Insurances Payable at the End of the Year of Death: Practice Problems and Solutions

When insurance policies pay at the end of the year of death, based solely on the number of complete years liked by the insured person prior to death. The random variable used in describing such policies is not the future-lifetime variable T, but rather the curate-future-lifetime random variable K. If the insured person dies in year k+1 of the insurance policy, then the curate-future-lifetime of the insured person is k. The present value of the benefit payment for such an insurance policy is z_k+1 and is expressed as follows.

z_k+1 = b_k+1v_k+1.

z_k+1 = Z is the present value, at policy issue, of the benefit payment.

b_k+1is the benefit function.

V_k+1is the discount function. v is the one-year discount factor by which a sum of money payable one year from now is multiplied to get its present value today. If the annual effective interest rate is r, then v = 1/(1+r).

Suppose we have an n-year term life insurance policy which pays one unit in benefits at the end of the year of death. The actuarial present value for this insurance policy is denoted as A¹_x:n¬ and can be found as follows:

E[Z] =A¹_x:n¬ = _k=0^n-1∑v^k+1*_kp_x*q_x+k

If δ_k+1 is the annual force of interest at time k+1, and the rule of moments holds for moments of E[Z]: E[Z^j] @ δ_k+1 = E[Z] @ jδ_k+1.

Thus, E[Z²] = ²A¹_x:n¬ = _k=0^n-1∑e^-2δ(k+1) *_kp_x*q_x+k

Var(Z) = ²A¹_x:n¬ - (A¹_x:n¬)²

The following recursion relation enables us to compute A¹_x:n¬when A¹_x+1:n-1¬is known:

A¹_x:n¬ = vq_x + vp_x*A¹_x+1:n-1¬

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 109-110.

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

Problem S3L33-1. 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 currently 0.06. Honorius the Triceratops is currently 2 years old and takes out a 3-year term life insurance policy paying a benefit of 1 at the end of the year of death. Find the actuarial present value of this policy.

Solution S3L33-1. We use the formula A¹_x:n¬ = _k=0^n-1∑v^k+1*_kp_x*q_x+k.

We know that δ = 0.06, so v = e^-0.06. We are given that x = 2 and n = 3.

We find _kp₂ = s(x + k)/s(x) = s(2 + k)/s(2) = e^-0.34(2+k)/e^-0.34(2) = e^-0.34k

We find q_2+k = 1 - s(3 + k)/s(2 + k) = 1 - e^-0.34.

Thus, A¹_2:3¬ = _k=0²∑ e^-0.06(k+1)*e^-0.34k*(1 - e^-0.34) =

e^-0.06*(1 - e^-0.34) + e^-0.12*e^-0.34*(1 - e^-0.34) + e^-0.18*e^-0.68*(1 - e^-0.34) =

(1 - e^-0.34)(e^-0.06 + e^-0.46 + e^-0.86) = A¹_2:3¬ = about 0.5753670393.

Problem S3L33-2. 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 currently 0.06. Honorius the Triceratops is currently 2 years old and takes out a 3-year term life insurance policy paying a benefit of 1 at the end of the year of death. Find the second moment of the present-value random variable for this policy.

Solution S3L33-2. We seek to find E[Z²] = ²A¹_x:n¬ = _k=0^n-1∑e^-2δ(k+1) *_kp_x*q_x+k for x = 2 and n = 3. We know from Solution S3L33-1 that v = e^-0.06, _kp₂ = e^-0.34k, and q_2+k = 1 - e^-0.34.

Thus, ²A¹_2:3¬ = _k=0²∑e^-0.12(k+1) * e^-0.34k *(1 - e^-0.34) =

e^-0.12*(1 - e^-0.34) + e^-0.24*e^-0.34*(1 - e^-0.34) + e^-0.36*e^-0.68*(1 - e^-0.34) =

(1 - e^-0.34)(e^-0.12 + e^-0.58 + e^-1.04) = ²A¹_2:3¬ = about 0.5188922456.

Problem S3L33-3. 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 currently 0.06. Honorius the Triceratops is currently 2 years old and takes out a 3-year term life insurance policy paying a benefit of 1 at the end of the year of death. Find the variance this policy.

Solution S3L33-3. We use the formula Var(Z) = ²A¹_x:n¬ - (A¹_x:n¬)².

From Solutions S3L33-1 and S3L33-2, we know that ²A¹_2:3¬ = 0.5188922456 and A¹_2:3¬ = 0.5753670393. Thus, Var(Z) = 0.5188922456 - (0.5753670393)² = Var(Z) = about 0.1878450157.

Problem S3L33-4. 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 currently 0.06. Rex the Triceratops is currently 3 years old and takes out a 2-year term life insurance policy paying a benefit of 1 at the end of the year of death. Find the actuarial present value of this policy. (Hint: This does not have to be done directly! A past solution in this section will help you.)

Solution S3L33-4. We use the recursion formula A¹_x:n¬ = vq_x + vp_x*A¹_x+1:n-1¬.

We already know from Solution S3L33-1 that A¹_2:3¬ = 0.5753670393.

Let x = 2 and n = 3. Then our desired value, A¹_3:2¬, is A¹_x+1:n-1¬ in this case.

We know that v = e^-0.06. Moreover, because s(x) = e^-0.34x is the survival function pertaining to an exponential distribution, it is the case that p_x = e^-0.34and q_x = (1 - e^-0.34).

We can rearrange our formula to solve directly for A¹_x+1:n-1¬:

(A¹_x:n¬ - vq_x)/(vp_x) = A¹_x+1:n-1¬

Thus, (0.5753670393 - e^-0.06(1 - e^-0.34))/( e^-0.4) = A¹_3:2¬ = about 0.4533991689.

Problem S3L33-5. Tibbar the Rabbit is deciding between two term life insurance policies, each of whose benefits is payable at the end of the year of death. Tibbar is currently 12 years old and knows that his probability of surviving to the end of the next year is 0.88. The policies he chooses between each pay one Golden Hexagon (GH) in benefits. Policy A is a 5-year policy that Tibbar would enter into immediately (at age 12). Policy A has an actuarial present value of 0.46. Policy B is a 4-year policy that Tibbar would enter into one year from now (at age 13). Policy B has an actuarial present value of 0.43. Find the annual force of interest.

Solution S3L33-5. We are given that p₁₂ = 0.88, and so q₁₂ = 1 - p₁₂ = 0.12. Moreover, we are given that A¹_12:5¬ = 0.46 and A¹_13:4¬ = 0.43. We need to find v.

We use the recursion formula A¹_x:n¬ = vq_x + vp_x*A¹_x+1:n-1¬ for x = 12 and n = 5.

We rearrange the formula thus:

A¹_x:n¬ = v(q_x + p_x*A¹_x+1:n-1¬)

v = A¹_x:n¬/(q_x + p_x*A¹_x+1:n-1¬)

Thus, v = 0.46/(0.12 + 0.88*0.43)

v = 0.922953451

e^-δ = 0.922953451

δ = -ln(0.922953451) = δ = about 0.080176478.

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