# AS 5102 Study Guide - Quiz Guide: Life Table, Term Life Insurance, Life Insurance

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AS 3501/5102 ACTUARIAL MODELING I
PROBLEM SET CH.4 SOLUTION
Note: The problem set contains optional practice questions from many sources. In particular, it includes
the Exam LTAM/MLC sample questions and textbook exercises. We call the table used in the MLC Exam as
the SOA Illustrative Life Table, and call the LTAM Exam Table as the Standard and Ultimate Life Table
(SULT).
1. LTAM Sample Qs: see the solutions posted by SOA.
2. LTAM Written Sample Qs: see the solutions posted by SOA.
3. MLC Sample Qs: see the solutions posted by SOA.
4. MLC Written Sample Qs: see the solutions posted by SOA.
5. Textbook Exercises: see the answers in the textbook.
6. You are given that a certain insurance contract issued to (x)will pay a beneﬁt of \$100,000 at the
moment of death if death occurs within 10 years. Otherwise, no beneﬁt will be paid.
(a) Identify the type of this insurance.
(b) Write down the present value random variable, Z, of the beneﬁt in this insurance.
(c) Derive an expression in terms of standard actuarial functions (notations) for the expected value of
Z.
(d) Derive an expression in terms of standard actuarial functions (notations) for the variance of Z.
Solution:
(a) This is a continuous 10-year term insurance life insurance with face amount \$100,000.
(b)
Z=(100,000 ·vTx= 100,000eδTx, Tx10
0, Tx>10
(c)
E[Z] = 100,000 ¯
A1
x:10
Note that all the actuarial notations we learned are for \$1 face amount.
(d)
V ar(Z) = E[Z2](E[Z])2= 100,00022¯
A1
x:10 ¯
A1
x:10 2
7. For a whole life insurance of \$1,000,000 on (30) with beneﬁts payable at the moment of death, you
are given:
µx=(0.03, x < 50
0.05, x 50
If δ= 0.05 and let Zbe the random variable of the present value of the insurance beneﬁt, calculate
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(b) V ar(Z)
Solution:
(a) We have two approaches to solve this problem. The ﬁrst approach is to use the basic deﬁnition.
Since
tp30 =(e0.03t, t < 20
20p30 ·t20p30 =e0.6·e0.05(t20), t 20
we have
E[Z] = 1,000,000 ¯
A30
= 1,000,000 Z20
0
eδt ·0.03e0.03tdt +Z
20
eδt0.05e0.6·e0.05(t20)dt
= 400,237.0647
Alternatively, we use the recursive formula:
E[Z] = 1,000,000 ¯
A30
= 1,000,000 ¯
A1
30:20 +20E30 ·¯
A50
= 1,000,000 0.03
0.03 + 0.05 1e(0.03+0.05)·20+e(0.03+0.05)·20 ·0.05
0.05 + 0.05
= 400,237.0647
Remark: APVs for continuous whole life/term life under the constant force of mortality law are
directly used:
¯
Ax=µ
µ+δand ¯
A1
x:n=µ
µ+δ1e(µ+δ)n
(b) The second moment can be directly obtained by replacing δwith 2δ, i.e.
E[Z2] = 1,000,0002·2¯
A30
= 1,000,00020.03
0.03 + 0.11e(0.03+0.1)·20+e(0.03+0.1)·20 ·0.05
0.05 + 0.1
= 2.3839 ×1011
Therefore, the variance is
V ar(Z) = E[Z2](E[Z])2= 7.8197 ×1010
8. Assume the force of mortality satisfy De Moivre’s Law with ω= 100, that is
µx=1
100 x,
for 0x < 100. Let i= 0.05, calculate
(a) A50
(b) 10|A40
(c) A1
50:20
(d) A(2)
50
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