Actuarial Science
AS 3429/9429 Long Term Actuarial Math II
Practice Final Exam Solution
1. Solution:
(a) Under the Equivalence Principle
75, 000A40:10 + 25%P + 9%P ä40:10 + 20 + 5ä40:10 + 100A40:10 = P ä40:10
Hence, the gross premium is
P = 75, 100A40:10 + 20 + 5ä40:10
0.91ä40:10 − 0.25 = 6505.2
where from the LTAM standard ultimate life table
A40:10 = 0.61494, ä40:10 = 8.0863.
(b) The loss at issue random variable
Lg0 =75, 100v min(K40+1,10) + 0.25P + 20 + 5ämin(K40+1,10) − 0.91P ämin(K40+1,10)
=75, 100vmin(K40+1,10) + 0.25P + 20 + (5− 0.91P )1− v min(K40+1,10)
d
=(75, 100− 5− 0.91P d
)vmin(K40+1,10) + 0.25P + 20 + 5− 0.91P
d
The probability of making a loss
P(Lg0 > 0) =P((75, 100− 5− 0.91P
d )vmin(K40+1,10) + 0.25P + 20 +
5− 0.91P d
> 0)
=P(vmin(K40+1,10) > − 0.25P + 20 + 5−0.91Pd
75, 100− 5−0.91Pd )
=P(e−δmin(K40+1,10) > 0.61494)
=P(min(K40 + 1, 10) < − ln(0.61494)
δ )
=P(min(K40 + 1, 10) < 9.9658) =P(K40 + 1 < 9.9658)
=P(K40 < 8.9658) = 9q40 = 1− 9p40 = 1− l49 l40
= 0.006578
(c) • Gross premium policy value
5V g =75, 100A45:5 + 5ä45:5 − 0.91P ä45:5
=75, 100× 0.78387− (0.91× 6505.2− 5)× 4.53862 = 32, 023.9
where
ä45:5 = ä45 − 5E45 ä55 = 17.8162− 0.77991× 17.0245 = 4.53862
and A45:5 = 1− dä45:5 = 1− 0.04762× 4.53862 = 0.78387.
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• Net premium policy value
5V n =75, 000A45:5 − Pn ä45:5
=75, 000× 0.78387− 5703.5356× 4.53862 = 32, 904.1
where
Pn = 75, 000A40:10
ä40:10 =
75, 000× 0.61494 8.0863
= 5703.5356
• FPT policy value
5V FPT =75, 000A45:5 − PFPT ä45:5
=75, 000× 0.78387− 6502.95× 4.53862 = 29275.8
where
ä41:9 = ä41 − 9E41 ä50 = 7.44458, A41:9 = 1− dä41:9 = 0.64549
PFPT = 75, 000A41:9
ä41:9 = 6502.95
*Numbers could be slightly different due to rounding.
2. Solution:
The loss at issue random variable for the portfolio is
L =
n∑ i=1
L0,i,
with mean E[L] = nE[L0,1] and variance V ar[L] = nV ar[L0,1], and n = 10, 000. For an individual policy,
L0,1 =150, 000v K50+1 + 20%P − 0.95P äK50+1
=150, 000vK50+1 + 0.2P − 0.95P 1− v K50+1
d
=
( 150, 000 +
0.95P
d
) vK50+1 + 0.2P − 0.95P
d
which gives E[L0,1] = 150, 000A50 + 0.2P − 0.95P ä50
and
V ar[L0,1] =
( 150, 000 +
0.95P
d
)2 (2A50 −A502)
Using normal approximation
P(L < 0) = P( L− E[L]√ V ar[L]
< 0− E[L]√ V ar[L]
) = 0.95 =⇒ − E[L]√ V ar[L]
= 1.645
i.e.,
− 10, 000(150, 000A50 + 0.2P − 0.95P ä50)( 150, 000 + 0.95Pd
)√ 10, 000(2A50 −A502)
= 1.645
which solves P = 1801.39.
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3. Solution:
(a) For j = 0, …, 14,
Lj =
1000vKx+j+1 − 3πäKx+j+1 , Kx+j = 0, 1, …, 14− j 1000vKx+j+1 − 3πä15−j − 2πv15−j äKx+j−(15−j)+1 , Kx+j = 15− j, …, 24− j 500vKx+j+1 − 3πä15−j − 2πv15−j äKx+j−(15−j)+1 , Kx+j = 25− j, …, 29− j 500vKx+j+1 − 3πä15−j − 2πv15−j ä15 , Kx+j = 30− j, 31− j, …
For j = 15, …, 24,
Lj =
1000vKx+j+1 − 2πäKx+j+1 , Kx+j = 0, 1, …, 24− j 500vKx+j+1 − 2πäKx+j+1 , Kx+j = 25− j, …, 29− j 500vKx+j+1 − 2πä15−j , Kx+j = 30− j, 31− j, …
For j = 25, …, 29
Lj =
{ 500vKx+j+1 − 2πäKx+j+1 , Kx+j = j, j + 1, …, 29 500vKx+j+1 − 2πä15−j , Kx+j = 30− j, 31− j, …
For j = 30, 31, … Lj = 500v
Kx+j+1, Kx+j = 0, 1, …
(b) Under the equivalence principle,
500A40 + 500A 1 40:25 = 2πä40:30 + πä40:15 ,
which implies that
π = 500A40 + 500A
1 40:25
2ä40:30 + ä40:15 =
500 (0.202) + 500 (0.114)
2 (13.414) + 10.005 = 4.2896.
Policy values at times 10, 20 and 40 are respectively given by
10V = 500A50 + 500A 1 50:15 −
( 2πä50:20 + πä50:5
) = 500 (0.316) + 500 (0.149)− 4.2896 (2 (10.806) + 4.386) = 120.98,
20V = 500A60 + 500A 1 60:5 − 2πä60:10
= 500 (0.458) + 500 (0.112)− 2 (4.2896) (6.929) = 225.55,
and 40V = 500A80 = 500 (0.744) = 372.
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4. Solution:
(a) The policy value at time 10 is
10V = 100, 200A70:10 = 100, 200× 0.63576 = 63703.152
Using the recursion formula
10V × (1 + 5%) = q70 × 100, 200 + p70 × 11V
which gives the policy value at time 11
11V = 66537.785
Hence the total profit or gain for this year (t = 10) is
nt(tV + Pt − eAt )× (1 + iAt )− [ dt(St+1 + E
A t+1) + nt+1 × t+1V
] =200(63703.152 + 0)(1 + 4%)− [2(100, 000 + 250) + 198(66537.785)] =− 124725.8 (loss)
(b) Expense (interest rate and mortality as expected)
200(10V )× (1 + 5%)− [200q70(100, 000 + 250) + 200p70 × 11V ] − {200(10V )× (1 + 5%)− [200q70(100, 000 + 200) + 200p70 × 11V ]}
=− 200q70 × 50 =− 104.13 (loss)
Interest (expenses as actual and mortality as expected)
200(10V )× (1 + 4%)− [200q70(100, 000 + 250) + 200p70 × 11V ] − {200(10V )× (1 + 5%)− [200q70(100, 000 + 250) + 200p70 × 11V ]}
=200(10V )(4%− 5%) =− 127406.304 (loss)
Mortality (expenses and interest as actual)
200(10V )× (1 + 4%)− [2(100, 000 + 250) + 198× 11V ] − {200(10V )× (1 + 4%)− [200q70(100, 000 + 250) + 200p70 × 11V ]}
=2784.629 (gain)
we can easily check −104.13− 127406.304 + 2784.629 = −124725.8.
5. Solution:
(a) From the CBD model, we know that logit(q(65, 2018)) follows a normal distribution with mean
µ = K (1) 2017 +c
(1) +(K (2) 2017 +c
(2))(x− x̄) = −3.2−0.02+(0.01+0.0006)(65−70) = −3.273
and variance
σ2 = σ2k1+(x−x̄) 2σ2k2+2(x−x̄)ρσk1σk2 = 0.03
2+(65−70)20.0052+2(65−70)×0.2×0.03×0.005 = 0.0352
hence the standard deviation is 0.035.
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(b) Let X = logit(q(65, 2018)) and Y = p(65, 2018). Hence
X = log 1− Y Y
or Y = 1 1+eX
= f(X), we see that f(x) is a decreasing function of x, since
f ′(x) = − e X
(1 + eX)2 < 0
Therefore, the median of Y will be related to the median of X
πY0.5 = f(π X 0.5) =
1
1 + eπ X 0.5
= 1
1 + e−3.273 = 0.96349
and the 95th percentile of Y will be related to the 5th percentile of X
πY0.95 = f(π X 0.05) =
1
1 + eπ X 0.05
= 1
1 + e−3.330575 = 0.96546
Note that since X ∼ N(µ, σ2) the median of X πX0.5 = µ = −3.273 and the 5th percentile of X πX0.05 = µ+ σΦ
−1(0.05) = −3.273− 1.645× 0.035 = −3.330575.
6. Solution:
(a) Summarize the data
i yi si ri si/ri 1 1 1 20 1/20 2 2 1 19 1/19 3 4 2 17 2/17 4 5 1 13 1/13 5 8 3 11 3/11 6 9 4 8 4/8 7 12 2 3 2/3
ymax = 15
(b) Kaplan–Meier Estimate of survival function is
S20(y) =
1, 0 ≤ y < 1, 1− 1/20 = 0.950, 1 ≤ y < 2, 0.950(1− 1/19) = 0.900, 2 ≤ y < 4, 0.900(1− 2/17) = 0.794, 4 ≤ y < 5, 0.794(1− 1/13) = 0.733, 5 ≤ y < 8, 0.733(1− 3/11) = 0.533, 8 ≤ y < 9, 0.533(1− 4/8) = 0.267, 9 ≤ y < 12, 0.267(1− 2/3) = 0.089, 12 ≤ y < 15.
Tail correction:
• Efron’s method: S20(y) = 0 for y ≥ 15.
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• Klein and Moeschberger’s method: S20(y) = 0.089 for 15 ≤ y < 22, and S20(y) = 0 for y ≥ 22. • Brown, Hollander and Korwar’s exponential tail correction: S20(y) = (0.089)
y/15 for y ≥ 15. Withe the change in value, we have ymax = y7 = 12 and S20(y) = 0 for y ≥ 12.
(c) Nelson–Aaloen estimates for cumulative hazard function and the corresponding esti- mated survival function are
Ĥ(y) =
0, 0 ≤ y < 1, 1/20 = 0.050, 1 ≤ y < 2, 0.050 + 1/19 = 0.103, 2 ≤ y < 4, 0.103 + 2/17 = 0.220, 4 ≤ y < 5, 0.220 + 1/13 = 0.297, 5 ≤ y < 8, 0.297 + 3/11 = 0.570, 8 ≤ y < 9, 0.570 + 4/8 = 1.070, 9 ≤ y < 12, 1.070 + 2/3 = 1.737, 12 ≤ y < 15.
Ŝ(y) =
e0 = 1, 0 ≤ y < 1, e−0.050 = 0.951, 1 ≤ y < 2, e−0.103 = 0.902, 2 ≤ y < 4, e−0.220 = 0.803, 4 ≤ y < 5, e−0.297 = 0.743, 5 ≤ y < 8, e−0.570 = 0.566, 8 ≤ y < 9, e−1.070 = 0.343, 9 ≤ y < 12, e−1.737 = 0.176, 12 ≤ y < 15.
Tail correction:
• Efron’s method: S20(y) = 0 for y ≥ 15. • Klein and Moeschberger’s method: S20(y) = 0.176 for 15 ≤ y < 22, and S20(y) = 0 for y ≥ 22. • Brown, Hollander and Korwar’s exponential tail correction: S20(y) = (0.176)
y/15 for y ≥ 15. (d) For the Kaplan–Meier estimator
V̂ ar[S20(2)] = [S20(2)] 2
[ 1
20(19) +
1
19(18)
] = 0.9002 × 0.00556 = 0.0045,
and for the Nelson–Aalen estimator,
V̂ ar[Ŝ(2)] = [Ŝ(2)]2 [
1(19)
203 +
1(18)
193
] = 0.9022 × 0.00500 = 0.004068.
(e) The 95% log-transform confidence interval for S(2) based on the Kaplan–Meier estima- tor is (
Ŝ(2) 1/U
, Ŝ(2) U )
= (0.65604, 0.97401)
with U = exp
( Φ−1( 1+α2 )×
√ V̂ ar[S(y)]
Ŝ(y)×ln Ŝ(y)
) = exp
( 1.96×
√ 0.0045
0.900×ln 0.900
) = 0.24994.
(f) The 95% log-transform confidence interval for S(2) based on the Nelson–Aalen estimator is (
exp ( −Ĥ(y)U
) , exp
( −Ĥ(y)/U
)) = (0.6733, 0.9735)
with U = exp
( Φ−1( 1+α2 )×
√ V̂ ar[H(y)]
Ĥ(y)
) = exp
( 1.96×
√ 0.00500
0.103
) = 3.84.
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