The Endowment Insurance-Annuity Identity

Why the Timing Match Matters

A discrete endowment insurance pays a benefit if death occurs during a specified term and also pays a benefit if the insured survives to the end of that term. A temporary life annuity-due pays while the insured is alive, with the first payment at the start of the term. These contracts have a useful relationship: with unit benefits and matching term and timing, their present values add to a simple constant after the annuity value is multiplied by the effective discount rate.

The timing conditions are essential. The insurance death benefit is paid at the end of the year of death, and the survival benefit is paid at the end of the term. The annuity-due pays at times $0,1,\ldots,n-1$, provided the insured is alive at each payment time. Thus, the annuity and insurance payments divide the same time interval in complementary ways.

Definition: For a unit-benefit discrete endowment insurance issued to age $x$ for $n$ years, $A_{x:\overline{n}|}$ is the expected present value of 1 paid at the end of the year of death if death occurs within the term, or 1 paid at time $n$ if the insured survives to time $n$. The matching temporary life annuity-due has value $\ddot{a}_{x:\overline{n}|}$: it pays 1 at each time $k=0,1,\ldots,n-1$ if the insured is alive at time $k$. The effective annual interest rate is $i$, the discount factor is $v=1/(1+i)$, and the effective annual discount rate is $d=1-v=i/(1+i)$.

Assume a fixed effective annual interest rate, annual discrete benefits, and a life-contingency model that assigns probabilities to survival and death over each year. Write ${}_kp_x$ for the probability that a life aged $x$ survives $k$ years, and ${}_kq_x=1-{}_kp_x$. The endowment insurance and annuity are both valued on the same mortality and interest basis. The identity does not directly apply if the benefits, payment dates, or terms differ.

The temporary annuity-due value is the sum of the expected present values of its possible payments. A payment at time $k$ is made only if the insured has survived $k$ years. The endowment insurance value can be written by separating death in each policy year from survival to the end of the term.

\ddot{a}_{x:\overline{n}|} =\sum_{k=0}^{n-1}v^k\,{}_kp_x

A_{x:\overline{n}|} =\sum_{k=0}^{n-1}v^{k+1}\,{}_kp_xq_{x+k} +v^n\,{}_np_x

Here $q_{x+k}$ is the probability of death during the next year conditional on being alive at age $x+k$. The final term in the insurance value is the survival benefit. The next argument establishes the identity path by path, which explains why it does not depend on the particular mortality probabilities.

Consider death in year $k+1$.
If death occurs between times $k$ and $k+1$, the annuity pays at times $0$ through $k$. The endowment insurance pays at time $k+1$.

Discount that path's payments.
The annuity payments have present value $1+v+\cdots+v^k$, and the insurance payment has present value $v^{k+1}$.

Use $d=1-v$.
Multiplying the annuity present value by $d$ gives $d(1+v+\cdots+v^k)=1-v^{k+1}$. Adding the insurance present value $v^{k+1}$ gives 1.

Check the survival path.
If the insured survives to time $n$, the annuity pays at times $0$ through $n-1$, and the endowment pays at time $n$. The same calculation gives $d(1+v+\cdots+v^{n-1})+v^n=1$.

Theorem: For unit benefits, a discrete endowment insurance and a temporary life annuity-due with the same term $n$, annual interest rate, and life-contingency basis satisfy $$ A_{x:\overline{n}|}+d\,\ddot{a}_{x:\overline{n}|}=1. $$ Equivalently, $A_{x:\overline{n}|}=1-d\,\ddot{a}_{x:\overline{n}|}$ and $\ddot{a}_{x:\overline{n}|}=(1-A_{x:\overline{n}|})/d$, when $d>0$.

The pathwise argument proves the result: for every possible death year or survival to the end of the term, the discounted endowment benefit plus $d$ times the discounted annuity payments equals 1. Taking expected values preserves that equality. This also explains why the mortality probabilities do not appear explicitly in the identity, even though they determine each contract's individual value.

Applying the Identity

Use the identity when one value is known or is easier to calculate than the other. First find the effective discount rate $d=i/(1+i)$; do not substitute $i$ for $d$. Then use the annuity-due value or endowment value on the same term and basis. A useful check is to calculate the insurance value directly from the death-year probabilities if those are available.

Worked Example: Calculating Endowment Insurance from a Temporary Annuity

A hypothetical policy covers a life aged $x$ for two years. The effective annual interest rate is $5\%$. The mortality assumption is $q_x=0.10$ and $q_{x+1}=0.20$. Calculate the unit-benefit endowment insurance value using the identity, then verify it from the benefit cash flows.

Method and assumption. The annuity-due pays at times 0 and 1 while the life is alive. The endowment pays at the end of the year of death or at time 2 on survival. The annual discount factor and discount rate are $v=1/1.05=20/21$ and $d=1-v=1/21$.

The one-year survival probability is ${}_1p_x=1-q_x=0.90$. Therefore,

\ddot{a}_{x:\overline{2}|} =1+v\,{}_1p_x =1+\frac{20}{21}(0.90) =1+\frac{6}{7} =\frac{13}{7}

Apply the identity:

A_{x:\overline{2}|} =1-\frac{1}{21}\left(\frac{13}{7}\right) =1-\frac{13}{147} =\frac{134}{147} \approx 0.911565

Direct check. Death in year 1 has probability $0.10$. Death in year 2 has probability $0.90(0.20)=0.18$, and survival through year 2 has probability $0.90(0.80)=0.72$. Thus,

A_{x:\overline{2}|} =(0.10)v+(0.18)v^2+(0.72)v^2 =\frac{2}{21}+\frac{24}{147}+\frac{96}{147} =\frac{134}{147} \approx 0.911565

The direct cash-flow value agrees with the identity. The final two terms share the same discount factor because both benefits are paid at time 2.

Worked Example: Calculating the Annuity from Endowment Insurance

A hypothetical three-year contract has a unit endowment benefit and is valued at an effective annual interest rate of $10\%$. Assume a constant one-year mortality probability $q_{x+k}=0.10$ for $k=0,1,2$. The endowment insurance value is $1030/1331$. Find the matching temporary annuity-due value and verify the insurance value directly.

Method and assumption. The survival probabilities to times 1 and 2 are $0.90$ and $0.90^2=0.81$. The interest and discount rates are $v=10/11$ and $d=1/11$. Rearrange the identity to solve for the annuity:

\ddot{a}_{x:\overline{3}|} =\frac{1-A_{x:\overline{3}|}}{d} =\frac{1-\frac{1030}{1331}}{\frac{1}{11}} =\frac{\frac{301}{1331}}{\frac{1}{11}} =\frac{301}{121} \approx 2.487603

Direct check of the insurance value. The probabilities of death in years 1, 2, and 3 are $0.10$, $0.90(0.10)=0.09$, and $0.81(0.10)=0.081$. The probability of survival through year 3 is $0.729$. Death in year 3 and survival to time 3 both produce a payment at time 3. Therefore,

A_{x:\overline{3}|} =(0.10)v+(0.09)v^2+(0.081+0.729)v^3 =\frac{121}{1331}+\frac{99}{1331}+\frac{810}{1331} =\frac{1030}{1331} \approx 0.773855

As a final check, $d\ddot{a}_{x:\overline{3}|}=(1/11)(301/121)=301/1331$, so $A_{x:\overline{3}|}+d\ddot{a}_{x:\overline{3}|}=(1030+301)/1331=1$.

Worked Example: Avoiding an Annuity-Immediate Timing Error

A hypothetical two-year policy has an effective annual interest rate of $5\%$, with $q_x=0.20$ and $q_{x+1}=0.25$. Calculate the endowment insurance value from the matching annuity-due, and show what goes wrong if an annuity-immediate is substituted into the identity.

Method and assumption. The matching annuity-due pays at times 0 and 1 if the life is then alive. The survival probabilities at those times are 1 and $0.80$. With $v=20/21$ and $d=1/21$,

\ddot{a}_{x:\overline{2}|} =1+0.80\left(\frac{20}{21}\right) =\frac{37}{21}

Thus the endowment value is

A_{x:\overline{2}|} =1-\frac{1}{21}\left(\frac{37}{21}\right) =1-\frac{37}{441} =\frac{404}{441} \approx 0.916100

A temporary annuity-immediate with two possible payments instead pays at times 1 and 2, provided the life is alive at those times. Its value is $0.80v+0.60v^2$, since survival to time 2 is $0.80(0.75)=0.60$. This equals $64/49$. Incorrectly substituting it gives $1-(1/21)(64/49)=965/1029\approx0.937804$, not the endowment value. The error is the timing mismatch: an annuity-immediate does not make the payment at time 0 that is included in the pathwise proof. Use the annuity-due with matching term and payment timing.

Direct check. The endowment value is $0.20v+0.20v^2+0.60v^2$, because death in year 2 has probability $0.80(0.25)=0.20$, and survival to time 2 has probability $0.60$. This is $4/21+160/441=244/441$? That sum would incorrectly count the first term: $0.20v=4/21=84/441$, and the time-2 benefits total $0.80v^2=320/441$. Hence the correct total is $404/441\approx0.916100$, agreeing with the identity.

Common Traps and Practical Checks

Using $i$ instead of $d$. The multiplier is $d=i/(1+i)=1-v$, not the effective interest rate $i$. At $i=5\%$, for example, $d=1/21$, not $0.05$.
Using an annuity-immediate. The identity requires an annuity-due with a payment at time 0 and remaining payments at times $1$ through $n-1$. Payment dates determine which identity applies.
Changing the term. The insurance and annuity must have the same term. A longer annuity includes payment opportunities that are not paired with the endowment benefits in the proof.
Forgetting the survival benefit. Endowment insurance pays at time $n$ on survival. Omitting that payment would leave term insurance, not endowment insurance.
Changing benefit amounts. The stated identity is for unit benefits. If benefits differ, scale the corresponding present values before applying a relationship.

A quick numerical check is to calculate both terms and confirm that $A_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}=1$, allowing only for rounding. If a result differs materially from 1, first check the value of $d$, the term, and whether the annuity is due. This matching relationship is useful for converting between insurance and annuity values on a common basis; it is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated annual discrete timing and unit benefits. Keep the term and payment timing matched.

A two-year endowment is valued at $5\%$ effective annual interest, with $q_x=0.20$ and $q_{x+1}=0.25$. Find the matching temporary annuity-due value and endowment insurance value. (Answer: $\ddot{a}_{x:\overline{2}|}=37/21\approx1.761905$, $A_{x:\overline{2}|}=404/441\approx0.916100$.)
A one-year endowment is valued at $8\%$ effective annual interest, with $q_x=0.30$. Find both values. (Answer: $\ddot{a}_{x:\overline{1}|}=1$, $A_{x:\overline{1}|}=25/27\approx0.925926$.)
A three-year endowment is valued at $4\%$ effective annual interest, with $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.10$. Find the temporary annuity-due value and the endowment insurance value. (Answer: $\ddot{a}_{x:\overline{3}|}=1+\frac{45}{52}+\frac{225}{338}\approx2.531065$, $A_{x:\overline{3}|}\approx0.902651$.)
A two-year endowment is valued at $10\%$ effective annual interest, with $q_x=0.10$ and $q_{x+1}=0.40$. Find both values using the identity. (Answer: $\ddot{a}_{x:\overline{2}|}=20/11\approx1.818182$, $A_{x:\overline{2}|}=101/121\approx0.834711$.)
A two-year endowment is valued at $4\%$ effective annual interest, with $q_x=0.15$ and $q_{x+1}=0.30$. Find the matching annuity-due value and endowment insurance value. (Answer: $\ddot{a}_{x:\overline{2}|}=189/104\approx1.817308$, $A_{x:\overline{2}|}=2515/2704\approx0.930104$.)

The Endowment Insurance-Annuity Identity

What You'll Learn

Why the Timing Match Matters