

2 22 13

0 2 20.060.09

0.20Var 0.20

0.03 1 1 1where 0.09 3 0.09

x xA AL

a

A a

δ

µµ δ µ δ

− −= = =

= = = = =+ +

Question #4 Key: B

( )( ) ( ) ( ) ( )0.1 60 0.08 60

602

0.005354

e es− −+

=

=

( )( )( ) ( )( )0.1 61 0.08 61

612

0.00492

e es− −+

=

=

600.004921 0.0810.005354

q = − =

Question #5 Key: B

( )2

For , 0.480

0.632580

50.6

F ωω

ω

ω

⎛ ⎞Ω = = ⎜ ⎟⎝ ⎠

=

=

For T(0) using De Moivre, ( )0.750.6

t tF tω

= = =

( )( )0.7 50.6 35.42t = = Question #6 Key: C

[ ] 1.81.8 0.63

E N mq q= = ⇒ = =

x ( )Nf x ( )NF x 0 0.064 0.064 1 0.288 0.352 2 0.432 0.784 3 0.216 1.000

First: 0.432 0.7 0.784< < so N = 2. Use 0.1 and 0.3 for amounts Second: 0.064 0.1 0.352< < so N = 1 Use 0.9 for amount Third: 0.432 0.5 0.784< < so N = 2 Use 0.5 and 0.7 for amounts Discrete uniform ( ) 0.2 , 1,2,3,4,5XF x x x⇒ = =

1

2

0.4 0.5 0.6 30.6 0.7 0.8 4

xx

< < ⇒ =< < ⇒ =

Aggregate claims = 3+4 = 7

Question #7 Key: C

( )1 20001

1 2000xE X x

x x

αθ θα θ

−⎡ ⎤⎛ ⎞∧ = − =⎢ ⎥⎜ ⎟− + +⎝ ⎠⎢ ⎥⎣ ⎦

x ( )E X x∧ ∞ 2000

250 222 2250 1059 5100 1437

( ) ( )( ) ( ) ( )( )

( ) ( )0.75 2250 250 0.95 5100

0.75 1059 222 0.95 2000 1437 1162.6

E X E X E X E X∧ − ∧ + − ∧

− + − =

The 5100 breakpoint was determined by when the insured’s share reaches 3600: 3600 = 250 + 0.25 (2250 – 250) + (5100 – 2250) Question #8 Key: D Since each time the probability of a heavy scientist is just half the probability of a success, the distribution is binomial with q = 0.6×0.5 = 0.3 and m = 8. ( ) ( ) ( ) ( )2 8 7 / 2 0.3 ^ 2 0.7 ^ 6 0.30f = × × × =

Question #9 Key: A

( ) ( ) ( ) 0.08 0.04 0.12xy x yt t tµ µ µ= + = + =

( ) ( )( )/ 0.5714x x xA t tµ µ δ= + =

( ) ( )( )/ 0.4y y yA t tµ µ δ= + =

( ) ( )( )/ 0.6667xy xy xyA t tµ µ δ= + =

( )( )1/ 5.556xy xya tµ δ= + =

0.5714 0.4 0.6667 0.3047x y xyxyA A A A= + − = + − =

Premium = 0.304762/5.556 = 0.0549 Question #10 Key: B

40 40 40/ 0.16132 /14.8166 0.0108878P A a= = =

42 42 42/ 0.17636 /14.5510 0.0121201P A a= = =

45 45 1 13.1121a a= − =

( )3 45 40 42 4542 3 1000 1000 1000E L K A P P a⎡ ⎤≥ = − −⎣ ⎦

( )( )201.20 10.89 12.12 13.112131.39

= − −

=

Many similar formulas would work equally well. One possibility would be

( )3 42 42 401000 1000 1000V P P+ − , because prospectively after duration 3, this differs from the normal benefit reserve in that in the next year you collect 401000P instead of 421000P .

Question #11 Key: E

For De Moivre’s Law: 2x

xe ω −=

1xk q

xω=

−

1 11 11

1

x xk k

x xkk b k b

xx

xx

A v q vx

aA

xAa

d

ω ω

ω

ω

ω

− − − −+ +

= =

−

= =−

=−−

=

∑ ∑

50 25 100 for typical annuitants

15 Assumed age = 70y

e

e y

ω= ⇒ =

= ⇒ =

3070

70

20

0.4588330

9.5607500000 52,297

aA

ab a b

= =

=

= ⇒ =

Question #12 Key: D

( ) ( ) ( ) ( )1 2' ' 0.8 0.7 0.56x x xp p pτ = = =

( )( )( )( )( )

( )1

1ln '

since UDD in double decrement tableln

xx x

x

pq q

pτ

τ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

( )( )

ln 0.80.44

ln 0.56

0.1693

⎡ ⎤= ⎢ ⎥⎣ ⎦

=

( )( )

( )

11

0.3 0.10.3 0.053

1 0.1x

xx

qqq τ+ = =

−

To elaborate on the last step:

( )10.3 0.1

Number dying from cause1 between 0.1 and 0.4

Number alive at 0.1xx x

qx+

⎛ ⎞⎜ ⎟+ +⎝ ⎠=

+

Since UDD in double decrement,

( ) ( ) ( )

( ) ( )( )10.3

1 0.1x x

x x

l q

l q

τ

τ τ=

−

Question #13 Key: B

non absorbing matrix 0.7 0.10.3 0.6

T⎛ ⎞

= ⎜ ⎟⎝ ⎠

, the submatrix excluding “Terminated”, which is an

absorbing state. 0.3 0.10.3 0.4

I T−⎛ ⎞

− = ⎜ ⎟−⎝ ⎠

( ) 1

0.4 0.14.44 1.110.09 0.09

0.3 0.3 3.33 3.330.09 0.09

I T −

⎛ ⎞⎜ ⎟ ⎛ ⎞

− = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎝ ⎠

Future costs for a healthy = 4.44 500 1.11 3000× + × = 5555 Question #14 Key: D

0.7 0.1 0.20.3 0.6 0.10 0 1

T⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

20.52 0.13 0.350.39 0.39 0.22

0 0 1T

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

Actuarial present value (A.P.V.) prem = 800(1 + (0.7 + 0.1) + (0.52 + 0.13)) = 1,960 A.P.V. claim = 500(1 + 0.7 + 0.52) + 3000(0 + 0.1 + 0.13) = 1800 Difference = 160

Question #15 Key: A Let 1 2,N N denote the random variable for # of claims for Type I and II in 2 years 1 2,X X denote the claim amount for Type I and II 1S = total claim amount for type I in 2 years 2S = total claim amount for Type II at time in 2 years 1 2S S S= + = total claim amount in 2 years

1S → compound poisson ( )1 12 6 12 0,1X Uλ = × = ∼

2S → compound poisson ( )2 22 2 4 0, 5X Uλ = × = ∼ ( ) ( )1 1 2 6 12E N Var N= = × =

( ) ( ) ( ) ( )( )1 1 1 12 0.5 6E S E N E X= = =

( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( )( )

21 1 1 1 1

21 012 12 0.5

124

Var S E N Var X Var N E X= +

−= +

=

( ) ( )2 2 2 2 4E N Var N= = × =

With formulas corresponding to those for 1S ,

( )254 102

E S = × =

( ) ( )2 2

25 0 54 4 33.3

12 2Var S

− ⎛ ⎞= × + =⎜ ⎟⎝ ⎠

( ) ( ) ( )1 2 6 10 16E S E S E S= + = + = Since 1S and 2S are independent,

( ) ( ) ( )1 2 4 33.3 37.3Var S Var S Var S= + = + =

( ) 16 2Pr 18 Pr 0.32739.3 37.3

SS⎛ ⎞−

> = > =⎜ ⎟⎝ ⎠

Using normal approximation ( ) ( )Pr 18 1 0.327

0.37S > = −Φ

=

Question #16 Key: D Since the rate of depletion is constant there are only 2 ways the reservoir can be empty sometime within the next 10 days. Way #1: There is no rainfall within the next 5 days Way #2 There is one rainfall in the next 5 days And it is a normal rainfall And there are no further rainfalls for the next five days Prob (Way #1) = Prob(0 in 5 days) = exp(-0.2*5) = 0.3679 Prob (Way #2) = Prob(1 in 5 days) × 0.8 × Prob(0 in 5 days) = 5*0.2 exp(-0.2* 5)* 0.8 * exp(-0.2* 5) = 1 exp(-1) * 0.8 * exp(-1) = 0.1083 Hence Prob empty at some time = 0.3679 + 0.1083 = 0.476 Question #17 Key: C Let X be the loss random variable, So ( )5X

+− is the claim random variable.

( ) 10 6.62.5 1

E X = =−

( )2.5 110 105 1

2.5 1 5 10

3.038

E X−⎡ ⎤⎛ ⎞ ⎛ ⎞∧ = −⎢ ⎥⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

=

( ) ( ) ( )5 5

6.6 3.0383.629

E X E X E X+

− = − ∧

= −=

Expected aggregate claims ( ) ( )5E N E X

+= −

( )( )5 3.62918.15

=

=

Question #18 Key: B A Pareto ( )2, 5α θ= = distribution with 20% inflation becomes Pareto with

2, 5 1.2 6α θ= = × =

In 2004, ( ) 6 62 1

E X = =−

( )

( ) ( ) ( )

2 16 610 1 3.752 1 10 6

10 10

E X

E X E X E X

−

+

⎛ ⎞⎛ ⎞∧ = − =⎜ ⎟⎜ ⎟⎜ ⎟− +⎝ ⎠⎝ ⎠− = − ∧

6 3.75 2.25= − = ( )

( )10 2.25LER 1 1 0.625

6E X

E X+

−= − = − =

Question #19 Key: A Let X = annual claims ( ) ( )( ) ( )( ) ( )( )0.75 3 0.15 5 0.1 7

3.7E X = + +

=

π = Premium ( ) 43.7 4.933

⎛ ⎞= =⎜ ⎟⎝ ⎠

Change during year 4.93 3 1.93= − = + with 0.75p = 4.93 5 0.07= − = − with 0.15p = 4.93 7 2.07= − = − with 0.10p = Since we start year 1 with surplus of 3, at end of year 1 we have 4.93, 2.93, or 0.93 (with associated probabilities 0.75, 0.15, 0.10). We cannot drop more then 2.07 in year 2, so ruin occurs only if we are at 0.93 after 1 and have a drop of 2.07.

( )( )Prob 0.1 0.1 0.01= =

Question #20 Key: E

( )

( )1

1

1

0.96ln 0.96 0.04082

0.04082 0.04082 0.01 0.03082

e µ λ

µ λµ λ

− +=

+ = − =

= − = − =

Similarly

( )2

1 2

ln 0.97 0.03046 0.01 0.020460.03082 0.02046 0.01 0.06128xy

µ λ

µ µ µ λ

= − − = − =

= + + = + + =

( ) ( )5 0.06128 0.30645 0.736xyp e e− −= = =

Question #21 Key: C

602

60

0.36913 0.05660

0.17741

A d

A

= =

=

and 2 260 60 0.202862A A− =

Expected Loss on one policy is ( ) 60100,000E L Ad dπ ππ ⎛ ⎞⎡ ⎤ = + −⎜ ⎟⎣ ⎦ ⎝ ⎠

Variance on one policy is ( ) ( )2

2 260 60Var 100,000L A A

dππ ⎛ ⎞⎡ ⎤ = + −⎜ ⎟⎣ ⎦ ⎝ ⎠

On the 10000 lives, [ ] ( )10,000E S E L π⎡ ⎤= ⎣ ⎦ and [ ] ( )Var 10,000 VarS L π⎡ ⎤= ⎣ ⎦

The π is such that [ ] [ ]0 / Var 2.326E S S− = since ( )2.326 0.99Φ =

60

2 260 60

10,000 100,0002.326

100 100,000

Ad d

A Ad

π π

π

⎛ ⎞⎛ ⎞− +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ =⎛ ⎞+ −⎜ ⎟⎝ ⎠

( )

( )

100 100,000 0.369132.326

100,000 0.202862

d d

d

π π

π

⎛ ⎞⎛ ⎞− +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ =⎛ ⎞+⎜ ⎟⎝ ⎠

0.63087 369130.004719

100,000d

d

π

π

−=

+

0.63087 36913 471.9 0.004719d dπ π− = =

36913 471.90.63087 0.00471959706

dπ +=

−=

59706 3379dπ = × =

Question #22 Key: C

( )( ) ( )1 0 1 1 75

75

1 10001.05 1000

V V i V V qq

π

π

= + + − + − ×

= −

Similarly,

( )( )

( )( )

( ) ( )( )

2 1 76

3 2 77

3 2 23 75 76 77

275 76 77

3 2

2

1.05 1000

1.05 1000

1000 1.05 1.05 1.05 1000 1.05 1000 1.05 1000 *

1000 1000 1.05 1.05

1.05 1.05 1.05

1000 1 1.05 0.05169 1.05 0.05647 0.06168

3.31

x

V V q

V V q

V q q q

q q q

π

π

π π π

π

= + × −

= + × −

= = + ⋅ + − × × − × × − ×

+ + +=

+ +

+ × + × +=

01251000 1.17796 355.87

3.310125×

= =

* This equation is algebraic manipulation of the three equations in three unknowns ( )1 2, ,V V π . One method – usually effective in problems where benefit = stated amount plus reserve, is to multiply the 1V equation by 21.05 , the 2V equation by 1.05, and add those two to the 3V equation: in the result, you can cancel out the 1V , and 2V terms. Or you can substitute the 1V equation into the 2V equation, giving 2V in terms of π , and then substitute that into the 3V equation.

Question #23 Key: D Actuarial present value (APV) of future benefits = ( ) ( )( ) 20.005 2000 0.04 1000 /1.06 1 0.005 0.04 0.008 2000 0.06 1000 /1.06= × + × + − − × + × 47.17 64.60111.77

= +=

APV of future premiums ( )1 1 0.005 0.04 /1.06 50⎡ ⎤= + − −⎣ ⎦

( )( )1.9009 5095.05

=

=

( )1 55 1 111.77 95.05 16.72E L K⎡ ⎤≥ = − =⎣ ⎦

Question #24 Key: D

0 200:20 20 0pe e e= +

[ ]050 25

3 1E Te = = =

−

[ ]3 1

0:2050 5020 1

3 1 50 20

12.245

E Te−⎛ ⎞⎛ ⎞= ∧ = −⎜ ⎟⎜ ⎟⎜ ⎟− +⎝ ⎠⎝ ⎠

=

( )3

20 0501 20 1 1

50 20

0.3644

Tp F⎛ ⎞⎛ ⎞= − = − −⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

=

20

20

25 12.245 0.3644

35

e

e

= +

=

Alternate approach: if losses are Pareto with 50 and 3θ α= = , then claim payments per payment with an ordinary deductible of 20 are Pareto with 50 20θ = + and 3α = .

Thus ( )( ) 50 2020 353 1

E T += =

−

This alternate approach was shown here for educational reasons: to reinforce the idea that many life contingent models and non-life models can have similar structure. We doubt many candidates would take that approach, especially since it involves specific properties of the Pareto distribution.

Question #25 Key: E Q P≥ since in Q you only test at intervals; surplus below 0 might recover before the next test.

In P, ruin occurs if you are ever below 0. R P≥ since you are less likely to have surplus below 0 in the first N years (finite horizon) than

forever. Add the inequalities

2Q R P+ ≥ Also (why other choices are wrong) S Q≥ by reasoning comparable to R P≥ . Same testing frequency in S and Q, but Q tests

forever. S R≥ by reasoning comparable to Q P≥ . Same horizon in S and R; R tests more frequently. S P≥ P tests more frequently, and tests forever. Question #26 Key: A This is a nonhomogeneous Poisson process with intensity function λ (t) = 3+3t, 0 2t≤ ≤ , where t is time after noon

( )( )

221

1

22

1

Average 3 31

332

7.5

t dtt dt

tt

λλ = = +

⎡ ⎤= +⎢ ⎥⎣ ⎦

=

∫∫

( )7.5 27.52 0.0156

2!ef−

= =

Question #27 Key: E X t Y tb g b g− is Brownian motion with initial value –2 and 2 0.5 1 1.5σ = + =

By formula 10.6, the probability that X t Y tb g b g− ≥ 0 at some time between 0 and 5 is

= × − ≥

= × −FHG

IKJ

LNMM

OQPP =

2 5 5 0

2 1 25 15

0 4652

Prob X Yb g b g

b gΦ.

.

The 2 in the numerator of ( )2

5 1.5 comes from ( ) ( )0 0 2X Y− = − ; the process needs to move 2

to reach ( ) ( ) 0X t Y t− ≥ .

Question #28 Key: B

80 84 80:842 2 2 280:84q q q q= + −

( ) ( )0.5 0.4 1 0.6 0.2 0.15 1 0.1= × × − + × × −

= 0.10136 Using new 82p value of 0.3

( ) ( )0.5 0.4 1 0.3 0.2 0.15 1 0.1× × − + × × − = 0.16118 Change = 0.16118 – 0.10136 = 0.06 Alternatively, 2 80 0.5 0.4 0.20p = × =

3 80 2 80 0.6 0.12p p= × =

2 84 0.20 0.15 0.03p = × =

3 84 2 84 0.10 0.003p p= × =

( )( )2 2 80 2 84 2 80 2 8480:84 since independent

0.20 0.03 0.20 0.03 0.224

p p p p p= + −

= + − =

( )( )3 3 80 3 84 3 80 3 8480:84

0.12 0.003 0.12 0.003 0.12264

p p p p p= + −

= + − =

2 32 80:84 80:84 80:84

0.224 0.12264 0.10136

q p p= −

= − =

Revised 3 80 0.20 0.30 0.06p = × =

( )( )3 80:84 0.06 0.003 0.06 0.003

0.06282

p = + −

=

2 80:84 0.224 0.06282 0.16118

change 0.16118 0.10136 0.06

q = − =

= − =

Question #29 Key: B

11

8.83 0.950481 9.29

xx x x x x

x

ee p p e pe+

+

= + ⇒ = = =+

221 ....x x xa vp v p= + + +

22:2

1 ...xxa v v p= + + +

:2 5.6459 5.60 0.0459x xxa a vq− = = − = ( )1 0.95048 0.0459v − =

0.9269v = 1 1 0.0789iv

= − =

Question #30 Key: A Let π be the benefit premium Let kV denote the benefit reserve a the end of year k. For any ( )( ) ( )25 1 25 1, 1n n n n nn V i q V p Vπ + + + ++ + = × + × 1n V+= Thus ( )( )1 0 1V V iπ= + +

( )( ) ( )( )( )2 1 21 1 1V V i i i sπ π π π= + + = + + + =

( )( ) ( )( )3 2 2 31 1V V i s i sπ π π π= + + = + + = By induction (proof omitted) n nV sπ= For 6035, nn V a= = (actuarial present value of future benefits; there are no future premiums)

60 35a sπ=

60

35

as

π =

20 20

6020

35

For 20,n V s

a ss

π= =

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

Alternatively, as above ( )( ) 11n nV i Vπ ++ + = Write those equations, for 0n = to 34n =

( )( )0 10 : 1V i Vπ+ + =

( )( )1 21: 1V i Vπ+ + =

( )( )

( )( )

2 3

34 35

2 : 1

34 : 1

V i V

V i V

π

π

+ + =

+ + =

Multiply equation k by ( )341 ki −+ and sum the results:

( )( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )

35 34 330 1 2 34

34 33 321 2 3 34 35

1 1 1 1

1 1 1 1

V i V i V i V i

V i V i V i V i V

π π π π+ + + + + + + + + + + + =

+ + + + + + + + +

For 1, 2, , 34,k = the ( )351 kkV i −+ terms in both sides cancel, leaving

( ) ( ) ( ) ( )35 35 340 351 1 1 1V i i i i Vπ ⎡ ⎤+ + + + + + + + =⎣ ⎦

Since 0 0V =

3535

60

s V

a

π =

=

(see above for remainder of solution) This technique, for situations where the death benefit is a specified amount (here, 0) plus the benefit reserve is discussed in section 8.3 of Bowers. This specific problem is Example 8.3.1.

Question #31 Key: B

( ) ( ) ( )t y t x t x t yxy

t x t y t x t y t x t y

q p x t q p y tt

q p p q p pµ µ

µ+ + +

=× + × + ×

For (x) = (y) = (50)

( ) ( )( )( )( ) ( )

( )( )( )( )( )( )( ) ( )

10.5 50 10 50 602 250:50

10.5 50 10.5 50 10.5 50

0.09152 0.91478 0.01376 2210.5 0.0023

2 0.09152 0.90848 2 0.90848

q p q

q p pµ

⋅= = =

⋅ + +

where

( ) ( )1122 60 61

10.5 5050

10.5 50 10.5 50

10 50

8,188,074 8,075,4030.90848

8,950,9011 0.091528,188,074 0.914788,950,901

l lp

lq p

p

++= = =

= − =

= =

( ) ( )10.5 50 10 50 6050 10.5p p qµ + = since UDD

Alternatively, ( ) 50 10 50 6010 tt p p p+ =

( ) ( ) ( )2 250:50 10 50 6010 tt p p p+ =

( ) ( ) ( )2 210 50 60 10 50 6010 50:50 2 t tt p p p p p+ = −

( ) ( ) ( )( ) ( )

2 210 50 60 10 50 60

210 50 60 10 50 60 60

2 1 1 since UDD

Derivative 2 2 1

p tq p tq

p q p tq q

= − − −

= − + −

Derivative at 10 10.5t+ = is

( )( ) ( ) ( )( )( )( )22 0.91478 0.01376 0.91478 1 0.5 0.01376 0.01376 0.0023− + − = −

( )( ) ( )

210.5 10.5 50 10.5 5050:50

2

2

2 0.90848 0.908480.99162

p p p= −

= −

=

µ (for any sort of lifetime) ( )0.00230.0023

0.99162

dpdtp

− − −= = =

Question #32 Key: E

( ) [ ]

( ) [ ]

( ) [ ]

( )

4

004

04

2 1

0

4 4

0

1 1( ) 2 Pr is the density of on 0, 4 .4 4

1 2 see note4

1 using formula from tables for the pgf of the Poisson4

1 1 14 413.4

i

iE W N i d

P d

e d

e e

λ

λ

λ λ λ

λ λ

λ

∞

=

−

⎡ ⎤= = ⎢ ⎥⎣ ⎦

=

=

= = −

=

∑∫

∫

∫

Note: the probability generating function (pgf) is ( )0

kk

kP Z p Z

∞

=

= ∑ so the integrand is ( )2P ,

or in this case ( )2P λ since λ is not known. Alternatively,

( )4

00

1( ) 2 Pr4

i

iE W N i dλ λ

∞

=

= =∑∫

4

00

1 24 !

i i

i

e di

λλ λ−∞

=

= ∑∫

( )4

00

214 !

i

i

ed

i

λ λλ

−∞

=

= ∑∫

We know ( )2

0

21

!

i

i

ei

λ λ−∞

=

=∑ since ( )2 2!

iei

λ λ−

is ( )f i for a Poisson with mean Zλ

so ( )2

0

2!

i

i

e e ei e

λ λλ

λ

λ− −∞

−=

= =∑

Thus ( )4

0

14

E W e dλ λ= ∫

( )4 4

0

1 1 14 413.4

e eλ= = −

=

Question #33 Key: A or E ( ) [ ] ( )( ) ( )2

2 / 3 1/ 4 2 / 4 3/ 2 2 / 3 9 / 4 3/ 2

Var 2 / 3 1/ 4 4 / 4 9 / 2 23/ 6

E S E X

S E X

λ

λ

= = + + = × =

⎡ ⎤= = + + =⎣ ⎦

So cumulative premium to time 2 is ( )2 3/ 2 1.8 23/ 6 10+ = , where the expression in

parentheses is the annual premium Times between claims are determined by 1/ log uλ− and are 0.43, 0.77, 1.37, 2.41 So 2 claims before time 2 (second claim is at 1.20; third is at 2.57) Sizes are 2, 3, 1, 3, where only the first two matter. So gain to the insurer is 10-(2+3) = 5 Note: since the problem did not specify that we wanted the gain or loss from the insurer’s viewpoint, we gave credit to answer A; a loss of 5 from the insured’s viewpoint.

Question #34 Key: C To get number of claims, set up cdf for Poisson:

x f(x) F(x) 0 0.135 0.135 1 0.271 0.406 2 0.271 0.677 3 0.180 0.857

0.80 simulates 3 claims.

( ) ( )( ) ( )2 1/ 21 500 / 500 , so 1 500 500F x x u x u −= − + = = − − 0.6 simulates 290.57 0.25 simulates 77.35 0.7 simulates 412.87 So total losses equals 780.79 Insurer pays ( )( ) ( )0.80 750 780.79 750 631+ − =

Question #35 Key: D

( ) ( ) ( ) ( ) ( ) ( )1 2 0.01 2.29 2.30x x xt t tτµ µ µ= + = + = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

2 22 10 0 2

50,000 50,000t t tt x x t x x t x xP P v p t dt v p t dt v p t dtτ τ τ τµ µ µ

∞= + +∫ ∫ ∫

2 20.1 2.3 0.1 2.3 0.1 2.30 0 2

2.29 50,000 0.01 50,000 2.3t t t t t tP P e e dt e e dt e e dt∞− − − − − −= × + × + ×∫ ∫ ∫

( ) ( ) ( )2 2.4 2 2.4 2 2.41 11 2.29 50000 0.01 2.32.4 2.4 2.4e e eP− − −⎡ ⎤ ⎡ ⎤− −

− × = × + ×⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

P = 11,194 Question #36 Key: D

( ) 0.001accidµ = ( ) 0.01totalµ = ( ) 0.01 0.001 0.009otherµ = − =

Actuarial present value ( )0.05 0.010

500,000 0.009t te e dt∞ − −= ∫

( )0.04 0.05 0.010

10 50,000 0.001t t te e e dt∞ − −+ ∫

0.009 0.001500,000 100,0000.06 0.02

⎡ ⎤= + =⎢ ⎥⎣ ⎦

Question #37 Key: B Variance 30

15 15x xv p q= Expected value 1515 xv p=

30 15

15 15 1515

15 15

0.065

0.065 0.3157x x x

x x

v p q v p

v q q

=

= ⇒ =

Since µ is constant

( )( )( )

1515

15

1

0.68430.9750.025

x x

x

x

x

q p

ppq

= −

=

=

=

Question #38 Key: E

( ) ( ) ( ) 1011 10

10 10

11 0 1000A A x

x x

i qV Vp p

+

+ +

+= + − ×

( ) ( ) ( ) 1011 10

10 10

12 1000B B B x

x x

i qV Vp p

π +

+ +

+= + − ×

( ) ( ) ( ) ( )11 11 10 10

10

11 2 A B A B B

x

iV V V V

pπ

+

+− − = − −

( ) ( )1.06101.35 8.36

1 0.004= −

−

98.97=

Question #39 Key: A

Actuarial present value Benefits ( )( )( ) ( )( )( )( )2 3

0.8 0.1 10,000 0.8 0.9 0.097 9,0001.06 1.06

= +

1,239.75=

( ) ( )( )

( )

20.8 0.8 0.9

1,239.75 11.06 1.06

2.3955517.53 518

P

PP

⎛ ⎞= + +⎜ ⎟

⎝ ⎠=

= ⇒

Question #40 Key: C

Event Prob Present Value 0x = ( )0.05 15

1x = ( )( )0.95 0.10 0.095= 15 20 /1.06 33.87+ =

2x ≥ ( )( )0.95 0.90 0.855= 215 20 /1.06 25 /1.06 56.12+ + = [ ] ( )( ) ( )( ) ( )( )0.05 15 0.095 33.87 0.855 56.12 51.95E X = + + =

( )( ) ( )( ) ( )( )2 2 22 0.05 15 0.095 33.87 0.855 56.12 2813.01E X⎡ ⎤ = + + =⎣ ⎦

[ ] ( ) ( ) ( )2 22 2813.01 51.95 114.2Var X E X E X= − = − =