Value Term and Whole-Life Benefits from Shared Mortality Data

Introduction

A term insurance value and a whole-life insurance value are not separate calculations when they use the same mortality basis. The whole-life benefit can be divided according to when death occurs: during the first $n$ years, or after the first $n$ years. The first part is the $n$-year term benefit; the second part is the value at issue of continuing whole-life coverage, provided the insured survives to the end of the term.

This division provides both a calculation method and a strong check. It is especially useful when a problem supplies mortality over an initial period and a continuation value at the end of that period. The key is to handle the boundary consistently: a death in year $n$ belongs to the term benefit, while continuation coverage applies only to a life who survives $n$ full years.

Discrete Death Benefits on an Annual Basis

Assume an annual effective interest rate $i$, with $v=1/(1+i)$. Assume death benefits are paid at the end of the policy year of death. For a life aged $x$, let ${}_kp_x$ be the probability of surviving $k$ years and let $q_{x+k}$ be the probability of dying during the following year, conditional on being alive at age $x+k$. Thus the probability of surviving $k$ years and then dying in the next year is ${}_kp_xq_{x+k}$. The mortality probabilities and interest rate must use compatible annual periods.

Definition: For a unit benefit payable at the end of the year of death, the discrete whole-life insurance value is $$ A_x=\sum_{k=0}^{\infty}v^{k+1}{}_kp_xq_{x+k}. $$ Define $T_{x:n}$ as the value at age $x$ of a unit $n$-year term insurance benefit, payable at the end of the year of death if death occurs within the first $n$ years: $$ T_{x:n}=\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}. $$

The term sum has $n$ possible payment years, indexed by $k=0,\ldots,n-1$. Its $k$th summand discounts a payment $k+1$ years from issue and weights it by the probability of surviving to the start of that year and dying in it. A benefit amount $B$, rather than a unit benefit, has value $B T_{x:n}$ or $B A_x$, because the expected present value scales linearly with the benefit.

Why the Complementary-Interval Check Works

The whole-life sum includes every possible year of death. Divide its terms into those with $k=0,\ldots,n-1$ and those with $k\ge n$. The first group is exactly the term value. For a death in year $n+j+1$, the insured must first survive $n$ years, then survive another $j$ years from age $x+n$, and then die in the next year. Conditional survival gives ${}_{n+j}p_x={}_np_x\,{}_jp_{x+n}$. Also, the payment discount separates as $v^{n+j+1}=v^n v^{j+1}$. Therefore the later-death terms together equal the probability of surviving $n$ years, discounted for $n$ years, multiplied by the whole-life value at age $x+n$.

\begin{aligned} A_x &=T_{x:n}+\sum_{j=0}^{\infty}v^{n+j+1}{}_np_x\,{}_jp_{x+n}q_{x+n+j}\\ &=T_{x:n}+v^n{}_np_x A_{x+n}. \end{aligned}

Theorem: On the same annual interest and mortality basis, a unit whole-life value satisfies $$ A_x=T_{x:n}+v^n{}_np_xA_{x+n}. $$ The continuation term is not simply $A_{x+n}$: it must be discounted back $n$ years and weighted by the probability of surviving to age $x+n$.

This identity requires the term and whole-life values to use the same benefit timing, interest rate, and mortality basis. It is an identity for death benefits, not a statement that the term contract pays a benefit on survival. If a contract pays a separate survival benefit, that benefit must be valued according to its own terms.

Worked Example: Constant Mortality and a Three-Year Term

A hypothetical insured has annual death probability $q=0.10$ at every age, so $p=0.90$, and the annual effective interest rate is $i=5\%$. Find the unit three-year term value and the unit whole-life value. Check the whole-life value by splitting deaths at the end of year 3.

Method. Use the discrete annual death-benefit sums. Here $v=1/1.05=20/21$, and $vp=(20/21)(0.90)=6/7$. The first-year discounted death term is $vq=(20/21)(0.10)=2/21$.

The three-year term value includes death in each of the first three years:

\begin{aligned} T_{x:3} &=vq+v^2pq+v^3p^2q\\ &=\frac{2}{21}\left(1+\frac{6}{7}+\frac{36}{49}\right)\\ &=\frac{2}{21}\left(\frac{127}{49}\right) =\frac{254}{1029} \approx 0.246841. \end{aligned}

For whole life, each additional year's discounted death term is multiplied by $vp=6/7$. The geometric series converges because $6/7<1$:

A_x=\frac{vq}{1-vp} =\frac{2/21}{1-6/7} =\frac{2}{3} \approx 0.666667.

The survival-continuation component after year 3 is $(vp)^3A_{x+3}$. Mortality is unchanged at the attained age, so $A_{x+3}=2/3$. Thus \[ (vp)^3A_{x+3}=\left(\frac{6}{7}\right)^3\frac{2}{3} =\frac{144}{343}\approx0.419825. \] The reconciliation is \[ \frac{254}{1029}+\frac{144}{343} =\frac{254+432}{1029} =\frac{2}{3}. \] The term value plus the continuation value equals the whole-life value. The common mistake avoided here is to use $A_{x+3}$ without discounting and without the probability of surviving three years.

Using Mortality That Changes by Year

When mortality varies by attained age, calculate each year’s death probability from the applicable survival probability and conditional $q$. Do not replace a sequence of annual probabilities with one $q$ unless the model explicitly assumes constant mortality. For the continuation part, use mortality beginning at age $x+n$, because that is the age at which the surviving insured enters the remaining coverage period.

Worked Example: Different Mortality in the Term Years

A hypothetical policy has a unit death benefit payable at the end of the year of death. The annual effective interest rate is $25\%$. The annual mortality probabilities are $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$. From age $x+3$ onward, assume $q=0.20$ every year. Find the three-year term value and the whole-life value.

Method. Here $v=1/1.25=0.80$. The survival probabilities to the starts of years 1, 2, and 3 are $1$, $0.90$, and $(0.90)(0.80)=0.72$. The three term contributions are:

\begin{aligned} \text{Year 1: }&v q_x=(0.80)(0.10)=0.08000,\\ \text{Year 2: }&v^2p_xq_{x+1}=(0.64)(0.90)(0.20)=0.11520,\\ \text{Year 3: }&v^3p_xp_{x+1}q_{x+2} =(0.512)(0.90)(0.80)(0.25)=0.09216. \end{aligned}

Therefore $T_{x:3}=0.08000+0.11520+0.09216=0.28736$. At age $x+3$, mortality is constant at $q=0.20$, so $p=0.80$ and the whole-life value from that age is \[ A_{x+3}=\frac{vq}{1-vp} =\frac{(0.80)(0.20)}{1-(0.80)(0.80)} =\frac{0.16}{0.36} =\frac{4}{9}. \] The probability of surviving three years is $0.90(0.80)(0.75)=0.54$. The discounted continuation value is therefore \[ v^3{}_3p_xA_{x+3}=(0.512)(0.54)\left(\frac{4}{9}\right)=0.12288. \] Consequently, \[ A_x=0.28736+0.12288=0.41024. \] The check passes because the term portion covers death in years 1 through 3, while the continuation portion covers only death after surviving those three years.

Worked Example: Reconcile Values for a Specified Benefit

A hypothetical insurer offers a $ \$50{,}000 $ whole-life death benefit. The annual effective interest rate is $25\%$. Mortality is $q_x=0.05$, $q_{x+1}=0.10$, and $q=0.20$ from age $x+2$ onward. Find the value of the first two years of term coverage, the value of the later-death component, and the whole-life value.

Method. With $v=0.80$, the first-year term contribution per dollar of benefit is $(0.80)(0.05)=0.04000$. The second-year contribution is $(0.80^2)(0.95)(0.10)=0.06080$. Thus the two-year term value per dollar is $0.10080$, and for the specified benefit it is \[ \$50{,}000(0.10080)=\$5{,}040.00. \]

At age $x+2$, the continuation whole-life value per dollar is $4/9$, as follows from constant $q=0.20$ and $v=0.80$. The probability of surviving two years is $(0.95)(0.90)=0.855$, and $v^2=0.64$. The later-death component per dollar at issue is \[ (0.64)(0.855)\left(\frac{4}{9}\right)=0.24320. \] For a $ \$50{,}000 $ benefit, this is $\$50{,}000(0.24320)=\$12{,}160.00$. The whole-life value is therefore \[ \$5{,}040.00+\$12{,}160.00=\$17{,}200.00. \]

As a check, the total value per dollar is $0.10080+0.24320=0.34400$, and $\$50{,}000(0.34400)=\$17{,}200.00$. A frequent error is to apply the continuation value to all policies rather than only those surviving the two-year term; doing so would omit both the survival weighting and the time discount.

Common Exam Traps and Practical Checks

The interval identity is simple, but its timing conventions are easy to misread. Before calculating, write down the benefit payment time, the years included in the term, and the survival probability to the continuation age. Then check that the death years in the two components do not overlap and leave no gap.

Off-by-one indexing: an $n$-year term includes deaths in years 1 through $n$, so its sum uses $k=0$ through $n-1$.
Wrong discount exponent: a death in year $k+1$ is discounted by $v^{k+1}$, not $v^k$, when the benefit is paid at the end of the year of death.
Missing survival to the boundary: continuation coverage begins only for lives surviving $n$ years, so its issue-age weight is $v^n{}_np_x$.
Changing the mortality basis: the $A_{x+n}$ used in the check must start at age $x+n$ and follow the same mortality assumptions as the tail of the original whole-life contract.
Mixing benefit timing: an annual end-of-year death benefit is not the same payment pattern as a benefit paid immediately at death. Use formulas that match the stated payment timing.

For numerical work, calculate the term contributions separately before summing. This makes it easier to locate a wrong $q$, survival product, or discount exponent. For a final reconciliation, compare $A_x$ with $T_{x:n}+v^n{}_np_xA_{x+n}$; a discrepancy usually points to an inconsistent boundary or basis.

Key takeaway: A discrete whole-life death-benefit value splits exactly into the value of deaths during the first $n$ years and the discounted, survival-weighted whole-life value for deaths after year $n$. This calculation chain is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use annual end-of-year death benefits and the stated annual effective interest rate. Answers are per unit benefit unless a dollar benefit is specified.

A life has constant $q=0.10$, and $i=0\%$. Find the value of a four-year term insurance. (Answer: $0.4000$)
A life has constant $q=0.20$, and $i=25\%$. Find the two-year term value. (Answer: $0.2624$)
At interest rate $i=25\%$, let $q_x=0.10$, $q_{x+1}=0.20$, and $q=0.20$ at all later ages. Find $T_{x:2}$, the discounted continuation value, and $A_x$. (Answer: $T_{x:2}=0.1952$, continuation value $=0.2048$, and $A_x=0.4000$)
At $i=25\%$, let $q_x=0.05$, $q_{x+1}=0.10$, and $q=0.20$ from age $x+2$ onward. Find the two-year term value and the whole-life value. (Answer: $T_{x:2}=0.1008$ and $A_x=0.3440$)
A whole-life value is $A_x=0.41024$, and the three-year term value on the same basis is $T_{x:3}=0.28736$. The annual effective interest rate is $25\%$, and ${}_3p_x=0.54$. Find $A_{x+3}$ using the interval identity. (Answer: $A_{x+3}=4/9\approx0.4444$)

What You'll Learn