Recovering Future Whole-Life Values from Term Insurance

Why a Whole-Life Value Can Reveal a Future Value

A whole-life insurance value at age $x$ combines two parts: the expected present value of benefits paid during the first $n$ years, and the expected present value of benefits paid later, provided the insured survives to age $x+n$. If the first part and the probability-discount factor for reaching year $n$ are known, the whole-life value can be rearranged to find the whole-life value at that future age.

The calculation concerns a discrete insurance with a unit benefit paid at the end of the year of death. Assume an effective annual interest rate $i$, and let $v=1/(1+i)$. Mortality is described by the one-year death probabilities $q_{x+k}$ and survival probabilities ${}_kp_x$. The same interest and mortality basis must be used for every value in the identity.

Definition: $A_x$ is the expected present value of a unit whole-life insurance benefit paid at the end of the year of death. $A^1_{x:n}$ is the expected present value of a unit term insurance benefit paid at the end of the year of death if death occurs within $n$ years. The pure endowment factor is ${}_nE_x=v^n\,{}_np_x$: the probability of surviving $n$ years, multiplied by the discount factor for $n$ years.

For example, the $k+1$st possible annual whole-life payment has present value $v^{k+1}$ if it is paid. Its probability of being paid is ${}_kp_xq_{x+k}$: the insured must survive $k$ years and then die during the next year. Thus $A_x$ sums these expected present values across all possible years of death. The term value includes only the first $n$ years of that sum.

The Decomposition and Its Rearrangement

The whole-life value can be split at duration $n$. The first component is the $n$-year term value. For the later component, the insured must first survive to age $x+n$; conditional on doing so, the remaining insurance has the same value as a whole-life policy issued at that age. Discounting and survival to year $n$ together give the factor ${}_nE_x$.

Theorem: For a unit-benefit discrete whole-life insurance and an $n$-year term insurance valued on the same interest and mortality basis, $$ A_x=A^1_{x:n}+{}_nE_xA_{x+n}. $$ If ${}_nE_x>0$, then $$ A_{x+n}=\frac{A_x-A^1_{x:n}}{{}_nE_x}. $$

The condition ${}_nE_x>0$ matters: division is possible only if there is a positive probability of surviving to duration $n$. For ordinary life-table values at a reachable age, this factor is positive. The rearrangement also requires matching benefit amounts and benefit timing. Values for a continuous insurance, or for a benefit paid at a different time, cannot be substituted without changing the decomposition.

Why the Future Component Has This Form

The identity follows by separating the terms in the whole-life present-value sum. The term portion is the sum for deaths during the first $n$ years. For every later possible year of death, factor out the discount and survival to year $n$; the terms left over are exactly the whole-life sum at age $x+n$.

A_x=\sum_{k=0}^{\infty}v^{k+1}\,{}_kp_xq_{x+k}

The first $n$ terms in this sum are $A^1_{x:n}$. For the tail, set $k=n+j$, where $j=0,1,2,\ldots$. The survival probabilities factor as ${}_{n+j}p_x={}_np_x\,{}_jp_{x+n}$, so

\begin{aligned} \sum_{j=0}^{\infty}v^{n+j+1}\,{}_{n+j}p_xq_{x+n+j} &=v^n\,{}_np_x \sum_{j=0}^{\infty}v^{j+1}\,{}_jp_{x+n}q_{x+n+j}\\ &={}_nE_xA_{x+n}. \end{aligned}

This tail argument depends on the correct survival interval after age $x+n$. In an alternative indexing where the first tail payment is labelled $j=1$, the probability of surviving the $j-1$ years after reaching age $x+n$ is ${}_{j-1}p_{x+n}$, not ${}_jp_{x+n-1}$. In fact,

{}_{n+j-1}p_x={}_np_x\,{}_{j-1}p_{x+n}.

The factorization also preserves the death probability $q_{x+n+j-1}$ for the year in which the benefit is paid. Using a survival probability over the wrong interval changes the tail sum and breaks the identity.

Recovering the Value Step by Step

Confirm the contract basis.
Check that the whole-life and term values both use unit benefits paid at the end of the year of death, and that they share the same interest and mortality assumptions.

Find the pure endowment factor.
Calculate ${}_nE_x=v^n\,{}_np_x$. It includes both the probability of surviving $n$ years and discounting over those years.

Remove the term value.
Calculate $A_x-A^1_{x:n}$. This is the part of the whole-life value attributable to payments after the term period.

Divide by the factor.
Divide the remaining value by ${}_nE_x$ to recover $A_{x+n}$. Check that the result is between 0 and 1 for a unit-benefit insurance.

Worked Calculations

Worked Example: Recovering a Future Value After Two Years

A hypothetical policy has a unit benefit paid at the end of the year of death. The effective annual interest rate is $i=25\%$, so $v=1/1.25=0.8$. The mortality probabilities for the first two years are $q_x=0.10$ and $q_{x+1}=0.20$. At ages $x+2$ and later, assume a constant annual death probability of $0.50$. Find $A_{x+2}$ by first calculating $A_x$ and $A^1_{x:2}$, then applying the rearranged identity. This example also checks the identity directly from the stated mortality assumptions.

Method. Calculate the first two term benefits and the pure endowment factor, then calculate the future whole-life value using the constant mortality assumption. Use the decomposition to check the result.

For $A^1_{x:2}$, the year-one benefit has probability $q_x=0.10$, and the year-two benefit has probability ${}_1p_xq_{x+1}=0.90(0.20)=0.18$. Therefore,

A^1_{x:2}=0.8(0.10)+0.8^2(0.18)=0.08+0.1152=0.1952.

The survival probability to year two is ${}_2p_x=(1-0.10)(1-0.20)=0.90(0.80)=0.72$. Hence

{}_2E_x=0.8^2(0.72)=0.64(0.72)=0.4608.

From age $x+2$, the annual death probability is $q=0.50$ each year, with survival probability $p=0.50$. The whole-life value is the sum of a geometric series:

A_{x+2}=\sum_{k=0}^{\infty}v^{k+1}p^kq =\frac{vq}{1-vp} =\frac{0.8(0.50)}{1-0.8(0.50)} =\frac{0.40}{0.60} =\frac{2}{3}.

Thus $A_x=0.1952+0.4608(2/3)=0.1952+0.3072=0.5024$. Rearranging recovers

A_{x+2}=\frac{0.5024-0.1952}{0.4608} =\frac{0.3072}{0.4608} =\frac{2}{3}.

Check. The recovered value agrees with the geometric-series value. A common mistake is to use $v^2=0.64$ alone as the denominator, giving $0.3072/0.64=0.48$. That omits the probability of surviving two years; the correct denominator is $v^2{}_2p_x=0.4608$.

Worked Example: A Three-Year Term and a Future Constant Mortality Rate

For a hypothetical unit-benefit policy, let $i=25\%$, so $v=0.8$. The death probabilities over the first three years are $q_x=0.20$, $q_{x+1}=0.25$, and $q_{x+2}=0.50$. At age $x+3$ and later, let the annual death probability remain $0.50$. Calculate $A^1_{x:3}$, ${}_3E_x$, and $A_x$, then recover $A_{x+3}$.

Method. Sum the first three expected discounted benefits for the term value. Then multiply the three-year survival probability by $v^3$ to obtain the pure endowment factor.

The survival probabilities before each of the three possible term benefits are $1$, $0.80$, and $0.80(0.75)=0.60$. Thus

\begin{aligned} A^1_{x:3} &=0.8(0.20)+0.8^2(0.80)(0.25)+0.8^3(0.60)(0.50)\\ &=0.16+0.128+0.1536\\ &=0.4416. \end{aligned}

Survival through all three years is $0.80(0.75)(0.50)=0.30$, and $v^3=0.8^3=0.512$. Therefore ${}_3E_x=0.512(0.30)=0.1536$. The future whole-life value, with $q=0.50$ at every future age, is

A_{x+3}=\frac{0.8(0.50)}{1-0.8(0.50)} =\frac{0.40}{0.60} =\frac{2}{3}.

The corresponding whole-life value at age $x$ is

A_x=0.4416+0.1536\left(\frac{2}{3}\right) =0.4416+0.1024 =0.5440.

Now solve for the future value:

A_{x+3}=\frac{0.5440-0.4416}{0.1536} =\frac{0.1024}{0.1536} =\frac{2}{3}.

Check. The result agrees with the future constant-mortality calculation. The third-year term benefit is discounted by $v^3$ and weighted by survival to the start of that year, $0.60$; using survival only to age $x+1$ would assign the wrong probability.

Worked Example: Recovering a Value from Exact Fractions

A hypothetical unit-benefit policy has $i=10\%$, so $v=10/11$. Let $q_x=0.10$ and $q_{x+1}=0.20$. From age $x+2$ onward, assume $q=0.25$ each year. Find $A_{x+2}$ using $A_x$, $A^1_{x:2}$, and ${}_2E_x$. Keep exact fractions during the calculation and report the final answer to four decimal places.

Method. Calculate the two-year term value and pure endowment factor, then form the whole-life value using the stipulated future mortality. Use the exact fractions to avoid introducing intermediate rounding error.

The first-year term contribution is $v(0.10)=1/11$. The probability of surviving the first year and dying in the second is $0.90(0.20)=0.18=9/50$. Its discounted contribution is $(100/121)(9/50)=18/121$. Therefore,

A^1_{x:2}=\frac{1}{11}+\frac{18}{121} =\frac{11}{121}+\frac{18}{121} =\frac{29}{121}.

The two-year survival probability is $0.90(0.80)=0.72=18/25$. Thus

{}_2E_x=\left(\frac{10}{11}\right)^2\frac{18}{25} =\frac{72}{121}.

For the future whole-life value, the annual survival probability is $0.75$. The geometric-series denominator is $1-v(0.75)=1-15/22=7/22$, while $vq=(10/11)(1/4)=5/22$. Therefore $A_{x+2}=(5/22)/(7/22)=5/7$. The whole-life value at age $x$ is consequently

A_x=\frac{29}{121}+\frac{72}{121}\left(\frac{5}{7}\right) =\frac{203}{847}+\frac{360}{847} =\frac{563}{847}.

Applying the rearrangement gives

A_{x+2} =\frac{\frac{563}{847}-\frac{29}{121}}{\frac{72}{121}} =\frac{\frac{360}{847}}{\frac{72}{121}} =\frac{5}{7} =0.7143.

Check. The exact recovered fraction is the same as the future geometric-series value. The final answer is rounded to four decimal places; no earlier quantity was rounded.

Consistency Checks and Common Traps

The rearrangement is simple, but its inputs have to match the decomposition. Before accepting a result, check both the policy basis and the numerical range. For a unit benefit, each expected present value is nonnegative and no greater than 1, because the benefit is at most 1 and is discounted. In particular, a recovered $A_{x+n}$ outside $[0,1]$ signals inconsistent inputs, a mismatched basis, or an arithmetic error.

Leaving out survival. The denominator is ${}_nE_x=v^n{}_np_x$, not $v^n$. The insured must both survive to duration $n$ and have the future value discounted back.
Subtracting the wrong term value. Subtract $A^1_{x:n}$, which covers exactly the first $n$ years, not a term value with a different duration.
Mixing benefit timing. The displayed identity assumes benefits paid at the end of the year of death. A continuous benefit or a different payment time changes the values and requires a matching identity.
Using inconsistent assumptions. If the whole-life value, term value, and pure endowment factor use different interest or mortality assumptions, the decomposition need not hold.

This method is useful when a future-age whole-life value is not given directly but the whole-life value at the starting age, a matching term value, and the survival-discount factor are available. It also provides a direct check: the whole-life value should equal the term component plus the future component weighted by the probability-discount factor.

Check Your Understanding

Use the unit-benefit decomposition. Unless a question gives an interest rate and mortality assumptions, treat the stated values as already calculated on the same discrete insurance basis.

Given $A_x=0.62$, $A^1_{x:4}=0.26$, and ${}_4E_x=0.45$, find $A_{x+4}$. (Answer: $0.8000$)
A matching three-year term value is $0.31$, the pure endowment factor is $0.40$, and $A_{x+3}=0.70$. Find $A_x$. (Answer: $0.5900$)
A hypothetical policy has $i=25\%$, so $v=0.8$, and first-two-year death probabilities $q_x=0.20$, $q_{x+1}=0.25$. Given $A_x=0.4800$, calculate $A^1_{x:2}$, ${}_2E_x$, and $A_{x+2}$. (Answer: $A^1_{x:2}=0.2880$, ${}_2E_x=0.3840$, $A_{x+2}=0.5000$)
Suppose $A_x=0.66$, $A^1_{x:5}=0.21$, and ${}_5E_x=0.50$. Recover $A_{x+5}$ and state whether it passes the unit-benefit range check. (Answer: $0.9000$; yes, it is in $[0,1]$)
A matching whole-life value is $0.72$ and ${}_6E_x=0.60$. If the six-year term value increases from $0.24$ to $0.27$, with the whole-life value and factor unchanged, what is the new recovered value $A_{x+6}$? (Answer: $0.7500$)

What You'll Learn