Term and Endowment Boundary Conventions

Why the Term Boundary Matters

In a term or endowment contract, the final policy year is a boundary between two different possible benefits. If the insured dies during that year, a death benefit may be payable at the end of the year. If the insured is alive at the term’s end, a survival benefit may instead be payable at that date. Confusing these events can omit a benefit, count one twice, or discount it for the wrong number of years.

The safest approach is to identify the event that triggers each payment before calculating its present value. For an $n$-year contract with annual benefit timing, the term death benefit applies to deaths in policy years 1 through $n$, while the survival benefit applies only if the insured survives to time $n$. Under the usual continuous future-lifetime model, death at exactly time $n$ has probability zero. If a contract or model allows positive probability of death at that exact instant, the contract wording must specify which benefit applies.

Definition: An $n$-year term insurance pays a death benefit if death occurs during the first $n$ policy years. With benefits paid at the end of the year of death, a death in policy year $k+1$ is paid at time $k+1$, for $k=0,\ldots,n-1$. A pure endowment pays a benefit at time $n$ only if the insured survives to that time. An $n$-year endowment insurance combines the term death benefit with a survival benefit payable at time $n$.

Write the Events Before the Formula

Let ${}_k p_x$ be the probability that a life aged $x$ survives $k$ years, and let $q_{x+k}$ be the probability of death within the following year, conditional on reaching age $x+k$. Thus, ${}_0p_x=1$, and the probability of death in policy year $k+1$ is ${}_k p_xq_{x+k}$. Let $v=1/(1+i)$, where $i$ is the effective annual interest rate.

For unit benefits, the present value of an $n$-year term insurance is conventionally denoted $A^{1}_{x:\overline{n}|}$. The superscript distinguishes the term death benefit from the survival benefit. The pure endowment value is ${}_nE_x$, and the endowment insurance value is $A_{x:\overline{n}|}$. These values are expected present values; multiply by the relevant benefit amount when it is not one.

A^{1}_{x:\overline{n}|} =\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}, \qquad {}_nE_x=v^n{}_np_x, \qquad A_{x:\overline{n}|}=A^{1}_{x:\overline{n}|}+{}_nE_x.

The term-insurance sum ends at $n-1$, because its last included death interval is policy year $n$, from time $n-1$ through time $n$. The payment for that death is discounted $n$ years. The pure endowment has a different event: survival to time $n$, with a payment at that time, also discounted $n$ years. The equal discount exponents do not make the events interchangeable.

Worked Example: Separate the Final Death Year From Survival

A hypothetical three-year endowment pays \$10,000 at the end of the year of death if death occurs during the term, and \$4,000 at time 3 if the insured survives to time 3. The annual survival probabilities for the first, second, and third years are $0.90$, $0.80$, and $0.75$, respectively. The effective annual interest rate is $5\%$. Find the expected present value of the benefits.

Method. Calculate the probability and payment time for each death year, then calculate the survival payment separately. This prevents the year-3 death benefit from being mistaken for the time-3 survival benefit.

The discount factor is $v=1/1.05=0.95238095$. The probabilities of death in policy years 1, 2, and 3 are, respectively, $0.10$, $0.90(0.20)=0.18$, and $0.90(0.80)(0.25)=0.18$. The probability of survival to time 3 is $0.90(0.80)(0.75)=0.54$.

The death-benefit present value is $10000[0.10v+0.18v^2+0.18v^3]$. Using $v^2=0.90702948$ and $v^3=0.86383760$, the bracketed value is $0.09523810+0.16326531+0.15549077=0.41399418$. Thus the death-benefit value is $\$4,139.94$.

The survival-benefit present value is $4000(0.54)v^3=4000(0.54)(0.86383760)=\$1,865.89$. Adding the two components gives $\$4,139.94+\$1,865.89=\$6,005.83$, rounded to the nearest cent.

Check. The three death probabilities sum to $0.46$, and the probability of survival to time 3 is $0.54$, so the outcomes partition the probability space. The common mistake is to treat the year-3 death benefit as a survival payment: the former has probability $0.18$, while the latter has probability $0.54$.

An Identity That Checks the Boundary

A temporary life annuity-due makes a useful independent check. Let $\ddot{a}_{x:\overline{n}|}$ denote the expected present value of unit payments at times $0,1,\ldots,n-1$, provided the life is alive at each payment time. Let $d=1-v$ be the effective annual discount rate. Under the same annual timing convention, the term insurance, temporary annuity-due, and pure endowment satisfy an identity.

Theorem: For annual death benefits paid at the end of the year of death and an $n$-year temporary life annuity-due, $A^{1}_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}+{}_nE_x=1$. Equivalently, $A_{x:\overline{n}|}=1-d\ddot{a}_{x:\overline{n}|}$.

To see why the identity holds, put $p_k={}_kp_x$. In the $k$th interval, the probability of death is $p_k-p_{k+1}={}_kp_xq_{x+k}$. Using this expression for the term insurance and the payment times for the annuity gives the following telescoping calculation.

\begin{aligned} A^{1}_{x:\overline{n}|} &=\sum_{k=0}^{n-1}v^{k+1}(p_k-p_{k+1}),\\ d\ddot{a}_{x:\overline{n}|} &=\sum_{k=0}^{n-1}(1-v)v^kp_k =\sum_{k=0}^{n-1}(v^k-v^{k+1})p_k,\\ A^{1}_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|} &=\sum_{k=0}^{n-1}(v^kp_k-v^{k+1}p_{k+1}) =1-v^np_n. \end{aligned}

Since $p_0=1$, the final sum cancels term by term except for its first and last terms. Adding ${}_nE_x=v^np_n$ gives 1. The endowment value is the sum of the term and pure endowment values, so rearranging also gives $A_{x:\overline{n}|}=1-d\ddot{a}_{x:\overline{n}|}$. The identity depends on consistent payment timing: the annuity is due, and the term death benefit is paid at each year-end.

Worked Example: Check a Two-Year Endowment With an Annuity Identity

A hypothetical contract has unit death benefit at the end of the year of death for two years and a unit survival benefit at time 2. The annual survival probability is $0.80$ in each year, and the effective annual interest rate is $25\%$. Calculate the endowment value directly and verify it with the annuity identity.

Direct calculation. Here $v=1/1.25=0.80$. Death in year 1 has probability $0.20$, and death in year 2 has probability $0.80(0.20)=0.16$. The term value is $0.20(0.80)+0.16(0.80^2)=0.16+0.1024=0.2624$.

Survival to time 2 has probability $0.80^2=0.64$, so the pure endowment value is $0.64(0.80^2)=0.4096$. The total endowment value is therefore $0.2624+0.4096=0.6720$.

Identity check. The temporary annuity-due value is $\ddot{a}_{x:\overline{2}|}=1+0.80(0.80)=1.64$: the first payment is certain at time 0, and the second is made at time 1 only if the life survives the first year. Also, $d=1-0.80=0.20$. Hence $1-d\ddot{a}_{x:\overline{2}|}=1-0.20(1.64)=1-0.328=0.672$, matching the direct value.

Check. The survival benefit is discounted for two years, not one, and the time-1 annuity payment is conditional on survival for one year. Using an annuity-immediate in this identity would change the payment dates and would not be a valid check.

Isolate the Last Covered Policy Year

A reliable boundary check is to compare an $n$-year term with an $(n-1)$-year term. The only additional outcome is death during policy year $n$. Its probability is ${}_{n-1}p_xq_{x+n-1}$, and its benefit is paid at time $n$. Therefore,

A^{1}_{x:\overline{n}|}-A^{1}_{x:\overline{n-1}|} =v^n{}_{n-1}p_xq_{x+n-1}.

This difference is a direct test of the upper summation limit: the new term is the death probability in year $n$, not the probability of surviving $n$ years. The survival probability belongs in the pure endowment term, ${}_nE_x=v^n{}_np_x$. Keeping these two expressions separate also helps when the death benefit and survival benefit have different amounts.

Worked Example: Add the Final Year to a Term Benefit

A hypothetical policy pays \$25,000 at the end of the year of death. The annual survival probability is $0.80$ in each year, and the effective annual interest rate is $25\%$. Find the value of the three-year term insurance, the increase over a two-year term, and the value of an additional \$7,000 survival benefit at time 3.

Method. First value the two covered death years, then add the death benefit for year 3 using the probability of reaching the start of that year and dying during it. Value the survival benefit using the probability of reaching time 3.

As before, $v=0.80$. The probabilities of death in years 1 and 2 are $0.20$ and $0.80(0.20)=0.16$. The two-year term value is $25000[0.20(0.80)+0.16(0.80^2)]=25000(0.2624)=\$6,560.00$.

The probability of death in year 3 is $0.80^2(0.20)=0.128$. Its discounted probability is $v^3(0.128)=0.512(0.128)=0.065536$. The increase in term value is $25000(0.065536)=\$1,638.40$, so the three-year term value is $\$6,560.00+\$1,638.40=\$8,198.40$.

The probability of survival to time 3 is $0.80^3=0.512$. The survival benefit value is $7000(0.512)(0.80^3)=7000(0.262144)=\$1,835.01$, rounded to the nearest cent. The combined value is $\$8,198.40+\$1,835.01=\$10,033.41$.

Check. The year-3 death benefit uses probability $0.80^2(0.20)$, because the life must survive only to the start of year 3 before dying in that year. The survival benefit uses probability $0.80^3$. Using $0.80^3(0.20)$ for the year-3 death probability incorrectly requires the life to survive through year 3 and then die in that same year.

Boundary-Year Errors to Avoid

Most errors in these calculations are not difficult probability errors; they are mismatches between an event, its payment date, and its discount factor. Before evaluating a formula, write down the policy-year intervals and the payment time for each benefit. In particular, check the final interval separately.

Leaving out the last death year. An $n$-year term includes death in policy year $n$. Its death-year index is $k=n-1$, and its payment occurs at time $n$.
Using the survival event for a death claim. The final-year death probability is ${}_{n-1}p_xq_{x+n-1}$, not ${}_np_x$. Survival to time $n$ triggers a pure endowment, not a term death benefit.
Discounting an end-of-year payment as if it were paid at the beginning of the year. A death in policy year $k+1$ has payment time $k+1$, so its discount factor is $v^{k+1}$.
Counting the boundary benefit twice. In an endowment, the death and survival benefits correspond to separate outcomes. Do not add both benefits for the same insured at time $n$ unless the contract explicitly provides both.

These checks are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus: they connect contract wording to event probabilities, payment times, and standard insurance-annuity identities. The central habit is simple: define the covered death years and the survival date first, then assign each benefit its own probability and discount factor.

Check Your Understanding

For each problem, identify the relevant boundary event before calculating. Round monetary answers to the nearest cent.

A two-year term pays \$5,000 at the end of the year of death. The annual survival probability is $0.90$ in each year, and $i=10\%$. Find the expected present value of the death benefit. (Answer: \$826.45)
A three-year pure endowment pays \$2,000 at time 3. The annual survival probability is $0.70$ in each year, and $i=0\%$. Find its expected present value. (Answer: \$686.00)
A three-year endowment pays \$1,000 at the end of the year of death and \$500 at time 3 if the insured survives. The annual survival probability is $0.80$ in each year, and $i=25\%$. Find the combined expected present value. (Answer: \$459.01)
A four-year term pays \$10,000 at the end of the year of death. The annual survival probability is $0.90$ in each year, and $i=5\%$. By how much does its value exceed the value of a three-year term? (Answer: \$599.75)

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