NOVEMBER 2004 COURSE 3 SOLUTIONS

NOVEMBER 2004 COURSE 3 SOLUTIONS Question #1 Key: D ∞ ∞ 1 dt 20 1 ⎛ 100 ⎞ 5 −0.07 t ⎤ ∞ = ⎜ = ⎟ ⎡⎣ −e ⎦ 0 20 ⎝ 7 ⎠ 7 2 ∞ 1 1 ∞ bt vt t px µ ( x + t...
Author: Myles Andrews
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NOVEMBER 2004 COURSE 3 SOLUTIONS

Question #1 Key: D ∞



1 dt 20 1 ⎛ 100 ⎞ 5 −0.07 t ⎤ ∞ = ⎜ = ⎟ ⎡⎣ −e ⎦ 0 20 ⎝ 7 ⎠ 7 2 ∞ 1 1 ∞ bt vt t px µ ( x + t ) dt = ∫ e0.12t e−0.16t e−0.05t dt = ∫ e−0.09t dt 0 20 20 0 1 ⎛ 100 ⎞ −0.09t ∞ 5 ⎤ = ⎜ ⎟ ⎡e ⎦0 = 9 20 ⎝ 9 ⎠ ⎣

E [ Z ] = ∫ bt v t t px µ ( x + t ) dt = ∫ e0.06 t e−0.08 t e−0.05 t 0

0

E ⎡⎣ Z 2 ⎤⎦ = ∫

( )



0

Var [ Z ] =

2

5 ⎛5⎞ − ⎜ ⎟ = 0.04535 9 ⎝7⎠

Question #2 Key: C

Let ns = nonsmoker and s = smoker k=

q xb + kg ns

pxb + kg

q xb +gk s

px( +)k

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 ns A x:2( ) =

ns v q x( ) +

1 ( 0.05) 1.02 A x:2( ) 1 s

v

1 1.02

ns px( )

v2

1 1.022 s qx( )

ns

ns

0.95 × 0.10 = 0.1403

+ v2

( 0.10 ) +

qx( +1)

1

(1.02 )2

s px( )

qx( +)1 s

0.90 × 0.20 = 0.2710

A1x:2 = weighted average = (0.75)(0.1403) + (0.25)(0.2710)

= 0.1730

s

Question #3 Key: A

P ( Ax ) = µ = 0.03 2

µ

0.03 2δ + µ 2δ + 0.03 ⇒ δ = 0.06

Ax = 0.20 =

Var ( 0 L ) = where A =

2

=

Ax − ( Ax )

(δ a )2 µ µ +δ

=

=

0.20 − ( 13 )

( 0.06 0.09 )

0.03 1 = 0.09 3

Question #4 Key: B − 0.1 60 − 0.08 60 e ( )( ) + e ( )( ) s ( 60 ) = 2 = 0.005354 − 0.1 61 − 0.08 61 e ( )( ) + e ( )( ) s ( 61) = 2 = 0.00492 0.00492 q60 = 1 − = 0.081 0.005354

2

2

2

a=

= 0.20 1 1 = µ + δ 0.09

Question #5 Key: B

⎛ω ⎞ For Ω, 0.4 = F (ω ) = ⎜ ⎟ ⎝ 80 ⎠ 0.6325 =

2

ω 80

ω = 50.6 t ω 50.6 t = ( 0.7 )( 50.6 ) = 35.42

For T(0) using De Moivre, 0.7 = F ( t ) =

t

=

Question #6 Key: C

x

1.8 = 0.6 3 fN ( x)

FN ( x )

0 1 2 3

0.064 0.288 0.432 0.216

0.064 0.352 0.784 1.000

E [ N ] = mq = 1.8 ⇒ q =

First: Second: Third:

0.432 < 0.7 < 0.784 so N = 2. 0.064 < 0.1 < 0.352 so N = 1 0.432 < 0.5 < 0.784 so N = 2

Discrete uniform ⇒ FX ( x ) = 0.2 x, x = 1, 2,3, 4,5 0.4 < 0.5 < 0.6 ⇒ x1 = 3 0.6 < 0.7 < 0.8 ⇒ x2 = 4 Aggregate claims = 3+4 = 7

Use 0.1 and 0.3 for amounts Use 0.9 for amount Use 0.5 and 0.7 for amounts

Question #7 Key: C

E ( X ∧ x) =

α −1 ⎤

θ ⎡ ⎛ θ ⎞ ⎢1 − α − 1 ⎢⎣ ⎜⎝ x + θ ⎟⎠

2000 x ⎥= ⎥⎦ x + 2000

x

E ( X ∧ x)

∞ 250 2250 5100

2000 222 1059 1437

0.75 ( E ( X ∧ 2250 ) − E ( X ∧ 250 ) ) + 0.95 ( E ( X ) − E ( X ∧ 5100 ) ) 0.75 (1059 − 222 ) + 0.95 ( 2000 − 1437 ) = 1162.6 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. f ( 2 ) = ( 8 × 7 / 2 ) × ( 0.3 ^ 2 ) × ( 0.7 ^ 6 ) = 0.30

Question #9 Key: A

µ xy ( t ) = µ x ( t ) + µ y ( t ) = 0.08 + 0.04 = 0.12 Ax = µ x ( t ) / ( µ x ( t ) + δ ) = 0.5714

(

)

Ay = µ y ( t ) / µ y ( t ) + δ = 0.4

(

)

Axy = µ xy ( t ) / µ xy ( t ) + δ = 0.6667

(

)

axy = 1/ µ xy ( t ) + δ = 5.556 Axy = Ax + Ay − Axy = 0.5714 + 0.4 − 0.6667 = 0.3047 Premium = 0.304762/5.556 = 0.0549

Question #10 Key: B

P40 = A40 / a40 = 0.16132 /14.8166 = 0.0108878 P42 = A42 / a42 = 0.17636 /14.5510 = 0.0121201 a45 = a45 − 1 = 13.1121 E ⎡⎣ 3 L K ( 42 ) ≥ 3⎤⎦ = 1000 A45 − 1000 P40 − 1000 P42 a45

= 201.20 − 10.89 − (12.12 )(13.1121) = 31.39

Many similar formulas would work equally well. One possibility would be 1000 3V42 + (1000 P42 − 1000 P40 ) , because prospectively after duration 3, this differs from the normal benefit reserve in that in the next year you collect 1000 P40 instead of 1000 P42 .

Question #11 Key: E

For De Moivre’s Law:

ex = k

ω−x 2

qx =

Ax = Ax = ax =

1 ω−x

ω − x −1



k =b

v k +1 k qx =

aω − x

ω−x 1 − Ax d

e50 = 25 ⇒ ω = 100 for typical annuitants e y = 15 ⇒ y = Assumed age = 70 A70 =

a30

= 0.45883 30 a70 = 9.5607 500000 = b a20 ⇒ b = 52, 297

1 ω − x −1 k +1 ∑v ω − x k =b

Question #12 Key: D τ 1 2 px( ) = p ' x( ) p ' x( ) = 0.8 ( 0.7 ) = 0.56

(1)

qx

( ) ( )

⎡ ln p ' (1) ⎤ x ⎥ q(τ ) since UDD in double decrement table =⎢ ⎢ ln p(τ ) ⎥ x x ⎢⎣ ⎦⎥ ⎡ ln ( 0.8 ) ⎤ =⎢ ⎥ 0.44 ⎣ ln ( 0.56 ) ⎦ = 0.1693 (1)

0.3 q x + 0.1

=

0.3qx( ) 1

= 0.053 τ 1 − 0.1qx( )

To elaborate on the last step: ⎛ Number dying from cause ⎞ ⎜ ⎟ 1 between x + 0.1 and x + 0.4 ⎠ (1) ⎝ 0.3 q x + 0.1 = Number alive at x + 0.1 Since UDD in double decrement, =

τ 1 l x( ) ( 0.3) qx( )

(

τ τ lx( ) 1 − 0.1qx( )

)

Question #13 Key: B

⎛ 0.7 0.1 ⎞ non absorbing matrix T = ⎜ ⎟ , the submatrix excluding “Terminated”, which is an ⎝ 0.3 0.6 ⎠ absorbing state. ⎛ 0.3 −0.1 ⎞ I −T = ⎜ ⎟ ⎝ −0.3 0.4 ⎠ ⎛ 0.4 ⎜ −1 ( I − T ) = ⎜ 0.09 ⎜ 0.3 ⎜ ⎝ 0.09

0.1 ⎞ 0.09 ⎟ ⎛ 4.44 1.1 1 ⎞ ⎟=⎜ ⎟ 0.3 ⎟ ⎝ 3.33 3.33 ⎠ ⎟ 0.09 ⎠

Future costs for a healthy = 4.44 × 500 + 1.1 1 × 3000 = 5555

Question #14 Key: D ⎛ 0.7 0.1 0.2 ⎞ ⎜ ⎟ T = ⎜ 0.3 0.6 0.1 ⎟ ⎜ 0 0 1 ⎟⎠ ⎝

⎛ 0.52 0.13 0.35 ⎞ ⎜ ⎟ T = ⎜ 0.39 0.39 0.22 ⎟ ⎜ 0 0 1 ⎟⎠ ⎝ 2

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 N1, N 2 denote the random variable for # of claims for Type I and II in 2 years X 1, X 2 denote the claim amount for Type I and II S1 = total claim amount for type I in 2 years S2 = total claim amount for Type II at time in 2 years S = S1 + S2 = total claim amount in 2 years

{S1} → compound poisson λ1 = 2 × 6 = 12 {S2} → compound poisson λ2 = 2 × 2 = 4

X 1 ∼ U ( 0, 1)

X 2 ∼ U ( 0, 5 )

E ( N1 ) = Var ( N1 ) = 2 × 6 = 12 E ( S1 ) = E ( N1 ) E ( X 1 ) = (12 )( 0.5 ) = 6

Var ( S1 ) = E ( N1 )Var ( X 1 ) + Var ( N1 ) ( E ( X 1 ) ) = (12 )

(1 − 0 ) + 12

2

(12 )( 0.5)2

=4 E ( N 2 ) = Var ( N 2 ) = 2 × 2 = 4

With formulas corresponding to those for S1 , 5 E ( S2 ) = 4 × = 10 2 Var ( S2 )

2 2 5 − 0) ( ⎛5⎞ = 4× +4 = 33.3

⎜ ⎟ 12 ⎝2⎠ E ( S ) = E ( S1 ) + E ( S 2 ) = 6 + 10 = 16

Since S1 and S2 are independent,

Var ( S ) = Var ( S1 ) + Var ( S 2 ) = 4 + 33.3 = 37.3

⎛ S − 16 ⎞ 2 > = 0.327 ⎟ Pr ( S > 18 ) = Pr ⎜ 37.3 ⎝ 39.3 ⎠ Using normal approximation Pr ( S > 18 ) = 1 − Φ ( 0.327 ) = 0.37

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 ( X − 5 )+ is the claim random variable. E(X ) =

10 = 6.6 2.5 − 1

2.5−1 ⎤ ⎛ 10 ⎞ ⎡ ⎛ 10 ⎞ E ( X ∧ 5) = ⎜ 1 − ⎢ ⎥ ⎟ ⎜ ⎟ ⎝ 2.5 − 1 ⎠ ⎣⎢ ⎝ 5 + 10 ⎠ ⎦⎥

= 3.038 E ( X − 5)+ = E ( X ) − E ( X ∧ 5) = 6.6 − 3.038 = 3.629 Expected aggregate claims = E ( N ) E ( X − 5 )+ = ( 5 )( 3.629 ) = 18.15

Question #18 Key: B

A Pareto (α = 2, θ = 5 ) distribution with 20% inflation becomes Pareto with

α = 2, θ = 5 ×1.2 = 6 6 =6 2 −1 2 −1 6 ⎛ ⎛ 6 ⎞ ⎞ E ( X ∧ 10 ) = ⎜1 − ⎜ ⎟ ⎟ = 3.75 2 − 1 ⎜⎝ ⎝ 10 + 6 ⎠ ⎟⎠ E ( X − 10 )+ = E ( X ) − E ( X ∧ 10 )

In 2004, E ( X ) =

= 6 − 3.75 = 2.25 E ( X − 10 )+ 2.25 LER = 1 − = 1− = 0.625 E(X ) 6

Question #19 Key: A

Let X = annual claims E ( X ) = ( 0.75 )( 3) + ( 0.15 )( 5 ) + ( 0.1)( 7 ) = 3.7

⎛ 4⎞ ⎝ ⎠ Change during year = 4.93 − 3 = +1.93 with p = 0.75 = 4.93 − 5 = −0.07 with p = 0.15 = 4.93 − 7 = −2.07 with p = 0.10

π = Premium = ( 3.7 ) ⎜ ⎟ = 4.93 3

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 − µ +λ 0.96 = e ( 1 )

µ1 + λ = − ln ( 0.96 ) = 0.04082 µ1 = 0.04082 − λ = 0.04082 − 0.01 = 0.03082 Similarly

µ 2 = − ln ( 0.97 ) − λ = 0.03046 − 0.01 = 0.02046 µ xy = µ1 + µ 2 + λ = 0.03082 + 0.02046 + 0.01 = 0.06128 5

− 5 0.06128 ) pxy = e ( ) ( = e −0.3064 = 0.736

Question #21 Key: C A60 = 0.36913 2

d = 0.05660

A60 = 0.17741

and

2

2 A60 − A60 = 0.202862

π⎞ π ⎛ Expected Loss on one policy is E ⎣⎡ L (π ) ⎦⎤ = ⎜100,000 + ⎟ A60 − d⎠ d ⎝ π⎞ ⎛ 2 Variance on one policy is Var ⎡⎣ L (π ) ⎤⎦ = ⎜100,000 + ⎟ 2 A60 − A60 d ⎝ ⎠ On the 10000 lives, E [ S ] = 10,000 E ⎡⎣ L (π ) ⎤⎦ and Var [ S ] = 10,000 Var ⎡⎣ L (π ) ⎤⎦ 2

(

)

The π is such that 0 − E [ S ] / Var [ S ] = 2.326 since Φ ( 2.326 ) = 0.99 ⎛π ⎛ π⎞ ⎞ 10,000 ⎜ − ⎜100,000 + ⎟ A60 ⎟ d⎠ ⎝d ⎝ ⎠ = 2.326 π⎞ ⎛ 2 100 ⎜100,000 + ⎟ 2 A60 − A60 d⎠ ⎝ ⎛π ⎛ π ⎞⎞ 100 ⎜ − ⎜100,000 + ⎟ ⎟ ( 0.36913) d ⎠⎠ ⎝d ⎝ = 2.326 π⎞ ⎛ ⎜100,000 + ⎟ ( 0.202862 ) d⎠ ⎝

0.63087

π d

− 36913

100,000 + 0.63087

π

π

= 0.004719

d

− 36913 = 471.9 = 0.004719

d 36913 + 471.9 = d 0.63087 − 0.004719 = 59706 π = 59706 × d = 3379

π

π d

Question #22 Key: C

1V

= ( 0V + π ) (1 + i ) − (1000 + 1V − 1V ) × q75 = 1.05π − 1000q75

Similarly, 2 V = ( 1V + π ) × 1.05 − 1000q76 3V

= ( 2V + π ) × 1.05 − 1000q77

(

)

1000 =3V = 1.053π + 1.052 ⋅ π + 1.05π − 1000 × q75 × 1.052 − 1000 × 1.05 × q76 − 1000 × q77 *

π= = =

(

1000 + 1000 1.052 q75 + 1.05q76 + q77

(1.05) + (1.05) 3

(

2

)

+ 1.05

1000 x 1 + 1.052 × 0.05169 + 1.05 × 0.05647 + 0.06168

)

3.310125 1000 × 1.17796 = 355.87 3.310125

* This equation is algebraic manipulation of the three equations in three unknowns

( 1V , 2V , π ) .

One method – usually effective in problems where benefit = stated amount plus reserve, is to multiply the 1V equation by 1.052 , 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 = = ( 0.005 × 2000 + 0.04 × 1000 ) /1.06 + (1 − 0.005 − 0.04 )( 0.008 × 2000 + 0.06 × 1000 ) /1.062 = 47.17 + 64.60 = 111.77

APV of future premiums = ⎡⎣1 + (1 − 0.005 − 0.04 ) /1.06 ⎤⎦ 50 = (1.9009 )( 50 ) = 95.05 E ⎡⎣ 1 L K ( 55 ) ≥ 1⎤⎦ = 111.77 − 95.05 = 16.72

Question #24 Key: D

e0 = e0:20 + 20 p0 e20

e0 = E [T ] =

e0:20

50 = 25 3 −1

50 ⎛ ⎛ 50 ⎞ = E [T ∧ 20] = ⎜1 − ⎜ ⎟ 3 − 1 ⎝⎜ ⎝ 50 + 20 ⎠

3−1 ⎞

⎟ ⎟ ⎠

= 12.245 ⎛ ⎛ 50 ⎞3 ⎞ 20 p0 = 1 − FT ( 20 ) = 1 − ⎜ 1 − ⎜ ⎟ ⎟⎟ ⎜ ⎝ ⎝ 50 + 20 ⎠ ⎠ = 0.3644 25 = 12.245 + 0.3644 e20

e20 = 35 Alternate approach: if losses are Pareto with θ = 50 and α = 3 , then claim payments per payment with an ordinary deductible of 20 are Pareto with θ = 50 + 20 and α = 3 . Thus E (T ( 20 ) ) =

50 + 20 = 35 3 −1

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

Q + R ≥ 2P 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 ≤ t ≤ 2 , where t is time after noon

∫1 λ ( t ) dt = Average λ = 2

1

∫1 ( 3 + 3t )dt 2

⎡ 3t 2 ⎤ = ⎢3t + ⎥ 2 ⎦ ⎣ = 7.5 e −7.5 7.52 = 0.0156 f ( 2) = 2!

2

1

Question #27 Key: E

bg bg

X t − Y t is Brownian motion with initial value –2 and σ 2 = 0.5 + 1 = 1.5

bg bg = 2 × Prob X b5g − Y b5g ≥ 0 LM F 2 I OP = 2 × 1 − ΦG MN H 5b15. g JK PQ = 0.4652

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

The 2 in the numerator of to reach X ( t ) − Y ( t ) ≥ 0 .

2 5 (1.5 )

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

Question #28 Key: B

2

q80:84 = 2 q80 + 2 q84 − 2 q80:84

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

= 0.10136 Using new p82 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 p80 = 0.5 × 0.4 = 0.20 3 p80 = 2 p80 × 0.6 = 0.12 2 p84 = 0.20 × 0.15 = 0.03 3 p84 = 2 p84 × 0.10 = 0.003 2 p80:84 = 2 p80 + 2 p84 − 2 p80 2 p84 since independent 3

p80:84

= 0.20 + 0.03 − ( 0.20 )( 0.03) = 0.224 = 3 p80 + 3 p84 − 3 p80 3 p84

= 0.12 + 0.003 − ( 0.12 )( 0.003) = 0.12264 2 q 80:84 = 2 p80:84 − 3 p80:84 = 0.224 − 0.12264 = 0.10136

Revised 3 p80 = 0.20 × 0.30 = 0.06 3

p 80:84 = 0.06 + 0.003 − ( 0.06 )( 0.003) = 0.06282 2 q 80:84 = 0.224 − 0.06282 = 0.16118

change = 0.16118 − 0.10136 = 0.06

Question #29 Key: B

ex = px + pxex +1 ⇒ px =

ex 8.83 = = 0.95048 1 + ex +1 9.29

ax = 1 + vpx + v 2 2 px + .... a

x:2

= 1 + v + v 2 2 p x + ...

ax:2 − ax = vq x = 5.6459 − 5.60 = 0.0459 v (1 − 0.95048 ) = 0.0459 v = 0.9269 1 i = − 1 = 0.0789 v

Question #30 Key: A

Let π be the benefit premium Let kV denote the benefit reserve a the end of year k.

For any n, ( nV + π ) (1 + i ) = ( q25+ n × n +1V + p25+ n × n+1V ) = Thus 1V = ( 0 V + π ) (1 + i ) 2V 3V

n +1V

= ( 1V + π )(1 + i ) = (π (1 + i ) + π ) (1 + i ) = π s2

(

)

= ( 2V + π )(1 + i ) = π s2 + π (1 + i ) = π s3

By induction (proof omitted) n V = π sn For n = 35, n V = a60 (actuarial present value of future benefits; there are no future premiums) a60 = π s35

π=

a60 s35

For n = 20,

20 V

= π s20 ⎛a ⎞ = ⎜ 60 ⎟ s20 ⎜s ⎟ ⎝ 35 ⎠

Alternatively, as above ( nV + π ) (1 + i ) = n+1V Write those equations, for n = 0 to n = 34 0 : ( 0V + π ) (1 + i ) = 1V 1: ( 1V + π )(1 + i ) = 2V

2 : ( 2V + π )(1 + i ) = 3V

34 : ( 34V + π ) (1 + i ) = 35V Multiply equation k by (1 + i )

34 − k

and sum the results:

( 0V + π ) (1 + i )35 + ( 1V + π )(1 + i )34 + ( 2V + π )(1 + i )33 + + ( 34V + π ) (1 + i ) = 34 33 32 + 34V (1 + i ) + 35V 1V (1 + i ) + 2V (1 + i ) + 3V (1 + i ) +

For k = 1, 2, 0V

, 34, the k V (1 + i )

35− k

(1 + i )35 + π ⎡⎣(1 + i )35 + (1 + i )34 +

terms in both sides cancel, leaving + (1 + i ) ⎤ = 35V ⎦

Since 0V = 0 π s35 = 35V = a60 (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

µ xy ( t ) =

t qy t t qx

px µ ( x + t ) + t qx t p y µ ( y + t )

× t p y + t px × t q y + t px × t p y

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

µ50:50 (10.5 ) = where 10.5

p50 =

10.5 q50 10

2

( l60 + l61 ) = 12 (8,188,074 + 8,075, 403) = 0.90848 8,950,901

l50

= 1 − 10.5 p50 = 0.09152

p50 =

10.5

1

( 10.5 q50 )( 10 p50 ) q60 ⋅ 2 ( 0.09152 )( 0.91478)( 0.01376 )( 2 ) = 0.0023 = 2 ( 10.5 q50 )( 10.5 p50 ) ⋅ 2 + ( 10.5 p50 ) ( 0.09152 )( 0.90848)( 2 ) + ( 0.90848)2

8,188,074 = 0.91478 8,950,901

p50 µ ( 50 + 10.5 ) = ( 10 p50 ) q60 since UDD

Alternatively, (10+t ) p50 = 10 p50 t p60 (10+t ) p50:50 = ( 10 p50 )

2

( t p60 )

2

(10+t ) p 50:50 = 2 10 p50 t p60 − ( 10 p50 )

2

( t p60 )

2

= 2 10 p50 (1 − tq60 ) − ( 10 p50 ) (1 − tq60 ) since UDD 2

2

Derivative = −2 10 p50 q60 + 2 ( 10 p50 ) (1 − tq60 ) q60 Derivative at 10 + t = 10.5 is 2

−2 ( 0.91478 )( 0.01376 ) + ( 0.91478 ) (1 − ( 0.5 )( 0.01376 ) ) ( 0.01376 ) = −0.0023 2

10.5

p 50:50 = 2 10.5 p50 − ( 10.5 p50 )

2

= 2 ( 0.90848 ) − ( 0.90848 )

2

= 0.99162

µ (for any sort of lifetime) =



dp dt = − ( −0.0023) = 0.0023 p 0.99162

Question #32 Key: E 1 ∞ E (W ) = ∫ ∑ 2i Pr ( N = i λ ) d λ 4 0 i =0 4

4

1 = ∫ P (2 λ ) dλ 40

[see note]

4

=

1 λ ( 2−1) e dλ 4 ∫0

(

⎡1 ⎤ ⎢ 4 is the density of λ on [ 0, 4] .⎥ ⎣ ⎦

[ using formula from tables for the pgf of the Poisson ]

)

1 λ 4 1 4 e = e −1 4 0 4 = 13.4 =



Note: the probability generating function (pgf) is P ( Z ) = ∑

k =0

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

pk Z k so the integrand is P ( 2 ) ,

Alternatively, 1 ∞ i E (W ) = ∫ ∑ 2 Pr ( N = i λ ) d λ 4 0 i =0 4

4 ∞

1 4

∫0 ∑ i =0

1 = 4

∫0 ∑ i =0

=

4

e − λ ( 2λ ) dλ i! i

e −2λ ( 2λ ) e −2λ ( 2λ ) = 1 since is f ( i ) for a Poisson with mean Z λ i! i! i



We know



2i e− λ λ i dλ i!

∑ i =0

e − λ ( 2λ ) e−λ = −2λ = eλ so ∑ i! e i =0 4 1 Thus E (W ) = ∫ eλ d λ 4 0 i



1 λ e 4 = 13.4 =

4 0

=

(

)

1 4 e −1 4

i

Question #33 Key: A or E E ( S ) = λ E [ X ] = 2 / 3 (1/ 4 + 2 / 4 + 3/ 2 ) = 2 / 3 × 9 / 4 = 3/ 2 Var ( S ) = λ E ⎡⎣ X 2 ⎤⎦ = 2 / 3 (1/ 4 + 4 / 4 + 9 / 2 ) = 23/ 6

(

)

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/ λ log u 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: f(x) 0.135 0.271 0.271 0.180

x 0 1 2 3

F(x) 0.135 0.406 0.677 0.857

0.80 simulates 3 claims. F ( x ) = 1 − ( 500 / ( x + 500 ) ) = u, so x = (1 − u ) 2

−1/ 2

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

500 − 500

Question #35 Key: D

µ x(τ ) ( t ) = µ x(1) ( t ) + µ x( 2 ) ( t ) = 0.01 + 2.29 = 2.30 2 2 ∞ τ τ τ τ 2 1 P = P ∫ vt t px( ) µ x( ) ( t ) dt + 50, 000 ∫ vt t px( ) µ x( ) ( t ) dt + 50, 000∫ vt t px( ) µ x( ) ( t ) dt 0 2 −0.1t −2.3t

P = P∫ e 0

e

0

2

2 −0.1t −2.3t

× 2.29dt + 50, 000∫ e 0

−2( 2.4 )

⎡ 1− e P ⎢1 − 2.29 × 2.4 ⎣⎢ P = 11,194

e

−2( 2.4 )

⎤ ⎡ 1− e ⎥ = 50000 ⎢ 0.01× 2.4 ⎦⎥ ⎣⎢



× 0.01dt + 50, 000 ∫ e−0.1t e−2.3t × 2.3dt 2

+ 2.3 ×

e

−2( 2.4 )

⎤ ⎥ 2.4 ⎦⎥

Question #36 Key: D

µ ( accid ) = 0.001 µ ( total ) = 0.01 µ ( other ) = 0.01 − 0.001 = 0.009 ∞

Actuarial present value = ∫ 500,000 e −0.05t e−0.01t ( 0.009 ) dt 0



+10 ∫ 50,000 e0.04t e−0.05t e −0.01t ( 0.001) dt 0

⎡ 0.009 0.001 ⎤ = 500, 000 ⎢ + = 100, 000 ⎣ 0.06 0.02 ⎥⎦

Question #37 Key: B

Expected value = v15 15 px

Variance = v30 15 px 15 qx

v30 15 px 15 qx = 0.065 v15 15 px v15 15qx = 0.065 ⇒ 15 qx = 0.3157 Since µ is constant 15 q x

( px )

15

(

= 1 − ( px )

15

)

= 0.6843

px = 0.975 qx = 0.025

Question #38 Key: E

(1)

( 2)

11V

A

11V

B

(1) − ( 2 )

=

=

(

10V

(

11V

10V

A

A

B

+0

) (p

1+ i)

+π B

x +10

) (p

− 11V B =

qx +10 × 1000 px +10



1+ i)

(

x +10

10V

A



qx +10 × 1000 px +10

− 10V B − π B

= (101.35 − 8.36 ) = 98.97

) (p

1+ i) x +10

(1.06 ) 1 − 0.004

Question #39 Key: A

Actuarial present value Benefits =

( 0.8)( 0.1)(10,000 ) + ( 0.8)( 0.9 )( 0.097 )( 9,000 )

1.062 = 1, 239.75

1.063

⎛ ( 0.8 ) ( 0.8 )( 0.9 ) ⎞ 1, 239.75 = P ⎜ 1 + + ⎟ 1.062 ⎠ ⎝ 1.06 = P ( 2.3955 ) P = 517.53 ⇒ 518

Question #40 Key: C

Event x=0

Prob ( 0.05)

Present Value 15

x =1

( 0.95)( 0.10 ) = 0.095 ( 0.95)( 0.90 ) = 0.855

15 + 20 /1.06 = 33.87

x≥2

15 + 20 /1.06 + 25 /1.062 = 56.12

E [ X ] = ( 0.05 )(15 ) + ( 0.095 )( 33.87 ) + ( 0.855 )( 56.12 ) = 51.95 E ⎡⎣ X 2 ⎤⎦ = ( 0.05 )(15 ) + ( 0.095 )( 33.87 ) + ( 0.855 )( 56.12 ) = 2813.01 2

( )

2

2

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

2