Endowment Insurance Recursions with Maturity Benefits

Why Maturity Changes the Boundary

An endowment insurance policy pays a death benefit if the insured dies during the term and a maturity benefit if the insured survives to the end. A reserve recursion must account for both possible payments. In particular, the last survival branch is not an ordinary continuation year: the policy ends and the maturity benefit is paid.

The key is to state exactly when each reserve is measured. This tutorial uses a reserve immediately before a premium due at the start of a policy year. At maturity, the boundary reserve is measured immediately before the maturity benefit is paid. With those conventions, the maturity benefit appears directly as the terminal reserve.

Definition: Consider an $n$-year endowment insurance issued at age $x$, with level annual premium $P$ payable at times $0,1,\ldots,n-1$ while the insured is alive. The policy pays death benefit $B$ at the end of the year of death and maturity benefit $M$ at time $n$ if the insured survives. For $k=0,\ldots,n-1$, let $V_k$ be the reserve at time $k$, immediately before the premium due then. Let $V_n$ be the reserve at time $n$, immediately before the maturity payment. Thus $V_n=M$. Assume a constant effective annual interest rate $i$, and write ${}_1q_{x+k}$ and ${}_1p_{x+k}=1-{}_1q_{x+k}$ for the one-year death and survival probabilities at age $x+k$.

The reserve convention matters. At time $k<n$, the premium due at that time has not yet been received, so it is added to $V_k$ before the year’s interest is earned. At the end of that year, death leads to payment of $B$; survival leads to the next reserve. At time $n$, the next reserve is $M$, because the maturity payment is still due. Immediately after that payment, the policy has ended and its reserve is zero.

The One-Year Endowment Recursion

Under the stated convention, the recursion is

(V_k+P)(1+i) = {}_1q_{x+k}B+{}_1p_{x+k}V_{k+1}, \qquad k=0,\ldots,n-1, \qquad V_n=M.

For $k=n-1$, the survival term is ${}_1p_{x+n-1}M$. The terminal benefit is therefore included even though there is no premium at time $n$. Solving for the current reserve gives the backward form

V_k = \frac{{}_1q_{x+k}B+{}_1p_{x+k}V_{k+1}}{1+i}-P.

The forward form is useful when the current reserve is known and a later reserve is required. If ${}_1p_{x+k}>0$, rearrangement gives

V_{k+1} = \frac{(V_k+P)(1+i)-{}_1q_{x+k}B}{{}_1p_{x+k}}.

The forward form requires a positive survival probability; if nobody can survive the year, there is no survival-branch reserve to infer. The backward form does not require division by the survival probability.

Derivation: At time $k$, the available amount is the reserve $V_k$ plus the premium $P$ collected immediately. After one year at effective rate $i$, this amount is $(V_k+P)(1+i)$. Death occurs with probability ${}_1q_{x+k}$ and requires payment $B$; survival occurs with probability ${}_1p_{x+k}$ and leaves the insurer with the next reserve $V_{k+1}$. Equating the accumulated amount to the probability-weighted amount required by these two branches gives the recursion. At the final time, the survival branch requires $M$, so $V_n=M$.

This is a cash-flow balance for a reserve measured before the current premium. A recursion written for a reserve measured after a premium would place the premium differently. Do not combine the premium timing from one convention with the reserve definition from another.

Working Backward from the Maturity Benefit

The terminal value $V_n=M$ makes backward calculation direct: begin with the maturity benefit, calculate $V_{n-1}$, and continue to the issue date. If $P$ is the net premium found by the equivalence principle, then $V_0=0$ at issue under this pre-premium convention. If a premium is prescribed rather than equivalence-priced, the resulting $V_0$ need not be zero.

Worked Example: Two-Year Endowment with a Prescribed Premium

A hypothetical two-year policy pays $B=\$10{,}000$ at the end of the year of death and $M=\$12{,}000$ at maturity. Annual premiums of $P=\$500$ are due at times 0 and 1 while the insured is alive. The effective annual interest rate is $i=5\%$. Assume ${}_1q_x=0.10$ and ${}_1q_{x+1}=0.20$. Calculate $V_1$ and $V_0$.

Method. Work backward from the maturity boundary. The survival probabilities are ${}_1p_x=1-0.10=0.90$ and ${}_1p_{x+1}=1-0.20=0.80$.

At time 2, immediately before the maturity payment, $V_2=M=\$12{,}000$. At time 1, death produces $\$10{,}000$, and survival leads to $V_2=\$12{,}000$. Therefore,

V_1 = \frac{0.20(10{,}000)+0.80(12{,}000)}{1.05}-500 = \frac{2{,}000+9{,}600}{1.05}-500 = 11{,}047.6190-500 = \$10{,}547.62.

At time 0, death leads to the death benefit and survival leads to $V_1$:

V_0 = \frac{0.10(10{,}000)+0.90(10{,}547.6190)}{1.05}-500 = \frac{1{,}000+9{,}492.8571}{1.05}-500 = 9{,}993.1972-500 = \$9{,}493.20.

Check. At time 1, $(V_1+P)(1+i)=(10{,}547.6190+500)(1.05)=\$11{,}600$, equal to $0.20(10{,}000)+0.80(12{,}000)=\$11{,}600$. At time 0, $(V_0+P)(1.05)=\$10{,}492.8571$, equal to $0.10(10{,}000)+0.90(10{,}547.6190)=\$10{,}492.8571$. The reserves satisfy both one-year balances. The common error avoided here is dropping the maturity benefit from the final survival branch.

Finding a Net Premium from the Initial Boundary

For a level net premium under the equivalence principle, the expected present value of future benefits equals the expected present value of premiums. Equivalently, using this tutorial’s reserve convention, the initial reserve is zero: $V_0=0$. For a one-year contract, the recursion makes the premium calculation especially transparent. For longer contracts, the same principle applies through the full sequence of annual survival probabilities and benefits.

Worked Example: Net Premium for a One-Year Endowment

A hypothetical one-year endowment issued at age $x$ pays $\$11{,}000$ at the end of the year if death occurs and $\$5{,}500$ at the end of the year if the insured survives. The annual premium is due at time 0. Assume $i=10\%$ and ${}_1q_x=0.20$. Find the net premium that gives $V_0=0$.

Method. The survival probability is ${}_1p_x=0.80$, and the terminal reserve is $V_1=M=\$5{,}500$. Substitute these values into the one-year recursion:

$$ (0+P)(1.10) = 0.20(11{,}000)+0.80(5{,}500) = 2{,}200+4{,}400 = 6{,}600. $$

Thus $P=6{,}600/1.10=\$6{,}000$. The death and survival benefits have expected value $\$6{,}600$ at time 1, and discounting that amount by $1.10$ gives the required time-0 premium.

Check. With $P=\$6{,}000$, the accumulated premium is $6{,}000(1.10)=\$6{,}600$, exactly the probability-weighted benefit at year-end. Hence the initial reserve is zero. The mistake avoided is treating the maturity benefit as payable only in addition to a survival reserve: at maturity the policy ends, so the survival branch is $M$, not $M$ plus another continuation value.

Forward Calculation and a Terminal-Value Check

The same recursion can be used forward if a reserve and premium are known. This is useful for checking a sequence calculated backward or for finding a later reserve from an earlier one. The final step must still be consistent with the maturity boundary; a forward calculation arriving at time $n$ should produce $V_n=M$.

Worked Example: Three-Year Reserves with Equal Annual Mortality

A hypothetical three-year endowment pays a death benefit of $B=\$8{,}000$ at the end of the year of death and a maturity benefit of $M=\$4{,}000$ at time 3. The annual premium is $P=\$1{,}000$. Assume $i=0\%$ and a one-year death probability of $0.50$ at each of the three attained ages. Calculate the reserves from $V_2$ back to $V_0$, then check one step forward.

Method. Each survival probability is $0.50$. Begin with $V_3=M=\$4{,}000$. For year 3, the backward recursion gives

V_2 = 0.50(8{,}000)+0.50(4{,}000)-1{,}000 = 4{,}000+2{,}000-1{,}000 = \$5{,}000.

For year 2, survival leads to $V_2=\$5{,}000$:

V_1 = 0.50(8{,}000)+0.50(5{,}000)-1{,}000 = 4{,}000+2{,}500-1{,}000 = \$5{,}500.

For year 1, survival leads to $V_1=\$5{,}500$:

V_0 = 0.50(8{,}000)+0.50(5{,}500)-1{,}000 = 4{,}000+2{,}750-1{,}000 = \$5{,}750.

Check. The forward recursion from $V_2$ to $V_3$ gives $(5{,}000+1{,}000)-0.50(8{,}000)=\$2{,}000$, which must be divided by the survival probability $0.50$, producing $V_3=\$4{,}000$, the stated maturity boundary. All monetary amounts are exact to the cent.

Common Boundary Error

A frequent mistake is to set $V_n=0$ because the policy has no reserve after it terminates. That statement is true after the maturity benefit has been paid, but the recursion needs the amount required immediately before that payment. In the first worked example, setting $V_2=0$ would incorrectly give

V_1^{\text{wrong}} = \frac{0.20(10{,}000)+0.80(0)}{1.05}-500 = \$1{,}404.76,

rather than the correct $V_1=\$10{,}547.62$. The incorrect calculation omits the $\$12{,}000$ maturity benefit, whose probability-weighted contribution in the final year is $0.80(12{,}000)=\$9{,}600$. Use $V_n=M$ when the terminal reserve is defined immediately before payment; use zero only for the reserve after the policy’s final benefit has been paid.

This distinction is central to a reliable calculation. Before applying a recursion, identify the reserve’s timing, the timing of the premium, and whether the terminal value is before or after the maturity payment. Then condition on death and survival for each year, including the final one.

Key takeaway: For a level-premium endowment with a death benefit $B$ and maturity benefit $M$, the pre-premium reserve satisfies $(V_k+P)(1+i)={}_1q_{x+k}B+{}_1p_{x+k}V_{k+1}$, with terminal boundary $V_n=M$. The maturity benefit is the final survival-branch liability, not a continuation reserve after the contract ends.

Check Your Understanding

Use the pre-premium reserve convention and show the death and survival branches in each calculation.

A one-year endowment has $B=\$6{,}000$, $M=\$3{,}000$, $P=\$1{,}000$, $i=5\%$, and ${}_1q_x=0.20$. Find $V_0$, rounded to the nearest cent. (Answer: $\$2{,}428.57$)
A two-year endowment has $B=\$8{,}000$, $M=\$4{,}000$, $P=\$500$, $i=0\%$, and death probability $0.25$ in each year. Find $V_1$ and $V_0$. (Answer: $V_1=\$4{,}500$, $V_0=\$4{,}875$)
A two-year endowment has $B=\$10{,}000$, $M=\$5{,}000$, $i=0\%$, ${}_1q_x=0.20$, and ${}_1q_{x+1}=0.25$. Find the level net premium payable at times 0 and 1. (Answer: $\$3{,}888.89$)
A two-year endowment has $B=\$6{,}000$, $P=\$1{,}000$, $i=0\%$, and death probability $0.50$ in each year. If $V_0=\$4{,}500$, infer the maturity benefit $M$. (Answer: $\$6{,}000$)
In the final year of an endowment, $B=\$20{,}000$, $M=\$12{,}000$, $P=\$8{,}000$, $i=4\%$, and ${}_1q_{x+n-1}=0.10$. Find $V_{n-1}$, rounded to the nearest cent. (Answer: $\$4{,}307.69$)

What You'll Learn