Fully Discrete Benefit Reserves: Practice Problems and Solutions

Let us suppose we have a life insurance policy that pays some benefits and is paid for via some benefit premiums.

Then the benefit reserve at time t is "the conditional expectation of the difference between the present value of future benefits and the present value of future benefit premiums, the conditional event being survivorship of the insured to time t." (Bowers et. al. 1997, p. 205).

Fully discrete benefit reserves occur in conjunction with fully discrete benefit premiums (see Section 52) and insurance policies paying benefits at the end of the year of death (see Section 34).

For a fully discrete benefit reserve on a whole life insurance policy on life (x) on the condition of that life's survival for k additional years, the value of the benefit reserve is denoted _kV_x and can be found via the following formula.

_kV_x = A_x+k - P_x*ä_x+k

We recall that P_x = A_x/ä_x, where A_x is the actuarial present value of a fully discrete whole life insurance policy and ä_x is the actuarial present value of a fully discrete whole life annuity-due. We recall also from Section 41 that ä_x = _k=0^∞Σv^k*_kp_x.

Alternatively, _kV_x = 1 - ä_x+k/ä_x.

Each of the above formulas may be useful in different circumstances.

For a fully discrete n-year endowment insurance policy with actuarial present value A_x:n¬, the benefit reserve at time k < n can be denoted as _kV_x:n¬ and can be found as follows:

_kV_x:n¬ = 1 -ä_x+k:n-k¬/ä_x:n¬.

Prospective formula: _kV_x:n¬ = A_x+k:n-k¬ - P_x:n¬*ä_x+k:n-k¬

Retrospective formula: _kV_x:n¬ = (1/_kE_x)(P_x:n¬* ä_x:k¬ - A¹_x:k¬)

Recall: ä_x:n¬ = _k=0^n-1Σv^k*_kp_x, P_x:n¬= A_x:n¬/ä_x:n¬, _kE_x = vⁿ*_np_x, and A¹_x:n¬ = _k=0^n-1∑v^k+1*_kp_x*q_x+k.

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. 205, 215-218.

Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3 - Fall 2006.

Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3 - Fall 2007.

Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3L - Spring 2008.

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

Problem S3L58-1. Iago owns a whole life insurance policy which pays a benefit of 1 whose benefits are payable at the end of the year of death. Iago is currently 35 years old and can purchase a discrete whole life annuity-due that pays a benefit of 1 at the beginning of each year for $65. If Iago survived for another 17 years, he would be able to purchase such a whole life annuity-due for $56. Find the fully discrete benefit reserve on Iago's insurance policy 17 years from now.

Solution S3L58-1. We want to find ₁₇V₃₅, which we will do using the formula _kV_x = 1 - ä_x+k/ä_x.

We are given x = 35, k = 17, ä₃₅ = 65, and ä₅₂ = 56. Thus, ₁₇V₃₅ = 1 - 56/65 = ₁₇V₃₅ = about 0.1384615385.

Problem S3L58-2. Roland is 95 years old and owns a whole life insurance policy that pays a benefit of 1 at the end of the year of death and whose benefit premium 0.19. In 29 years, the actuarial present value of Roland's policy will be 0.89, and Roland will be able to purchase can purchase a discrete whole life annuity-due that pays a benefit of 1 at the beginning of each year for $5. Find the fully discrete benefit reserve on Roland's policy 29 years from now.

Solution S3L58-2. We want to find ₂₉V₉₅. We use the formula _kV_x = A_x+k - P_x*ä_x+k. We are given A₁₂₄ = 0.89, P₉₅ = 0.19, and ä₁₂₄ = 5. Thus, ₂₉V₉₅ = 0.89 - 0.19*5 = ₂₉V₉₅ = -0.06.

(Note: Nothing theoretically precludes benefit reserves from being negative. This problem is one example where a negative benefit reserve indeed occurs. A negative benefit premium simply means that the present value of future benefit premiums is greater than the present value of future benefits, under the condition that the insured life survives during the specified time period.)

Problem S3L58-3. Similar to Question 37 from the Casualty Actuarial Society's Fall 2006 Exam 3. The following probabilities are associated with the lives of rabbit-bears aged 5.

_k│q₅ = (k + 1)/15 for k = 0, 1, 2, 3, 4. The annual effective interest rate is 0.09. Reabtibbar the Rabbit-Bear is 5 years old has a whole life insurance policy that pays a benefit of 1 at the end of the year of death. Find the fully discrete benefit reserve on Reabtibbar's policy 3 years from now.

Solution S3L58-3. We use the formula _kV_x = 1 - ä_x+k/ä_x.

The above probabilities of death in each of the years 0, 1, 2, 3, 4 add to 1, indicating that Reabtibbar will not survive past 5 years from now. Thus, to find ä₅, we only need to consider payments in years 0 through 4. Thus, we use the formula ä_x = _k=0^∞Σv^k*_kp_x, which, applied to our case, is as follows.

ä₅ = 1 + v*p₅ + v²*₂p₅ + v³*₃p₅+ v⁴*₄p₅

ä₅ = 1 + (1/1.09)(1 - 1/15) + (1/1.09)²(1 - 3/15) + (1/1.09)³(1 - 6/15) + (1/1.09)⁴(1 - 10/15)

ä₅ = 3.229064933

To find ä₅₊₃ = ä₈ we only need to consider payments in years 3 and 4.

ä₈ = 1 + v*p₈. We find p₈ = (1 - 10/15)/(1 - 6/15) = (5/9).

Thus, ä₈ = 1 + (1/1.09)(5/9) = 1.509683996.

Thus, ₃V₅ = 1 - ä₈/ä₅ = 1 - 1.509683996/3.229064933 = ₃V₅ = about 0.5324702267.

Problem S3L58-4. Similar to Question 24 from the Casualty Actuarial Society's Spring 2008 Exam 3L. Arachne is currently 78 years old and owns a 7-year fully discrete endowment insurance policy that pays a benefit of 1. You know the following: ä_78:7¬ = 5.9. Find the ₆V_78:7¬, the fully discrete benefit reserve for this policy six years from now.

Solution S3L58-4. We use the formula _kV_x:n¬ = 1 - ä_x+k:n-k¬/ä_x:n¬. In our case, ₆V_78:7¬ = 1 - ä_84:1¬/ä_78:7¬. We only need to consider what the value of ä_84:1¬ might be. We use the formula ä_x:n¬ = _k=0^n-1Σv^k*_kp_x, so

ä_84:1¬ = _k=0^1-1Σv^k*_kp₈₄ = _k=0⁰Σv^k*_kp₈₄ = v⁰*₀p₈₄ = 1*1 = 1.

Thus, ₆V_78:7¬ = 1 - 1/5.9 = ₆V_78:7¬ = about 0.8305084746.

Note: Exam Question 24 from the Spring 2008 Exam 3L is slightly defective. It only makes sense if the value asked for is ₄V_35:5¬, the fully discrete benefit reserve, rather than the fully continuous benefit reserve.

Problem S3L58-5. Similar to Question 39 from the Casualty Actuarial Society's Fall 2007 Exam 3.

A given 30-year endowment insurance policy on the life of a 40-year-old white panther pays a benefit of $10000 and has a benefit premium of $20. You know that the one-year discount factor v is 0.95 and that p₄₀ = 0.88 for white panthers. Find the fully discrete benefit reserve one year from now (₁V_40:30¬) for this policy.

Solution S3L58-5. We use the retrospective formula _kV_x:n¬ = (1/_kE_x)(P_x:n¬* ä_x:k¬ - A¹_x:k¬)

In our case, ₁V_40:30¬ = (1/₁E₄₀)(P_40:30¬* ä_40:1¬ - A¹_40:1¬)

We use the formula _kE_x = vⁿ*_np_x to find ₁E₄₀ = v*p₄₀ = 0.95*0.88 = ₁E₄₀ = 0.836.

Since the formulas we use assume a benefit of 1 for the insurance policy, we need to scale our value of the benefit and the benefit premium by (1/10000) to make them fit the conditions of the formulas.

Thus, P_40:30¬ = 20/10000 = P_40:30¬ = 0.002.

Moreover, we use the formula A¹_x:n¬ = _k=0^n-1∑v^k+1*_kp_x*q_x+k to find

A¹_40:1¬ = _k=0⁰∑v^k+1*_kp₄₀*q_40+k = v*₀p₄₀*q₄₀ = 0.95*1*(1 - p₄₀) = 0.95(1 - 0.88) = A¹_40:1¬ = 0.114.

We use the formula ä_x:n¬ = _k=0^n-1Σv^k*_kp_x to find

ä_40:1¬ = _k=0^1-1Σv^k*_kp₄₀ = _k=0⁰Σv^k*_kp₄₀ = v⁰*₀p₄₀ = 1*1 = 1.

Thus, we have ₁V_40:30¬ = (1/0.836)(0.002*1 - 0.114) = -0.1339712919.

But we need to scale our value of ₁V_40:30¬ by 10000 to have it match the size of the policy's benefit.
Thus, our final answer is -0.1339712919*1000 = about -$1339.71.

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