Exam-Style Questions on Life Insurance Policies

Before we attempt some exam-style questions pertaining to the material covered thus far in this study guide, it will be useful to know some special properties that hold under the assumption of uniform distributions of deaths (UDD) for fractional ages.

Under UDD for fractional ages, it is the case that

μ_x+t*_tp_x = q_xand

_tq_x+s = _tq_x/(1 - _sq_x)

The problems in this section were designed to be similar to problems from past versions of the Casualty Actuarial Society's Exam 3L and the Society of Actuaries' Exam MLC. They use original exam questions as their inspiration - and the specific inspiration for each problem is cited so as to give students a chance to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L36-1.

Similar to Question 21 from the Casualty Actuarial Society's Spring 2008 Exam 3L.

Two whole life insurance policies for 60-year-old yak-rabbits are as follows:

Policy J assumes that the lives of yak-rabbits follow an exponential distribution with mean 8.

Policy K assumes that the lives of yak-rabbits follow a uniform distribution, and no yak-rabbit survives beyond age 90.

The annual force of interest in Yak-Rabbitland is 0.05. What is the absolute value of the difference between Policy J and Policy K?

Solution S3L36-1. We use the formula Ā_x = ₀^∞∫v^t*_tp_x*μ_x(t)dt = ₀^∞∫v^t*f_T(t)dt

For Policy J, an exponential distribution of lives implies that μ_x(t) = 1/λ (where λ is the mean of the exponential distribution) = 1/8 = 0.125 for all x.

Likewise, _tp_x = e^-0.125t for all x.

We find v^t = e^-0.05t. Thus, Ā₆₀ = ₀^∞∫0.125*e^-0.05t*e^-0.125tdt = ₀^∞∫0.125*e^-0.175tdt =

(-125/175)e^-0.175t│₀^∞ = 125/175 = Ā₆₀ = 0.714285714 for Policy J.

For Policy K, since it is assumed that a 60-year-old yak-rabbit has at most 30 years left ti live and deaths are uniformly distributed, it follows that f_T(t) = 1/30.

Thus, Ā₆₀ = ₀³⁰∫(1/30)e^-0.05tdt = (-20/30)e^-0.05t│₀³⁰ = (2/3)(1 - e^-0.05*30) = about 0.517913227 for Policy K.

Thus, the absolute value of the difference between Policy J and Policy K is

0.714285714 - 0.517913227 = about 0.196372487

Problem S3L36-2.

Similar to Question 22 from the Casualty Actuarial Society's Spring 2008 Exam 3L.

For velociraptors, the force of mortality is a constant 0.12, and the force of interest is 0.09. For a velociraptor aged x, a whole life insurance policy has been issued. What is the 46^th percentile of the distribution of the present value of this policy?

Solution S3L36-2. The present-value random variable of this policy is Z. We recall from Section 26 that F_Z(z) = 1 - F_T(ln(z)/ln(v)). Since the force of mortality given here is constant, we know that the lives of velociraptors are exponentially distributed with mean 1/0.12. Thus, F_T(t) = 1 - e^-0.12t. Thus,

1 - F_T(ln(z)/ln(v)) = e^{-0.12(ln(z)/ln(v))}.

Since the force of interest is 0.09, we have v = e^-0.09, so ln(v) = -0.09. Thus,

e^{-0.12(ln(z)/ln(v))} = e^{-0.12(ln(z)/-0.09)} = e^((4/3)ln(z)) = e^{(ln(z^(4/3)))} = F_Z(z) = z^4/3. We want to find z such that z^4/3 = 0.46. This is z = 0.46^3/4 = about 0.558558125.

Problem S3L36-3.

Similar to Question 30 from the Casualty Actuarial Society's Fall 2007 Exam 3.

The survival of white bulls follows a Weibull distribution, such that the force of mortality is µ_x = kx⁴.

You know that for white bulls, ₅q₁ = 0.64535. Find _2│2q₃.

Solution S3L36-3. We first need to find the value of k. We know that a Weibull survival distribution looks as follows: s(x) = exp[-uxⁿ⁺¹], where u = k/(n + 1). Moreover, a Weibull force of mortality is of the formµ_x = kxⁿ,so we know here that n = 4 and u = k/5.

We are given that ₅q₁ = 1 - s(6)/s(1) = 0.64535 and 0.35465 = s(6)/s(1)

We find s(1) = exp[-(k/5)1⁵] = exp[-(k/5)]

We find s(6) = exp[-(k/5)6⁵]

Thus, 0.35465 = exp[-(k/5)6⁵]/exp[-(k/5)]

0.35465 = exp[-(k/5)(6⁵ - 1)]

ln(0.35465) = -(k/5)(6⁵ - 1)

-5ln(0.35465)/(6⁵ - 1) = k = about 0.000666639158.

Now we can find _2│2q₃ = (s(5) - s(7))/s(3) =

(exp[-(k/5)5⁵] - exp[-(k/5)7⁵])/exp[-(k/5)3⁵] =

(exp[-(0.000666639158/5)5⁵] - exp[-(0.000666639158/5)7⁵])/

exp[-(0.000666639158/5)3⁵] = (0.659251965 - 0.106369025)/0.968120551 = _2│2q₃ = about 0.571088941.

Problem S3L36-4. Similar to Question 13 from the Casualty Actuarial Society's Fall 2006 Exam 3.

This question uses the Illustrative Life Table, which can be found here.

Using the Illustrative Life Table and assuming a uniform distribution of deaths for fractional ages, determine the truth or falsehood of each of these statements.

(a) μ_34.5*_0.5p₃₄ = q₃₄.

(b) _0.5p₃₄*_0.3q_34.5 = 0.00057

(c) _0.3q_34.7< _0.3q_34.5

Solution S3L36-4.

Under a uniform distribution of deaths for fractional ages, it is the case that

μ_x+t*_tp_x = q_x. (a) is the expression of this statement for x = 34 and t = 0.5. Thus, (a) is true.

We examine statement (b):

_0.5p₃₄*_0.3q_34.5 = _0.5p₃₄*(1 - _0.3p_34.5) = _0.5p₃₄ - _0.8p₃₄ = _0.8q₃₄ - _0.5q₃₄ = 0.3q₃₄(by the UDD assumption).

From the Illustrative Life Table, 1000q₃₄ = 1.90, so 0.3q₃₄ = 1.90*0.3/1000 = 0.00057, so (b) is true.

Under the UDD assumption, _tq_x+s = _tq_x/(1 - _sq_x).

Thus, for x = 34, s = 0.5, t = 0.3, _0.3q_34.5 = _0.3q₃₄/(1 - _0.5q₃₄).

For x = 34, s = 0.7, t = 0.3, _0.3q_34.5 = _0.3q₃₄/(1 - _0.7q₃₄).

We know that _0.7q₃₄ >_0.5q₃₄, since _0.7q₃₄ = _0.5q₃₄+ _0.2q_34.5.

Thus, (1 - _0.7q₃₄) < (1 - _0.5q₃₄)

Thus, _0.3q₃₄/(1 - _0.7q₃₄) >_0.3q₃₄/(1 - _0.5q₃₄), so

_0.3q_34.7 >_0.3q_34.5 and (c) is false.

Problem S3L36-5. Similar to Question 33 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Red rodents exhibit a constant force of mortality of 0.1, and the annual force of interest among red rodents is 0.03. Tnedor the Red Rodent has a 7-year deferred whole life insurance paying a benefit of 1 Golden Hexagon (GH) upon death. Find the variance Var(Z) of the present-value random variable for this policy.

Solution S3L36-5. We use the formula
_m│Ā_x= _m^∞∫v^t*_tp_x*μ_x(t)dt to findthe actuarial present value of this policy:

A constant force of mortality corresponds to an exponential distribution with mean

1/μ_x

So μ_x(t) = 0.1 and _tp_x = e^-0.1t. Moreover, we can find v^t = e^-0.03t.

Thus, _7│Ā_x= ₇^∞∫e^-0.03t*e^-0.1t*0.1dt = ₇^∞∫0.1e^-0.13tdt = (-10/13)e^-0.13t│₇^∞ = (10/13)e^-0.13*7 = _7│Ā_x= 0.309634018.

Now we find the second moment of the present value for this policy:

²_7│Ā_x= ₇^∞∫e^-0.03*2t*e^-0.1t*0.1dt = ₇^∞∫0.1e^-0.16tdt = (-10/16)e^-0.16t│₇^∞ = (5/8)e^-0.16*7 = ²_7│Ā_x= 0.203924872.

Now we use the formula Var(Z) = ²_m│Ā_x - (_m│Ā_x) ² = 0.203924872 - 0.309634018² = Var(Z) = 0.108051647.

See other sections of The Actuary's Free Study Guide for Exam 3L.