Fully Continuous Benefit Premiums for Pure Endowments and Exam-Style Questions on Life Insurance and Life Annuities

This section will be a mix of exam-style questions on previously covered material and questions covering new material.

We continue to examine fully continuous benefit premiums under the equivalence principle for various kinds of life insurance policies.

For an n-year pure endowment which pays a benefit of 1 to life x, the fully continuous benefit premium P^-(_nE_x) can be found as follows.

P^-(_nE_x) = _nE_x/ ā_x:n¬

Recall: _nE_x = vⁿ*_np_x and ā_x:n¬ = ₀ⁿ∫v^t*_tp_x*dt.

Moreover, ā_x:n¬ = (1 - e^-(δ+μ)n)/(μ + δ) for constant force of mortality μ and constant force of interest δ.

In Section 37, we introduced the formula

ā_x = ā_x:1¬ + vp_x*ā_x+1

An analogous formula holds for any number of years n:

ā_x = ā_x:n¬ + vⁿ*_np_x*ā_x+n

But vⁿ*_np_x is the same as _nE_x. Thus,

ā_x = ā_x:n¬ + _nE_x*ā_x+n

The recursion relation

A¹_x:n¬ = vq_x + vp_x*A¹_x+1:n-1¬ from Section 34 can also be generalized to m time periods:

A¹_x:n¬ = A¹_x:m¬+ v^m*_mp_x*A¹_x+m:n-m¬

Moreover, if one is given a life table and required to calculate the actuarial present values of term life insurance policies, the following identity is useful.

A¹_x:m¬ = _n=0^mΣvⁿ(_n-1│q_x)

Moreover, the following relationship holds between the actuarial present values of term and whole life insurance policies.

A_x = A¹_x:m¬ + v^m*_mp_x*A_x+m

Some of 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.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 173.

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

Problem S3L45-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is 0.09. Otto the Triceratops is currently 7 years old has a 2-year pure endowment, which pays a benefit of 1 Triceratops Currency Unit (TCU). Under the equivalence principle, what is the annual fully continuous level benefit premium that Hasdrubal will pay for this policy?

Solution S3L45-1. We use the formula P^-(_nE_x) = _nE_x/ ā_x:n¬.

Since triceratopses exhibit a constant force of mortality, we have μ_x(t) = 0.34 and therefore ā_x:n¬ = (1 - e^-(δ+μ)n)/(μ + δ) For n = 2, ā_7:2¬ = (1 - e^-(0.43)2)/(0.43) =

(1 - e^-0.86)/(0.43) = about 1.341483529 TCU.

Moreover, since triceratops lifetimes are exponentially distributed, we have _tp_x = e^-0.34t.

Moreover, v^t = e^-0.07t.

Thus, ₂E₇ = vⁿ*_np_x = e^-0.07*2e^-0.34*2 = e^-0.86.

Thus, P(₂E₇) = e^-0.86/1.341483529 = P(₂E₇) = about 0.3154433677 TCU.

Problem S3L45-2.

Similar to Question 38 from the Casualty Actuarial Society's Fall 2007 Exam 3.

You know the following about the actuarial present values of continuous whole life annuities and a 5-year pure endowment.

1) ₅E₄ = 0.35

2) ā₄ = 3.25

3) ā₉ = 7.776

Find ā_4:5¬, the value of a 5-year continuous temporary life annuity for life (4).

Solution S3L45-2. We use the formula ā_x = ā_x:n¬ + _nE_x*ā_x+n, which we rearrange thus:

ā_x:n¬ = ā_x - _nE_x*ā_x+n

Thus, ā_4:5¬ = 3.25 - 0.35*7.776 = ā_4:5¬ = 0.5284.

Problem S3L45-3.

Similar to Question 34 from the Casualty Actuarial Society's Fall 2006 Exam 3.

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

Mortality occurs according to the Illustrative Life Table. There is a 6-year term life insurance policy on a life currently aged 54.

For the next three years, the annual effective interest rate will be 0.07.

Thereafter, the annual effective interest rate will be 0.09.

Find A¹_54:6¬, the actuarial present value of term life insurance policy in question.

Solution S3L45-3.

Since the interest rates are different for different time periods, we will need to use the formula A¹_x:n¬ = A¹_x:m¬+ v^m*_mp_x*A¹_x+m:n-m¬.

Here, A¹_54:6¬ = A¹_54:3¬+ v³*₃p₅₇*A¹_57:3¬.

The "v" in this formula is the three-year discount factor using the interest rate 0.09, the rate that takes effect 3 years from now. Thus, v³ = 1/1.09³

Moreover, we can use the Illustrative Life Table to find

₃p₅₇ = l₆₀/l₅₇ = 8188074/8479908 = 0.9655852398.

Thus, v³*₃p₅₇ = 0.9655852398/1.09³ = 0.7456089708.

Hence, we are left with A¹_54:6¬ = A¹_54:3¬+ 0.7456089708*A¹_57:60¬.

Now we find A¹_54:3¬ using the formula A¹_x:m¬ = _n=0^mΣvⁿ(_n-1│q_x).

Thus, A¹_54:3¬ = vq₅₄ + v²(_1│q₅₄) + v³(_2│q₅₄)

A¹_54:3¬ = 0.00824/1.09 + (1- 0.00824)(0.00896)/1.09² + (1- 0.00824)(1 - 0.00896)(0.00975)/1.09³ = A¹_54:3¬ = 0.0224387938.

Therefore, A¹_54:6¬ = 0.0224387938 + 0.7456089708*A¹_57:3¬.

Now we need to find A¹_57:3¬. We use the formula A_x = A¹_x:m¬ + v^m*_mp_x*A_x+m.

In this case,

A₅₇ = A¹_57:3¬ + v³*₃p₅₇*A₆₀.

A¹_57:3¬ = A₅₇ - v³*₃p₅₇*A₆₀

We already know that v³*₃p₅₇ = 0.7456089708.

From the Illustrative Life Table, we know that A₅₇ = 0.32984 and A₆₀ = 0.36913.

Thus, A¹_57:3¬ = 0.32984 - 0.7456089708*0.36913 = A¹_57:3¬ = 0.0546133606.

Therefore, A¹_54:6¬ = 0.0224387938 + 0.7456089708*0.0546133606 =

A¹_54:6¬ = about 0.0631590054.

Problem S3L45-4.

Similar to Question 39 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Real Risky Insurance Co. collects a premium P at the beginning of the year. It also has a surplus of $56 at the beginning of the year. All funds that the company has at the beginning of the year are invested at an annual effective interest rate of 20%.

The following losses may occur at the end of the year:

Loss of $0 with probability 0.2

Loss of $89 with probability 0.5

Loss of $78000 with probability 0.3.

What is the smallest premium P that Real Risky Insurance Co. can charge in order to keep its probability of ruin during this year at most 30%?

Note: This question requires only a basic knowledge of probability theory and interest rates. The rest is just a matter of thinking the problem through and applying the theory to the given context. Try solving the problem yourself before you examine the solution here.

Solution S3L45-4. To keep the probability of ruin during this year at most 30%, the company needs to make sure that a loss of $89 will not bankrupt it. (A loss of $0 will already not bankrupt it.) Thus, it needs to have at least $89 in funds at the end of the year.

The company has $(56 + P) to invest at 20% interest.

Thus, at the lowest value of P, 89 = 1.2(56 + P), and so P = 89/1.2 - 56 =

P = about $18.166666667

Problem S3L45-5.

Similar to Question 39 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Ridiculously Risky Insurance Co. collects a premium of $40 at the beginning of the year. It also has a surplus of $56 at the beginning of the year. Because of excessive risk-taking in this world, all financial markets have collapsed, and thus the annual effective interest rate is 0.

The following losses may occur at the end of the year:

Loss of $0 with probability 0.2

Loss of $89 with probability 0.5

Loss of $100 with probability 0.3.

What is the probability that Ridiculously Risky Insurance Co. will avoid bankruptcy for the next 2 years?

Note: This question requires only a basic knowledge of probability theory and interest rates. The rest is just a matter of thinking the problem through and applying the theory to the given context. Try solving the problem yourself before you examine the solution here.

Solution S3L45-5.

During the first year, the company has 56 + 40 = $96 to pay losses with.

It will only be bankrupted during the first year if its loss is $100.

If the year 1 loss is 0, then at the beginning of year 2, the company will collect $40 and will have $136, thus rendering it solvent in the event of any loss in year 2.

This is one component of the desired probability and is equal to 0.2.

If the year 1 loss is 89, then the company will have 96 - 89 + 40 = $47 at the beginning of year 2. It then will only survive year 2 if the year 2 loss is 0.

This is the second component of the desired probability and is equal to 0.5*0.2 = 0.1.

Thus, the total probability of the company surviving for the next two years is 0.2 + 0.1 = 0.3.

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