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SOCIETY OF ACTUARIES
EXAM M ACTUARIAL MODELS
EXAM M SAMPLE SOLUTIONS
Copyright 2005 by the Society of Actuaries
Some of the questions in this study note are taken from past SOA examinations.
M-09-05
PRINTED IN U.S.A.
Question #1
Key: E
2
q30:34 = 2 p30:34 − 3 p30:34
2
p30 = ( 0.9 )( 0.8 ) = 0.72
2
p34 = ( 0.5 )( 0.4 ) = 0.20
2
p30:34 = ( 0.72 )( 0.20 ) = 0.144
2
p30:34 = 0.72 + 0.20 − 0.144 = 0.776
3
p30 = ( 0.72 )( 0.7 ) = 0.504
3
p34 = ( 0.20 )( 0.3) = 0.06
3
p30:34 = ( 0.504 )( 0.06 ) = 0.03024
3
p30:34 = 0.504 + 0.06 − 0.03024
= 0.53376
2
q30:34 = 0.776 − 0.53376
= 0.24224
Alternatively,
2 q30:34 = 2 q30 + 2 q34 − 2 q30:34
b
g
= 2 p30q32 + 2 p34q36 − 2 p30:34 1 − p32:36
= (0.9)(0.8)(0.3) + (0.5)(0.4)(0.7) – (0.9)(0.8)(0.5)(0.4) [1-(0.7)(0.3)]
= 0.216 + 0.140 – 0.144(0.79)
= 0.24224
Alternatively,
2
q30:34 = 3 q30 × 3 q34 − 2 q30 × 2 q34
= (1 − 3 p30 )(1 − 3 p34 ) − (1 − 2 p30 )(1 − 2 p34 )
= (1 − 0.504 )(1 − 0.06 ) − (1 − 0.72 )(1 − 0.20 )
= 0.24224
(see first solution for
2
p30 ,
2
p34 ,
3
p30 ,
3
p34 )
Question #2
Key: E
1000 Ax = 1000 ⎡ Ax1:10 + 10 Ax ⎤
⎣
⎦
∞
10
= 1000 ⎡ ∫ e −0.04t e −0.06t (0.06)dt + e−0.4e−0.6 ∫ e−0.05t e−0.07t (0.07)dt ⎤
⎢⎣ 0
⎥⎦
0
∞
10
= 1000 ⎡ 0.06∫ e −0.1t dt + e −1 (0.07) ∫ e−0.12t dt ⎤
0
0
⎣⎢
⎦⎥
−0.10 t 10
−0.12 t ∞ ⎤
⎡
= 1000 ⎢ 0.06 ⎡ − e0.10 ⎤ + e −1 (0.07) ⎡ − e0.12 ⎤ ⎥
⎣
⎦0
⎣
⎦0 ⎦
⎣
−1 ⎡
−1.2 ⎤ ⎤
⎡1 − e −1 ⎤ + 0.07
= 1000 ⎡ 0.06
⎦ 0.12 e ⎣1 − e ⎦ ⎦
⎣ 0.10 ⎣
= 1000 ( 0.37927 + 0.21460 ) = 593.87
Because this is a timed exam, many candidates will know common results for constant
force
and constant interest without integration.
For example Ax1:10 =
10 E x
Ax =
µ
µ +δ
=e
(1 − 10 Ex )
−10 ( µ +δ )
µ
µ +δ
With those relationships, the solution becomes
1000 Ax = 1000 ⎡⎣ Ax1:10 + 10 Ex Ax +10 ⎤⎦
(
)
⎡⎛
0.06 ⎞
0.07 ⎞ ⎤
−( 0.06+ 0.04 )10
− 0.06+ 0.04 )10 ⎛
= 1000 ⎢⎜
+e (
⎟ 1− e
⎜
⎟⎥
⎝ 0.07 + 0.05 ⎠ ⎦
⎣⎝ 0.06 + 0.04 ⎠
(
)
= 1000 ⎡( 0.60 ) 1 − e −1 + 0.5833 e −1 ⎤⎥
⎣
⎦
= 593.86
Question #3
Key: A
⎧c ( 400 − x ) x < 400
B=⎨
x ≥ 400
0
⎩
100 = E ( B ) = ci400 − cE ( X ∧ 400 )
300 ⎞
⎛
= ci400 − ci300 ⎜1 −
⎟
⎝ 300 + 400 ⎠
4⎞
⎛
= c ⎜ 400 − 300i ⎟
7⎠
⎝
100
c=
= 0.44
228.6
Question #4
Key: C
Let N = # of computers in department
Let X = cost of a maintenance call
Let S = aggregate cost
Var ( X ) = ⎡⎣Standard Deviation ( X ) ⎤⎦ = 2002 = 40,000
2
( )
E X 2 = Var ( X ) + ⎡⎣ E ( X ) ⎤⎦
2
= 40,000 + 802 = 46, 400
E ( S ) = N × λ × E ( X ) = N × 3 × 80 = 240 N
( )
Var ( S ) = N × λ × E X 2 = N × 3 × 46, 400 = 139, 200 N
We want 0.1 ≥ Pr ( S > 1.2 E ( S ) )
⎛ S − E (S )
0.2 E ( S )
≥ Pr ⎜
>
139, 200 N
⎝ 139, 200 N
2
⎛ 1.282 × 373.1 ⎞
N ≥⎜
⎟ = 99.3
48
⎝
⎠
⎞ 0.2 × 240 N
≥ 1.282 = Φ ( 0.9 )
⎟⇒
N
373.1
⎠
Question #5
Key: B
µ x(τ ) = µ x(1) + µ x( 2 ) + µ x( 3) = 0.0001045
t
τ
px( ) = e −0.0001045t
∞
τ
1
APV Benefits = ∫ e −δ t 1,000,000 t px ( ) µ x( ) dt
0
∞ −δ t
+∫ e
0
τ
2
500,000 t px ( ) µ x( ) dt
∞
τ
3
+ ∫ e −δτ 200,000 t px ( ) µ x( ) dt
0
=
1,000,000 ∞ −0.0601045t
500,000 ∞ −0.0601045t
250,000 ∞ −0.0601045t
e
dt +
e
dt +
e
dt
∫
∫
2,000,000 0
250,000 0
10,000 ∫0
= 27.5 (16.6377 ) = 457.54
Question #6
Key: B
APV Benefits
1
= 1000 A40:20
APV Premiums = π a40:20 +
∞
+
∑ k E401000vq40+k
k = 20
∞
∑ k E401000vq40+k
k = 20
Benefit premiums ⇒ Equivalence principle ⇒
1
1000 A40:20
+
∞
∞
∑ k E401000vq40+k = π a40:20 + ∑ k E401000vq40+k
k = 20
20
π
1
= 1000 A40:
20
=
/ a40:20
161.32 − ( 0.27414 )( 369.13)
14.8166 − ( 0.27414 )(11.1454 )
= 5.11
1
While this solution above recognized that π = 1000P40:20
and was structured to take
advantage of that, it wasn’t necessary, nor would it save much time. Instead, you could
do:
APV Benefits = 1000 A40 = 161.32
∞
APV Premiums =π a40:20 + 20 E40 ∑ k E60 1000vq60+ k
k =0
= π a40:20 + 20 E40 1000 A60
= π ⎡⎣14.8166 − ( 0.27414 )(11.1454 ) ⎤⎦ + ( 0.27414 )( 369.13)
= 11.7612π + 101.19
11.7612π + 101.19 = 161.32
161.32 − 101.19
π=
= 5.11
11.7612
Question #7
Key: C
ln (1.06 )
( 0.53) = 0.5147
0.06
1 − A70 1 − 0.5147
=
= 8.5736
a70 =
d
0.06 /1.06
⎛ 0.97 ⎞
a69 = 1 + vp69a70 = 1 + ⎜
⎟ ( 8.5736 ) = 8.8457
⎝ 1.06 ⎠
A70 = δ A70 =
i
( )
= α ( 2 ) a69 − β ( 2 ) = (1.00021)( 8.8457 ) − 0.25739
a69
2
= 8.5902
( m − 1) works well (is closest to the exact answer,
m
Note that the approximation ax( ) ≅ ax −
2m
1
only off by less than 0.01). Since m = 2, this estimate becomes 8.8457 − = 8.5957
4
Question #8
Key: C
The following steps would do in this multiple-choice context:
1.
From the answer choices, this is a recursion for an insurance or pure
endowment.
2.
Only C and E would satisfy u(70) = 1.0.
1+ i
u ( k − 1)
3.
It is not E. The recursion for a pure endowment is simpler: u ( k ) =
pk −1
4.
Thus, it must be C.
More rigorously, transform the recursion to its backward equivalent,
u ( k − 1) in terms of u ( k ) :
⎛ q ⎞ ⎛ 1+ i ⎞
u ( k ) = − ⎜ k −1 ⎟ + ⎜
⎟ u ( k − 1)
p
p
⎝ k −1 ⎠ ⎝ k −1 ⎠
pk −1u ( k ) = − qk −1 + (1 + i ) u ( k − 1)
u ( k − 1) = vqk −1 + vpk −1 u ( k )
This is the form of (a), (b) and (c) on page 119 of Bowers with x = k − 1 . Thus, the
recursion could be:
Ax = vqx + vpx Ax +1
or
A1x: y − x = vq x + vp x Ax1+1: y − x −1
or
Ax: y − x = vqx + vpx Ax +1: y − x −1
Condition (iii) forces it to be answer choice C
u ( k − 1) = Ax fails at x = 69 since it is not true that
(
)
A69 = vq69 + vp69 (1)
u ( k − 1) = A1x: y − x fails at x = 69 since it is not true that
(
)
1
A69:1
= vq69 + vp69 (1)
u ( k − 1) = Ax: y − x is OK at x = 69 since
(
)
A69:1 = vq69 + vp69 (1)
Note: While writing recursion in backward form gave us something exactly like page
119 of Bowers, in its original forward form it is comparable to problem 8.7 on page 251.
Reasoning from that formula, with π h = 0 and bh +1 = 1 , should also lead to the correct
answer.
Question #9
Key: A
You arrive first if both (A) the first train to arrive is a local and (B) no express arrives in
the 12 minutes after the local arrives.
P ( A) = 0.75
Expresses arrive at Poisson rate of ( 0.25 )( 20 ) = 5 per hour, hence 1 per 12 minutes.
e −110
= 0.368
0!
A and B are independent, so
P ( A and B ) = ( 0.75 )( 0.368 ) = 0.276
f ( 0) =
Question #10
Key: E
d = 0.05 → v = 0.095
At issue
(
49
)
(
)
A40 = ∑ v k +1 k q40 = 0.02 v1 + ... + v50 = 0.02v 1 − v50 / d = 0.35076
k =0
and a40 = (1 − A40 ) / d = (1 − 0.35076 ) / 0.05 = 12.9848
so P40 =
E
(
10 L
1000 A40 350.76
=
= 27.013
a40
12.9848
)
K ( 40 ) ≥ 10 = 1000 A50
Revised
Revised
− P40a50
= 549.18 − ( 27.013)( 9.0164 ) = 305.62
where
Revised
A50
and
24
= ∑ v k +1 k q50
k =0
Revised
a50
=
Revised
(1 − A
Revised
50
(
)
(
)
= 0.04 v1 + ... + v 25 = 0.04v 1 − v 25 / d = 0.54918
) / d = (1 − 0.54918) / 0.05 = 9.0164
Question #11
Key: E
Let NS denote non-smokers and S denote smokers.
The shortest solution is based on the conditional variance formula
(
)
(
Var ( X ) = E Var ( X Y ) + Var E ( X Y )
)
Let Y = 1 if smoker; Y = 0 if non-smoker
1 − AxS
S
E aT Y = 1 = ax =
(
)
δ
1 − 0.444
= 5.56
0.1
1 − 0.286
= 7.14
Similarly E aT Y = 0 =
0.1
=
(
( (
E E aT Y
)
) ) = E ( E ( aT 0 ) ) × Prob ( Y=0 ) + E ( E ( aT 1) ) × Prob ( Y=1)
= ( 7.14 )( 0.70 ) + ( 5.56 )( 0.30 )
= 6.67
( (
E ⎡⎢ E aT Y
⎣
))
( (
2⎤
(
)
(
)
2
2
⎥⎦ = 7.14 ( 0.70 ) + 5.56 ( 0.30 )
= 44.96
) ) = 44.96 − 6.672 = 0.47
E ( Var ( aT Y ) ) = ( 8.503)( 0.70 ) + ( 8.818 )( 0.30 )
Var E aT Y
= 8.60
( )
Var aT = 8.60 + 0.47 = 9.07
Alternatively, here is a solution based on
( )
Var(Y ) = E Y 2 − ⎡⎣ E (Y ) ⎤⎦ , a formula for the variance of any random variable. This
can be
2
( )
transformed into E Y 2 = Var (Y ) + ⎡⎣ E (Y ) ⎤⎦ which we will use in its conditional form
(( )
E aT
2
)
(
2
)
(
)
NS = Var aT NS + ⎡ E aT NS ⎤
⎣
⎦
( )
Var ⎡⎣ aT ⎤⎦ = E ⎡ aT
⎢⎣
2⎤
(
− E ⎡⎣ aT ⎤⎦
⎥⎦
)
2
2
E ⎡⎣ aT ⎤⎦ = E ⎡⎣ aT S⎤⎦ × Prob [S] + E ⎡⎣ aT NS⎤⎦ × Prob [ NS]
= 0.30axS + 0.70axNS
=
(
0.30 1 − AxS
) + 0.70 (1 − A )
NS
x
0.1
0.1
0.30 (1 − 0.444 ) + 0.70 (1 − 0.286 )
=
= ( 0.30 )( 5.56 ) + ( 0.70 )( 7.14 )
0.1
= 1.67 + 5.00 = 6.67
( )
E ⎡ aT
⎢⎣
2⎤
= E ⎡⎣ aT 2 S⎤⎦ × Prob [S] + E ⎡⎣ aT 2 NS⎤⎦ × Prob [ NS]
⎥⎦
( (
) (
)
2
= 0.30 Var aT S + E ⎡⎣ aT S⎤⎦ ⎞⎟
⎠
(
(
)
(
+0.70 Var aT NS + E aT NS
)
2
)
2
2
= 0.30 ⎡8.818 + ( 5.56 ) ⎤ + 0.70 ⎡8.503 + ( 7.14 ) ⎤
⎣
⎦
⎣
⎦
11.919 + 41.638 = 53.557
Var ⎡⎣ aT ⎤⎦ = 53.557 − ( 6.67 ) = 9.1
2
Alternatively, here is a solution based on aT =
1 − vT
δ
⎛1 v ⎞
Var aT = Var ⎜ − ⎟
⎝δ δ ⎠
T
( )
⎛ − vT ⎞
= Var ⎜
⎟ since Var ( X + constant ) = Var ( X )
⎝ δ ⎠
=
( ) since Var ( constant × X ) = constant
Var vT
δ2
2
=
Ax − ( Ax )
δ2
2
× Var ( X )
2
which is Bowers formula 5.2.9
( )
This could be transformed into 2Ax = δ 2 Var aT + Ax2 , which we will use to get
2
Ax NS and 2AxS .
Ax = E ⎡⎣ v 2T ⎤⎦
2
= E ⎡⎣ v 2T NS⎤⎦ × Prob ( NS) + E ⎡⎣ v 2T S⎤⎦ × Prob ( S )
2
= ⎡⎢δ 2 Var aT NS + Ax NS ⎤⎥ × Prob ( NS)
⎣
⎦
2
+ ⎡⎢δ 2Var aT S + AxS ⎤⎥ × Prob ( S)
⎣
⎦
(
) (
(
) ( )
)
= ⎡⎣( 0.01)( 8.503) + 0.2862 ⎤⎦ × 0.70
+ ⎡⎣( 0.01)( 8.818 ) + 0.4442 ⎤⎦ × 0.30
= ( 0.16683)( 0.70 ) + ( 0.28532 )( 0.30 )
= 0.20238
Ax = E ⎡⎣ vT ⎤⎦
= E ⎡⎣ vT NS⎤⎦ × Prob ( NS) + E ⎡⎣ vT S⎤⎦ × Prob ( S)
= ( 0.286 )( 0.70 ) + ( 0.444 )( 0.30 )
= 0.3334
Ax − ( Ax )
2
( )
Var aT =
=
2
δ2
0.20238 − 0.33342
= 9.12
0.01
Question #12
Key: A
To be a density function, the integral of f must be 1 (i.e., everyone dies eventually). The
solution is written for the general case, with upper limit ∞ . Given the distribution of
f 2 ( t ) , we could have used upper limit 100 here.
Preliminary calculations from the Illustrative Life Table:
l50
= 0.8951
l0
l40
= 0.9313
l0
∞
50
∞
0
0
50
1 = ∫ fT ( t ) dt = ∫ k f1 ( t ) dt + ∫ 1.2 f 2 ( t )dt
= k∫
50
0
∞
f1 ( t ) dt + 1.2∫ f 2 ( t )dt
50
= k F1 ( 50 ) + 1.2 ( F2 ( ∞ ) − F2 ( 50 ) )
= k (1 − 50 p0 ) + 1.2 (1 − 0.5 )
= k (1 − 0.8951) + 0.6
k=
1 − 0.6
= 3.813
1 − 0.8951
For x ≤ 50, FT ( x ) = ∫ 3.813 f1 ( t ) dt = 3.813F1 ( x )
x
0
⎛ l ⎞
FT ( 40 ) = 3.813 ⎜ 1 − 40 ⎟ = 3.813 (1 − 0.9313) = 0.262
l0 ⎠
⎝
⎛ l ⎞
FT ( 50 ) = 3.813 ⎜ 1 − 50 ⎟ = 3.813 (1 − 0.8951) = 0.400
l0 ⎠
⎝
10
p40 =
1 − FT ( 50 ) 1 − 0.400
=
= 0.813
1 − FT ( 40 ) 1 − 0.262
Question #13
Key: D
Let NS denote non-smokers, S denote smokers.
Prob (T < t ) = Prob (T < t NS) × Prob ( NS ) + P rob (T < t S) × P rob ( S )
(
)
(
)
= 1 − e −0.1t × 0.7 + 1 − e −0.2t × 0.3
= 1 − 0.7e −0.1t − 0.3 e−0.2t
S ( t ) = 0.3e−0.2 t + 0.7e−0.1t
Want tˆ such that 0.75 = 1 − S ( tˆ ) or 0.25 = S ( tˆ )
(
ˆ
ˆ
Substitute: let x = e −0.1t
0.3 x 2 + 0.7 x − 0.25 = 0
ˆ
)
ˆ 2
0.25 = 0.3e−2t + 0.7e −0.1t = 0.3 e−0.1t
+ 0.7e−0.1t
ˆ
This is quadratic, so x =
−0.7 ± 0.49 + ( 0.3)( 0.25 ) 4
2 ( 0.3)
x = 0.3147
e −0.1t = 0.3147
ˆ
so tˆ = 11.56
Question #14
Key: D
The modified severity, X*, represents the conditional payment amount given that a
payment occurs. Given that a payment is required (X > d), the payment must be
uniformly distributed between 0 and c ⋅ (b − d ) .
The modified frequency, N*, represents the number of losses that result in a payment.
b−d
The deductible eliminates payments for losses below d, so only 1 − Fx ( d ) =
of
b
losses will require payments. Therefore, the Poisson parameter for the modified
b−d
. (Reimbursing c% after the deductible affects only the
frequency distribution is λ ⋅
b
payment amount and not the frequency of payments).
Question #15
Key: C
Let N = number of sales on that day
S = aggregate prospective loss at issue on those sales
K = curtate future lifetime
N ∼ Poisson ( 0.2*50 )
0L
= 10,000v K +1 − 500aK +1
500 ⎞ K +1 500
⎛
0 L = ⎜ 10,000 +
⎟v −
d ⎠
d
⎝
⇒ E [ N ] = Var [ N ] = 10
⇒ E [ 0 L ] = 10,000 A65 − 500a65
2
500 ⎞ ⎡ 2
2⎤
⎛
⇒ Var [ 0 L ] = ⎜ 10,000 +
⎟ ⎣ A65 − ( A65 ) ⎦
d ⎠
⎝
S = 0 L1 + 0 L2 + ... + 0 LN
E [ S ] = E [ N ] ⋅ E [ 0 L]
Var [ S ] = Var [ 0 L ] ⋅ E [ N ] + ( E [ 0 L ]) ⋅ Var [ N ]
2
⎛
0 − E [S ] ⎞
⎟
Pr ( S < 0 ) = Pr ⎜ Z <
⎜
⎟
Var
S
[
]
⎝
⎠
Substituting d = 0.06/(1+0.06), 2 A65 = 0.23603, A65 = 0.43980 and a65 = 9.8969 yields
E [ 0 L ] = −550.45
Var [ 0 L ] = 15,112,000
E [ S ] = −5504.5
Var [ S ] = 154,150,000
Std Dev (S) = 12,416
⎛ S + 5504.5 5504.5 ⎞
Pr ( S < 0 ) = Pr ⎜
<
12, 416 ⎟⎠
⎝ 12,416
= Pr ( Z < 0.443)
= 0.67
With the answer choices, it was sufficient to recognize that:
0.6554 = φ ( 0.4 ) < φ ( 0.443) < φ ( 0.5 ) = 0.6915
By interpolation, φ ( 0.443) ≈ ( 0.43) φ ( 0.5 ) + ( 0.57 ) φ ( 0.4 )
= ( 0.43)( 0.6915 ) + ( 0.57 )( 0.6554 )
= 0.6709
Question #16
Key: A
1000 P40 =
A40 161.32
=
= 10.89
a40 14.8166
⎛ a ⎞
⎛ 11.1454 ⎞
1000 20V40 = 1000 ⎜ 1 − 60 ⎟ = 1000 ⎜ 1 −
⎟ = 247.78
⎝ 14.8166 ⎠
⎝ a40 ⎠
( 20V + 5000 P40 ) (1 + i) − 5000q60
21V =
P60
=
( 247.78 + (5)(10.89) ) × 1.06 − 5000 ( 0.01376 ) = 255
1 − 0.01376
[Note: For this insurance, 20V = 1000 20V40 because retrospectively, this is identical to
whole life]
Though it would have taken much longer, you can do this as a prospective reserve.
The prospective solution is included for educational purposes, not to suggest it
would be suitable under exam time constraints.
1000 P40 = 10.89 as above
1
1000 A40 + 4000 20 E40 A60:5
= 1000 P40 + 5000 P40 × 20 E40 a60:5 + π
20 E40 × 5 E60
a65
1
where A60:5
= A60 − 5 E60 A65 = 0.06674
a40:20 = a40 − 20 E40 a60 = 11.7612
a60:5 = a60 − 5 E60 a65 = 4.3407
1000 ( 0.16132 ) + ( 4000 )( 0.27414 )( 0.06674 ) =
= (10.89 )(11.7612 ) + ( 5 )(10.89 )( 0.27414 )( 4.3407 ) + π ( 0.27414 )( 0.68756 )( 9.8969 )
161.32 + 73.18 − 128.08 − 64.79
1.86544
= 22.32
π=
Having struggled to solve for π , you could calculate
above)
calculate 21V recursively.
20V
1
= 4000 A60:5
+ 1000 A60 − 5000 P40 a60:5 − π 5 E60 a65
20 V
prospectively then (as
= ( 4000 )( 0.06674 ) + 369.13 − ( 5000 )( 0.01089 )( 4.3407 ) − ( 22.32 )( 0.68756 )( 9.8969 )
= 247.86 (minor rounding difference from 1000 20V40 )
Or we can continue to
21V
=
1
5000 A61:4
where
4 E61
=
21V
prospectively
+ 1000 4 E61 A65 − 5000 P40 a61:4 − π 4 E61 a65
l65 4 ⎛ 7,533,964 ⎞
v =⎜
⎟ ( 0.79209 ) = 0.73898
l61
⎝ 8,075, 403 ⎠
1
A61:4
= A61 − 4 E61 A65 = 0.38279 − 0.73898 × 0.43980
a61:4
= 0.05779
= a61 − 4 E61 a65 = 10.9041 − 0.73898 × 9.8969
= 3.5905
21V
= ( 5000 )( 0.05779 ) + (1000 )( 0.73898 )( 0.43980 )
− ( 5 )(10.89 )( 3.5905 ) − 22.32 ( 0.73898 )( 9.8969 )
= 255
Finally. A moral victory. Under exam conditions since prospective benefit reserves
must equal retrospective benefit reserves, calculate whichever is simpler.
Question #17
Key: C
Var ( Z ) = 2 A41 − ( A41 )
2
A41 − A40 = 0.00822 = A41 − (vq40 + vp40 A41 )
= A41 − (0.0028 /1.05 + ( 0.9972 /1.05 ) A41 )
⇒ A41 = 0.21650
2
(
A41 − 2 A40 = 0.00433 = 2 A41 − v 2 q40 + v 2 p40 2 A41
(
)
= 2 A41 − (0.0028 /1.052 + 0.9972 /1.052
2
A41 = 0.07193
Var ( Z ) = 0.07193 − 0.216502
= 0.02544
)
2
A41 )
Question #18
Key: D
This solution looks imposing because there is no standard notation. Try to focus on the
big picture ideas rather than starting with the details of the formulas.
Big picture ideas:
1.
We can express the present values of the perpetuity recursively.
2.
Because the interest rates follow a Markov process, the present value (at time t )
of the future payments at time t depends only on the state you are in at time t ,
not how you got there.
3.
Because the interest rates follow a Markov process, the present value of the
future payments at times t1 and t2 are equal if you are in the same state at times
t1 and t2 .
Method 1: Attack without considering the special characteristics of this transition matrix.
Let sk = state you are in at time k ( thus sk = 0, 1 or 2 )
Let Yk = present value, at time k , of the future payments.
Yk is a random variable because its value depends on the pattern of discount factors,
which are random. The expected value of Yk is not constant; it depends on what state
we are in at time k.
Recursively we can write
Yk = v × (1 + Yk +1 ) , where it would be better to have notation that indicates the v’s are not
constant, but are realizations of a random variable, where the random variable itself has
different distributions depending on what state we’re in. However, that would make the
notation so complex as to mask the simplicity of the relationship.
Every time we are in state 0 we have
E ⎡⎣Yk sk = 0 ⎤⎦ = 0.95 × 1 + E ⎡⎣Yk +1 sk = 0 ⎤⎦
(
( {(
)
))
(
= 1⎤⎦ ) × Pr ob ( s
= 2 ⎤⎦ ) × Pr ob ( s
)
= 1 s = 0])
= 2 s = 0])}
= 0.95 × 1 + E ⎡⎣Yk +1 sk +1 = 0 ⎤⎦ × Pr ob s k+1 = 0 sk = 0]
(
= ( E ⎡⎣Y
+ E ⎡⎣Yk +1 sk +1
k +1
(
sk +1
= 0.95 × 1 + E ⎡⎣Yk +1 sk +1 = 1⎤⎦
k +1
k +1
)
k
k
That last step follows because from the transition matrix if we are in state 0, we always
move to state 1 one period later.
Similarly, every time we are in state 2 we have
E ⎡⎣Yk sk = 2 ⎤⎦ = 0.93 × 1 + E ⎡⎣Yk +1 sk = 2 ⎤⎦
(
(
= 0.93 × 1 + E ⎡⎣Yk +1 sk +1 = 1⎤⎦
)
)
That last step follows because from the transition matrix if we are in state 2, we always
move to state 1 one period later.
Finally, every time we are in state 1 we have
E ⎡⎣Yk sk = 1⎤⎦ = 0.94 × 1 + E ⎡⎣Yk +1 sk = 1⎤⎦
( {
(
)
})
= 0.94 × 1 + E ⎡⎣Yk +1 sk +1 = 0 ⎤⎦ × Pr ⎡⎣ sk +1 = 0 sk = 1⎤⎦ + E ⎡⎣Yk +1 sk +1 = 2⎤⎦ × Pr ⎡⎣ sk +1 = 2 sk = 1⎤⎦
( {
})
= 0.94 × 1 + E ⎡⎣Yk +1 sk +1 = 0 ⎤⎦ × 0.9 + E ⎡⎣Yk +1 sk +1 = 2 ⎤⎦ × 0.1 . Those last two steps follow
from the fact that from state 1 we always go to either state 0 (with probability 0.9) or
state 2 (with probability 0.1).
Now let’s write those last three paragraphs using this shorter notation:
xn = E ⎡⎣Yk sk = n ⎤⎦ . We can do this because (big picture idea #3), the conditional
expected value is only a function of the state we are in, not when we are in it or how we
got there.
x0 = 0.95 (1 + x1 )
x1 = 0.94 (1 + 0.9 x0 + 0.1x2 )
x2 = 0.93 (1 + x1 )
That’s three equations in three unknowns. Solve (by substituting the first and third into
the second) to get x1 = 16.82 .
That’s the answer to the question, the expected present value of the future payments
given in state 1.
The solution above is almost exactly what we would have to do with any 3 × 3 transition
matrix. As we worked through, we put only the non-zero entries into our formulas. But
if for example the top row of the transition matrix had been ( 0.4 0.5 0.1) , then the first
of our three equations would have become x0 = 0.95 (1 + 0.4 x0 + 0.5 x1 + 0.1x2 ) , similar in
structure to our actual equation for x1 . We would still have ended up with three linear
equations in three unknowns, just more tedious ones to solve.
Method 2: Recognize the patterns of changes for this particular transition matrix.
This particular transition matrix has a recurring pattern that leads to a much quicker
solution. We are starting in state 1 and are guaranteed to be back in state 1 two steps
later, with the same prospective value then as we have now.
Thus,
E [Y ] = E ⎡⎣Y first move is to 0 ⎤⎦ × Pr [ first move is to 0] + E ⎡⎣Y first move is to 2 ⎤⎦ × Pr [first move is to 2 ]
(
)
(
= 0.94 × ⎡ 1 + 0.95 × (1 + E [Y ] ⎤ × 0.9 + ⎡ 0.94 × ⎡ 1 + 0.93 × (1 + E [Y ]) × 0.1⎤
⎣
⎦
⎣
⎣
⎦
(Note that the equation above is exactly what you get when you substitute x0 and x2
into the formula for x1 in Method 1.)
= 1.6497 + 0.8037 E [Y ] + 0.1814 + 0.0874 E [Y ]
E [Y ] =
1.6497 + 0.1814
(1 − 0.8037 − 0.0874 )
= 16.82
Question #19
Key: E
The number of problems solved in 10 minutes is Poisson with mean 2.
If she solves exactly one, there is 1/3 probability that it is #3.
If she solves exactly two, there is a 2/3 probability that she solved #3.
If she solves #3 or more, she got #3.
f(0) = 0.1353
f(1) = 0.2707
f(2) = 0.2707
⎛1⎞
⎛ 2⎞
P = ⎜ ⎟ ( 0.2707 ) + ⎜ ⎟ ( 0.2707 ) + (1 − 0.1353 − 0.2707 − 0.2707 ) = 0.594
⎝ 3⎠
⎝ 3⎠
Question #20
Key: D
µ x(τ ) = µ x(1) ( t ) + µ x( 2 ) ( t )
τ
2
= 0.2 µ x( ) ( t ) + µ x( ) ( t )
2
τ
⇒ µ x( ) ( t ) = 0.8µ x( ) ( t )
1
− ∫ 0.2 k t
'1
'1
qx( ) = 1 − px( ) = 1 − e 0
k
3
2
dt
= 1− e
−0.2
k
3
= 0.04
⇒ ln (1 − 0.04 ) / ( −0.2 ) = 0.2041
k = 0.6123
( 2)
2 qx
=∫
2
0 t
2
τ
2
px( ) µ x( ) dt = 0.8∫
0
(
τ
τ
= 0.8 2 qx( ) = 0.8 1 − 2 px( )
t
τ
τ
px( ) µ x( ) ( t ) dt
)
− ∫ µ x (t ) d t
(τ )
0
2 px = e
2
2
− kt
= e ∫0
=
2
dt
−8 k
e 3
−( 8 ) ( 0.6123)
3
e
=
= 0.19538
( 2)
2 qx
= 0.8 (1 − 0.19538 ) = 0.644
Question #21
Key: A
k
k ∧3
f(k)
f ( k ) × ( k ∧ 3)
f ( k ) × (k ∧ 3) 2
0
1
2
3+
0
1
2
3
0.1
(0.9)(0.2) = 0.18
(0.72)(0.3) = 0.216
1-0.1-0.18-0.216 = 0.504
0
0.18
0.432
1.512
2.124
0
0.18
0.864
4.536
5.580
E ( K ∧ 3) = 2.124
(
E ( K ∧ 3)
2
) = 5.580
Var ( K ∧ 3) = 5.580 − 2.1242 = 1.07
Note that E [ K ∧ 3] is the temporary curtate life expectancy, ex:3 if the life is age x.
Problem 3.17 in Bowers, pages 86 and 87, gives an alternative formula for the variance,
basing the calculation on k px rather than k q x .
Question #22
Key: E
f ( x) = 0.01, 0 ≤ x ≤ 80
= 0.01 − 0.00025( x − 80) = 0.03 − 0.00025 x,
80
120
0
80
E ( x) = ∫ 0.01x dx + ∫
( 0.03x − 0.00025x ) dx
2
0.01x 2 80 0.03 x 2 120 0.00025 x3
+
−
2 0
2 80
3
= 32 + 120 − 101.33 = 50.66667
=
120
80
E ( X − 20) + = E ( X ) − ∫ x f ( x ) dx − 20(1 − ∫
20
20
0
0
f ( x ) dx)
(
0.01x 2 20
= 50.6667 −
− 20 1 − 0.01x
2 0
= 50.6667 − 2 − 20(0.8) = 32.6667
Loss Elimination Ratio = 1 −
32.6667
= 0.3553
50.6667
20
0
)
80 < x ≤ 120
Question #23
Key: D
Let q64 for Michel equal the standard q64 plus c. We need to solve for c.
Recursion formula for a standard insurance:
20V45
= ( 19V45 + P45 ) (1.03) − q64 (1 − 20V45 )
Recursion formula for Michel’s insurance
20V45
= ( 19V45 + P45 + 0.01) (1.03) − ( q64 + c ) (1 − 20V45 )
The values of
19 V45
and
20V45
are the same in the two equations because we are told
Michel’s benefit reserves are the same as for a standard insurance.
Subtract the second equation from the first to get:
0 = − (1.03) (0.01) + c(1 − 20V45 )
c=
(1.03) ( 0.01)
(1 − 20V45 )
0.0103
1 − 0.427
= 0.018
=
Question #24
Key: B
K is the curtate future lifetime for one insured.
L is the loss random variable for one insurance.
LAGG is the aggregate loss random variables for the individual insurances.
σ AGG is the standard deviation of LAGG .
M is the number of policies.
⎛ π⎞
L = v K +1 − π aK +1 = ⎜ 1 + ⎟ v K +1 − π
d
⎝ d⎠
(1 − Ax )
E [ L ] = ( Ax − π ax ) = Ax − π
d
⎛ 0.75095 ⎞
= 0.24905 − 0.025 ⎜
⎟ = −0.082618
⎝ 0.056604 ⎠
⎛ π⎞
Var [ L ] = ⎜ 1 + ⎟
⎝ d⎠
2
(
2
2
)
(
)
0.025 ⎞
2
⎛
Ax − Ax2 = ⎜ 1 +
⎟ 0.09476 − ( 0.24905 ) = 0.068034
⎝ 0.056604 ⎠
E [ LAGG ] = M E [ L ] = −0.082618M
Var [ LAGG ] = M Var [ L ] = M (0.068034) ⇒ σ AGG = 0.260833 M
⎡L
− E [ LAGG ] − E ( LAGG ) ⎤
>
Pr [ LAGG > 0] = ⎢ AGG
⎥
σ AGG
σ AGG ⎦
⎣
⎛
0.082618M ⎞
≈ Pr ⎜ N (0,1) >
⎟
⎜
⎟
M
0.260833
(
)
⎝
⎠
0.082618 M
0.260833
⇒ M = 26.97
⇒ 1.645 =
⇒ minimum number needed = 27
Question #25
Key: D
Death benefit:
1 − v K +1
Z1 = 12,000
for K = 0,1, 2,...
d
Z 2 = Bv K +1
for K = 0,1, 2,...
New benefit:
Z = Z1 + Z 2 = 12,000
Annuity benefit:
1 − v K +1
+ Bv K +1
d
12,000 ⎛
12,000 ⎞ K +1
=
+⎜B−
⎟v
d
d ⎠
⎝
2
(
)
12,000 ⎞
⎛
K +1
Var( Z ) = ⎜ B −
⎟ Var v
d ⎠
⎝
12,000
Var ( Z ) = 0 if B =
= 150,000 .
0.08
In the first formula for Var ( Z ) , we used the formula, valid for any constants a and b and
random variable X,
Var ( a + bX ) = b 2Var ( X )
Question #26
Key: B
First restate the table to be CAC’s cost, after the 10% payment by the auto owner:
Towing Cost, x
72
90
144
p(x)
50%
40%
10%
Then E ( X ) = 0.5*72 + 0.4*90 + 0.1*144 = 86.4
( )
E X 2 = 0.5*722 + 0.4*902 + 0.1*1442 = 7905.6
Var ( X ) = 7905.6 − 86.42 = 440.64
Because Poisson, E ( N ) = Var ( N ) = 1000
E ( S ) = E ( X ) E ( N ) = 86.4*1000 = 86, 400
Var( S ) = E ( N )Var( X ) + E ( X )2 Var( N ) = 1000* 440.64 + 86.42 *1000 = 7,905,600
⎛ S − E ( S ) 90,000 − 86, 400 ⎞
>
Pr( S > 90,000) + Pr ⎜
⎟⎟ = Pr( Z > 1.28) = 1 − Φ (1.28) = 0.10
⎜ Var( S )
7,905,600
⎝
⎠
Since the frequency is Poisson, you could also have used
Var ( S ) = λ E X 2 = (1000 )( 7905.6 ) = 7,905,600
( )
That way, you would not need to have calculated Var ( X ) .
Question #27
Key: C
(
)
− d /θ
E ( X ∧ d ) θ 1− e
LER =
=
= 1 − e− d /θ
θ
E(X )
0.70 = 1 − e− d /θ ⇒ − d = θ log 0.30
− d new = θ log(1 − LER new )
4
Hence θ log (1 − LER new ) = − d new = θ log 0.30
3
log (1 − LER new ) = −1.6053
Last year
Next year:
(1 − LER new ) = e−1.6053 = 0.20
LER new = 0.80
Question #28
Key: E
E ( X ) = e(d ) S (d ) + E ( X ∧ d )
[Klugman Study Note, formula 3.10]
62 = e 40 × 40 p0 + E (T ∧ 40 )
62 = (e 40 )(0.6) + 40 − (0.005)(402 )
= 0.6 e 40 + 32
e40 =
( 62 − 32 ) = 50
0.6
The first equation, in the notation of Bowers, is e0 = e40 × 40 p0 + e0:40 . The corresponding
formula, with i > 0 , is a very commonly used one:
ax = ax:n + n Ex ax + n
Question #29
Key: B
d = 0.05 ⇒ v = 0.95
Step 1 Determine px from Kevin’s work:
608 + 350vpx = 1000vqx + 1000v 2 px ( px +1 + qx +1 )
608 + 350 ( 0.95 ) px = 1000 ( 0.95 ) (1 − px ) + 1000 ( 0.9025 ) px (1)
608 + 332.5 px = 950 (1 − px ) + 902.5 px
px = 342 / 380 = 0.9
Step 2 Calculate 1000 Px:2 , as Kira did:
608 + 350 ( 0.95 )( 0.9 ) = 1000 Px:2 ⎡⎣1 + ( 0.95 )( 0.9 ) ⎤⎦
[ 299.25 + 608] = 489.08
1000 Px:2 =
1.855
The first line of Kira’s solution is that the actuarial present value of Kevin’s benefit
premiums is equal to the actuarial present value of Kira’s, since each must equal the
actuarial present value of benefits. The actuarial present value of benefits would also
have been easy to calculate as
(1000 )( 0.95 )( 0.1) + (1000 ) 0.952 ( 0.9 ) = 907.25
(
)
Question #30
Key: E
Because no premiums are paid after year 10 for (x), 11Vx = Ax +11
Rearranging 8.3.10 from Bowers, we get
10V
=
h +1V
=
( hV + π h ) (1 + i ) − bh+1qx+ h
( 32,535 + 2,078 ) × (1.05) − 100,000 × 0.011 = 35,635.642
px + h
0.989
( 35,635.642 + 0 ) × (1.05) − 100,000 × 0.012 = 36,657.31 = A
11V =
x +11
0.988
Question #31
Key: B
x⎞
⎛
For De Moivre’s law where s ( x ) = ⎜ 1 − ⎟ :
⎝ ω⎠
ω−x
t ⎞
⎛
and t px = ⎜ 1 −
⎟
2
⎝ ω −x⎠
105 − 45
°
e 45 =
= 30
2
105 − 65
e°65 =
= 20
2
°
ex =
°
e 45:65 = ∫
=
40
0
t
p45:65dt = ∫
40 60 − t
0
FG
H
60
×
40 − t
dt
40
1
60 + 40 2 1 3
t + t
60 × 40 × t −
60 × 40
2
3
IJ
K
40
0
= 1556
.
e 45:65 = e 45 + e 65 − e 45:65
= 30 + 20 − 1556
. = 34
In the integral for e45:65 , the upper limit is 40 since 65 (and thus the joint status also)
can survive a maximum of 40 years.
Question #32
Answer: E
bg bg bg
− d − e / 100i
=
µ 4 = −s' 4 / s 4
4
1 − e4 / 100
=
e4 / 100
1 − e4 / 100
=
e4
100 − e4
= 1202553
.
Question # 33
Answer: A
bi g
ig
τ g L ln px′ O
b
b
bτ g LM ln e− µ OP
qx = qx M
=
q
P
x
M ln p bτ g P M ln e-µb g P
bi g
N
x
Q
N
τ
µb g
τg
b
= q x × bτ g
µ
i
µ x bτ g = µ x b1g + µ x b 2 g + µ x b 3g = 15
.
q xbτ g = 1 − e − µ bτ g = 1 − e −1.5
=0.7769
b0.7769gµ
q xb 2 g =
µ bτ g
= 0.2590
b2g
=
b0.5gb0.7769g
15
.
Q
Question # 34
Answer: D
22
A 60 = v 3
×
B
B
× q 60 + 2
live
then die
2 years in year 3
+ v4
3p 60
×
live
3 years
+
B
pay at end
of year 3
pay at end
of year 4
=
2 p 60
×
q60+ 3
then die
in year 4
1
1 − 0.09gb1 − 011
. gb013
. g+
. gb1 − 013
. gb015
. g
b
b1 − 0.09gb1 − 011
. g
. g
b103
b103
1
3
4
= 019
.
Question # 35
Answer: B
a x = a x:5 +5 E x a x +5
a x:5
5 Ex
1 − e −0.07b5g
=
= 4.219 , where 0.07 = µ + δ for t < 5
0.07
= e −0.07b5g = 0.705
a x +5 =
1
= 12.5 , where 0.08 = µ + δ for t ≥ 5
0.08
b
gb g
∴ a x = 4.219 + 0.705 12.5 = 13.03
Question # 36
Answer: E
The distribution of claims (a gamma mixture of Poissons) is negative binomial.
b g d c hi b g
Var b N g = E dVar c N Λ hi + Var d E c N Λ hi
= E b Λ g + Var b Λ g = 6
rβ = 3
rβ b1 + β g = 6
b1 + β g = 6 / 3 = 2 ; β = 1
E N = E∧ E N Λ = E∧ Λ = 3
∧
∧
∧
∧
rβ = 3
r=3
.
b g = 0125
.
b g = 01875
Pr obbat most 1g = p + p
p0 = 1 + β
rβ
p1 =
1+ β
−r
r +1
0
1
= 0.3125
Question # 37
Answer: A
bg b g b g
Var b S g = E b N g × Var b X g + E b X g × Var b N g
E S = E N × E X = 110 × 1101
,
= 121110
,
2
= 110 × 702 + 11012 × 750
= 909,689,750
bg
Std Dev S = 30,161
b
g
g
c b
= Pr b Z < −0.70g = 0.242
h
Pr S < 100,000 = Pr Z < 100,000 − 121110
,
/ 30,161 where Z has standard normal
distribution
Question # 38
Answer: C
This is just the Gambler’s Ruin problem, in units of 5,000 calories.
Each day, up one with p = 0.45 ; down 1 with q = 0.55
Will Allosaur ever be up 1 before being down 2?
e1 − b0.55 / 0.45g 2j = 0.598
P =
e1 − b0.55 / 0.45g 3j
∧
∧
2
Or, by general principles instead of applying a memorized formula:
Let P1 = probability of ever reaching 3 (15,000 calories) if at 1 (5,000 calories).
Let P2 = probability of ever reaching 3 (15,000 calories) if at 2 (10,000 calories).
From either, we go up with p = 0.45 , down with q = 0.55
P(reaching 3) = P(up) × P(reaching 3 after up) + P(down) × P(reaching 3 after down)
P2 = 0.45 × 1 + 0.55 × P1
P1 = 0.45 × P2 + 0.55 × 0 = 0.45 × P2
P2 = 0.45 + 0.55 × P1 = 0.45 + 0.55 × 0.45 × P2 = 0.45 + 0.2475 P2
P2 = 0.45 / 1 − 0.2475 = 0.598
b
g
Here is another approach, feasible since the number of states is small.
Let states 0,1,2,3 correspond to 0; 5,000; 10,000; ever reached 15,000 calories. For
purposes of this problem, state 3 is absorbing, since once the allosaur reaches 15,000
we don’t care what happens thereafter.
The transition matrix is
LM1
MM 00.55
MN 0
0
0
0
0.45
0.55 0
0
0
OP
PP
PQ
0
0
0.45
1
Starting with the allosaur in state 2;
[0
0
1
0]
at inception
[0
0.55
0
0.45]
after 1
[0.3025
0
0.2475
0.45]
after 2
[0.3025
0.1361
0
0.5614] after 3
[0.3774
0
0.0612
0.5614] after 4
[0.3774
0.0337
0
0.5889] after 5
[0.3959
0
0.0152
0.5889] after 6
By this step, if not before, Prob(state 3) must be converging to 0.60. It’s already closer
to 0.60 than 0.57, and its maximum is 0.5889 + 0.0152
Question # 39
Answer: D
b gb gb g
Per 10 minutes, find coins worth exactly 10 at Poisson rate 0.5 0.2 10 = 1
Per 10 minutes,
f
f
f
f
b0g = 0.3679
b1g = 0.3679
.
b2g = 01839
b3g = 0.0613
bg
bg
bg
bg
F 0 = 0.3679
F 1 = 0.7358
F 2 = 0.9197
F 3 = 0.9810
Let Period 1 = first 10 minutes; period 2 = next 10.
Method 1, succeed with 3 or more in period 1; or exactly 2, then one or more in period 2
P = 1 − F 2 + f 2 1 − F 0 = 1 − 0.9197 + 01839
.
1 − 0.3679
b gh b gc
c
b gh b
g b
gb
g
= 01965
.
bg
bg bg
b gb
Pr ob = F 1 = 0.7358
Pr ob = f 2 f 0
= 01839
.
0.3679 = 0.0677
Pr ob = 1 − 0.7358 − 0.0677
= 01965
.
Method 2: fail in period 1 if < 2;
fail in period 2 if exactly 2 in period 1, then 0;
Succeed if fail neither period;
g
(Method 1 is attacking the problem as a stochastic process model; method 2 attacks it
as a ruin model.)
Question # 40
Answer: D
Use Mod to designate values unique to this insured.
b
g
b
g b gb g
= 1000b0.36933 / 111418
.
.
g = 3315
a60 = 1 − A60 / d = 1 − 0.36933 / 0.06 / 106
.
= 111418
.
1000 P60 = 1000 A60 / a60
d
i
Mod
Mod
A60Mod = v q60
A61 =
+ p60
b
gb
g
1
.
01376
+ 0.8624 0.383 = 0.44141
.
106
d
b
i
g
a Mod = 1 − A60Mod / d = 1 − 0.44141 / 0.06 / 106
. = 9.8684
d
Mod
E 0 LMod = 1000 A60Mod − P60a60
i
b
= 1000 0.44141 − 0.03315 9.8684
g
= 114.27
Question # 41
Answer: D
The prospective reserve at age 60 per 1 of insurance is A60 , since there will be no
future premiums. Equating that to the retrospective reserve per 1 of coverage, we have:
s40:10
A60 = P40
+ P50Mod s50:10 − 20 k40
10 E50
A
A60 = 40 ×
a40
0.36913 =
a40:10
10 E40 10 E50
+
P50Mod
a50:10
10 E50
−
1
A40
:20
20 E40
016132
.
7.70
7.57
0.06
×
+ P50Mod
−
14.8166 0.53667 0.51081
0.51081 0.27414
b
gb
g
0.36913 = 0.30582 + 14.8196 P50Mod − 0.21887
1000 P50Mod = 19.04
Alternatively, you could equate the retrospective and prospective reserves at age 50.
Your equation would be:
A50 −
P50Mod
a50:10
1
A40 a40:10 A40:10
=
×
−
a40 10 E40 10 E40
1
where A40
= A40 −10 E40 A50
:10
b
gb
g
= 016132
.
− 0.53667 0.24905
= 0.02766
d
.
7.70
0.02766
×
−
ib g 14016132
.8166 0.53667 0.53667
.
1000
014437
b gb
g = 19.07
=
0.24905 − P50Mod 7.57 =
1000 P50Mod
7.57
Alternatively, you could set the actuarial present value of benefits at age 40 to the
actuarial present value of benefit premiums. The change at age 50 did not change the
benefits, only the pattern of paying for them.
A40 = P40 a40:10 + P50Mod
016132
.
=
10 E40
a50:10
.
FG 016132
IJ b7.70g + d P ib0.53667gb7.57g
H 14.8166K
b1000gb0.07748g = 19.07
=
Mod
50
1000 P50Mod
4.0626
Question # 42
Answer: A
d xb 2 g = q xb 2 g × lxbτ g = 400
b g
d xb1g = 0.45 400 = 180
q x′ b 2 g =
d xb 2 g
lxbτ g − d xb1g
=
400
= 0.488
1000 − 180
px′b 2 g = 1 − 0.488 = 0.512
Note: The UDD assumption was not critical except to have all deaths during the year so
that 1000 - 180 lives are subject to decrement 2.
Question #43
Answer: D
Use “age” subscripts for years completed in program. E.g., p0 applies to a person newly
hired (“age” 0).
Let decrement 1 = fail, 2 = resign, 3 = other.
Then q0′b1g = 1 4 , q1′ b1g = 15 , q2′b1g = 1 3
q0′b 2 g =
1
q ′b 3g = 1
5,
q1′b 2 g =
3,
q ′b3g = 1
10 ,
0
1
1
b
9,
gb
q2′b 2 g =
1
8
q ′b 3g = 1
2
gb
4
g
This gives p0bτ g = 1 − 1 / 4 1 − 1 / 5 1 − 1 / 10 = 0.54
b gb gb g
pb g = b1 − 1 / 3gb1 − 1 / 8gb1 − 1 / 4g = 0.438
.
So 1b g = 200, 1b g = 200 b0.54g = 108 , and 1b g = 108 b0.474g = 512
p1bτ g = 1 − 1 / 5 1 − 1 / 3 1 − 1 / 9 = 0.474
τ
2
τ
τ
0
τ
1
2
q2b1g = log p2′b1g / log p2bτ g q2bτ g
c h / logb0.438g 1 − 0.438
= b0.405 / 0.826gb0.562g
q2b1g = log
2
3
= 0.276
d2b1g = l2bτ g q2b1g
= 512
. 0.276 = 14
b gb
g
Question #44
Answer: C
Let:
N = number
X = profit
S = aggregate profit
subscripts G = “good”, B = “bad”, AB = “accepted bad”
c hb60g = 40
= c hc hb60g = 10
λG =
λ AB
2
3
1
2
1
3
(If you have trouble accepting this, think instead of a heads-tails
rule, that the application is accepted if the applicant’s government-issued identification
number, e.g. U.S. Social Security Number, is odd. It is not the same as saying he
automatically alternates accepting and rejecting.)
b g b g
b g
b g b g
Var SG = E N G × Var X G + Var N G × E X G
2
b gb g b gd i
Var b S g = E b N g × Var b X g + Var b N g × E b X g
= b10gb90,000g + b10gb−100g = 1,000,000
S and S are independent, so
Var b S g = Var b S g + Var b S g = 4,000,000 + 1,000,000
= 40 10,000 + 40 3002 = 4,000,000
2
AB
AB
AB
AB
AB
2
G
AB
G
AB
= 5,000,000
If you don’t treat it as three streams (“goods”, “accepted bads”, “rejected bads”), you
can compute the mean and variance of the profit per “bad” received.
λ B = 13 60 = 20
c hb g
b g b g
= 90,000 + b−100g = 100,000
d i
If all “bads” were accepted, we would have E X 2 B = Var X B + E X B
2
2
Since the probability a “bad” will be accepted is only 50%,
E X B = Prob accepted × E X B accepted + Prob not accepted × E X B not accepted
b g
b
g c
h
= b0.5gb −100g + b0.5gb0g = −50
E d X i = b0.5gb100,000g + b0.5gb0g = 50,000
b
g c
2
B
Likewise,
Now Var S B = E N B × Var X B + Var N B × E X B
b g b g b g b g b g
= b20gb47,500g + b20gd50 i = 1,000,000
2
SG and S B are independent, so
bg
b g
b g
Var S = Var SG + Var S B = 4,000,000 + 1,000,000
= 5,000,000
2
h
Question # 45
Answer: C
Let N = number of prescriptions then S = N × 40
bg
n
0
1
2
3
∞
b g
bg
fN n
0.2000
0.1600
0.1280
0.1024
FN n
0.2000
0.3600
0.4880
0.5904
bg
1− FN n
0.8000
0.6400
0.5120
0.4096
c b gh
E N = 4 = ∑ 1− F j
j =0
b
g
E S − 80
b
b
+
g
E S − 120
g
∞
b gh
L
O
= 40 × M∑ c1 − F b j gh − ∑ c1 − F b j ghP
MN
PQ
= 40b4 − 144
. g = 40 × 2.56 = 102.40
= 40 × E N − 2
+
+
j =2
∞
1
j =0
j =0
b
g
c
= 40 × ∑ 1 − F j
∞
c b gh
L
O
= 40 × M∑ c1 − F b j gh − ∑ c1 − F b j ghP
MN
PQ
= 40b4 − 1952
. g = 40 × 2.048 = 8192
.
= 40 × E N − 3
+
= 40 × ∑ 1 − F j
j =3
∞
2
j =0
j =0
Since no values of S between 80 and 120 are possible,
+ b100 − 80g × E b S − 120g
g = b120 − 100g × E bS − 80g 120
b
E S − 100
+
+
+
= 92.16
Alternatively,
b
g
E S − 100
∞
+
b
g bg
bg
bg
bg
= ∑ 40 j − 100 f N j + 100 f N 0 + 60 f N 1 + 20 f N 2
j =0
b
g
(The correction terms are needed because 40 j − 100 would be negative for j = 0, 1, 2;
we need to add back the amount those terms would be negative)
∞
bg
∞
b g b gb
g b gb g b
gb g
= 40∑ j × f N j − 100∑ f N j + 100 0.200 + 016
.
60 + 0128
.
20
j=0
b g
j=0
= 40 E N − 100 + 20 + 9.6 + 2.56
= 160 − 67.84 = 92.16
Question #46
Answer: B
10 E30:40 = 10 p30 10
d p v id p v ib1 + i g
= b E gb E gb1 + i g
= b0.54733gb0.53667 gb179085
.
g
p40 v10 =
10
10
30
10
10
10
40
10
10
30
10
40
= 0.52604
The above is only one of many possible ways to evaluate
10
p30
10
should give 0.52604
a30:40:10 = a30:40 −10 E30:40 a30+10:40+10
b
g b
gb
g
.
= b13.2068g − b0.52604gb114784
g
= a30:40 − 1 − 0.52604 a40:50 − 1
= 7.1687
Question #47
Answer: A
Equivalence Principle, where π is annual benefit premium, gives
d
b g
1000 A35 + IA
π=
d
35
i
× π = a xπ
1000 A35
1000 × 0.42898
=
− 616761
(1199143
.
.
)
a35 − IA 35
b gi
428.98
.
582382
= 73.66
=
We obtained a35 from
a35 =
1 − A35 1 − 0.42898
=
= 1199143
.
0.047619
d
p40 v10 , all of which
Question #48
Answer: C
Time until arrival = waiting time plus travel time.
Waiting time is exponentially distributed with mean 1 λ . The time you may already have
been waiting is irrelevant: exponential is memoryless.
You: E (wait) = 201 hour = 3 minutes
E (travel) = 0.25 16 + 0.75 28 = 25 minutes
E (total) = 28 minutes
b gb g b gb g
Co-worker:
E (wait) = 15 hour = 12 minutes
E (travel) = 16 minutes
E (total) = 28 minutes
Question #49
Answer: C
µ xy = µ x + µ y = 014
.
µ
0.07
Ax = Ay =
=
= 0.5833
µ + δ 0.07 + 0.05
µ xy
014
.
014
.
1
1
Axy =
=
=
= 0.7368 and a xy =
=
= 5.2632
µ xy + δ 014
µ xy + δ 014
. + 0.05 019
.
. + 0.05
P=
Axy
a xy
=
Ax + Ay − Axy
a xy
=
b
g
2 0.5833 − 0.7368
= 0.0817
5.2632
Question #50
Answer: E
b V + P gb1 + ig − q b1− V g= V
b0.49 + 0.01gb1 + ig − 0.022b1 − 0.545g = 0.545
b1 + ig = b0.545gb1 − 0.022g + 0.022
20 20
20
40
21 20
21 20
0.50
= 111
.
b V + P gb1 + ig − q b1− V g= V
. g − q b1 − 0.605g = 0.605
b0.545+.01gb111
21 20
20
41
22 20
22 20
41
q41 =
0.61605 − 0.605
0.395
= 0.028
Question #51
Answer: E
1000 P60 = 1000 A60 / a60
b
gb
g
= 1000bq + p A g / b106
. +p a g
= c15 + b0.985gb382.79gh / c106
. + b0.985gb10.9041gh = 33.22
= 1000 v q60 + p60 A61 / 1 + p60 v a61
60
60
61
60
61
Question #52
Answer: E
Method 1:
In each round,
N = result of first roll, to see how many dice you will roll
X = result of for one of the N dice you roll
S = sum of X for the N dice
b g b g
b g b g
E b S g = E b N g * E b X g = 12.25
Var b S g = E b N gVar b X g + Var b N g E b X g
= b 35
. gb2.9167 g + b 2.9167 gb 35
.g
E X = E N = 35
.
Var X = Var N = 2.9167
2
2
= 45.938
Let S1000 = the sum of the winnings after 1000 rounds
b g
Stddevb S g = sqrt b1000 * 45.938g = 214.33
E S1000 = 1000 *12.25 = 12,250
1000
After 1000 rounds, you have your initial 15,000, less payments of 12,500, plus winnings
of S1000 .
Since actual possible outcomes are discrete, the solution tests for continuous outcomes
greater than 15000-0.5. In this problem, that continuity correction has negligible impact.
b
g
Pr 15000 − 12500 + S1000 > 14999.5 =
cb
g
b
g
h
= Pr S1000 − 12250 / 214.33 > 14999.5 − 2500 − 12250 / 214.33 =
b g
= 1 − Φ 117
. = 012
.
Method 2
Realize that you are going to determine N 1000 times and roll the sum of those 1000
N’s dice, adding the numbers showing.
Let N1000 = sum of those N’s
b
g
b g b gb g
E N1000 = 1000 E N = 1000 35
. = 3500
b
g
Var N1000 = 1000Var ( N ) = 2916.7
b g b g b g b gb g
Var b S g = E b N g Var b X g + Var b N g E b X g
= b3500gb2.9167g + b2916.7gb35
. g = 45.938
E S1000 = E N1000 E X = 3500 35
. = 12.250
2
1000
1000
1000
2
b g
Stddev S1000 = 214.33
Now that you have the mean and standard deviation of S1000 (same values as method
1), use the normal approximation as shown with method 1.
Question #53
Answer: B
FG
H
IJ
K
0.25 = ba + bg × 0.25 ⇒ a + b = 1
F bI
F bI
01875
.
= G a + J × 0.25 ⇒ G 1 − J × 0.25 = 01875
.
H 2K
H 2K
pk = a +
b
pk −1
k
b = 0.5
a = 0.5
FG
H
p3 = 0.5 +
IJ
K
0.5
× 01875
.
= 0125
.
3
Question #54
Answer: B
Transform these scenarios into a four-state Markov chain, where the final disposition of
rates in any scenario is that they decrease, rather than if rates increase, as what is
given.
0
from year t – 3
to year t – 2
Decrease
from year t – 2
to year t – 1
Decrease
Probability that year t will
decrease from year t - 1
0.8
1
Increase
Decrease
0.6
2
Decrease
Increase
0.75
3
Increase
Increase
0.9
State
LM0.80
0.60
Transition matrix is M
MM0.00
N0.00
0.00
0.00
0.75
0.90
0.20
0.40
0.00
0.00
OP
PP
PQ
0.00
0.00
0.25
010
.
P002 + P012 = 0.8 * 0.8 + 0.2 * 0.75 = 0.79
For this problem, you don’t need the full transition matrix. There are two cases to
consider. Case 1: decrease in 2003, then decrease in 2004; Case 2: increase in 2003,
then decrease in 2004.
For Case 1: decrease in 2003 (following 2 decreases) is 0.8; decrease in 2004
(following 2 decreases is 0.8. Prob(both) = 0.8 × 0.8 = 0.64
For Case 2: increase in 2003 (following 2 decreases) is 0.2; decrease in 2004 (following
a decrease, then increase) is 0.75. Prob(both) = 0.2 × 0.75 = 0.15
Combined probability of Case 1 and Case 2 is 0.64 + 0.15 = 0.79
Question #55
Answer: B
lx = ω − x = 105 − x
⇒t P45 = l45+ t / l45 = 60 − t / 60
Let K be the curtate future lifetime of (45). Then the sum of the payments is 0 if K ≤ 19
and is K – 19 if K ≥ 20 .
F 60 − K IJ × 1
60 K
b40 + 39+...+1g = b40gb41g = 13.66
=
60
2b60g
20 a45 =
∑ 1 × GH
60
K = 20
Hence,
c
h
c
Prob K − 19 > 13.66 = Prob K > 32.66
b g
= ProbbT ≥ 33g
= Prob K ≥ 33 since K is an integer
= 33p45 =
= 0.450
l78 27
=
l45 60
h
Question #56
Answer: C
Ax =
2
Ax =
d IAi
∞
E
0s x
zd
∞
0
= 0.4
µ +δ
x
= 0.25 → µ = 0.04
µ + 2δ
µ
z
=
µ
=
z
∞
0 s
Ax ds
Ax ds
ib g
e −0.1s 0.4 ds
F −e I
= b0.4 gG
H 01. JK
−0.1s
∞
=
0
0.4
=4
01
.
Alternatively, using a more fundamental formula but requiring more difficult integration.
c IA h
x
=
=
z
z
∞
0
∞
0
bg
b0.04g e
t t px µ x t e − δ t dt
t e −0.04 t
= 0.04
z
∞
0
−0.06 t
dt
t e −0.1t dt
(integration by parts, not shown)
∞
1
−t
e −0.1 t
= 0.04
−
0
01
. 0.01
0.04
=
=4
0.01
FG
H
IJ
K
Question #57
Answer: E
Subscripts A and B here just distinguish between the tools and do not represent ages.
ο
We have to find e AB
ο
eA =
ο
eB =
ο
FG 1 − t IJ dt = t − t
H 10K
20
z
z FGH
z FGH
e AB =
10
0
7
2 10
1−
0
7
ο
IJ
K
t
t2
dt = t −
7
14
1−
t
7
7
0
= 49 −
0
49
= 35
.
14
IJ FG 1 − t IJ dt = z FG 1 − t − t + t IJ dt
K H 10K
H 10 7 70K
7
0
t2 t2
t3
=t− − +
20 14 210
7
0
= 7−
49 49 343
−
+
= 2.683
20 14 210
ο
ο
ο
= 10 − 5 = 5
ο
e AB = e A + e B − e AB
= 5 + 35
. − 2.683 = 5817
.
2
Question #58
Answer: A
bg
µ bxτ g t = 0100
.
+ 0.004 = 0104
.
t
pxbτ g = e −0.104 t
Actuarial present value (APV) = APV for cause 1 + APV for cause 2.
z
b
5
z
g
b
5
g
2000 e −0.04 t e −0.104 t 0100
.
dt + 500,000 e −0.04 t e −0.104 t 0.400 dt
0
c b g
0
b
= 2000 010
. + 500,000 0.004
=
ghz e
5 −0.144 t
0
dt
2200
1 − e −0.144b5g = 7841
.
0144
e
j
Question #59
Answer: A
R = 1 − px = q x
S = 1 − px
× eb − k g since
d
e z
−
1
0
bg i
µ x t + k dt
=e z
=e z
1
bg
z
bg
e z
1
− µ x t dt − k dt
0
1
0
1
− µ x t dt − k dt
0
So S = 0.75R ⇒ 1 − px × e − k = 0.75q x
1 − 0.75q x
px
1 − qx
px
ek =
=
1 − 0.75q x 1 − 0.75q x
e− k =
k = ln
LM 1 − q OP
N1 − 0.75q Q
x
x
0
Question #60
Answer: E
b g
β = mean = 4;
pk = β k / 1 + β
b
P N =n
n
g
k +1
x
f b1g x
bg
f b2g x
bg
f b 3g x
bg
0
0.2
0
0
0
0
1
0.16
1
0.25
0
0
2
0.128
2
0.25
0.0625
0
3
0.1024
3
0.25
0.125
0.0156
bg
f b g b x g = f b x g; the claim amount distribution for a single claim
f b g b x g = ∑ e f b g b j gj x f b x − j g
f b k g x = probability that, given exactly k claims occur, that the aggregate amount is x.
1
x
k
k −1
j =0
bg
b
bg
possible claim size 1, f b g b x g = 0 for k > x
f b0g = 0.2
f b1g = 016
. * 0.25 = 0.04
f b2g = 016
. * 0.25 + 0128
. * 0.0625 = 0.048
f b3g = 016
. * 0.25 + 0128
. * 0125
.
+ 01024
.
* 0.0156 = 0.0576
F b3g = 0.2 + 0.04 + 0.048 + 0.0576 = 0.346
g
f s x = ∑ P N = k × f b k g x ; upper limit of sum is really ∞ , but here with smallest
x
k =0
k
s
s
s
s
s
Question #61
Answer: E
Let L = incurred losses; P = earned premium = 800,000
L
Bonus = 015
. × 0.60 −
× P if positive
P
= 015
. × 0.60 P − L if positive
= 015
. × 480,000 − L if positive
FG
H
IJ
K
g
b
b
g
= 015
. × c480,000 − b L ∧ 480,000gh
E (Bonus) = 0.15 (480,000– E c L ∧ 480,000gh
From Appendix A.2.3.1
= 0.15{480,000 – [500,000 × (1 – (500,000 / (480,000+500,000)))]}
= 35,265
Question #62
Answer: D
A281:2 =
z
3V
b g
d
i
FG IJ
H K
1
.
= 0.05827
1 − e −2δ = 0.02622 since δ = ln 106
72δ
71
= 19303
.
= 1+ v
72
=
a28:2
2 −δ t
e 1 72 dt
0
= 500,000 A281:2 − 6643 a28:2
= 287
Question #63
Answer: D
bg
Let Ax and a x be calculated with µ x t and δ = 0.06
bg
Let Ax and a x be the corresponding values with µ x t
increased by 0.03 and δ decreased by 0.03
*
*
ax =
1 − Ax
δ
ax = ax
=
0.4
= 6.667
0.06
*
LM Proof: a
N
*
x
z
z
z
=
=
=
∞
0
e z
z
e
∞
0
d
e z
t
−
0
bg
i
µ x s + 0.03 ds −0.03t
bg
t
− µ x s ds −0.03t −0.03t
0
t
bg
e
e
∞ − 0 µ x s ds −0.06 t
0
e
e
dt
dt
dt
= ax
Ax* = 1 − 0.03 a x * = 1 − 0.03 a x
b gb
= 1 − 0.03 6.667
g
= 0.8
Question #64
Answer: A
bulb ages
Year
0
1
0
1
2
3
10000
0
1000
9000
100+2700
900
280+270+3150
2
3
0
0
6300
0
0
0
The diagonals represent bulbs that don’t burn out.
E.g., of the initial 10,000, (10,000) (1-0.1) = 9000 reach year 1.
(9000) (1-0.3) = 6300 of those reach year 2.
Replacement bulbs are new, so they start at age 0.
At the end of year 1, that’s (10,000) (0.1) = 1000
At the end of 2, it’s (9000) (0.3) + (1000) (0.1) = 2700 + 100
At the end of 3, it’s (2800) (0.1) + (900) (0.3) + (6300) (0.5) = 3700
1000 2800 3700
+
+
.
105
105
. 2 105
. 3
= 6688
Actuarial present value =
#
replaced
1000
2800
3700
Question #65
Key: E
Model Solution:
e25:25 =
=
z
z
15
0 t
p25dt +15 p25
15 −.04 t
0
e
10
0t
p40 dt
FG z IJ e
H
Kz
i + e LMN.051 d1 − e
dt + e
d
z
−
15
0
10 −.05t
.04 ds
0
1
− .60
1 − e−.60
.04
.
= 112797
+ 4.3187
= 15.60
=
dt
−.50
iOPQ
Question #66
Key: C
Model Solution:
5
p 60 +1 =
e1 − q je1 − q jb1 − q gb1 − q gb1 − q g
= b0.89gb0.87gb0.85gb0.84gb0.83g
60 +1
60 + 2
63
64
65
= 0.4589
Question # 67
Key: E
Model Solution:
12.50 = a x =
Ax =
µ
1
⇒ µ + δ = 0.08 ⇒ µ = δ = 0.04
µ +δ
= 0.5
µ +δ
µ
1
2
Ax =
=
µ + 2δ 3
e j
Var aT =
2
Ax − Ax2
δ2
1
−
1
= 3 4 = 52.083
0.0016
S.D. = 52.083 = 7.217
Question # 68
Key: D
Model Solution:
v = 0.90 ⇒ d = 010
.
Ax = 1 − da x = 1 − 010
. 5 = 0.5
b gb g
Benefit premium π =
=
5000 Ax − 5000vqx
ax
b5000gb0.5g − 5000b0.90gb0.05g = 455
5
a x +10
ax
a
0.2 = 1 − x +10 ⇒ a x +10 = 4
5
10Vx
= 1−
b gb g
−πa
= b5000gb0.6g − b455gb4g = 1180
Ax +10 = 1 − da x +10 = 1 − 010
.
4 = 0.6
10V
= 5000 Ax +10
x +10
Question #68
Key: D
Model Solution:
v is the lowest premium to ensure a zero % chance of loss in year 1 (The present value of the
payment upon death is v, so you must collect at least v to avoid a loss should death occur).
Thus v = 0.95.
bg
b g
2
E Z = vqx + v 2 px qx +1 = 0.95 × 0.25 + 0.95 × 0.75 × 0.2
= 0.3729
b g
d i
b g
2
4
E Z2 = v 2qx + v 4 px qx +1 = 0.95 × 0.25 + 0.95 × 0.75 × 0.2
= 0.3478
b g d i c b gh
Var Z = E Z2 − E Z
2
b
g
2
= 0.3478 − 0.3729 = 0.21
Question # 70
Key: D
Model Solution:
Severity
after increase
Severity
after increase and
deductible
0
20
80
200
60
120
180
300
Expected payment per loss = 0.25 × 0 + 0.25 × 20 + 0.25 × 80 + 0.25 × 200
= 75
Expected payments = Expected number of losses × Expected payment per loss
= 75 × 300
= 22,500
Question # 71
Key: A
Model Solution:
E (S) = E (N) E (X) = 50 × 200 = 10,000
2
Var S = E N Var X + E X Var N
bg b g b g b g b g
= b50gb400g + d200 ib100g
2
= 4,020,000
b
g
FG 8,000 − 10,000 IJ
4,020,000 K
H
= Prb Z < −0.998g ≅ 16%
Pr S < 8,000 = Pr Z <
Question #72
Key: A
Model Solution:
Let Z be the present value random variable for one life.
Let S be the present value random variable for the 100 lives.
bg
E Z = 10
= 10
z
∞ δt µ t
e e µ dt
5
µ
δ +µ
e − bδ + µ g 5
= 2.426
FG µ IJ e b g
H 2δ + µ K
F 0.04IJ de i = 11233
= 10 G
.
H 0.16K
Var b Z g = E d Z i − c E b Z gh
d i
− 2δ + µ 5
E Z 2 = 102
−0.8
2
2
2
= 11233
− 2.4262
.
= 5.348
bg
bg
VarbSg = 100 Varb Zg = 534.8
E S = 100 E Z = 242.6
F − 242.6
= 1645
.
→ F = 281
534.8
Question #73
Key: D
Model Solution:
Prob{only 1 survives} = 1-Prob{both survive}-Prob{neither survives}
b
ge
= 1− 3p50 × 3p 50 − 1− 3p50 1− 3p 50
b
gb
gb
gb
j
gb
g b
gb
gb
= 1 − 0.9713 0.9698 0.9682 0.9849 0.9819 0.9682 − 1 − 0.912012 1 − 0.93632
= 0.912012
g
0.936320
= 0.140461
Question # 74
Key: C
Model Solution:
The tyrannosaur dies at the end of the first day if it eats no scientists that day. It dies at the end
of the second day if it eats exactly one the first day and none the second day. If it does not die
by the end of the second day, it will have at least 10,000 calories then, and will survive beyond
2.5.
b g bg b g
b gb g
Prob (ruin) = f 0 + f 1 f 0
= 0.368 + 0.368 0.368
= 0.503
e −110
= 0.368
0!
e −111
f 1 =
= 0.368
1!
bg
bg
since f 0 =
Question #75
Key: B
Model Solution:
Let X = expected scientists eaten.
For each period, E X = E X dead × Prob already dead + E X alive × Prob alive
b
b g
g
b g
= 0 × Prob dead + E X alive × Prob alive
Day 1, E X1 = 1
e −1 01
Prob dead at end of day 1 = f 0 =
= 0.368
0!
Day 2, E X 2 = 0 × 0.368 + 1 × 1 − 0.368 = 0.632
Prob (dead at end of day 2) = 0.503
[per problem 10]
b
g bg
b
g
b g
b
g
Day 2.5, E X 2.5 = 0 × 0.503 + 0.5 × 1 − 0.503 = 0.249
where E X 2.5 alive = 0.5 since only
1
2
day in period.
E X = E X1 + E X 2 + E X 2.5 = 1 + 0.632 + 0.249 = 1881
.
E 10,000 X = 18,810
Question # 76
Key: C
Model Solution:
This solution applies the equivalence principle to each life. Applying the equivalence principle
to the 100 life group just multiplies both sides of the first equation by 100, producing the same
result for P.
b
b
g
g
b10gb0.03318g + b10gb1 − 0.03318gb0.03626g + Pb1 − 0.03318gb1 − 0.03626g
P=
APV Prems = P = APV Benefits = 10q70 v + 10 p70q71v 2 + Pp70 p71v 2
.
108
. 2
108
= 0.3072 + 0.3006 + 0.7988 P
0.6078
P=
= 3.02
0.2012
. 2
108
(APV above means Actuarial Present Value).
Question #77
Key: E
Model Solution:
One approach is to recognize an interpretation of formula 7.4.11 or exercise 7.17a:
Level benefit premiums can be split into two pieces: one piece to provide term
insurance
for n years; one to fund the reserve for those who survive.
If you think along those lines, you can derive formula 7.4.11:
Px = Px1:n + Px:n1 nVx
And plug in to get
b
gb
g
0.090 = Px1:n + 0.00864 0.563
Px1:n = 0.0851
Another approach is to think in terms of retrospective reserves. Here is one such solution:
nVx
e
j
a
= eP − P j
E
= Px − Px1:n sx:n
x:n
1
x
x:n
e
= Px − Px1:n
n
x
jP
a x:n
1
x:n
a x:n
eP − P j
=
eP j
1
x
x:n
1
x:n
e
j
0.563 = 0.090 − Px1:n / 0.00864
b
gb
g
Px1:n = 0.090 − 0.00864 0.563
= 0.0851
Question #78
Key: A
Model Solution:
b g
δ = ln 1.05 = 0.04879
Ax =
=
=
z
z
ω −x
bg
px µ x t e −δt dt
0
t
ω −x
1
e −δt dt for DeMoivre
ω−x
0
1
a
ω − x ω −x
From here, many formulas for
10 V
c A h could be used.
40
One approach is:
Since
A50 =
A40 =
a50
50
a60
60
FG
H
F1− A
=G
H δ
IJ
K
IJ = 1387
K .
=
18.71
1 − A50
= 12.83
= 0.3742 so a50 =
50
δ
=
19.40
= 0.3233 so a40
60
40
.3233
= 0.02331
c h 01387
.
V c A h = A − P c A ha = 0.3742 − b0.02331gb12.83g = 0.0751.
so P A40 =
10
40
50
40
50
Question #79
Key: D
Model Solution:
b g
bg
F 0.03 IJ × 0.70 + FG 0.6 IJ × 0.30
=G
H 0.03 + 0.08 K
H 0.06 + 0.08 K
Ax = E v T b x g = E v T b x g NS × Prob NS + E v T b x g S × Prob S
= 0.3195
Similarly, 2 Ax =
F
H
Var aT
I
b gK =
2
FG 0.03 IJ × 0.70 + FG 0.06 IJ × 0.30 = 01923
.
.
H 0.03 + 0.16K
H 0.06 + 0.16K
Ax − Ax2
x
δ2
=
01923
.
− 0.31952
..
= 141
0.082
Question #80
Key: B
Model Solution:
Let S denote aggregate losses before deductible.
E S = 2 × 2 = 4, since mean severity is 2.
e −2 20
.
, since must have 0 number to get aggregate losses = 0.
fS 0 =
= 01353
0!
bg
F e 2 I FG 1IJ = 0.0902, since must have 1 loss whose size is 1 to get aggregate losses = 1.
f b1g = G
H 1! JK H 3K
E b S ∧ 2g = 0 × f b0g + 1 × f + 2 × c1 − f b0g − f b1gh
.
.
= 0 × 01353
+ 1 × 0.0902 + 2 × b1 − 01353
− 0.0902g
−2
S
S
S
S
S
.
= 16392
b
E S −2
g
+
= E S − E S ∧2
.
= 4 − 16392
= 2.3608
Question #81
Key: D
Model Solution:
Poisson processes are separable. The aggregate claims process is therefore equivalent to two
independent processes, one for Type I claims with expected frequency
one for Type II claims.
Let
S I = aggregate Type I claims.
N I = number of Type I claims.
X I = severity of a Type I claim (here = 10).
b g
b g
Since X I = 10, a constant , E X I = 10; Var X I = 0.
b g b g b g b g b g
= b1000gb0g + b1000gb10g
Var S I = E N I Var X I + Var N I E X I
2
= 100,000
2
FG 1IJ b3000g = 1000 and
H 3K
bg
b g
b g
2,100,000 = 100,000 + Var b S g
Var b S g = 2 ,000,000
Var S = Var S I + Var S II since independent
II
II
Question #82
Key: A
Model Solution:
5
bτ g = p′b1g p′b2g
p50
5 50 5 50
FG 100 − 55IJ e b gb g
H 100 − 50K
= b0.9gb0.7788g = 0.7009
− 0.05 5
=
Similarly
10
FG 100 − 60IJ e b g b g
H 100 − 50 K
= b0.8gb0.6065g = 0.4852
bτ g =
p50
− 0.05 10
bτ g = pbτ g − pbτ g = 0.7009 − 0.4852
5 5 q50
5 50
10 50
= 0.2157
Question #83
Key: C
Model Solution:
Only decrement 1 operates before t = 0.7
b g = b0.7g q′b1g = b0.7gb010
. g = 0.07
′1
0.7 q40
40
Probability of reaching t = 0.7
since UDD
is
1-0.07 = 0.93
Decrement 2 operates only at t = 0.7, eliminating 0.125 of those who reached 0.7
b gb
g
b2g = 0.93 0125
q40
.
= 011625
.
Question #84
Key: C
Model Solution:
e
j
π 1+ 2 p80v 2 = 1000 A80 +
FG
H
π 1+
0.83910
1.062
πvq80 πv 3 2 p80q82
+
2
2
IJ = 665.75 + πFG 0.08030 + 0.83910 × 0.09561IJ
K
. g
2b1.06g
H 2b106
K
3
b
g
b
πb167524
.
g = 665.75
g
π 174680
.
= 665.75 + π 0.07156
π = 397.41
3,284 ,542
= 0.83910
3,914 ,365
Where 2 p80 =
b
gb
g
Or 2 p80 = 1 − 0.08030 1 − 0.08764 = 0.83910
Question #85
Key: E
Model Solution:
At issue, actuarial present value (APV) of benefits
=
z
z
∞
0
bt v t t p65 µ
∞
65
bt gdt
d id
bg
i
= 1000 e0.04 t e −0.04 t t p65 µ 65 t dt
0
= 1000
z
∞
p
0 t 65
bg
µ 65 t dt = 1000 ∞ q65 = 1000
APV of premiums = π a65 = π
FG 1 IJ = 16.667π
H 0.04 + 0.02K
Benefit premium π = 1000 / 16.667 = 60
2V
=
=
z
z
∞
0
∞
0
b g
b2+u v u u p67 µ 65 2 + u du − π a67
b g b gb
g
1000 e0.04b 2+uge −0.04u u p67 µ 65 2 + u du − 60 16.667
= 1000e0.08
z
∞
0 u
b g
p67 µ 65 2 + u du − 1000
= 1083.29 ∞ q67 − 1000 = 1083.29 − 1000 = 83.29
Question #86
Key: B
Model Solution:
(1)
a x:20 = a x:20 − 1+ 20 Ex
(2)
a x:20 =
(3)
Ax:20 =
(4)
Ax = A1x:20 + 20 Ex Ax +20
1 − Ax:20
d
A1x:20
+ Ax:201
b gb g
0.28 = A1x:20 + 0.25 0.40
A1x:20 = 018
.
Now plug into (3):
Now plug into (2):
Now plug into (1):
Ax:20 = 018
. + 0.25 = 0.43
a x:20 =
b
1 − 0.43
= 1197
.
0.05 / 105
.
g
a x:20 = 1197
. − 1 + 0.25 = 1122
.
Question #87
Key: A
Model Solution:
E N = EΛ E N Λ = EΛ Λ = 2
Var N = E Λ Var N Λ + VarΛ E N Λ
= E Λ Λ + VarΛ Λ = 2 + 4 = 6
Distribution is negative binomial.
b g
rβ = 2 and rβ 1 + β = 6
1+ β = 3
β=2
rβ = 2 r = 1
Per supplied tables:
p1 =
rβ 1
b g
1! 1+ β
r +1
=
b1gb2g = 0.22
b1gb3g
2
Alternatively, if you don’t recognize that N will have a negative binomial distribution, derive
gamma density from moments (hoping α is an integer).
Mean = θα = 2
Var = E Λ2 − E Λ
2
e
j
= θ2 α 2 + α − θ2α 2
= θ2α = 4
θ2α 4
= =2
θα 2
θα
=1
α=
θ
θ=
p1 =
z
∞
0
c p λ h f bλ gdλ = z
1
∞
0
e− λ λ1
1!
z
bλ / 2ge b
λ Γ b1g
− λ /2
g
dλ
α
1 ∞ −2
=
λe dλ
2 0
Integrate by parts; not shown
F
GH
3
3
1 2 − 2λ 4 − 2λ
=
− λe
− e
2 3
9
=
2
= 0.22
9
I
JK
∞
0
Question #88
Key: C
Model Solution:
Limited expected value =
z c1 − Fbxghdx = z d0.8e
1000
1000
0
0
−0.02 x
i
d
+ 0.2e−0.001x dx = −40e−0.02 x − 200e−0.001x
i
1000
0
= 40 + 126.4
= 166.4
Question #89
Key: E
Model Solution:
M = Initial state matrix = 1 0 0 0
LM0.20
0.50
T = One year transition matrix = M
MM0.75
N1.00
0.80
0
0
0.50
0
0
0
0
OP
P
0.25P
P
0 Q
0
0
M × T = 0.20 0.80 0 0
.
0.40 0
b M × T g × T = 0.44 016
.
cb M × T g × T h × T = 0.468 0.352 0.08 010
Probability of being in state F after three years = 0.468.
d
ib g
Actuarial present value = 0.468v 3 500 = 171
Notes:
1. Only the first entry of the last matrix need be calculated (verifying that the four sum
to 1 is useful “quality control.”)
2. Compare this with solution 23. It would be valid to calculate T 3 here, but advancing
M one year at a time seems easier.
Question #90
Key: B
Model Solution:
Let Yi be the number of claims in the ith envelope.
b g
Let X 13 be the aggregate number of claims received in 13 weeks.
b
g b
g b g b g
E Y = b1 × 0.2g + b4 × 0.25g + b9 × 0.4g + b16 × 015
. g = 7.2
E X b13g = 50 × 13 × 2.5 = 1625
Var X b13g = 50 × 13 × 7.2 = 4680
. g
ProbmXb13g ≤ Zr = 0.90 = Φb1282
R X b13g − 1625 ≤ 1282
U
⇒ Prob S
. V
T 4680
W
X b13g ≤ 1712.7
E Yi = 1 × 0.2 + 2 × 0.25 + 3 × 0.4 + 4 × 015
. = 2.5
2
i
b g
Note: The formula for Var X 13 took advantage of the frequency’s being Poisson.
The more general formula for the variance of a compound distribution,
2
Var S = E N Var X + Var N E X , would give the same result.
bg b g b g
b gb g
Question #91
Key: E
Model Solution:
b g ω −1 60 = 75 −1 60 = 151
b60g = ω ′ −1 60 = 151 × 53 = 251 ⇒ ω ′ = 85
µ M 60 =
µF
t
10
t
= 1−
25
t
M
p65
= 1−
t
F
p60
Let x denote the male and y denote the female.
ex
ey
e xy
cmean for uniform distribution over b0,10gh
= 12 .5 c mean for uniform distribution over b0,25gh
FG1 − t IJ FG1 − t IJ ⋅ dt
=
H 10K H 25K
F1 − 7 t + t I ⋅ dt
=
GH 50 250JK
F 7 t + t I = 10 − 7 × 100 + 1000
= Gt −
100
750
H 100 750JK
=5
z
z
10
0
2
10
0
3
2
10
0
= 10 − 7 +
exy = ex + e y − exy = 5 +
25 13 30 + 75 − 26
− =
= 1317
.
2
3
6
Question #92
Key: B
Model Solution:
1
µ
=
µ+δ 3
1
µ
2
Ax =
=
µ + 2δ 5
Ax =
c h
P Ax = µ = 0.04
F Pc A hI A − A
Var b Lg = G 1 +
j
H δ JK e
F 0.04 IJ FG 1 − FG 1IJ IJ
= G1 +
H 0.08 K H 5 H 3K K
F 3I F 4 I
=G J G J
H 2 K H 45K
2
x
2
2
x
x
2
2
=
1
5
4 13
=
3 3
2
Question #93
Key: B
Model Solution:
Mean excess loss =
=
b g b
b g b
g
b g
E X − E X ∧ 100
1 − F 100
331 − 91
= 300
0.8
g
b g
E X = E X ∧ 1000 since F 1000 = 10
.
Question #94
Key: E
Model Solution:
b
g
Expected insurance benefits per factory = E X − 1 +
= 0.2 × 1 + 01
. × 2 = 0.4.
Insurance premium = (1.1) (2 factories) (0.4 per factory) = 0.88.
Let R = retained major repair costs, then
bg
f b1g = 2 × 0.4 × 0.6 = 0.48
f b2g = 0.6 = 0.36
f R 0 = 0.42 = 016
.
R
2
R
b gb g
. 3 , if positive
Dividend = 3 − 0.88 − R − 015
= 167
. − R , if positive
b
g b gb
g b gb
g b gb g
E Dividend = 016
. 167
. − 0 + 0.48 167
. − 1 + 0.36 0 = 0.5888
b
g
[The (0.36)(0) term in the last line represents that with probability 0.36, 167
. − R is
negative so the dividend is 0.]
Question #95
Key: A
Model Solution:
E X =
αθ
4α
=
= 8 ⇒ 4α = 8α − 8
α −1 α −1
α =2
bg
F 6 =1−
FG θ IJ
H 6K
α
=1−
FG 4 IJ
H 6K
2
= 0.555
bg
bg
s 6 = 1 − F 6 = 0.444
Question #96
Key: B
Model Solution:
ex = px + 2 px + 3 px + ... = 1105
.
b g
Annuity = v 3 3 p x 1000 + v 4 4 p x × 1000 × 104
. + ...
=
∞
. g
∑1000b104
k −3 k
v
k px
k =3
= 1000v 3
∞
∑ k px
k =3
b
g
FG 1 IJ
H 104
. K
= 1000v 3 ex − 0.99 − 0.98 = 1000
Let π = benefit premium.
e
j
π 1 + 0.99v + 0.98v 2 = 8072
2.8580π = 8072
π = 2824
3
× 9.08 = 8072
Question #97
Key B
Model Solution:
b g
b ge
1
π a30:10 = 1000 A30 + P IA 30
+ 10π
:10
π=
10
A30
j
1000 A30
b g
a30:10 − IA
=
1
30:10
− 1010 A30
b
g
b
1000 0102
.
7.747 − 0.078 − 10 0.088
g
102
6.789
= 15.024
=
Test Question: 98
Key: E
For de Moivre’s law,
FG1 − t IJ dt
H ω − 30K
L t OP
= Mt −
N 2bω − 30g Q
e 30 =
z
ω − 30
0
2
ω − 30
0
=
ω − 30
2
100 − 30
= 35
2
Prior to medical breakthrough
ω = 100 ⇒ e30 =
After medical breakthrough
e ′ 30 = e 30 + 4 = 39
so
e30
′ = 39 =
ω ′ − 30
2
⇒ ω ′ = 108
Test Question: 99
0L
= 100,000v 2.5 − 4000a3
Key: A
@5%
= 77,079
Test Question: 100
Key: C
Ε Ν = ΕΛ Ε Ν Λ = ΕΛ Λ = 2
Var Ν = Ε Λ Var Ν Λ + VarΛ E N Λ
= Ε Λ Λ + VarΛ Λ = 2 + 2 = 4
Distribution is negative binomial (Loss Models, 3.3.2)
Per supplied tables
mean = rβ = 2
b g
b1 + β g = 2
Var = rβ 1 + β = 4
β =1
rβ = 2
r=2
From tables
2 3 4 13
r r +1 r + 2 β3
4
.
=
=
= 0125
p3 =
5
r +3
32
3! 2
3! 1 + β
1000 p3 = 125
b gb g
b g
b gb gb g
Test Question: 101
Key: E
b gb g
b gb g b gb g b gb g
= b0.6gb1g + b0.2gb25g + b0.2gb100g = 25.6
E N = Var N = 60 0.5 = 30
E X = 0.6 1 + 0.2 5 + 0.2 10 = 3.6
E X2
Var X = 25.6 − 3.62 = 12.64
For any compound distribution, per Loss Models
c
Var S = E N Var X + Var N E X
d i
h
2
= (30) (12.64) + (30) 3.62
= 768
For specifically Compound Poisson, per Probability Models
Var S = λt E X 2 = (60) (0.5) (25.6) = 768
Alternatively, consider this as 3 Compound Poisson processes (coins worth 1; worth 5;
2
worth 10), where for each Var X = 0 , thus for each Var S = Var N E X .
Processes are independent, so total Var is
b g
bg b g
Var = b60gb0.5gb0.6g1 + b60gb0.5gb0.2g5 + b60gb0.5gb0.2gb10g
2
2
= 768
Test Question: 102
1000
20
20Vx
= 1000 Ax +20 =
=
a x +20 =
Key: D
1000
d
20
19Vx + 20 Px
. g − q b1000g
ib106
x +19
px +19
. g − 0.01254b1000g
b342.03 + 13.72gb106
= 36918
.
0.98746
1 − 0.36918
.
= 111445
.
0.06 / 106
b
so 1000 Px +20 = 1000
g
Ax +20 36918
.
=
= 331
.
a x +20 111445
.
2
Test Question: 103
k
pxbτ g = e
−
z
k
0
µ bxτ g t dt
bg
Key: B
z b gb g
F b g b g IJ
= Ge z
H
K
= b p g where
=e
−
k
0
2 µ x1 t dt
k
−
0
µ x1 t dt
2
2
k
10
is from Illustrative Life Table, since µ b1g follows I.L.T.
6,616,155
= 0.80802
8,188,074
6,396,609
=
= 0.78121
8,188,074
p60 =
11 p60
10
k px
x
b τ g = p b τ g − p bτ g
q60
10 60 11 60
=
b
10
p60
g −b
2
11 p60
g
2
from I.L.T.
= 0.80802 2 − 0.781212 = 0.0426
Test Question: 104
Ps =
1
− d , where s can stand for any of the statuses under consideration.
as
as =
1
Ps + d
1
= 6.25
. + 0.06
01
1
=
= 8.333
0.06 + 0.06
ax = a y =
a xy
Key: C
a xy + a xy = a x + a y
a xy = 6.25 + 6.25 − 8.333 = 4.167
Pxy =
1
− 0.06 = 018
.
4.167
Test Question: 105
z
Key: A
b
g
g j = 48
d 0bτ g = 1000 e − b µ + 0.04 gt µ + 0.04 dt
1
0
= 1000 1 − e − b µ + 0.04
e
e − b µ + 0.04 g = 0.952
b
µ + 0.04 = − ln 0.952
g
= 0.049
µ = 0.009
z
b
g
d 3b1g = 1000 e −0.049 t 0.009 dt
4
3
= 1000
0.009 − b 0.049 gb 3g − b 0.049 gb 4 g
e
−e
= 7.6
0.049
e
j
Test Question: 106
Key: B
bg
This is a graph of lx µ x .
bg
b
g
µ x would be increasing in the interval 80,100 .
The graphs of lx px , lx and lx2 would be decreasing everywhere.
The graph shown is comparable to Figure 3.3.2 on page 65 of Actuarial Mathematics
Test Question: 107
Key: A
Using the conditional mean and variance formulas:
c h
E N = EΛ N Λ
d c hj
d c hj
Var N = VarΛ E N Λ + E Λ Var N Λ
Since N, given lambda, is just a Poisson distribution, this simplifies to:
bg
bg
E N = EΛ Λ
Var N = VarΛ Λ + E Λ Λ
bg
bg
We are given that E N = 0.2 and Var N = 0.4 , subtraction gives Var Λ = 0.2
Test Question: 108
Key: B
N = number of salmon
X = eggs from one salmon
S = total eggs.
E(N) = 100t
Var(N) = 900t
bg b gb g
Var b S g = E b N gVar b X g + E b X gVar b N g = 100t ⋅ 5 + 25 ⋅ 900t = 23,000t
F S − 500t > 10,000 − 500t IJ =.95 ⇒
Pb S > 10,000g = PG
H 23,000t 23,000t K
E S = E N E X = 500t
2
10,000 − 500t = −1645
.
⋅ 23000 t = −250 t
40 − 2t = − t
d ti
2
2
− t − 40 = 0
1 ± 1 + 320
= 4.73
4
t = 22.4
round up to 23
t=
Test Question: 109
Key: A
2
A P V (x’s benefits) = ∑ v k +1bk +1 k px q x + k
k =0
b g
b gb g
b gb gb g
= 1000 300v 0.02 + 350v 2 0.98 0.04 + 400v 3 0.98 0.96 0.06
= 36,829
Test Question: 110
Key: E
π denotes benefit premium
= APV future benefits - APV future premiums
1
0.6 =
− π ⇒ π = 0.326
108
.
. − q65 10
10V + π 108
11V =
p65
19 V
b
=
gb g b gb g
. g − b010
. gb10g
b5.0 + 0.326gb108
1 − 010
.
=5.28
Test Question: 111
Key: C
X = losses on one life
E X = 0.3 1 + 0.2 2 + 01
. 3
b gb g b gb g b gb g
=1
S = total losses
E S = 3E X = 3
b g
c b gh
= E S − b1gc1 − f b0gh
= 3 − b1gd1 − 0.4 i
E S − 1 + = E S − 1 1 − Fs 0
s
3
= 3 − 0.936
= 2.064
Test Question: 112
Key: A
1180 = 70a30 + 50a40 − 20a30:40
b gb g b gb g
1180 = 70 12 + 50 10 − 20a30:40
a30:40 = 8
a30:40 = a30 + a40 − a30:40 = 12 + 10 − 8 = 14
100a30:40 = 1400
Test Question: 113
z
∞
bg
a = a¬ f t dt =
t
=
z
Key: B
∞ 1 − e −0.05t
1
0.05
0.05
zb
∞
te
−t
1
te − t dt
Γ2
bg
i
− te −1.05t dt
LM b g FG
IJ
H
K
N
1 L F 1 I O
=
.
M1 − G . JK PP = 185941
0.05 MN H 105
Q
1
1
t
=
− t + 1 e−t +
+
e −1.05t
2
0.05
105
.
105
.
2
20,000 × 185941
.
= 37,188
Test Question: 114
bg
Key: C
b g
LM OP b g
N Q
2
p k −1
k
2
= 0+ p k −1
k
pk =
Thus an (a, b, 0) distribution with a = 0, b = 2.
Thus Poisson with λ = 2 .
e −2 2 4
p4 =
4!
= 0.09
bg
OP
Q
∞
0
Test Question: 115
Key: B
By the memoryless property, the distribution of amounts paid in excess of 100 is still
exponential with mean 200.
With the deductible, the probability that the amount paid is 0 is
F 100 = 1 − e −100/ 200 = 0.393.
b g
Thus the average amount paid per loss is (0.393) (0) + (0.607) (200) = 121.4
The expected number of losses is (20) (0.8) = 16.
The expected amount paid is (16) (121.4) = 1942.
Test Question: 116
Key: D
Let M = the force of mortality of an individual drawn at random; and T = future lifetime
of the individual.
Pr T ≤ 1
n
= E Pr T ≤ 1 M
=
=
=
z
zz
zd
∞
0
0 0
0
bg
Pr T ≤ 1 M = µ f M µ dµ
2 1
2
s
1
2
µe − µ t dt dµ
1 − e−µ
= 0.56767
i 21 du = 21 d2 + e
−2
i 21 d1 + e i
−1 =
−2
Test Question: 117
Key: E
b gb g b gb g
E N = b0.8g1 + b0.2gb4g = 16
.
. − 12
. = 016
.
Var b N g = 16
E N = 0.8 1 + 0.2 2 = 12
.
2
2
E X = 70 + 100 = 170
bg
b
g
E S = E N E X = 12
. b170g = 204
VarbSg = E N Varb Xg + E X Varb N g = 12
. b78,100g + 170 b016
. g = 98,344
Std dev b S g = 98,344 = 313.6
Var X = E X 2 − E X
2
= 7000 + 100,000 − 1702 = 78,100
2
2
So B = 204 + 314 = 518
Test Question: 118
Key: D
Let π = benefit premium
Actuarial present value of benefits =
= 0.03 200,000 v + 0.97 0.06 150,000 v 2 + 0.97 0.94 0.09 100,000 v 3
b gb
g b gb gb
= 5660.38 + 7769.67 + 6890.08
= 20,32013
.
Actuarial present value of benefit premiums
= a x:3 π
b gb g
= 1 + 0.97v + 0.97 0.94 v 2 π
= 2.7266 π
20,32013
.
π =
= 7452.55
2.7266
1V
=
. g − b200,000gb0.03g
b7452.55gb106
1 − 0.03
= 1958.46
Initial reserve, year 2 = 1V + π
= 1958.56 + 7452.55
= 9411.01
Test Question: 119
Key: A
g b gb gb gb
g
Let π denote the premium.
b g
L = bT v T − π aT = 1+ i
T
× v T − π aT
= 1 − π aT
E L = 1 − π ax = 0
⇒π =
⇒ L = 1 − π aT = 1 −
=
b
g
aT
ax
=
1
ax
d
δ ax − 1 − v T
i
δ ax
v T − 1 − δ ax
v T − Ax
=
δ ax
1 − Ax
Test Question: 120
Key: D
(0, 1)
(1, 0.9)
(1.5, 0.8775)
(2, 0.885)
tp1
1
2
t
1
p1 = (1 − 01
. ) = 0.9
2
p1 = 0.9 1 − 0.05 = 0.855
b gb
g
b
g
since uniform, 1.5 p1 = 0.9 + 0.855 / 2
= 0.8775
e1:1.5 = Area between t = 0 and t = 15
.
=
FG 1 + 0.9 IJ b1g + FG 0.9 + 0.8775IJ b0.5g
H 2 K H 2 K
= 0.95 + 0.444
= 1394
.
Alternatively,
e11: .5 =
=
=
z
z
zb
1.5
0 t
p1dt
1
z
0.5
p1dt +1p1 x
t
0
0
1
0
g
p2 dx
1 − 01
. t dt + 0.9
= t − 0.12t
2
1
0
z
0.5
0
b1 − 0.05xgdx
+ 0.9 x − 0.052 x
0.5
2
0
= 0.95 + 0.444 = 1394
.
Test Question: 121
Key: A
b g
10,000 A63 112
. = 5233
A63 = 0.4672
Ax +1 =
A64 =
b g
Ax 1 + i − qx
px
. g − 0.01788
b0.4672gb105
1 − 0.01788
= 0.4813
A65 =
. g − 0.01952
b0.4813gb105
1 − 0.01952
= 0.4955
Single contract premium at 65 = (1.12) (10,000) (0.4955)
= 5550
b1 + ig
2
=
5550
5233
i=
5550
− 1 = 0.02984
5233
Test Question: 122
Key: B
Original Calculation (assuming independence):
µ x = 0.06
µ y = 0.06
µ xy = 0.06 + 0.06 = 012
.
µx
0.06
Ax =
=
= 0.54545
µ x + δ 0.06 + 0.05
µy
0.06
Ay =
=
= 0.54545
µ y + δ 0.06 + 0.05
µ xy
012
.
Axy =
=
= 0.70588
µ xy + δ 012
. + 0.05
Axy = Ax + Ay − Axy = 0.54545 + 0.54545 − 0.70588 = 0.38502
Revised Calculation (common shock model):
µ x = 0.06, µ Tx *b x g = 0.04
µ y = 0.06, µ Ty*b y g = 0.04
µ xy = µ Tx *b x g + µ Ty*b y g + µ Z + 0.04 + 0.04 + 0.02 = 010
.
µx
0.06
Ax =
=
= 0.54545
µ x + δ 0.06 + 0.05
µy
0.06
Ay =
=
= 0.54545
µ y +δ 0.06 + 0.05
µ xy
010
.
Axy =
=
= 0.66667
µ xy + δ 010
. + 0.05
Axy = Ax + Ay − Axy = 0.54545 + 0.54545 − 0.66667 = 0.42423
Difference = 0.42423 − 0.38502 = 0.03921
Test Question: 123
Key: E
Treat as three independent Poisson variables, corresponding to 1, 2 or 3 claimants.
rate1 = 6
rate 2 = 4
rate 3 = 2
LM= 1 × 12OP
N 2 Q
Var1 = 6
Var2 = 16
= 4 × 22
Var3 = 18
total Var = 6 + 16 + 18 = 40, since independent.
Alternatively,
d i
E X2 =
12 2 2 32 10
+
+
=
2
3 6
3
For compound Poisson, Var S = E N E X 2
b
= 12
Test Question: 124
z
3
0
bg
gFGH 103IJK = 40
Key: C
bg
λ t dt = 6 so N 3 is Poisson with λ = 6.
b
g g
P is Poisson with mean 3 (with mean 3 since Prob yi < 500 = 0.5
P and Q are independent, so the mean of P is 3, no matter what the value of Q is.
Test Question: 125
Key: A
At age x:
Actuarial Present value (APV) of future benefits =
APV of future premiums =
FG 4 a IJ π
H5 K
FG 1 A IJ 1000
H5 K
x
x
1000
4
A25 = π a25 by equivalence principle
5
5
1000 A25
1
8165
.
=π ⇒π = ×
= 1258
.
4 a25
4 16.2242
10V
= APV (Future benefits) – APV (Future benefit premiums)
1000
4
A35 − π a35
5
5
4
1
15.3926
.
= 128.72 − 1258
5
5
= 10.25
=
b
g b gb
g
Test Question: 126
Let
Key: E
Y = present value random variable for payments on one life
S = ∑ Y = present value random variable for all payments
E Y = 10a40 = 148166
.
Var Y = 10
2
d
d
2
2
A40 − A40
d
i
2
ib
g
2
= 100 0.04863 − 016132
106
.
. / 0.06
2
= 70555
.
E S = 100 E Y = 14,816.6
Var S = 100 Var Y = 70,555
Standard deviation S = 70,555 = 265.62
By normal approximation, need
E [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (265.62)
= 15,254
Test Question: 127
Key: B
e
5 A30 − 4 A301 :20
Initial Benefit Prem =
Where
j
5a30:35 − 4a30:20
b
g b
g b
g
=
− 4 0.02933
5 010248
.
5 14.835 − 4 11959
.
=
0.5124 − 011732
0.39508
.
=
= 0.015
74.175 − 47.836
26.339
e
b
g
j
1
A30
= A30:20 − A30:201 = 0.32307 − 0.29374 = 0.02933
:20
and
a30:20 =
1 − A30:20
d
=
1 − 0.32307
= 11959
.
0.06
106
.
FG IJ
H K
Comment: the numerator could equally well have been calculated as A30 + 4 20 E30 A50
= 0.10248 + (4) (0.29374) (0.24905)
= 0.39510
Test Question: 128
0.75
Key: B
b gb g
px = 1 − 0.75 0.05
= 0.9625
0.75
b gb g
p y = 1 − 0.75 .10
= 0.925
0.75 qxy = 1− 0.75 pxy
b p gd p i since independent
= 1- b0.9625gb0.925g
= 1−
0.75
x
0.75
y
= 01097
.
Test Question:
N = number of physicians
X = visits per physician
129 Key:
A
E(N) = 3
E(X) = 30
Var (N) = 2
Var (X) = 30
S = total visits
E(S) = E (N) E(X) = 90
Var(S) = E(N) Var(X) + E2 X Var(N) =
= 3 ⋅ 30 + 900 ⋅ 2 = 1890
Standard deviation (S) = 43.5
bg
Pr(S>119.5) = Pr
FG S − 90 > 119.5 − 90 IJ = 1 − Φb0.68g Course 3: November 2000
H 435.
K
435
.
Test Question:
130 Key:
A
The person receives K per year guaranteed for 10 years ⇒ Ka10 = 8.4353K
The person receives K per years alive starting 10 years from now ⇒10 a40 K
b
g
*Hence we have 10000 = 8.4353+10 E40a50 K
Derive
10 E40 :
1
+
A40 = A4010
:
10 E40
Derive a50 =
=
b
10 E40
1
A40 − A4010
:
A50
=
gA
50
0.30 − 0.09
= 0.60
0.35
1 − A50 1 − 0.35
=
= 16.90
.04
d
104
.
Plug in values:
10,000 = 8.4353 + 0.60 16.90 K
b gb
c
gh
= 18.5753K
K = 538.35
Test Question:
131 Key:
STANDARD: e25:11 =
MODIFIED:
p25
z
11
0
=e z
1
FG1 − t IJ dt = t − t
H 75K
2 × 75
− 0.1ds
0
D
2
= e −.1 = 0.90484
11
0
= 101933
.
z
1
e2511
: =
+ p25
p dt
0 t 25
z FGH
z FGH
10
0
1−
IJ
K
t
dt
74
IJ
K
F t I
1− e
=
+ e Gt −
01
.
H 2 × 74 JK
= 0.95163 + 0.90484b9.32432g = 9.3886
=
z
1 −0.1t
0
e
dt + e−0.1
−0.1
10
0
−0.1
1−
t
dt
74
2
10
0
Difference
=0.8047
Test Question:
132
Key: B
Comparing B & D: Prospectively at time 2, they have the same future benefits. At issue,
B has the lower benefit premium. Thus, by formula 7.2.2, B has the higher reserve.
Comparing A to B: use formula 7.3.5. At issue, B has the higher benefit premium. Until
time 2, they have had the same benefits, so B has the higher reserve.
Comparing B to C: Visualize a graph C* that matches graph B on one side of t=2 and
matches graph C on the other side. By using the logic of the two preceding paragraphs,
C’s reserve is lower than C*’s which is lower than B’s.
Comparing B to E: Reserves on E are constant at 0.
Test Question:
133 Key:
C
Since only decrements (1) and (2) occur during the year, probability of reaching the end
of the year is
p60
′b1g × p60
′b 2g = 1 − 0.01 1 − 0.05 = 0.9405
b
gb
g
Probability of remaining through the year is
p60
. = 0.84645
′b1g × p60
′b 2g × p60
′b3g = 1 − 0.01 1 − 0.05 1 − 010
b
gb
gb
Probability of exiting at the end of the year is
b3g = 0.9405 − 0.84645 = 0.09405
q60
g
Test Question:
134 Key:
E
b g
b g
b g
b g
bg
Var b S g = Eb N g Var b X g + E b X g Var b N g =
=.7 × 16 + 4 × 121
. = 16.04
Standard Devb S g = 4
Eb S g + 2 × Standard Devb S g = 14
. + 2 × 4 = 9.4
Since there are no possible values of S between 0 and 10,
Prb S > 9.4g = 1 − Prb S = 0g
E N =.7
Var N = 4×.2 + 9×.1−.49 = 121
.
E X =2
Var X = 100×.2 − 4 = 16
E S = 2×.7 = 14
.
2
= 1−.7−.2×.82 −.1×.83 =.12
Test Question:
135 Key:
APV of regular death benefit =
=
zb
∞
0
gd
ib
gd
D
zb
∞
0
gd ib
gd i
100000 e-δ t 0.008 e- µ t dt
i
100000 e-0.06t 0.008 e-0.008t dt
b
g
= 100000 0.008 / 0.06 + 0.008 = 11,764.71
APV of accidental death benefit =
=
zb
30
0
gd
ib
gd
i
zb
100000 e-0.06t 0.001 e-0.008t dt
= 100 1 − e-2.04 / 0.068 = 1,279.37
Total APV = 11765 + 1279 = 13044
30
0
gd ib
gd i
100000 e-δ t 0.001 e- µ t dt
Test Question:
136 Key:
b gb
g b gb
g
g b gb
g
B
l 60 +.6 = .6 79,954 + .4 80,625
= 80,222.4
b gb
l 60 +1.5 = .5 79,954 + .5 78,839
= 79,396.5
0.9 q 60 +.6
80222.4 − 79,396.5
80,222.4
= 0.0103
=
P0 = 111 = 9.0909%
Test Question:
z
z
137 Key:
bg
5 −λ
1
5 0 e dλ
bg
5
−λ
1
5 0 λe dλ
P0 =
P1 =
b
=
1
5
=
d− e i
−λ
1
5
d − λe
5
0
−λ
=
1
5
A
.
d1 − e i = 01987
− e− λ
−5
i
5
0
=
1
5
.
d1 − 6e i = 01919
−5
g
P N ≥ 2 = 1−.1987−.1919 =.6094
Test Question:
bτ g = qb1g + qb2g = 0.34
q40
40
40
1
2
= 1 − p40
′b g p40
′b g
0.34 = 1 − 0.75 p40
′ b 2g
p40
′ b 2g = 0.88
q40
. =y
′ b 2g = 012
q41
′ b 2g = 2 y = 0.24
138 Key:
A
b gb g
l b g = 2000b1 − 0.34gb1 − 0.392g = 803
bτ g = 1 − 0.8 1 − 0.24 = 0.392
q41
τ
42
Test Question:
139 Key:
C
b g
Pr L π ' > 0 < 0.5
Pr 10,000 v K +1 − π ' a K +1 > 0 < 0.5
From Illustrative Life Table,
47
p30 = 0.50816 and
48
p30 =.47681
Since L is a decreasing function of K, to have
Pr L π ′ > 0 < 0.5 means we must have L π ′ ≤ 0 for K ≥ 47.
b g
b g
b g
Highest value of L π ′ for K ≥ 47 is at K = 47.
b g
L π ′ at K = 47 = 10,000 v 47+1 − π ′ a47+1
b g
= 609.98 − 16.589π ′
b
g
L π ′ ≤ 0 ⇒ 609.98 − 16.589 π ′ ≤ 0
⇒ π′ >
609.98
= 36.77
16.589
Test Question:
140
Key: B
b g
Prb K = 1g = p − p = 0.9 − 0.81 = 0.09
Prb K > 1g = p = 0.81
E bY g = .1 × 1+.09 × 187
. +.81 × 2.72 = 2.4715
E dY i = .1 × 1 +.09 × 187
. +.81 × 2.72 = 6.407
VARbY g = 6.407 − 2.4715 = 0.299
Pr K = 0 = 1 − px = 01
.
1 x
2 x
2 x
2
2
2
2
2
Test Question:
141 Key:
D
Let X be the occurrence amount, Y = max(X-100, 0) be the amount paid.
E[X] = 1,000
2
Var[X] = 1,000
P(X>100) = exp(-100/1,000) = .904837
b
g
The distribution of Y given that X>100, is also exponential with mean 1,000
(memoryless property).
So Y is
RS 0 with prob .095163
Texponential mean 1000 with prob .904837
E Y =.095163 × 0+.904837 × 1,000 = 904.837
b
g
E Y 2 =.095163 × 0+.904837 × 2 × 1,000 = 1,809,675
b
g
2
Var Y = 1,809,675 − 904.837 = 990,944
2
Alternatively, think of this as a compound distribution whose frequency is Bernoulli with
p = .904837, and severity is exponential with mean 1,000.
2
Var = Var N × E X + Var X × E N = p 1 − p 1,000,000 + p 1,000,000
b gb
g b
g
Test Question:
142 Key:
B
b g b g c A−A h
2 2
In general Var L = 1 + δP
c h
Here P Ax =
x
2
x
1
1
− δ = −.08 =.12
ax
5
F .12 I
So Var b Lg = G 1 + J c A − A h =.5625
H .08K
F b.12gIJ c A − A h
and Var b L *g = G 1 +
H .08 K
b1 + g b0.5625g =.744
So Var b L *g =
b1 + g
E L * = A −.15a = 1 − a bδ +.15g = 1 − 5b.23g = −.15
E L * + Var b L *g =.7125
2
2
2
x
x
2
5
4
2
x
2
x
15 2
8
12 2
8
x
x
x
Test Question:
143 Key:
C
Serious claims are reported according to a Poisson process at an average rate of 2 per
month. The chance of seeing at least 3 claims is (1 – the chance of seeing 0, 1, or 2
claims).
b g
b g
b g bg b g
P 3 + ≥ 0.9 is the same as P 0,1,2 ≤ 01
. is the same as P 0 + P 1 + P 2 ≤ 01
.
d
i
01
. ≥ e − λ + λe− λ + λ2 / 2 e− λ
The expected value is 2 per month, so we would expect it to be at least 2 months
λ=4 .
Plug in and try
e −4 + 4e −4 + 42 / 2 e−4 =.238, too high, so try 3 months λ = 6
b
g
e −6 + 6e −6
d i
+ d6 / 2ie
2
b
−6
g
=.062, okay. The answer is 3 months.
[While 2 is a reasonable first guess, it was not critical to the solution. Wherever you
start, you should conclude 2 is too few, and 3 is enough].
Test Question:
144 Key:
B
Let l0bτ g = number of students entering year 1
superscript (f) denote academic failure
superscript (w) denote withdrawal
subscript is “age” at start of year; equals year - 1
p0bτ g = 1 − 0.40 − 0.20 = 0.40
l2bτ g = 10 l2bτ g q2b f g ⇒ q2b f g = 01
.
b
g
q2b w g = q2bτ g − q2b f g = 10
. − 0.6 − 01
. = 0.3
l1bτ gq1b f g = 0.4 l1bτ g 1 − q1b f g − q1b wg
e
j
q1b f g = 0.4 1 − q1b f g − 0.3
e
q1b f g =
j
0.28
= 0.2
14
.
p1bτ g = 1 − q1b f g − q1b w g = 1 − 0.2 − 0.3 = 0.5
b g = q b w g + p b τ g q b w g + p b τ g p bτ g q b w g
w
3 q0
0
0
1
0
1
2
b gb g b gb gb g
= 0.2 + 0.4 0.3 + 0.4 0.5 0.3
= 0.38
Test Question:
b
e25 = p25 1 + e26
145 Key:
g
N
M
since same µ
e26
= e26
N
p25
=e z
=e z
−
1
0
1
bg c h
M
µ 25
t + 0.1 1− t dt
bg
z
bg
e z
c h
1
M
− µ 25
t dt − 0.1 1− t dt
0
=e z
1
0
c h
1
M
− µ 25
t dt − 0.1 1− t dt
0
0
L F
M MN H
= p25 e
2
− 0.1 t − t2
I OP1
K Q0
M
= e−0.05 p25
b
N
N
e25
= p25
1 + e26
g
b
M
= e−0.05 p25
1 + e26
b
g
gb g
M
= 0.951 e25
= 0.951 10.0 = 9.5
D
Test Question:
146 Key:
D
b g
F c1 − A hI = 10,000,000
= 100b10,000gG
H δ JK
E YAGG = 100E Y = 100 10,000 a x
x
b10,000g δ1 c A − A h
b10,000g b0.25g − b016
. g = 50,000
=
δ
σ Y = Var Y =
2
2
b
2
x
x
2
g
σ AGG = 100σ Y = 10 50,000 = 500,000
0.90 = Pr
LM F − E Y
N σ
AGG
>0
AGG
⇒ 1282
=
.
OP
Q
F − E YAGG
σ AGG
F = 1282
. σ AGG + E YAGG
b
g
F = 1282
.
500,000 + 10,000,000 = 10,641,000
Test Question:
147 Key:
C
Expected claims under current distribution = 500
θ = parameter of new distribution
X = claims
E X =θ
bonus =.5 × 500 − X ∧ 500
b g
FG
H
FG
H
E (claims + bonus) = θ +.5 500 − θ 1 −
IJ IJ = 500
500 + θ K K
FG 500 IJ = 250
2 H 500 + θ K
2b500 + θ gθ − 500θ = 250b500 + θ g ⋅ 2
θ−
θ
1000θ + θ 2 ⋅ 2 − 500θ = 2 × 250 × 500 + 500θ
θ = 250 × 500 = 354
θ
Test Question:
148
b DAg
1
80:20
Key: E
eb g j
= 20vq80 + vp80 DA
bg
1
8119
:
b g
b g
b g
b g b gb g
2+.9b12.225g
=
= 12.267
20 .2
.8
1
+
DA 81
:19
106
.
106
.
13 106
. −4
1
∴ DA 8119
=
= 12.225
:
.8
DA801 :20 = 20 v .1 + v .9 12.225
q80 =.2
13 =
q80 =.1
106
.
Test Question:
149 Key:
B
Let T denote the random variable of time until the college graduate finds a job
Let N t , t ≥ 0 denote the job offer process
m bg
r
Each offer can be classified as either
Type I - - accept with probability p ⇒ N1 t
R|
S|Type II T
m b gr
- reject with probability b1 − pg ⇒ mN bt gr
By proposition 5.2, mN bt gr is Poisson process with λ = λ ⋅ p
p = Prb w > 28,000g = Prbln w > ln 28,000g
.
10.24 − 1012
. I
F ln w − 1012
= Prbln w > 10.24g = Pr G
>
JK = 1 − Φb1g
H 012
.
012
.
2
1
1
= 01587
.
λ1 = 01587
.
× 2 = 0.3174
T has an exponential distribution with θ =
b
g
bg
Pr T > 3 = 1 − F 3
=
−3
3.15
e
= 0.386
1
= 315
.
.3174
Test Question:
t
LM
N
px = exp −
z
150 Key:
OP
Q
b
A
ds
= exp ln 100 − x − s
0 100 − x − s
t
g
t
0
=
100 − x − t
100 − x
e50:60 = e50 + e60 − e50:60
e50
e60
LM
OP = 25
=z
N
Q
t O
40 − t
1 L
40t − P = 20
dt =
=z
M
40
40 N
2Q
F 50 − t IJ FG 40 − t IJ dt = z 1 d2000 − 90t + t idt
=z G
H 50 K H 40 K
2000
I = 14.67
1 F
t
2000
45
t
t
=
−
+
2000 GH
3 JK
50
50 50 − t
0
t2
1
50t −
dt =
50
50
2
0
2 40
40
0
0
e50:60
40
40
0
0
3
2
e50:60 = 25 + 20 − 14.67 = 30.33
40
0
2
Test Question:
151 Key:
b gb
g b gb
A
g
UDD ⇒ l21 = 0.8 53,488 + 0.2 17,384 = 46,267.2
z z bb
∞
b g
Mrl 21 = e21 =
∞
t p21
0
= ∑ areas
b gb g
1.000
1
Ο
Ο2
4
3
Ο
Ο
xi
21
Test: 152
g
.
= 2.751 + 1228
+ 0.361 + 0.072
= 4.412
S X i S 21
0.375731
.115611
.028897
0
g
S 21 + t
dt
S 21
25
30
35
Key: Question
D
µ ′n1 = E N = 25
µ x 2 = Var X = 675
µ N 2 = Var N = 25
E X = 50
E S = E X E N = 25 × 50 = 1250
Var S = E N Var X + Var N E X
2
= 25 × 675 + 25 × 2500 = 79,375
40
Standard Deviation S = 79,375 = 28174
.
b
b
g
b
g
b g
g
Pr S > 2000 = Pr S − 1250 / 28174
. > 2000 − 1250 / 28174
. = 1 − Φ 2.66
Test Question:
b g
Var b Λ g
Var 0 L
0
153
Key: E
b g
b g since VarbΛ g = 0
= v bb − V g p q
b10,000 − 3,209g b0.00832gb0.99168g
=
= Var Λ 0 + v 2 Var Λ 1
2
2
1 1
50 50
2
103
. 2
= 358664.09
b g
Var Λ 1
b
= v b2 − 2V
g
2
p50q51 p51
b10,000 − 6,539g b0.99168gb0.00911gb0.99089g
=
2
103
. 2
= 101075.09
b g
Var 0 L
= 358664.09 +
101075.09
= 453937.06
103
. 2
Alternative solution:
π = 10,000 v − 2V50:3 = 9708.74 − 6539 = 3169.74
R|10,000 v − π a = 6539
|
L = S10,000 v − π a = 3178.80
||10,000 v − π a = −83.52
T
1
2
0
2
3
3
for
K=0
for
K =1
for
K >1
b g
Prb K = 1g = p q = b0.99168gb0.00911g = 0.0090342
Prb K > 1g = 1 − Prb K = 0g − Prb K = 1g = 0.98265
Varb Lg = E L − E L = E L
since π is benefit premium
= 0.00832 × 6539 + 0.00903 × 3178.80 + 0.98265 × b−83.52g
Pr K = 0 = q50 = 0.00832
50 51
2
2
0
0
2
0
0
2
2
2
= 453,895 [difference from the other solution is due to rounding]
Test Question:
154 Key:
C
Let π denote the single benefit premium.
π = 30 a35 + π A351 :30
π=
30
a35
eA
=
35:30
1
1 − A35
:30
j
1
a65
− A35
:30
1−
=
1
A35
:30
=
b.21−.07g9.9
b1−.07g
1.386
.93
= 1.49
=
Test Question:
0.4 p0
=.5 = e z
−
0. 4
0
e F +e2 x jdx
LM e22x OP.4
N Q0
=e
0.8
−.4 F −F e 2 −1 I
H
K
=e
−.4 F −
.5 = e−.4 F −.6128
bg
⇒ ln .5 = −.4 F −.6128
⇒ −.6931 = −.4F −.6128
⇒ F = 0.20
155 Key:
E
Test Question:
156 Key:
C
bgbg
F 2000 IJ × LM1 − 2000 OP = 2000 × FG1 − 2IJ = 2000 × 3 = 1200
E X ∧ 3000 = G
H 1 K N b3000 + 2000g Q
H 5K
5
E X = 2000 1! / 1! = 2000
So the fraction of the losses expected to be covered by the reinsurance
2000 − 1200
= 0.4.
is
2000
The expected ceded losses are 4,000,000 ⇒ the ceded premium is 4,400,000.
Test Question:
157 Key:
E
X 2002 = 105
. × X 2001
so:
F x IJ = 1 − LM
FG
H 105
. K
Nbx
L
= 1− M
Nx
2002
2000
. + 2000
2002 / 105
2100
2002 + 2100
OP
Q
OP
gQ
2
2
This is just another Pareto distribution with α = 2,θ = 2100.
E X 2002 = 2100.
and
FG 2100 IJ × LM1 − FG 2100 IJ OP
H 1 K MN H b3000 + 2100gK PQ
L 3000 OP = 1235
= 2100 × M
N 5100 Q
E X 2002 ∧ 3000 =
So the fraction of the losses expected to be covered by the reinsurance is
2100 − 1235
= 0.412.
2100
The total expected losses have increased to 10,500,000, so
C2002 = 11
. × 0.412 × 10,500,000 = 4,758,600
And
C2002 4,758,600
=
= 108
.
C2001 4,400,000