n-Year Term Insurance With Constant Mortality

Introduction

An n-year term insurance pays a death benefit only if the insured dies within the specified coverage period. To value it, account for two things in each covered policy year: the probability the insured is still alive at the start of that year, and the time value of a benefit paid at the end of the year. Under constant mortality, the calculation follows a repeated pattern.

This tutorial considers a discrete contract with a level death benefit and no payment at the end of the term if the insured survives. The result is an actuarial present value (APV): the expected present value of the future benefit, using a specified interest rate and mortality assumption.

Definition: Suppose the conditional probability of death during each covered policy year, given survival to its start, is the same value $q$. Write $p=1-q$ for the corresponding annual probability of survival. An n-year level-benefit term insurance with benefit $B$ pays $B$ at the end of the policy year of death if death occurs in one of the first $n$ years; otherwise, it pays nothing. Assume a fixed effective annual interest rate $i$, with $v=1/(1+i)$, and no benefit other than the stated death benefit.

Survival, Death, and Discounting by Policy Year

For a life insured at age $x$, constant annual mortality means the conditional one-year death probability in each covered year is $q$. Thus, the probability of surviving $k$ years is ${}_kp_x=p^k$. For death to occur in policy year $k$, the insured must first survive $k-1$ years and then die during year $k$. The probability of that event is $p^{k-1}q$.

The benefit for death in policy year $k$ is paid at time $k$, so its present-value factor is $v^k$. The contribution of that death year to the APV is therefore $B p^{k-1}qv^k$. Adding the contributions for all covered years gives the finite sum below.

\operatorname{APV}_n = B\sum_{k=1}^{n}p^{k-1}qv^k = Bqv\sum_{j=0}^{n-1}(pv)^j.

Here $\operatorname{APV}_n$ denotes the APV of the specified n-year contract. The second expression follows by factoring out $Bqv$ and setting $j=k-1$. The sum has $n$ terms: $1,(pv),\ldots,(pv)^{n-1}$. If $pv\ne1$, the geometric-series formula gives an equivalent closed form.

\operatorname{APV}_n = Bqv\frac{1-(pv)^n}{1-pv}.

The finite sum is the direct formula and is valid without needing to divide by $1-pv$. If $pv=1$, use the finite sum, which is $Bqvn$. In the usual case with positive mortality or positive interest, $pv<1$, so the quotient form applies.

Why Each Term Has This Form

The death-in-year probability and discount factor can be justified directly. For year $k$, surviving each of the first $k-1$ years has probability $p^{k-1}$, by multiplying the conditional annual survival probabilities. Given survival to the start of year $k$, death in that year has probability $q$. Hence the probability of death in year $k$ is $p^{k-1}q$. Since the benefit is paid at time $k$, its present value in that event is $Bv^k$. Multiplying the event probability by the benefit's present value gives $Bp^{k-1}qv^k$, and summing across the mutually exclusive covered death years gives the APV.

This reasoning also explains the term boundary: there are exactly $n$ covered death years. Surviving the full $n$ years does not create a payment under this contract. The mortality assumption is constant only over the years being valued; if the annual probabilities differ by year, the repeated geometric factor $pv$ no longer represents every term.

Worked Example: Three-Year Coverage at a Five Percent Interest Rate

A hypothetical policy pays $B=\$10{,}000$ at the end of the policy year of death, provided death occurs within three years. The conditional annual death probability is $q=0.10$ in each covered year, and the effective annual interest rate is $i=0.05$. Find the APV.

Method and assumptions. Use the death-year sum for a discrete benefit paid at each year-end. The probability of surviving a covered year is $p=1-0.10=0.90$, and the discount factor is $v=1/1.05=20/21$.

The probabilities of death in years 1, 2, and 3 are respectively $0.10$, $0.90(0.10)=0.09$, and $0.90^2(0.10)=0.081$. Discount each at the end of its own year:

\begin{aligned} \operatorname{APV}_3 &=10{,}000\left(0.10v+0.09v^2+0.081v^3\right)\\ &=10{,}000\left(\frac{2}{21}+\frac{4}{49}+\frac{24}{343}\right)\\ &=10{,}000\left(\frac{98+84+72}{1029}\right)\\ &=10{,}000\left(\frac{254}{1029}\right)\\ &=\$2{,}468.42. \end{aligned}

The fractions follow from $0.10v=2/21$, $0.09v^2=4/49$, and $0.081v^3=24/343$. The death probability within the term is $0.10+0.09+0.081=0.271$, so the undiscounted expected benefit is $\$10{,}000(0.271)=\$2{,}710$. The lower APV, $\$2{,}468.42$, is consistent with discounting positive benefits paid in the future. This method avoids both omitting survival to later years and discounting the benefit to the wrong time.

Using the Geometric-Series Form

When mortality and interest are constant, each successive death-year term is the previous term multiplied by $pv$. For example, the ratio of the year-2 contribution to the year-1 contribution is $pv$, because the year-2 event requires one additional year of survival and its benefit is discounted one additional year. This repeated ratio is why the geometric-series form is efficient, especially when the term is longer.

Worked Example: Four-Year Coverage Using the Geometric Series

A hypothetical four-year term policy has benefit $B=\$50{,}000$. The conditional annual death probability is $q=0.08$, and the effective annual interest rate is $i=0.04$. Calculate the APV.

Method and assumptions. The annual survival probability is $p=0.92=23/25$, and $v=1/1.04=25/26$. Therefore $pv=23/26$, while $qv=(2/25)(25/26)=1/13$.

There are four covered death years, so the geometric sum has powers 0 through 3:

\begin{aligned} \operatorname{APV}_4 &=50{,}000\left(\frac{1}{13}\right) \left[1+\frac{23}{26} +\left(\frac{23}{26}\right)^2 +\left(\frac{23}{26}\right)^3\right]\\ &=50{,}000\left(\frac{1}{13}\right) \left(\frac{17{,}576+15{,}548+13{,}754+12{,}167}{17{,}576}\right)\\ &=50{,}000\left(\frac{59{,}045}{228{,}488}\right)\\ &=\$12{,}920.81. \end{aligned}

The denominator $17{,}576=26^3$ puts the four terms over a common denominator; dividing the sum by 13 gives $59{,}045/228{,}488$. As a check, the probability of death within four years is $1-0.92^4=0.28360704$, giving an undiscounted expected benefit of $\$14{,}180.35$. The APV is lower because payments are discounted. The method also avoids adding a fifth term: survival through year 4 produces no benefit.

Worked Example: A Higher Interest Rate and a Four-Year Term

A hypothetical policy pays $\$12{,}500$ at the end of the year of death, if death occurs within four years. The conditional annual death probability is $q=0.20$, and the effective annual interest rate is $i=0.25$. Find the APV.

Method and assumptions. Here $p=0.80$ and $v=1/1.25=0.80$, so $pv=0.64$ and $qv=0.16$. Apply the four-term geometric sum:

\begin{aligned} \operatorname{APV}_4 &=12{,}500(0.16)\left(1+0.64+0.64^2+0.64^3\right)\\ &=2{,}000(1+0.64+0.4096+0.262144)\\ &=2{,}000(2.311744)\\ &=\$4{,}623.49. \end{aligned}

The undiscounted probability of death within the four-year term is $1-0.80^4=0.5904$, so the expected benefit without discounting is $\$12{,}500(0.5904)=\$7{,}380$. The APV is lower, as expected with a positive interest rate. Rounding is to the nearest cent only at the final step.

Common Timing and Coverage Errors

A frequent error is to use the annual death probability $q$ in every year without also requiring survival to that year. For the first example, this incorrect calculation would be:

10{,}000(0.10)(v+v^2+v^3)\approx \$2{,}723.25.

It overstates the APV because it treats the insured as exposed to the same death probability in years 2 and 3 regardless of whether the insured survived the preceding years. The correct year-$k$ death probability is $p^{k-1}q$, not simply $q$. A related timing error is to discount the year-$k$ benefit by $v^{k-1}$; that would value a year-end payment as though it were paid one year earlier. For a benefit paid at the end of policy year $k$, use $v^k$.

Policy year	Probability of death in that year	Present-value factor for its benefit
1	$q$	$v$
2	$pq$	$v^2$
$k$	$p^{k-1}q$	$v^k$
$n$	$p^{n-1}q$	$v^n$

Keep the coverage boundary separate from the payment timing. The final covered death occurs in year $n$ and is paid at time $n$. Do not include a death benefit for year $n+1$, or a maturity benefit at time $n$, unless the contract explicitly provides one. Those are different payment conditions from the term insurance defined here.

Key takeaway: For a level-benefit discrete n-year term insurance under constant conditional annual mortality, the year-$k$ contribution is $Bp^{k-1}qv^k$. There are $n$ covered death years, with benefits paid at their respective year-ends and no benefit for survival through the term.

Check Your Understanding

For each problem, assume the stated annual death probability applies conditionally in every covered year, and benefits are paid at the end of the year of death. Round money to the nearest cent.

A hypothetical two-year term policy has benefit $\$12{,}000$, annual death probability $q=0.05$, and annual effective interest rate $i=0$. Find its APV. (Answer: $\$1{,}170.00$)
A three-year policy pays $\$8{,}000$. If $q=0.25$ and $i=0.25$, calculate the APV using the geometric-series form. (Answer: $\$3{,}136.00$)
A five-year term benefit is $\$40{,}000$, with $q=0.10$ and $i=0$. Find the APV and distinguish it from the undiscounted benefit amount. (Answer: $\$16{,}380.40$)
A hypothetical four-year policy has benefit $\$16{,}000$, $q=0.25$, and $i=0.25$. Find the APV for four years, then calculate how much larger it is than the APV for the same policy limited to two years. (Answer: Four-year APV $\$6{,}963.20$; increase $\$1{,}843.20$)
A three-year policy has benefit $\$24{,}000$, $q=0.10$, and $i=0.05$. Calculate its APV, showing the probability and discount factor for each covered death year. (Answer: $\$5{,}924.20$)

Policy year	Probability of death in that year	Present-value factor for its benefit
1	\(q\)	\(v\)
2	\(pq\)	\(v^2\)
\(k\)	\(p^{k-1}q\)	\(v^k\)
\(n\)	\(p^{n-1}q\)	\(v^n\)

What You'll Learn