Pure Endowment and Temporary Annuity Complements

Introduction

Over a fixed term, an individual is either alive at each scheduled annuity payment time, dies during one of the covered years, or survives to the end of the term. These outcomes give a useful relationship among a temporary annuity-due, term insurance, and a pure endowment. The relationship is especially useful as a calculation check: with consistent benefit timing and assumptions, their appropriately weighted present values sum to one.

The key is that the annuity value is not added to the two insurance values by itself. It is multiplied by the annual discount rate $d$. Omitting that factor is a common error because the identity concerns a discounted change in the annuity value, not the annuity value alone.

Definition: Consider a life aged $x$, a positive integer term $n$, and a constant effective annual interest rate $i$. Let $v=1/(1+i)$ and $d=1-v$. The temporary annuity-due pays 1 at times $0,1,\ldots,n-1$ when the life is then alive. The $n$-year term insurance pays 1 at the end of the year of death if death occurs within the term. The $n$-year pure endowment pays 1 at time $n$ if the life survives to that time. All values below are expected present values at time 0.

The Three Values and Their Timing

Write ${}_kp_x$ for the probability that a life aged $x$ survives at least $k$ years, with ${}_0p_x=1$. For $k=0,\ldots,n-1$, the probability of surviving $k$ years and then dying in the following year is ${}_kp_xq_{x+k}$. This is a probability of death during a particular year, not merely the conditional probability $q_{x+k}$.

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

The superscript $1$ identifies term insurance in this notation. For the annuity, a payment at time $k$ is made only if the life is alive then, and its discount factor is $v^k$. For term insurance, death in year $k+1$ results in a payment at time $k+1$, so its discount factor is $v^{k+1}$. The pure endowment pays only at time $n$, and only on survival to time $n$.

The annual discount rate satisfies $d=1-v=i/(1+i)$. The identity in this tutorial assumes the same fixed annual interest rate for each payment and the discrete timing just described. If death benefits are paid immediately at death, or the annuity pays at a different frequency, these formulas do not directly describe those cash flows.

Why the Values Partition One

The relationship follows by combining the value of each payment pattern and canceling adjacent terms. First use the survival/death partition for each year:

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

This holds because survival through year $k$ is divided into two disjoint outcomes: death during the next year or survival through that year. Also, $d=1-v$, so multiplying each annuity term by $d$ replaces its discount factor $v^k$ with the difference $v^k-v^{k+1}$.

Theorem: Under the assumptions above, the temporary annuity-due, $n$-year term insurance, and $n$-year pure endowment satisfy $$ d\ddot{a}_{x:\overline{n}|}+A^1_{x:\overline{n}|}+{}_nE_x=1. $$ Equivalently, $A^1_{x:\overline{n}|}=1-d\ddot{a}_{x:\overline{n}|}-{}_nE_x$.

To see the cancellation explicitly, expand each term. The annuity contribution is $\sum_{k=0}^{n-1}(v^k-v^{k+1}){}_kp_x$. The insurance contribution, after using the survival/death partition, is $\sum_{k=0}^{n-1}v^{k+1}({}_kp_x-{}_{k+1}p_x)$. The terms involving $v^{k+1}{}_kp_x$ cancel. What remains telescopes from the initial unit value to the survival value at time $n$, which the pure endowment supplies.

\begin{aligned} d\ddot{a}_{x:\overline{n}|}+A^1_{x:\overline{n}|}+{}_nE_x &=\sum_{k=0}^{n-1}(v^k-v^{k+1}){}_kp_x\\ &\quad+\sum_{k=0}^{n-1}v^{k+1}({}_kp_x-{}_{k+1}p_x) +v^n{}_np_x\\ &=\sum_{k=0}^{n-1}v^k{}_kp_x -\sum_{k=0}^{n-1}v^{k+1}{}_{k+1}p_x +v^n{}_np_x\\ &={}_0p_x=1. \end{aligned}

In the last line, every term from $v^1{}_1p_x$ through $v^{n-1}{}_{n-1}p_x$ appears once with a positive sign and once with a negative sign; the final negative term is canceled by the pure endowment. Only $v^0{}_0p_x=1$ remains. This derivation also shows why a pure endowment is essential: without it, the cancellation leaves an uncanceled terminal survival term.

Worked Applications

Worked Example: Using Unconditional Death Probabilities

A hypothetical two-year contract is valued at $i=10\%$. The probability of death during year 1 is $0.20$, and the probability of death during year 2, measured from issue, is also $0.20$. Survival to time 2 is $0.60$. Calculate the two-year term insurance, pure endowment, and temporary annuity-due values, then verify the identity.

Method and probability interpretation. The given year-2 death probability is unconditional from issue. Therefore, survival through year 1 is $0.80$, and the conditional probability of death in year 2 given survival to year 1 is $0.20/0.80=0.25$. For the term insurance formula, use the unconditional probabilities of death in each year: $0.20$ and $0.20$.

Interest factors. $v=1/1.10=10/11$, $v^2=100/121$, and $d=1-v=1/11$.

Term insurance. Discount each unconditional death probability to the end of its year:

A^1_{x:\overline{2}|} =0.20\left(\frac{10}{11}\right) +0.20\left(\frac{100}{121}\right) =\frac{22}{121}+\frac{20}{121} =\frac{42}{121} =0.3471.

Pure endowment. The survival probability to time 2 is $0.60$:

{}_2E_x=0.60\left(\frac{100}{121}\right) =\frac{60}{121} =0.4959.

Temporary annuity-due. The time-0 payment is certain, and the time-1 payment is made with probability $0.80$:

\ddot{a}_{x:\overline{2}|} =1+0.80\left(\frac{10}{11}\right) =\frac{19}{11} =1.7273.

Check. The weighted annuity value is $d\ddot{a}_{x:\overline{2}|}=(1/11)(19/11)=19/121=0.1570$. Then \[ \frac{19}{121}+\frac{42}{121}+\frac{60}{121} =\frac{121}{121}=1. \] The final values are rounded to four decimals. The calculation avoids treating the second-year unconditional death probability as though it were the conditional $q_{x+1}$.

Worked Example: Checking a Three-Year Contract

A hypothetical three-year contract uses $i=25\%$. The conditional annual survival probabilities are $p_x=0.75$, $p_{x+1}=0.80$, and $p_{x+2}=0.50$. Calculate all three values and verify the relationship.

Method. First find survival to each payment time and the unconditional probability of death in each covered year. Here $v=1/1.25=0.8$, $d=0.2$, and the discount factors are $1$, $0.8$, $0.64$, and $0.512$.

The survival probabilities are ${}_1p_x=0.75$, ${}_2p_x=0.75(0.80)=0.60$, and ${}_3p_x=0.60(0.50)=0.30$. The unconditional death probabilities in years 1, 2, and 3 are $0.25$, $0.75(0.20)=0.15$, and $0.60(0.50)=0.30$, respectively.

Temporary annuity-due:

\ddot{a}_{x:\overline{3}|} =1+0.75(0.8)+0.60(0.64) =1+0.60+0.384 =1.984.

Term insurance:

A^1_{x:\overline{3}|} =0.25(0.8)+0.15(0.64)+0.30(0.512) =0.200+0.096+0.1536 =0.4496.

Pure endowment:

{}_3E_x=0.30(0.512)=0.1536.

Check. The weighted annuity value is $0.2(1.984)=0.3968$, and $0.3968+0.4496+0.1536=1.0000$. Each death probability is weighted by the probability of reaching the year in which the death occurs; using only the conditional mortality rates as the death probabilities would give an incorrect insurance value.

Worked Example: Recovering the Annuity Value

A hypothetical two-year model has $i=5\%$, conditional death probabilities $q_x=0.10$ and $q_{x+1}=0.20$, and unit benefits under the timing conventions above. Calculate the term insurance and pure endowment values, then recover the temporary annuity-due value from the identity. Check it directly.

Method. The survival probabilities are ${}_1p_x=0.90$ and ${}_2p_x=0.90(0.80)=0.72$. The unconditional death probabilities are $0.10$ in year 1 and $0.90(0.20)=0.18$ in year 2. The interest factors are $v=20/21$, $v^2=400/441$, and $d=1/21$.

Term insurance:

A^1_{x:\overline{2}|} =0.10\left(\frac{20}{21}\right) +0.18\left(\frac{400}{441}\right) =\frac{2}{21}+\frac{72}{441} =\frac{114}{441} =\frac{38}{147} =0.2585.

Pure endowment:

{}_2E_x=0.72\left(\frac{400}{441}\right) =\frac{288}{441} =\frac{32}{49} =0.6531.

Recover the annuity value. Rearranging the identity gives $d\ddot{a}_{x:\overline{2}|}=1-A^1_{x:\overline{2}|}-{}_2E_x$. Thus:

\ddot{a}_{x:\overline{2}|} =\frac{1-\frac{38}{147}-\frac{96}{147}}{1/21} =\frac{13/147}{1/21} =\frac{13}{7} =1.8571.

Direct check. The annuity-due value is $1+0.90(20/21)=1+18/21=13/7$, as recovered. Also, $d\ddot{a}_{x:\overline{2}|}=13/147$, and $13/147+38/147+96/147=1$. Final decimal values are rounded to four decimals. The method avoids solving for the annuity value before recognizing that the identity can recover it directly.

Common Exam Trap and Practical Use

The most frequent error is to write $\ddot{a}_{x:\overline{n}|}+A^1_{x:\overline{n}|}+{}_nE_x=1$. This is wrong: the temporary annuity-due is not multiplied by the discount rate in that expression. For the first worked example, that incorrect calculation gives $1.7273+0.3471+0.4959=2.5703$, not 1. The correction is to use $d\ddot{a}_{x:\overline{n}|}$: $0.090909\ldots(1.7273\ldots)+0.3471\ldots+0.4959\ldots=1$, with exact fractions shown in the example.

A second trap is mixing conditional and unconditional probabilities. In an insurance sum, the probability assigned to death in year $k+1$ is ${}_kp_xq_{x+k}$. If a problem instead gives an unconditional probability of death in that year, use that probability directly; do not multiply it by survival to year $k$ a second time. Conversely, a stated conditional $q_{x+k}$ must be multiplied by the probability of reaching that year.

The identity is useful for checking a calculation and for recovering one value when the other two are known. It is also a compact way to organize a fixed-term problem: set the payment timing, calculate the survival and death probabilities, use the matching discount factors, and then apply the identity. A disagreement with 1 often points to a timing mismatch, an incorrectly interpreted mortality probability, or a missing factor $d$.

Check Your Understanding

For each problem, use annual discrete timing and unit benefits unless stated otherwise. Values in the answers are rounded to four decimals.

A one-year contract has $i=10\%$ and probability of death during the year $0.30$. Find the term insurance, pure endowment, and temporary annuity-due values, and check the identity. (Answer: $A^1=0.2727,\ {}_1E_x=0.6364,\ \ddot{a}_{x:\overline{1}|}=1.0000;\ d\ddot{a}+A^1+{}_1E_x=1.0000$.)
A two-year contract has $i=20\%$, with conditional survival probabilities $p_x=0.80$ and $p_{x+1}=0.75$. Calculate its three values and verify the identity. (Answer: $\ddot{a}_{x:\overline{2}|}=1.6667,\ A^1_{x:\overline{2}|}=0.3056,\ {}_2E_x=0.4167;\ d\ddot{a}+A^1+{}_2E_x=1.0000$.)
A three-year model has $i=0\%$ and conditional death probability $0.10$ in each year. Calculate the temporary annuity-due, term insurance, and pure endowment values. (Answer: $\ddot{a}_{x:\overline{3}|}=2.7100,\ A^1_{x:\overline{3}|}=0.2710,\ {}_3E_x=0.7290;\ d\ddot{a}+A^1+{}_3E_x=1.0000$.)
A two-year contract has $i=10\%$, $q_x=0.25$, and $q_{x+1}=0.20$. Calculate all three values and verify the relationship. (Answer: $\ddot{a}_{x:\overline{2}|}=1.6818,\ A^1_{x:\overline{2}|}=0.3512,\ {}_2E_x=0.4959;\ d\ddot{a}+A^1+{}_2E_x=1.0000$.)
A two-year contract has $i=5\%$. The unconditional probabilities of death in years 1 and 2 are $0.12$ and $0.22$, respectively, and survival to time 2 is $0.66$. Find the conditional death probability in year 2, then calculate the three values and check the identity. (Answer: $q_{x+1}=0.2500,\ \ddot{a}_{x:\overline{2}|}=1.8381,\ A^1_{x:\overline{2}|}=0.3138,\ {}_2E_x=0.5986;\ d\ddot{a}+A^1+{}_2E_x=1.0000$.)

What You'll Learn