Endowment Insurance Recursion and Maturity Boundary

Why the Maturity Boundary Matters

An n-year endowment insurance provides a benefit if the insured dies during the term and also provides a benefit if the insured survives to the end of the term. Its one-year recursion therefore has two possible first-year outcomes: death, which produces a benefit, and survival, which leaves a shorter endowment contract in force. The value assigned when no term remains is essential: at that point, the maturity benefit is due.

This boundary distinguishes endowment insurance from term insurance. A term contract with no time remaining has no future benefit, but a unit endowment contract with no time remaining has a unit maturity payment due immediately. Applying the wrong boundary changes the value of the whole contract.

Definition: For $n\geq 1$, $A_{x:\overline{n}|}$ denotes the expected present value at age $x$ of a unit n-year endowment insurance. The contract pays 1 at the end of the year of death if death occurs within the n-year term, or pays 1 at the end of year n if the insured survives to that time. For zero remaining years, define $A_{x:\overline{0}|}=1$: the maturity payment is due immediately.

Assume annual benefits are paid at the end of the year of death or at the maturity date, whichever applies; the two benefits are not both paid. Let $i$ be a fixed effective annual interest rate and $v=1/(1+i)$. Let $q_x$ be the probability that a life aged $x$ dies within the next year, and $p_x=1-q_x$ the probability of surviving that year. Probabilities at later ages are defined in the same way. This recursion uses annual, discrete benefit timing and the one-year mortality probabilities; it does not apply unchanged to a continuous benefit paid at the moment of death or to unequal death and maturity benefits.

The One-Year Endowment Recursion

Condition on what happens in the first year. Death in that year has probability $q_x$ and produces a unit payment at time 1. Survival has probability $p_x$; at time 1, the remaining contract is an $(n-1)$-year endowment insurance at age $x+1$. Its value at time 1 must be discounted one year to obtain a value at age $x$.

Theorem: For $n\geq 1$, under the assumptions above, $$ A_{x:\overline{n}|}=v q_x+v p_x A_{x+1:\overline{n-1}|}, \qquad A_{x:\overline{0}|}=1. $$

The first term is the present value of the first-year death benefit, weighted by its probability. The second term is the probability of surviving the first year multiplied by the discounted value of the remaining endowment contract. The boundary value is not discounted: when the remaining term reaches zero, the maturity benefit is payable at that instant.

Separate first-year outcomes.
Death within the first year has probability $q_x$; survival to age $x+1$ has probability $p_x$.

Value the death outcome.
The unit death benefit is paid at time 1, so its expected present value is $v q_x$.

Value the survival outcome.
Conditional on survival, the remaining contract has value $A_{x+1:\overline{n-1}|}$ at time 1. Its age-$x$ contribution is $v p_x A_{x+1:\overline{n-1}|}$.

Apply the maturity boundary.
Continue reducing the remaining term until it is zero; then use $A_{x+n:\overline{0}|}=1$.

Why the Recursion Gives the Endowment Value

The recursion follows from the expected present value of the possible benefits. If death occurs in year $k+1$, where $k=0,\ldots,n-1$, the probability of reaching age $x+k$ is ${}_kp_x$, and the conditional probability of dying during the following year is $q_{x+k}$. The benefit is paid at time $k+1$. If the insured survives all n years, the maturity benefit is paid at time n. Thus the value can also be written as

A_{x:\overline{n}|} = \sum_{k=0}^{n-1} v^{k+1}{}_kp_x q_{x+k} + v^n{}_np_x.

Here ${}_0p_x=1$, and ${}_kp_x$ is the probability of surviving k years from age $x$. The sum accounts for the mutually exclusive death-in-year outcomes; the final term accounts for survival to maturity. To see how this expression yields the recursion, separate the $k=0$ death term, $v q_x$. Every other death term and the maturity term requires survival through the first year. After that survival, each is a benefit in the remaining $(n-1)$-year endowment contract. Discounting its value at time 1 gives $v p_x A_{x+1:\overline{n-1}|}$. At $n=0$, the sum is empty and the maturity term is $v^0{}_0p_x=1$, which confirms the boundary.

Worked Example: Recursing Through a Three-Year Endowment

A hypothetical insurer values a unit three-year endowment insurance issued at age $x$. The effective annual interest rate is $i=25\%$, so $v=1/1.25=0.8$. The annual mortality probabilities are $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$; therefore the corresponding survival probabilities are $0.90$, $0.80$, and $0.75$. Find the value at issue. Use backward recursion from the maturity boundary.

Method. Start with zero remaining years at age $x+3$, where the maturity payment is due, and apply the recursion one year at a time.

With one year remaining at age $x+2$, the recursion uses the unit boundary:

A_{x+2:\overline{1}|} =0.8(0.25)+0.8(0.75)(1) =0.20+0.60 =0.80.

With two years remaining at age $x+1$:

A_{x+1:\overline{2}|} =0.8(0.20)+0.8(0.80)(0.80) =0.16+0.512 =0.672.

With three years remaining at age $x$:

A_{x:\overline{3}|} =0.8(0.10)+0.8(0.90)(0.672) =0.08+0.48384 =0.56384.

The unit endowment value is $0.56384$, or $0.56$ if rounded to two decimal places. To check, the discounted contributions for death in years 1, 2, and 3 are $0.08$, $0.1152$, and $0.09216$. The probability of surviving all three years is $0.90(0.80)(0.75)=0.54$, so the discounted maturity contribution is $0.8^3(0.54)=0.27648$. Their sum is $0.08+0.1152+0.09216+0.27648=0.56384$, agreeing with the recursion. The common mistake avoided is using a zero boundary at maturity, which would omit the final benefit.

Using the Recursion to Recover a Mortality Probability

When an endowment value and a shorter-duration value are known, the recursion can be rearranged to find an unknown one-year death probability. This is valid when the interest rate and benefit pattern are known, and the shorter contract really is the surviving policy after one year.

Worked Example: Finding the First-Year Death Probability

A hypothetical two-year unit endowment insurance is issued at age $x$. The effective annual interest rate is $25\%$, so $v=0.8$. The one-year endowment value at age $x+1$ is $0.8$, and the two-year value at age $x$ is $0.704$. Find $q_x$, the first-year death probability, and check that it is a valid probability.

Method. Substitute the known values into $A_{x:\overline{2}|}=v q_x+v(1-q_x)A_{x+1:\overline{1}|}$, then solve for $q_x$.

$$ 0.704=0.8q_x+0.8(1-q_x)(0.8) $$

Expand the right-hand side:

$$ 0.704=0.8q_x+0.64-0.64q_x =0.64+0.16q_x. $$

Subtract $0.64$ from both sides and divide by $0.16$:

q_x=\frac{0.704-0.64}{0.16} =\frac{0.064}{0.16} =0.40.

Thus $q_x=0.40$, and $p_x=1-q_x=0.60$. The probability is between 0 and 1, as required. Substitution checks the result: $0.8(0.40)+0.8(0.60)(0.8)=0.32+0.384=0.704$. The common mistake avoided is treating the known one-year value as an age-$x$ value; it is the value at time 1 and must be multiplied by both $p_x$ and $v$.

The Boundary Is Part of the Contract

The value at zero remaining years is determined by the benefits still payable, not by a general rule that every insurance value becomes zero when its term expires. For unit endowment insurance, the insured has survived to the maturity date, so the unit maturity benefit is due then. In value terms at that date, its present value is $1$.

This is different from a term insurance that pays only on death during its stated term. At the end of that term, there is no remaining term-insurance benefit, so its zero-term value is zero. The distinction is about the contract's benefit definition, not about how a recursion is algebraically arranged.

Worked Example: Correcting a Zero-Boundary Error

A hypothetical two-year unit endowment insurance has an effective annual interest rate of $25\%$, so $v=0.8$. Its death probabilities are $q_x=0.10$ in the first year and $q_{x+1}=0.20$ in the second year, with survival probabilities $0.90$ and $0.80$. Find the correct value and compare it with the value obtained by mistakenly setting the zero-remaining-term boundary to zero.

Correct method. At age $x+1$, one year remains. The second-year death benefit and the maturity benefit together cover the two possible outcomes, so use $A_{x+2:\overline{0}|}=1$:

A_{x+1:\overline{1}|} =0.8(0.20)+0.8(0.80)(1) =0.16+0.64 =0.80.

Then the value at issue is

A_{x:\overline{2}|} =0.8(0.10)+0.8(0.90)(0.80) =0.08+0.576 =0.656.

Incorrect calculation. If the boundary is wrongly set to zero, the one-year value at age $x+1$ is incorrectly calculated as $0.8(0.20)+0.8(0.80)(0)=0.16$. This then gives the incorrect issue value $0.8(0.10)+0.8(0.90)(0.16)=0.1952$.

The error removes the maturity payment for anyone surviving to the end of year 2. A direct check of the correct value gives a first-year death contribution of $0.08$, a second-year death contribution of $0.8^2(0.90)(0.20)=0.1152$, and a maturity contribution of $0.8^2(0.90)(0.80)=0.4608$. Their sum is $0.08+0.1152+0.4608=0.656$, confirming the correct recursion. The common mistake is to reuse the zero-term boundary for term insurance without checking which benefits remain payable.

Practical Checks

When evaluating a recursive endowment value, keep the duration and the payment time visible at every step. The subscript age advances by one when the remaining term decreases by one. The one-year discount factor applies both to a death benefit payable at the end of the first year and to the value of the surviving contract at that time.

At each step, check that the death and survival probabilities sum to 1: $q_x+p_x=1$.
At zero remaining years, use a unit value for a unit maturity benefit.
For a positive term, check that the recursion includes both the first-year death contribution and the survival continuation value.
When possible, verify the result by adding the discounted probabilities of death in each year and survival to maturity.

The recursion is a valuation identity for the specified benefit pattern and assumptions. It does not by itself include premiums, expenses, or other contract features. If a contract changes the death benefit, maturity benefit, or payment timing, those details must be incorporated into its valuation rather than assumed to follow this unit-benefit formula.

Check Your Understanding

For each problem, use the stated annual interest and mortality assumptions, and pay particular attention to the maturity boundary.

A unit one-year endowment insurance has $i=25\%$ and $q_x=0.30$. Find its value using the one-year recursion and the zero-remaining-term boundary. (Answer: $0.80$)
A unit two-year endowment insurance has $i=25\%$, $q_x=0.10$, and $q_{x+1}=0.25$. Calculate its value at issue by backward recursion. (Answer: $0.656$)
At $i=25\%$, a unit two-year endowment has value $0.704$ at age $x$, and its one-year endowment value at age $x+1$ is $0.80$. Find $q_x$. (Answer: $0.40$)
A unit three-year endowment has $i=25\%$ and death probabilities $q_x=0.20$, $q_{x+1}=0.10$, and $q_{x+2}=0.30$. Calculate its value at issue by recursion. (Answer: $0.57984$)
A hypothetical five-year endowment pays a benefit of $\$25{,}000$ at maturity if the insured survives, and has no remaining term at the maturity date. What is the value of the maturity payment at that date, before any discounting? (Answer: $\$25{,}000$)

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