Whole-Life Insurance Split at a Fixed Duration

Why Split a Whole-Life Insurance Value?

A whole-life insurance benefit can be valued by considering every possible year of death. Sometimes it is more useful to divide those possibilities at a fixed duration \(n\): deaths during the first \(n\) years are valued as term insurance, while deaths after year \(n\) are valued from the insured’s age at that point. This split can make a calculation shorter and can reveal which part of a contract drives its value.

The split is an identity for present values, not a change to the insurance benefit. In particular, surviving to year \(n\) does not itself produce a payment under the whole-life insurance considered here. This topic is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Consider a discrete whole-life insurance issued at age \(x\), with a unit benefit paid at the end of the year of death. Let \(A_x\) be its expected present value at effective annual interest rate \(i\). For a positive integer \(n\), let \(A^1_{x:n}\) be the expected present value of an \(n\)-year term insurance with the same benefit and timing, and let \({}_nE_x=v^n\,{}_np_x\) be the pure endowment factor, where \(v=1/(1+i)\). Then \(A_{x+n}\) is the value of the same discrete whole-life insurance issued at age \(x+n\).

The Fixed-Duration Identity

Assume a fixed effective annual interest rate, a benefit of one unit payable at the end of the year of death, and a life whose annual survival probabilities are those specified by the mortality model. The identity is

A_x=A^1_{x:n}+{}_nE_xA_{x+n}.

The term insurance component accounts for death in one of the first \(n\) policy years. The second component accounts for survival through \(n\) years followed by a later death. The factor \({}_nE_x\) discounts for \(n\) years and weights by the probability of surviving those years; it does not represent a benefit payable at year \(n\) in the whole-life contract.

Result: Under the assumptions above, \(A^1_{x:n}=\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}\), and \({}_nE_x=v^n{}_np_x\). Thus the identity separates the whole-life value into deaths during the first \(n\) years and deaths after survival to year \(n\).

Why the Identity Holds

The proof is a partition of the possible death years. Let \(k\) count completed years lived after age \(x\), so death in the \((k+1)\)st policy year has probability \({}_kp_xq_{x+k}\) and produces a payment discounted by \(v^{k+1}\). For the whole-life insurance, sum these discounted probabilities over all \(k\geq 0\). The first \(n\) terms are exactly the \(n\)-year term insurance value:

A_x=\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k} +\sum_{k=n}^{\infty}v^{k+1}{}_kp_xq_{x+k}.

For the second sum, write \(k=n+j\). Survival to age \(x+n\) has probability \({}_np_x\). Conditional on reaching that age, the remaining death-year probabilities are those for a whole-life insurance issued at age \(x+n\). The discount factor also separates:

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

Combining the two sums proves the identity. The argument relies on consistent benefit timing and on using the same interest and mortality assumptions in every component. It does not apply unchanged if, for example, a maturity benefit is added at year \(n\), the benefit amount varies by death year, or the payment timing differs.

Calculation Method

Fix the split duration.
Identify the first \(n\) policy years and the age \(x+n\) reached at the split.

Value deaths before the split.
For each year \(k+1\), multiply the probability of death in that year by \(v^{k+1}\), then sum for \(k=0,\ldots,n-1\).

Value survival to the split.
Calculate \({}_nE_x=v^n{}_np_x\). This combines survival to year \(n\) and discounting to that time.

Attach the future whole-life value.
Multiply \({}_nE_x\) by \(A_{x+n}\), then add the term value. Check that the result is consistent with the whole-life death-year probabilities.

Worked Example: Splitting at the End of Two Years

A hypothetical discrete whole-life policy pays one unit at the end of the year of death. The effective annual interest rate is \(i=5\%\). Mortality is \(q_x=0.10\), \(q_{x+1}=0.20\), and the whole-life value at age \(x+2\) is \(A_{x+2}=0.35\). Calculate \(A_x\) by splitting at \(n=2\). The method is to value the first two death years as term insurance, then multiply the future value by the two-year pure endowment factor.

First, \(v=1/1.05=0.95238095\) and \(v^2=0.90702948\). Death in year 1 has probability \(0.10\). Survival through year 1 is \(p_x=1-0.10=0.90\), so death in year 2 has probability \(0.90(0.20)=0.18\). Therefore,

A^1_{x:2}=0.10v+0.18v^2 =0.10(0.95238095)+0.18(0.90702948) =0.09523810+0.16326531 =0.25850340.

The probability of surviving both years is \(0.90(0.80)=0.72\). Hence

{}_2E_x=v^2{}_2p_x=0.90702948(0.72)=0.65306122, \qquad {}_2E_xA_{x+2}=0.65306122(0.35)=0.22857143.

Adding the two parts gives \(A_x=0.25850340+0.22857143=0.48707483\), or 0.4871 to four decimal places. As a check, the first two death-year contributions are \(0.09523810\) and \(0.16326531\), and all later contributions together equal \(0.22857143\) by the stated future value. Their sum matches the split calculation. A common mistake is to use the survival probability \(0.72\) without discounting; that would omit the factor \(v^2\) from the future component.

Using the Identity to Find an Unknown

The split can also turn a whole-life value into an equation for an unknown mortality probability. In that use, be careful to distinguish the probability of dying in a year from the probability of surviving that year. The future whole-life value is multiplied by the probability of surviving to the split age, not by the probability of dying during the final year of the term.

Worked Example: Solving for a Second-Year Mortality Probability

A hypothetical discrete whole-life insurance has unit benefit payable at the end of the year of death and effective annual interest \(i=5\%\). Its value at age \(x\) is \(A_x=0.50\). The first-year mortality probability is \(q_x=0.08\), the second-year mortality probability is an unknown \(r=q_{x+1}\), and \(A_{x+2}=0.35\). Find \(r\). Use the split at \(n=2\), keeping the year-2 death probability separate from survival through year 2.

The probability of death in year 1 is \(0.08\). The probability of death in year 2 is \(0.92r\), and the probability of surviving two years is \(0.92(1-r)\). With \(v=1/1.05\), the identity gives

0.50=v(0.08)+v^2(0.92)\bigl[r+0.35(1-r)\bigr].

The first term is \(v(0.08)=0.95238095(0.08)=0.07619048\). Also, \(v^2(0.92)=0.90702948(0.92)=0.83446712\). Thus

\begin{aligned} r+0.35(1-r) &=\frac{0.50-0.07619048}{0.83446712}\\ &=\frac{0.42380952}{0.83446712}\\ &\approx 0.5078804. \end{aligned}

Since \(r+0.35(1-r)=0.35+0.65r\),

r=\frac{0.5078804-0.35}{0.65}\approx 0.2429.

The result is a valid probability. Substitution checks it: \(0.35+0.65(0.2429)=0.507885\), and \(0.07619048+0.83446712(0.507885)\approx0.5000\). The common mistake is to omit the survival factor \(0.92\) before year 2, or to apply \(r\) to the future value as though survival were \(r\). The year-2 death and the survival-to-age-\(x+2\) cases are distinct events.

Checking With Several Death Years

A direct death-year calculation is a useful check, especially when the split duration is more than one year. If mortality is supplied year by year, calculate each death probability from survival up to that year; do not treat the annual mortality probabilities as unconditional probabilities from age \(x\).

Worked Example: A Three-Year Split With Changing Mortality

A hypothetical discrete whole-life policy pays one unit at the end of the year of death. Let \(i=4\%\), \(q_x=0.05\), \(q_{x+1}=0.10\), \(q_{x+2}=0.20\), and \(A_{x+3}=0.40\). Calculate \(A_x\) by splitting at \(n=3\), then check by identifying the first three unconditional death probabilities.

Here \(v=1/1.04=0.96153846\), \(v^2=0.92455621\), and \(v^3=0.88899636\). The probabilities of death in years 1, 2, and 3 are respectively \(0.05\), \(0.95(0.10)=0.095\), and \(0.95(0.90)(0.20)=0.171\). Therefore,

\begin{aligned} A^1_{x:3} &=0.05v+0.095v^2+0.171v^3\\ &=0.04807692+0.08783284+0.15201838\\ &=0.28792814. \end{aligned}

Survival through three years is \(0.95(0.90)(0.80)=0.684\). The pure endowment factor and future component are

{}_3E_x=v^3(0.684)=0.88899636(0.684)=0.60807351, \qquad {}_3E_xA_{x+3}=0.60807351(0.40)=0.24322940.

Thus \(A_x=0.28792814+0.24322940=0.53115754\), or 0.5312 to four decimal places. The check is that the first-three-year contributions sum to \(0.28792814\), while all deaths after year 3 contribute \(0.24322940\). Together, they reproduce the whole-life value. The key calculation is the year-3 death probability: it is \(0.171\), not \(0.20\), because the insured must first survive the preceding two years.

Common Traps and Practical Checks

The decomposition is simple, but its pieces have different meanings. The term insurance value covers death during a specified interval. The pure endowment factor combines survival and discounting to the end of that interval. The future whole-life value is conditional on reaching age \(x+n\). Keeping these roles separate prevents several common errors.

Using an unconditional annual mortality probability. For death in year \(k+1\), use \({}_kp_xq_{x+k}\), not just \(q_{x+k}\).
Forgetting discounting to the split date. The future component uses \(v^n{}_np_xA_{x+n}\), not \({}_np_xA_{x+n}\).
Adding a survival benefit that is not in the contract. The identity multiplies the future insurance value by a pure endowment factor; it does not add a separate pure endowment payment to the whole-life policy.
Changing benefit timing or amount without adjusting the formula. The stated identity assumes a unit benefit paid at the end of the year of death in all components.

A reliable check is to confirm that the term component includes exactly the first \(n\) death years, that the survival probability in \({}_nE_x\) is \({}_np_x\), and that the future insurance value is evaluated at age \(x+n\). If the underlying contract has a maturity benefit or a different payment timing, identify that feature before applying this identity.

Key takeaway: For a discrete unit-benefit whole-life insurance, splitting after \(n\) years gives \(A_x=A^1_{x:n}+{}_nE_xA_{x+n}\). The first term values deaths before the split; the second values survival to the split followed by a later death.

Check Your Understanding

For each problem, use the stated discrete benefit timing and keep survival, death, and discounting factors distinct.

A hypothetical policy pays one unit at the end of the year of death. Let \(i=4\%\), \(q_x=0.06\), \(q_{x+1}=0.12\), and \(A_{x+2}=0.30\). Use a two-year split to calculate \(A_x\). (Answer: \(A_x\approx0.3914\))
For a hypothetical unit-benefit policy, \(i=5\%\), \(q_x=0.04\), \(q_{x+1}=0.15\), and \(A_{x+2}=0.25\). Calculate the first-two-year term value, the future component, and \(A_x\). (Answer: \(A^1_{x:2}\approx0.1687\), future component \(\approx0.1850\), and \(A_x\approx0.3537\))
A hypothetical discrete whole-life value is \(A_x=0.42\), with \(i=4\%\), \(q_x=0.05\), and \(A_{x+2}=0.30\). If \(r=q_{x+1}\), solve for \(r\) using the two-year split. (Answer: \(r\approx0.1763\))
For a hypothetical unit-benefit contract, \(i=5\%\), \(q_x=0.10\), \(q_{x+1}=0.20\), and \(A_x=0.60\). Use a two-year split to recover \(A_{x+2}\). (Answer: \(A_{x+2}=251/480\approx0.5229\))
A hypothetical discrete policy has \(i=4\%\), \(q_x=0.10\), \(q_{x+1}=0.20\), \(q_{x+2}=0.25\), and \(A_{x+3}=0.40\). Use a three-year split to find \(A_x\). (Answer: \(A_x\approx0.6146\))

What You'll Learn