Endowment Insurance and Temporary Annuity Identity

Why the Endowment–Annuity Identity Works

An endowment insurance combines two possible benefits: a death benefit if death occurs within a fixed term, and a maturity benefit if the insured survives to the end of that term. A temporary life annuity-due pays while the insured is alive, but only during the same fixed term. With matching amounts and payment timing, these products fit together exactly: their discounted values add to one when the annuity value is multiplied by the effective discount rate.

The identity is useful for calculating an unknown value and for checking a calculation by an independent method. Its conditions matter. In particular, the endowment benefits must be payable at the end of the year of death or at the end of the term, and the annuity payments must be due at the beginning of each year while the insured is alive.

Definition: Consider a life aged $x$, an integer term $n$, and an effective annual interest rate $i$. Write $v=1/(1+i)$ and $d=1-v=i/(1+i)$. An $n$-year endowment insurance paying 1 at the end of the year of death if death occurs within $n$ years, or 1 at time $n$ if the insured survives, has actuarial present value $A_{x:\overline{n}|}$. The matching $n$-year temporary life annuity-due pays 1 at times $0,1,\ldots,n-1$ provided the insured is alive at each payment time; its actuarial present value is $\ddot{a}_{x:\overline{n}|}$.

Use annual survival probabilities for the life aged $x$. Here, ${}_kp_x$ is the probability of surviving $k$ years from age $x$, and ${}_kp_xq_{x+k}$ is the probability of surviving $k$ years and then dying in the following year. The convention is ${}_0p_x=1$. Under these assumptions, the present values are

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

The first expression includes one term for each possible year of death within the term and one maturity-benefit term. The annuity expression includes a payment at time 0, followed by a payment at each later integer time before $n$, weighted by the probability that the insured is alive then.

The Identity and Its Justification

Theorem: For a unit $n$-year endowment insurance and a unit $n$-year temporary life annuity-due with the benefit and payment timing defined above, $$ A_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}=1. $$

The identity can be proved by considering the present values in each possible outcome, rather than by manipulating sums of probabilities. Suppose death occurs during policy year $k$, where $1\leq k\leq n$. The endowment insurance pays 1 at time $k$, with present value $v^k$. The annuity pays at times $0$ through $k-1$, since the insured is alive at all those payment times. The present value of those annuity payments is a geometric sum.

\sum_{j=0}^{k-1}v^j = \frac{1-v^k}{1-v} = \frac{1-v^k}{d}.

Multiplying this annuity present value by $d$ gives $1-v^k$. Adding the endowment insurance present value $v^k$ therefore gives 1. If the insured survives to time $n$, the insurance pays its maturity benefit at time $n$, with present value $v^n$. The annuity pays at times $0$ through $n-1$, whose present value multiplied by $d$ is $1-v^n$. Again, the two amounts sum to 1. Thus the equality holds in every possible outcome. Taking expected present values gives the stated identity.

This proof also shows why matching timing is essential. If the insurance pays immediately on death, or the annuity pays at the end rather than the beginning of each year, the state-by-state present values used in the argument change. The identity in this form then does not apply.

Worked Applications

Worked Example: Calculate Both Values from Annual Survival Probabilities

A hypothetical two-year endowment insurance covers a life aged $x$. The effective annual interest rate is 5%. The probabilities of surviving the first and second policy years, conditional on being alive at the start of each year, are $p_x=0.90$ and $p_{x+1}=0.80$. Calculate the endowment insurance and matching temporary annuity-due values, then check the identity.

Method. First find the discount factors and the probabilities of the possible benefit outcomes. Then calculate both actuarial present values using their respective payment times.

The annual discount factor and discount rate are

v=\frac{1}{1.05}=\frac{20}{21}, \qquad d=1-v=\frac{1}{21}.

The first-year death probability is $q_x=1-0.90=0.10$. The probability of surviving one year and dying in the second is $0.90(1-0.80)=0.18$. The probability of surviving both years is $0.90(0.80)=0.72$. Therefore,

\begin{aligned} A_{x:\overline{2}|} &=0.10v+0.18v^2+0.72v^2\\ &=0.10\left(\frac{20}{21}\right) +0.90\left(\frac{20}{21}\right)^2\\ &=\frac{2}{21}+\frac{360}{441}\\ &=\frac{42}{441}+\frac{360}{441} =\frac{134}{147} \approx 0.911565. \end{aligned}

The annuity pays 1 at time 0 with certainty and 1 at time 1 if the insured survives the first year:

\ddot{a}_{x:\overline{2}|} =1+0.90v =1+0.90\left(\frac{20}{21}\right) =\frac{13}{7} \approx 1.857143.

Now check the identity:

A_{x:\overline{2}|}+d\ddot{a}_{x:\overline{2}|} = \frac{134}{147}+\frac{1}{21}\cdot\frac{13}{7} = \frac{134}{147}+\frac{13}{147} =1.

The result is exact, not just a consequence of rounding. A common mistake is to multiply the annuity value by $i=0.05$ instead of by $d=1/21$; the identity uses the effective discount rate, not the effective interest rate.

Worked Example: Check the Identity Across All Three Benefit Outcomes

A hypothetical three-year contract has an effective annual interest rate of 4%. The conditional annual survival probabilities are 0.90, 0.80, and 0.75, in order. Verify the identity by calculating the endowment insurance and annuity-due values. Report values to six decimal places.

Method. Enumerate deaths in each of the three years and survival to maturity. Use the same integer-time benefit schedule for the annuity and insurance.

Here $v=1/1.04=25/26$, and $d=1/26$. The probabilities of death in 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 surviving three years is $0.90(0.80)(0.75)=0.54$. Hence,

\begin{aligned} A_{x:\overline{3}|} &=0.10v+0.18v^2+0.18v^3+0.54v^3\\ &=0.10\left(\frac{25}{26}\right) +0.18\left(\frac{25}{26}\right)^2 +0.72\left(\frac{25}{26}\right)^3\\ &\approx 0.096154+0.166420+0.640077\\ &\approx 0.902651. \end{aligned}

The annuity pays at time 0, at time 1 if the life survives one year, and at time 2 if the life survives two years. The two-year survival probability is $0.90(0.80)=0.72$, so

\begin{aligned} \ddot{a}_{x:\overline{3}|} &=1+0.90v+0.72v^2\\ &=1+0.90\left(\frac{25}{26}\right) +0.72\left(\frac{25}{26}\right)^2\\ &\approx 1+0.865385+0.665680\\ &\approx 2.531065. \end{aligned}

Using unrounded values for the check,

A_{x:\overline{3}|}+\frac{1}{26}\ddot{a}_{x:\overline{3}|} \approx 0.902651+0.097349=1.000000.

The rounded values confirm the identity to six decimal places. The common mistake avoided here is omitting the maturity benefit for the 0.54 probability of surviving three years; that benefit is part of an endowment insurance.

Worked Example: Recover the Interest Rate from the Two Values

For a hypothetical two-year contract, annual conditional survival probabilities are $p_x=0.80$ and $p_{x+1}=0.50$. A unit endowment insurance has value $A_{x:\overline{2}|}=0.672$, and the matching unit annuity-due has value $\ddot{a}_{x:\overline{2}|}=1.64$. Find the implied effective annual interest rate and verify the values.

Method. Rearrange the identity to find the effective discount rate, convert that rate to the effective interest rate, and then recompute the contract values using the stated survival probabilities.

The identity gives

d=\frac{1-A_{x:\overline{2}|}}{\ddot{a}_{x:\overline{2}|}} =\frac{1-0.672}{1.64} =\frac{0.328}{1.64} =0.20.

Since $d=1-v$, the discount factor is $v=0.80$. From $v=1/(1+i)$,

i=\frac{1}{v}-1=\frac{1}{0.80}-1=0.25.

To verify, the death probabilities in years 1 and 2 are $0.20$ and $0.80(0.50)=0.40$; the maturity survival probability is also $0.80(0.50)=0.40$. Thus,

\begin{aligned} A_{x:\overline{2}|} &=0.20(0.80)+0.40(0.80)^2+0.40(0.80)^2\\ &=0.160+0.256+0.256=0.672,\\ \ddot{a}_{x:\overline{2}|} &=1+0.80(0.80)=1.64. \end{aligned}

Finally, $0.672+0.20(1.64)=1$, as required. The conversion step matters: $d=0.20$ is not an interest rate of 20%; it corresponds to $i=25\%$.

Using the Identity Correctly

For a positive effective annual interest rate, the identity can be rearranged in either direction:

A_{x:\overline{n}|}=1-d\ddot{a}_{x:\overline{n}|}, \qquad \ddot{a}_{x:\overline{n}|}=\frac{1-A_{x:\overline{n}|}}{d}.

These are relationships between actuarial present values for the same life, term, interest assumption, and unit benefit schedule. They do not mean the two products have identical cash flows: one pays a single benefit, while the other pays a sequence of possible annuity payments. Rather, after weighting the annuity value by $d$, their present values complement one another.

The identity also scales with the benefit amount. If both benefits are multiplied by a constant $B$, their present values are multiplied by $B$, giving

B A_{x:\overline{n}|}+dB\ddot{a}_{x:\overline{n}|}=B.

For example, this scaling can be used to check a calculation involving a benefit of several thousand dollars without changing the underlying identity. The unit-benefit version is usually simplest for valuation; apply the benefit amount after finding the unit value.

If $i=0$, then $d=0$, and the rearranged formula dividing by $d$ cannot be used. In that case, the endowment insurance pays exactly 1 at some time no later than $n$, so its present value is 1; the identity $A_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}=1$ still holds. This special case distinguishes the identity itself from the algebraic step of dividing by $d$.

Common Exam Traps

Using $i$ instead of $d$. The correct multiplier is $d=1-v=i/(1+i)$. For a 5% effective annual interest rate, $d=0.05/1.05$, not $0.05$.
Using the wrong annuity timing. The identity requires an annuity-due with payments at times $0$ through $n-1$, conditional on survival at each payment time.
Leaving out the maturity benefit. An endowment insurance pays at time $n$ if the insured survives the term. A term insurance alone does not include that benefit.
Mixing terms or benefit amounts. Both values must use the same term and unit benefit basis. A longer annuity or a different insurance benefit cannot be inserted unchanged.
Rounding too early. Keep exact discount factors or full precision until the final reported values, especially when checking whether the sum is one.

Key takeaway: With annual discrete timing, a unit $n$-year endowment insurance and its matching temporary life annuity-due satisfy $A_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}=1$. The state-by-state proof depends on the insurance benefit being paid at the end of the year of death or at maturity, and the annuity being paid at the beginning of each year.

Check Your Understanding

For each problem, use the stated annual benefit and payment timing. Answers are given after each question for checking.

A one-year endowment insurance has effective annual interest rate 6% and first-year survival probability 0.88. Find its unit value and the matching one-year temporary annuity-due value. (Answer: $A_{x:\overline{1}|}=0.9434$, $\ddot{a}_{x:\overline{1}|}=1$.)
A two-year contract has effective annual interest rate 5% and conditional survival probabilities 0.70 and 0.80. Calculate the endowment insurance value and verify the identity using the temporary annuity-due. (Answer: $A_{x:\overline{2}|}=406/441\approx0.9206$, $\ddot{a}_{x:\overline{2}|}=5/3$, and $A+d\ddot{a}=1$.)
For a three-year contract, the conditional annual survival probabilities are 0.90, 0.80, and 0.50, and $i=10\%$. Calculate the temporary annuity-due value, then use the identity to find the endowment insurance value. (Answer: $\ddot{a}_{x:\overline{3}|}=292/121\approx2.4132$, $A_{x:\overline{3}|}=1039/1331\approx0.7806$.)
A two-year contract has $i=5\%$, with conditional survival probabilities 0.80 and 0.50. An analyst uses the temporary annuity-immediate value instead of the annuity-due value. Calculate the endowment insurance value and the annuity-immediate value, and explain why they cannot be substituted directly into the identity. (Answer: $A_{x:\overline{2}|}=404/441\approx0.9161$; the annuity-immediate value is $496/441\approx1.1247$, but the identity requires the annuity-due value $37/21\approx1.7619$.)
A unit endowment insurance has value 0.84 at $i=5\%$. Find the matching temporary annuity-due value, then find the two present values when the common benefit amount is $12,500. (Answer: $\ddot{a}_{x:\overline{n}|}=3.36$; the insurance value is $10,500.00 and $d$ times the annuity value at that benefit amount is $2,000.00, totaling $12,500.00.)

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