Temporary Term Insurance and Temporary Annuity Identity

Why These Temporary Values Are Connected

A temporary term insurance pays only if death occurs during a specified term. A temporary life annuity-due pays only while the insured is alive, and only during that same term. Their payment times differ: the insurance benefit is paid at the end of the year of death, while the annuity payments are made at the beginnings of the years in which the insured is alive. A precise identity connects their expected present values and the value of surviving to the end of the term.

This connection is useful for checking calculations and moving between insurance and annuity values. Its validity depends on matching the term, interest basis, survival model, and payment timing. In particular, it uses an effective annual discount rate, not the effective annual interest rate.

Definition: Consider a life aged $x$, an integer term $n$, annual effective interest rate $i$, discount factor $v=1/(1+i)$, and annual effective discount rate $d=1-v=i/(1+i)$. The $n$-year end-of-year-of-death term insurance, denoted $A^1_{x:n}$, pays 1 at the end of the year of death if death occurs within the next $n$ years; it pays nothing otherwise. The $n$-year temporary life annuity-due, denoted $\ddot{a}_{x:n}$, pays 1 at times $0,1,\ldots,n-1$, provided the insured is alive at the payment time. The pure endowment value is ${}_nE_x=v^n\,{}_np_x$, the expected present value of 1 at time $n$ if the insured survives to that time.

Here ${}_kp_x$ is the probability that a life aged $x$ survives $k$ years, and ${}_kq_x=1-{}_kp_x$ is the probability of death within $k$ years. In particular, the conditional probability of death during year $k+1$, given survival to its start, is $q_{x+k}$. These probabilities may vary by year; constant mortality is not required.

The Temporary Insurance–Annuity Identity

Theorem: Under the timing and interest assumptions in the definition, $$ A^1_{x:n}+d\,\ddot{a}_{x:n}=1-{}_nE_x. $$ Equivalently, the insurance value is $A^1_{x:n}=1-{}_nE_x-d\,\ddot{a}_{x:n}$.

The actuarial present value of the term insurance is the sum of the discounted probabilities of death in each policy year. The temporary annuity-due value is the sum of the discounted probabilities of being alive at each payment time:

A^1_{x:n}=\sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}, \qquad \ddot{a}_{x:n}=\sum_{k=0}^{n-1}v^k{}_kp_x.

These expressions assume a unit benefit or payment. For another amount, multiply the corresponding value by that amount. They also assume level annual effective interest and benefits paid at the specified integer times. The formula does not directly apply to continuous insurance benefits, payments made at the end of each year of survival, or a contract that pays a maturity benefit as well as death benefits.

Why the Identity Holds

The key is to compare the annuity’s discounted survival terms with the insurance’s discounted death terms. For each year $k+1$, the probability of surviving to time $k+1$ is ${}_{k+1}p_x={}_kp_x(1-q_{x+k})$. Thus ${}_kp_xq_{x+k}={}_kp_x-{}_{k+1}p_x$. Multiplying by the discount factor for the end of that year gives the insurance term.

v^{k+1}{}_kp_xq_{x+k} = v^{k+1}{}_kp_x-v^{k+1}{}_{k+1}p_x.

Meanwhile, since $d=1-v$, the corresponding annuity contribution is $d v^k{}_kp_x=v^k{}_kp_x-v^{k+1}{}_kp_x$. Adding the insurance and annuity contributions for year $k+1$ cancels the intermediate term $v^{k+1}{}_kp_x$:

v^{k+1}{}_kp_xq_{x+k}+d v^k{}_kp_x = v^k{}_kp_x-v^{k+1}{}_{k+1}p_x.

Summing over all $n$ years leaves a telescoping sum: each negative survival term at an intermediate time cancels the positive term from the next year. Only the initial value 1 and the discounted probability of surviving to time $n$ remain.

\begin{aligned} A^1_{x:n}+d\,\ddot{a}_{x:n} &=\sum_{k=0}^{n-1}\left(v^k{}_kp_x-v^{k+1}{}_{k+1}p_x\right)\\ &=1-v^n{}_np_x\\ &=1-{}_nE_x. \end{aligned}

The initial term is 1 because ${}_0p_x=1$. The final term is the pure endowment value, not simply the probability of survival: survival to time $n$ is discounted for $n$ years.

Applying the Identity

A reliable calculation begins by fixing the time units and payment conditions. The term insurance counts death in policy years 1 through $n$; the annuity-due includes a payment at time 0 and stops before time $n$. The discount factor for a payment at time $k$ is $v^k$, so the annual discount rate used in the identity is $d=1-v$.

Worked Example: Calculate and Verify Both Temporary Values

A hypothetical three-year contract covers a life aged $x$. The effective annual interest rate is $i=25\%$, and the conditional probability of survival through each of the next three years is $0.9$. Find $A^1_{x:3}$, $\ddot{a}_{x:3}$, and ${}_3E_x$, then verify the identity. The death benefit is paid at the end of the year of death, and each annuity payment is 1 at the beginning of a year if the life is then alive.

Method. First find the interest quantities and survival probabilities. Then sum the annual present values and check the identity. This avoids using the interest rate in place of the discount rate.

Interest and survival values. The discount factor and discount rate are $v=1/1.25=0.8$ and $d=1-0.8=0.2$. The successive survival probabilities are ${}_0p_x=1$, ${}_1p_x=0.9$, ${}_2p_x=0.9^2=0.81$, and ${}_3p_x=0.9^3=0.729$. The death probability conditional on survival is $0.1$ in each year.

Term insurance. The three year-specific contributions are $0.8(1)(0.1)=0.08$, $0.8^2(0.9)(0.1)=0.0576$, and $0.8^3(0.81)(0.1)=0.041472$. Therefore $A^1_{x:3}=0.08+0.0576+0.041472=0.179072$.

Temporary annuity-due. Payments can occur at times 0, 1, and 2. Their expected present values are $1$, $0.8(0.9)=0.72$, and $0.8^2(0.81)=0.5184$. Hence $\ddot{a}_{x:3}=1+0.72+0.5184=2.2384$.

Pure endowment and check. The pure endowment value is ${}_3E_x=0.8^3(0.729)=0.373248$. The left side is $0.179072+0.2(2.2384)=0.179072+0.44768=0.626752$. The right side is $1-0.373248=0.626752$, so the two sides agree.

Worked Example: Recover Term Insurance from the Annuity and Survival Values

For a hypothetical two-year contract, $i=25\%$, ${}_1p_x=0.9$, and ${}_2p_x=0.72$. The benefit is 1 at the end of the year of death during the term; the annuity pays 1 at each beginning of a year in which the life is alive. Calculate the term insurance value using the identity and confirm it directly.

Method. Use the survival probabilities to calculate the temporary annuity-due and pure endowment, then solve the identity for insurance. The direct death-year calculation provides a check and guards against shifting a benefit to the wrong time.

Here $v=0.8$ and $d=0.2$. The annuity-due has payments at time 0 and time 1, so $\ddot{a}_{x:2}=1+v{}_1p_x=1+0.8(0.9)=1.72$. The pure endowment is ${}_2E_x=v^2{}_2p_x=0.64(0.72)=0.4608$.

Rearrange the identity: $A^1_{x:2}=1-{}_2E_x-d\ddot{a}_{x:2}$. Substitution gives $A^1_{x:2}=1-0.4608-0.2(1.72)=1-0.4608-0.344=0.1952$.

For a direct check, the first-year death probability is $1-0.9=0.1$, so its contribution is $0.8(0.1)=0.08$. The conditional second-year survival probability is ${}_2p_x/{}_1p_x=0.72/0.9=0.8$, making the second-year conditional death probability $0.2$. Its contribution is $0.8^2(0.9)(0.2)=0.1152$. The total is $0.08+0.1152=0.1952$, as obtained from the identity.

Worked Example: Infer a Later Survival Probability

A hypothetical two-year term contract has $i=25\%$, temporary annuity-due value $\ddot{a}_{x:2}=1.6$, and end-of-year-of-death term insurance value $A^1_{x:2}=0.296$. The annuity payments and insurance benefit follow the timings in the definition. Given that $p_x=0.75$, find $p_{x+1}$ and the two-year pure endowment value.

Method. Use the annuity-due sum to identify the first-year survival probability, then use the insurance identity to obtain the pure endowment. This avoids treating the annuity value alone as if it specified survival through the full term.

The discount factor is $v=0.8$. The two-year temporary annuity-due is $1+v p_x=1+0.8(0.75)=1.6$, consistent with the supplied value. From the identity, $1-{}_2E_x=A^1_{x:2}+d\ddot{a}_{x:2}$, where $d=0.2$. Thus $1-{}_2E_x=0.296+0.2(1.6)=0.296+0.32=0.616$, so ${}_2E_x=0.384$.

Since ${}_2E_x=v^2{}_2p_x$, the two-year survival probability is ${}_2p_x=0.384/0.64=0.6$. Also ${}_2p_x=p_xp_{x+1}$, so $p_{x+1}=0.6/0.75=0.8$.

Check directly: first-year death contributes $0.8(0.25)=0.2$, and second-year death contributes $0.8^2(0.75)(0.2)=0.096$. Their sum is $0.296$, the given insurance value. The inferred second-year survival probability and pure endowment are consistent with the contract data.

Timing Checks and Common Traps

The identity is especially useful as a consistency check, but only when its components use the same term and payment conventions. In the annuity-due, there is a payment at time 0 and no payment at time $n$. In the term insurance, death during year $n$ is covered and the benefit is discounted $n$ years if paid at the end of that year. The terminal survival value accounts for lives that reach time $n$ without a term death benefit.

Using $i$ instead of $d$. The coefficient is $d=1-v=i/(1+i)$, not $i$. In the first worked example, using $i=0.25$ would give $0.179072+0.25(2.2384)=0.738672$, which does not equal $1-{}_3E_x=0.626752$. Using $d=0.2$ gives the correct equality.
Including an annuity payment at time $n$. The temporary annuity-due stops at time $n-1$. A payment at time $n$ would require survival to the end of the term and would change the value.
Using the wrong death-benefit timing. The sum for $A^1_{x:n}$ discounts a death in year $k+1$ by $v^{k+1}$. A benefit paid immediately at death is a different contract value.
Omitting the pure endowment term. The right side is $1-{}_nE_x$, not 1. The value ${}_nE_x$ is the discounted probability of surviving to the term end.

For a practical check, write down the payment times before substituting values. Then calculate $v$, $d$, each survival probability needed, and the pure endowment using the same interest and mortality assumptions. The equality should hold to the precision of the inputs. This identity is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated end-of-year-of-death benefit and temporary annuity-due timing. Answers are given for checking after you have worked through the steps.

A hypothetical two-year contract has $i=25\%$ and conditional survival probability $0.8$ in each year. Find $A^1_{x:2}$, $\ddot{a}_{x:2}$, and ${}_2E_x$, and verify the identity. (Answer: $A^1_{x:2}=0.2624$, $\ddot{a}_{x:2}=1.64$, ${}_2E_x=0.4096$; both sides equal $0.5904$.)
A life has $i=25\%$, ${}_1p_x=0.9$, and ${}_2p_x=0.81$. Calculate the two-year temporary annuity-due and pure endowment, then use the identity to find the term insurance value. (Answer: $\ddot{a}_{x:2}=1.72$, ${}_2E_x=0.5184$, and $A^1_{x:2}=0.1376$.)
For a one-year contract, $i=5\%$ and the one-year death probability is $0.1$. Calculate the end-of-year term insurance value and verify the identity, including the one-year pure endowment. (Answer: $A^1_{x:1}=0.0952$, $\ddot{a}_{x:1}=1$, ${}_1E_x=0.8571$, and both sides equal $0.1429$, with probabilities and values rounded to four decimals.)
A two-year contract has $i=25\%$, $p_x=0.75$, $\ddot{a}_{x:2}=1.6$, and $A^1_{x:2}=0.296$. Find $p_{x+1}$ and ${}_2E_x$, and check by calculating the insurance value from the annual death probabilities. (Answer: $p_{x+1}=0.8$, ${}_2E_x=0.384$; the direct insurance value is $0.2+0.096=0.296$.)
A three-year contract has $i=25\%$, ${}_1p_x=0.9$, ${}_2p_x=0.81$, and ${}_3p_x=0.729$. Its term insurance value is $0.179072$. Find the temporary annuity-due value by the identity and verify it from its payment-time terms. (Answer: $\ddot{a}_{x:3}=2.2384$; the payment-time sum is $1+0.72+0.5184$.)

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