Term Insurance Values Across Nested Durations

Why Nested Durations Matter

A term insurance contract may cover a person for several years, while a calculation may provide values only for shorter durations. The longer contract can be valued by separating deaths during an initial period from deaths during the remaining period. The second part must be valued from the age at which that remaining coverage begins, and then weighted for the chance and cost of reaching that point.

This tutorial treats discrete annual term insurance: a unit benefit is payable at the end of the year of death if death occurs within the stated term. This timing matters. The identities below use the same benefit amount, annual effective interest rate, and mortality assumptions throughout; they are not formulas for continuous insurance benefits.

Definition: Let $A^1_{x:n}$ denote the actuarial present value, at age $x$, of a unit death benefit payable at the end of the year of death if death occurs within the next $n$ years. There is no benefit if the insured survives the full term. Here $n$ is a nonnegative integer, $i$ is the effective annual interest rate, and $v=1/(1+i)$. In particular, $A^1_{x:0}=0$.

For integer-age annual mortality notation, $q_{x+k}$ is the probability that a life aged $x+k$ dies within the next year, and ${}_kp_x$ is the probability of surviving from age $x$ to age $x+k$. Thus ${}_0p_x=1$, and ${}_{k+1}p_x={}_kp_x(1-q_{x+k})$. A death in policy year $k+1$ has probability ${}_kp_xq_{x+k}$ and its unit benefit is discounted for $k+1$ years.

A^1_{x:n}=\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}.

The Nested-Duration Identity

Suppose a term of $n+m$ years is divided at the end of year $n$. A death benefit can be paid either during the first $n$ years or during the next $m$ years, but not both. The first part is already valued by $A^1_{x:n}$. The second part is valued at age $x+n$, conditional on being alive then. To express that conditional value at age $x$, multiply it by the probability of surviving $n$ years and discount it $n$ years.

Theorem: For nonnegative integer durations $n$ and $m$, under the same annual effective interest rate and consistent mortality assumptions for both ages, $$ A^1_{x:n+m}=A^1_{x:n}+v^n{}_np_xA^1_{x+n:m}. $$

The factor $v^n{}_np_x$ is essential: it is the present value at age $x$ of one unit payable at time $n$ only if the insured survives to that time. The later term value $A^1_{x+n:m}$ is conditional on being alive at age $x+n$, so it does not already include the probability of reaching that age from $x$.

Why the Identity Holds

The formula follows by partitioning the possible death years. In the full $n+m$-year term, death in one of the first $n$ years contributes exactly the terms in $A^1_{x:n}$. For a death $j+1$ years after the split, where $0\leq j\leq m-1$, the probability from age $x$ is ${}_np_x{}_jp_{x+n}q_{x+n+j}$. The benefit is paid at time $n+j+1$. Therefore the contribution of that death year is

v^{n+j+1}{}_np_x{}_jp_{x+n}q_{x+n+j} = v^n{}_np_x\left(v^{j+1}{}_jp_{x+n}q_{x+n+j}\right).

Summing the expression in parentheses over $j=0,\ldots,m-1$ gives $A^1_{x+n:m}$. Adding the first $n$ years' contributions proves the identity. This argument also shows that if $n=0$, the first term is zero and the identity reduces to $A^1_{x:m}=A^1_{x:m}$; if $m=0$, the later term is zero.

Value the initial block.
Calculate $A^1_{x:n}$ for deaths in the first $n$ years.

Value the later block at its starting age.
Use $A^1_{x+n:m}$, conditional on survival to age $x+n$.

Bring the later block back to age $x$.
Multiply by both ${}_np_x$ and $v^n$, then add the initial-block value.

Worked Applications

Worked Example: Splitting a Five-Year Term

A hypothetical policy pays 1 at the end of the year of death during a five-year term. The effective annual interest rate is $i=5\%$. The annual death probabilities at ages $x,x+1,x+2,x+3,x+4$ are $0.10,0.12,0.15,0.18,0.20$, respectively. Find the five-year value by splitting after two years, and verify it by direct calculation.

Method. Calculate the first two years at age $x$, calculate the remaining three years conditional on survival to $x+2$, and apply the split identity. This keeps conditional and unconditional death probabilities distinct.

The discount factor is $v=1/1.05=0.9523809524$. Survival probabilities to the start of each year are $1,0.90,0.792,0.6732,0.552024$. The five unconditional death probabilities are consequently $0.10,0.108,0.1188,0.121176,0.1104048$.

For the first two years, $A^1_{x:2}=0.10v+0.108v^2$. The discounted contributions are $0.0952380952$ and $0.0979591837$, so $A^1_{x:2}=0.1931972789$, or $0.193197$ to six decimal places.

At age $x+2$, the remaining annual death probabilities are $0.15,0.18,0.20$. Thus $A^1_{x+2:3}=0.15v+0.85(0.18)v^2+0.85(0.82)(0.20)v^3$. The three contributions are $0.1428571429$, $0.1387755102$, and $0.1204189612$. Their sum is $0.4020516143$.

Survival to the split is ${}_2p_x=0.90(0.88)=0.792$, and $v^2=0.9070294785$. Therefore the later block contributes $v^2{}_2p_xA^1_{x+2:3}=0.9070294785(0.792)(0.4020516143)=0.288820756$, or $0.288821$ to six decimal places. The split value is $0.1931972789+0.288820756=0.482018035$, which rounds to $0.482018$.

Direct check. Discounting the five unconditional death probabilities gives contributions $0.0952380952,0.0979591837,0.1026239075,0.0996917948,$ and $0.0865050535$. Their sum is $0.4820180347$, or $0.482018$ to six decimal places. The split and direct values agree.

Worked Example: Combining a Short Term with a Later Term

A hypothetical four-year contract has $i=25\%$, so $v=0.8$. The annual death probabilities at ages $x,x+1,x+2,x+3$ are $0.20,0.25,0.25,$ and $5/24$. Find its value by combining a two-year term value with the conditional two-year value beginning at age $x+2$.

Method. Value each two-year block at its own starting age, then use the survival-discount factor for the later block. The fractional death probability $5/24$ is a valid probability and is retained exactly to avoid premature rounding.

The first block is $A^1_{x:2}=0.20(0.8)+0.80(0.25)(0.8^2)=0.16+0.128=0.288$. Survival to age $x+2$ is ${}_2p_x=0.80(0.75)=0.60$.

For the later block, conditional on reaching age $x+2$, $A^1_{x+2:2}=0.25(0.8)+0.75(5/24)(0.8^2)=0.20+0.10=0.30$. The factor bringing this value back two years is $v^2{}_2p_x=0.64(0.60)=0.384$. Its contribution is $0.384(0.30)=0.1152$. Hence $A^1_{x:4}=0.288+0.1152=0.4032$.

Check. Directly, the year-three death probability from age $x$ is $0.60(0.25)=0.15$, giving present value $0.15(0.8^3)=0.0768$. The year-four death probability is $0.60(0.75)(5/24)=0.09375$, giving present value $0.09375(0.8^4)=0.0384$. These later contributions total $0.1152$; adding the first-block value $0.288$ confirms $0.4032$.

Worked Example: Recovering a Conditional Later-Term Value

A hypothetical five-year term is valued at $0.569152$ at age $x$, with $i=0$. The death probabilities in the first two years are $0.12$ and $0.20$. Survival to age $x+2$ is therefore $0.704$. Find the conditional three-year term value at age $x+2$. The later annual death probabilities are $0.15,0.10,$ and $0.20$.

Method. First remove the initial two-year block from the total value. Divide the remaining value by the probability of surviving to the split (the discount factor is 1 because $i=0$).

The first two-year value is $A^1_{x:2}=0.12+0.88(0.20)=0.296$. The part of the five-year value attributable to the later block is $0.569152-0.296=0.273152$. The nested-duration identity gives $0.273152=1^2(0.704)A^1_{x+2:3}$, so $A^1_{x+2:3}=0.273152/0.704=0.388$.

Check. At age $x+2$, the three-year term value calculated from the stated mortality probabilities is $0.15+0.85(0.10)+0.85(0.90)(0.20)=0.15+0.085+0.153=0.388$. The original total is also recovered: $0.296+0.704(0.388)=0.569152$.

Common Exam Traps and Practical Checks

The most common error is to add a later conditional value without adjusting for survival and discounting. In the second example, the incorrect calculation $0.288+0.30=0.588$ treats the insured as certain to survive to the later block and treats its value as if it were already measured at age $x$. The correct calculation is $0.288+(0.8^2)(0.60)(0.30)=0.4032$. The survival factor is not optional, and the discount exponent is the number of years from the original valuation date to the split.

A second trap is confusing the mortality probabilities conditional on reaching a year with death probabilities from the original age. For example, $q_{x+2}=0.15$ is the conditional probability of death during the first year of the later block. From age $x$, the probability of that death is ${}_2p_xq_{x+2}$. Multiplying by survival twice, or not at all, changes the value.

Before using the identity, check that both blocks use the same benefit amount and timing, that the later value is conditional on being alive at the split, and that $n$ is measured in the same annual units as the interest and mortality assumptions. If the later block has a different benefit or payment timing, its value cannot simply be inserted as $A^1_{x+n:m}$ without adjusting the model and benefit definition.

Key takeaway: A term insurance value over $n+m$ years is the initial $n$-year value plus the later $m$-year value conditional on survival to the split, multiplied by $v^n{}_np_x$. This identity is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify whether the later term value is conditional on survival to the split before applying the nested-duration identity.

A hypothetical two-year term has $i=5\%$, with annual death probabilities $0.10$ and $0.20$. Find its value by splitting after one year. (Answer: $0.2585$, to four decimal places)
At $i=0$, a hypothetical three-year term has annual death probabilities $0.30,0.10,$ and $0.20$. Split after the first year and find the full term value. (Answer: $0.4960$)
A hypothetical four-year term has $i=10\%$. Its two-year value is $0.25$, survival to year two is $0.72$, and the conditional two-year value starting then is $0.40$. Find the four-year value. (Answer: $0.4880$, to four decimal places)
A hypothetical five-year term has $i=5\%$, $A^1_{x:2}=0.30$, ${}_2p_x=0.72$, and $A^1_{x:5}=0.55$. Find $A^1_{x+2:3}$. (Answer: $0.3828$, to four decimal places)
At $i=0$, annual death probabilities for a hypothetical three-year term are $0.10,0.20,$ and $0.25$. Find the full value by splitting after two years, and verify the later-block contribution. (Answer: $0.4600$; the later-block contribution is $0.1800$)

What You'll Learn