Term Insurance from Temporary Annuity-Due Values

Why a Temporary Annuity Helps Value Term Insurance

A discrete $n$-year term insurance pays a benefit if death occurs within a specified term. Its value can be calculated by summing the discounted benefits for each possible year of death. When a temporary annuity-due value and a pure endowment value are available, however, a finite-term complement identity provides another route. The identity is useful both for calculation and for checking that the term, timing, and survival probabilities have been handled consistently.

The key is to match the cash-flow definitions. The temporary annuity-due pays at the start of each year while the life is alive, for at most $n$ payments. The term insurance pays at the end of the year of death, only if death occurs within those $n$ years. The pure endowment pays at time $n$ if the life survives to then.

Definition: Assume a life aged $x$, a constant effective annual interest rate $i$, and annual mortality probabilities. Write $v=1/(1+i)$. A unit $n$-year discrete term insurance pays 1 at the end of the year of death if death occurs within $n$ years; its actuarial present value is $A^1_{x:\overline{n}|}$. The temporary life annuity-due value is $$ \ddot{a}_{x:\overline{n}|}=\sum_{k=0}^{n-1}v^k\,{}_kp_x, $$ and the unit pure endowment value is $$ {}_nE_x=v^n\,{}_np_x. $$ Here ${}_kp_x$ is the probability that a life aged $x$ survives $k$ years, with ${}_0p_x=1$. The term insurance benefit is paid at year-end, while the annuity payments are made at times $0,1,\ldots,n-1$.

The Finite-Term Complement Identity

For a level unit benefit paid at the end of the year of death, the value from a temporary annuity-due and a pure endowment is

A^1_{x:\overline{n}|}=1-d\,\ddot{a}_{x:\overline{n}|}-{}_nE_x, \qquad d=1-v=\frac{i}{1+i}.

This is a finite-term identity. It uses the same interest and survival assumptions for all three values, and it depends on the stated benefit timings. In particular, it is not a formula for a benefit paid immediately at death or for an annuity paid at the end of each year.

Write the term insurance as a sum over death years.
If death occurs in year $k+1$, its probability is ${}_kp_xq_{x+k}$, and the unit benefit is discounted $k+1$ years. Thus $A^1_{x:\overline{n}|}=\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}$.

Use the one-year survival/death partition.
Since $q_{x+k}=1-p_{x+k}$, the probability of surviving $k$ years and then dying in the next year is ${}_kp_x-{}_{k+1}p_x$. Substitute this difference into the sum.

Collect the discounted survival terms.
The death-year sum becomes $\sum_{k=0}^{n-1}v^{k+1}{}_kp_x-\sum_{j=1}^{n}v^j{}_jp_x$. The first sum is $v\ddot{a}_{x:\overline{n}|}$; the second ends with the survival term $v^n{}_np_x={}_nE_x$.

Apply the discount identity.
Subtracting the sums, or equivalently expanding $d\ddot{a}_{x:\overline{n}|}=(1-v)\ddot{a}_{x:\overline{n}|}$, gives $A^1_{x:\overline{n}|}+{}_nE_x=1-d\ddot{a}_{x:\overline{n}|}$. Rearranging gives the complement identity.

The derivation explains why both complementary values are needed. The term insurance covers deaths before time $n$, while the pure endowment represents survival to time $n$; the annuity-due value supplies the discounted survival terms that connect those outcomes. The identity also gives a useful sign check: for nonnegative values, $d\ddot{a}_{x:\overline{n}|}+{}_nE_x$ cannot exceed 1 if the inputs are consistent with the stated model.

Worked Example: Recovering a Three-Year Term Value

A hypothetical insurer values a unit three-year term benefit paid at the end of the year of death. The effective annual interest rate is $i=5\%$. The conditional probabilities of surviving each successive year are $p_x=0.9$, $p_{x+1}=0.8$, and $p_{x+2}=0.5$. Calculate the insurance value from the temporary annuity-due and pure endowment values, then check it by summing the death-year benefits.

Method and assumptions. Use the finite-term complement identity. The conditional survival probabilities give ${}_1p_x=0.9$, ${}_2p_x=0.9(0.8)=0.72$, and ${}_3p_x=0.9(0.8)(0.5)=0.36$. The discount factor is $v=1/1.05=20/21$, so $d=1-v=1/21$.

The temporary annuity-due value is

\ddot{a}_{x:\overline{3}|} =1+\frac{20}{21}(0.9)+\left(\frac{20}{21}\right)^2(0.72) =1+\frac{6}{7}+\frac{32}{49} =\frac{123}{49}.

The pure endowment value is

{}_3E_x=\left(\frac{20}{21}\right)^3(0.36) =\frac{8000}{9261}\cdot\frac{9}{25} =\frac{2880}{9261}.

Substitute these values into the identity:

A^1_{x:\overline{3}|} =1-\frac{1}{21}\left(\frac{123}{49}\right)-\frac{2880}{9261} =1-\frac{1107}{9261}-\frac{2880}{9261} =\frac{5274}{9261} =0.569484936\ldots.

Rounded to six decimal places, the insurance value is 0.569485.

Direct check. The conditional death probabilities are $0.1$, $0.2$, and $0.5$. The discounted death benefits are therefore

\begin{aligned} A^1_{x:\overline{3}|} &=\frac{20}{21}(0.1) +\left(\frac{20}{21}\right)^2(0.9)(0.2) +\left(\frac{20}{21}\right)^3(0.72)(0.5)\\ &=\frac{2}{21}+\frac{8}{49}+\frac{2880}{9261}\\ &=\frac{882+1512+2880}{9261} =\frac{5274}{9261} =0.569484936\ldots. \end{aligned}

The direct sum agrees, and both calculations round to 0.569485 to six decimal places. A common mistake is to truncate the decimal to 0.569484; the exact fraction shows that the next digit after the sixth decimal place is 9, so the correct rounded value is 0.569485.

Worked Example: Using a Conditional Mortality Schedule

A hypothetical four-year term policy pays 1 at the end of the year of death. The annual effective interest rate is $25\%$. The conditional probabilities of surviving the next year at successive attained ages are $0.8$, $0.75$, $0.5$, and $0.5$. Find the unit term insurance value using the complement identity, and verify it directly.

Method and assumptions. First form the cumulative survival probabilities: ${}_1p_x=0.8$, ${}_2p_x=0.8(0.75)=0.6$, ${}_3p_x=0.6(0.5)=0.3$, and ${}_4p_x=0.3(0.5)=0.15$. Here $v=1/1.25=0.8$ and $d=0.2$.

The temporary annuity-due value is

\ddot{a}_{x:\overline{4}|} =1+(0.8)(0.8)+(0.8)^2(0.6)+(0.8)^3(0.3) =1+0.64+0.384+0.1536 =2.1776.

The pure endowment is ${}_4E_x=(0.8)^4(0.15)=0.4096(0.15)=0.06144$. Thus

A^1_{x:\overline{4}|} =1-(0.2)(2.1776)-0.06144 =1-0.43552-0.06144 =0.50304.

Direct check. The conditional death probabilities are $0.2$, $0.25$, $0.5$, and $0.5$. The four discounted death benefits are $0.8(0.2)=0.16$, $0.8^2(0.8)(0.25)=0.128$, $0.8^3(0.6)(0.5)=0.1536$, and $0.8^4(0.3)(0.5)=0.06144$. Their sum is $0.16+0.128+0.1536+0.06144=0.50304$, as required. The calculation avoids treating the cumulative survival probability $0.6$ at year 2 as the conditional probability of surviving the third year; that conditional probability is $0.5$.

Scaling the Unit Value to a Policy Benefit

The identity gives the actuarial present value for a unit benefit. If a level benefit $B$ is paid at the end of the year of death, its value is $B A^1_{x:\overline{n}|}$, because each possible payment amount and its present value are multiplied by $B$. This scaling applies to a level benefit; a benefit that varies by year requires year-specific benefit amounts in the direct sum.

Worked Example: Valuing a Two-Year Benefit

A hypothetical policy pays \$30,000 at the end of the year of death if death occurs within two years. The effective annual interest rate is $5\%$, and the conditional survival probabilities are $p_x=0.8$ and $p_{x+1}=0.75$. Find the actuarial present value of the benefit using a temporary annuity-due value and a pure endowment value.

Method and assumptions. The cumulative survival probabilities are ${}_1p_x=0.8$ and ${}_2p_x=0.8(0.75)=0.6$. With $v=20/21$, the discount rate is $d=1/21$.

The two-payment temporary annuity-due value is

\ddot{a}_{x:\overline{2}|} =1+\frac{20}{21}(0.8) =1+\frac{16}{21} =\frac{37}{21}.

The pure endowment value is

{}_2E_x =\left(\frac{20}{21}\right)^2(0.6) =\frac{400}{441}\cdot\frac{3}{5} =\frac{240}{441}.

The unit term insurance value is

A^1_{x:\overline{2}|} =1-\frac{1}{21}\left(\frac{37}{21}\right)-\frac{240}{441} =\frac{441-37-240}{441} =\frac{164}{441} =0.371882086\ldots.

Multiplying by the \$30,000 benefit gives

\$30{,}000\left(\frac{164}{441}\right) =\$11{,}156.4625\ldots =\boxed{\$11{,}156.46}.

The amount is rounded to the nearest cent. As a check, the conditional death probabilities are $0.2$ in year 1 and $0.25$ in year 2. The unit value from direct summation is $(20/21)(0.2)+(20/21)^2(0.8)(0.25)=4/21+80/441=164/441$. Multiplication by \$30,000 gives the same result. The common mistake avoided here is multiplying the benefit by the annuity value rather than by the term insurance value.

Conditions and Common Exam Traps

Before applying the complement identity, check that all inputs refer to the same life, term, and interest basis. In particular, the temporary annuity-due must have payments at times $0$ through $n-1$, and the pure endowment must pay at time $n$ on survival. The term insurance benefit must be payable at the end of the year of death and must have the same level amount for each covered year.

Using an annuity-immediate value. An annuity-immediate pays at times $1$ through $n$, not $0$ through $n-1$. Its value cannot be substituted for $\ddot{a}_{x:\overline{n}|}$ without first converting it to the matching annuity-due value.
Omitting the pure endowment. The formula is $1-d\ddot{a}_{x:\overline{n}|}-{}_nE_x$, not merely $1-d\ddot{a}_{x:\overline{n}|}$. The survival-to-$n$ value is a separate term.
Using $i$ instead of $d$. The required rate is $d=1-v=i/(1+i)$, not the effective interest rate $i$.
Changing the benefit timing. A benefit paid at the moment of death has a different discount pattern. The identity here is for a benefit paid at the end of the year of death.
Mixing conditional and cumulative survival probabilities. The annuity and pure endowment use ${}_kp_x$, the probability of surviving $k$ years from age $x$. A year-specific death probability uses ${}_kp_xq_{x+k}$, not $q_{x+k}$ alone.

A practical check is to calculate the term value both ways when the mortality schedule is available: use the complement identity, then sum the discounted death benefits. Agreement helps detect an indexing error, a rate-conversion error, or a missing survival term.

Key takeaway: For a unit, level discrete term benefit paid at the end of the year of death, use $A^1_{x:\overline{n}|}=1-d\ddot{a}_{x:\overline{n}|}-{}_nE_x$. Match the annuity, pure endowment, and insurance term and payment timings before substituting values.

Check Your Understanding

For each problem, use the stated year-end benefit convention. Answers are given so you can check your calculation.

A two-year unit term insurance has $i=5\%$, with conditional survival probabilities $p_x=0.9$ and $p_{x+1}=0.8$. Calculate the temporary annuity-due value, the pure endowment value, and the term insurance value. (Answer: $\ddot{a}_{x:\overline{2}|}=13/7$, ${}_2E_x=32/49$, $A^1_{x:\overline{2}|}=38/147\approx0.258503$.)
A four-year unit term insurance has $i=25\%$, and conditional survival probabilities $0.8, 0.75, 0.5, 0.5$. Calculate its value using the complement identity. (Answer: $0.50304$.)
A three-year unit term insurance has $i=6.25\%$, with conditional survival probabilities $0.8, 0.75, 0.5$. Find the cumulative survival probabilities, then calculate the term insurance value using the complement identity. (Answer: ${}_1p_x=0.8$, ${}_2p_x=0.6$, ${}_3p_x=0.3$, and $A^1_{x:\overline{3}|}=3024/4913\approx0.615510$, to six decimal places.)
A hypothetical two-year policy pays \$25,000 at the end of the year of death. The effective annual interest rate is $0\%$, and the conditional survival probabilities are $0.9$ and $0.8$. Find the actuarial present value. (Answer: $ \$16{,}000.00$.)

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