One-Year Recursions with Nonunit Death Benefits

Why the Death Benefit Changes the Recursion

A one-year insurance recursion splits the possible outcomes during the coming year into death and survival. With a unit death benefit, the death term is the probability of death during the year. With a constant benefit $B$, that term must include the amount paid on death: probability multiplied by benefit. The continuation term, however, is still the value of the insurance conditional on survival, provided that continuation value already reflects the same benefit $B$.

The distinction matters in both whole-life and term insurance. It is easy to multiply the first-year death term by $B$ and then accidentally multiply the continuation value by $B$ again. The recursion is correct only when both terms are expressed in consistent monetary units.

Definition: Assume an effective annual interest rate $i$, with $v=1/(1+i)$. Let $q_x={}_1q_x$ be the probability that a life aged $x$ dies during the next year, and let $p_x={}_1p_x=1-q_x$ be the probability of surviving that year. A level death benefit $B$ is paid at the end of the year of death. $A_x^{(B)}$ denotes the expected present value of a whole-life death benefit of $B$, conditional on being alive at age $x$. $A_{x:\overline{n}|}^{(B)}$ denotes the corresponding expected present value for an $n$-year term insurance: it pays $B$ for death within the term and nothing if the insured survives to the end.

These are benefit present values, not premium or reserve calculations: no premiums, expenses, or other cash flows are included. The annual timing assumption is essential. The formulas below apply to benefits paid at the end of the year of death; a benefit paid immediately at death requires a different timing treatment.

Deriving the Whole-Life Recursion

At age $x$, either death occurs during the first year or the insured survives to age $x+1$. In the first case, $B$ is paid at time 1. In the second case, the future whole-life insurance has value $A_{x+1}^{(B)}$ at time 1. Both first-year outcomes are discounted back one year.

A_x^{(B)}=v\left(q_xB+p_xA_{x+1}^{(B)}\right).

The equation is a conditional expectation. The death branch has probability $q_x$ and time-1 payment $B$; the survival branch has probability $p_x$ and time-1 continuation value $A_{x+1}^{(B)}$. Multiplying their probability-weighted values by $v$ converts the time-1 amounts to values at age $x$.

Theorem: For a constant death benefit $B$, the first-year death term in the whole-life recursion is $q_xB$. The continuation term is $p_xA_{x+1}^{(B)}$, where the continuation value is already based on benefit $B$. Thus the entire probability-weighted sum, not just one of its terms, is discounted by $v$.

The recursion also follows directly from the unit-benefit value. For unit benefit, the standard whole-life recursion is $A_x=v(q_x+p_xA_{x+1})$. Every possible death payment for benefit $B$ is $B$ times the corresponding payment for unit benefit. Expected present values are linear in the payments, so

A_x^{(B)}=B A_x.

Substituting $A_{x+1}^{(B)}=B A_{x+1}$ in the nonunit recursion gives $v(q_xB+p_xB A_{x+1})=Bv(q_x+p_xA_{x+1})=BA_x$. This confirms that the recursion and benefit scaling agree. It is valid because the benefit is the same fixed amount in every year of death; it does not justify multiplying a continuation value that is already benefit-adjusted by $B$ again.

Worked Example: Whole-Life Value with a Nonunit Benefit

A hypothetical whole-life policy pays $B=\$10{,}000$ at the end of the year of death. The effective annual interest rate is $i=4\%$. At age $x$, $q_x=0.10$, so $p_x=0.90$. The whole-life benefit present value conditional on survival to age $x+1$ is $A_{x+1}^{(B)}=\$4{,}000$. Find the value at age $x$.

Method. Use the nonunit-benefit whole-life recursion. The stated continuation value is already in dollars for a $\$10{,}000$ benefit, so it must not be multiplied by $B$ again.

First calculate the discount factor: $v=1/(1.04)=0.96153846\ldots$. The first-year expected death payment, measured at time 1, is $q_xB=0.10(\$10{,}000)=\$1{,}000$. The survival-weighted continuation value at time 1 is $p_xA_{x+1}^{(B)}=0.90(\$4{,}000)=\$3{,}600$.

A_x^{(B)} =\frac{\$1{,}000+\$3{,}600}{1.04} =\frac{\$4{,}600}{1.04} =\$4{,}423.0769\ldots

Rounded to the nearest cent, $A_x^{(B)}=\$4{,}423.08$. As a check, discounting each branch separately gives $\$1{,}000/1.04=\$961.5385\ldots$ and $\$3{,}600/1.04=\$3{,}461.5385\ldots$; their sum is $\$4{,}423.0769\ldots$, the same value.

Term Insurance and Its Boundary

For an $n$-year term insurance, the same split applies during a year when coverage remains. If death occurs during the year, the benefit is paid; if the insured survives, the contract continues with one fewer year of coverage. Once the term has expired, there is no remaining death benefit under this contract.

A_{x:\overline{n}|}^{(B)} =v\left(q_xB+p_xA_{x+1:\overline{n-1}|}^{(B)}\right), \qquad n\geq 1,

A_{x:\overline{0}|}^{(B)}=0.

The zero boundary is a consequence of the product definition: with no coverage time left, there can be no payment under this term insurance. In particular, a one-year term value is $A_{x:\overline{1}|}^{(B)}=vq_xB$. There is no survival payment at the end of the year. A contract that also pays on survival to maturity is a different benefit design and does not use this zero-maturity-payment boundary.

Worked Example: Two-Year Term Insurance

A hypothetical two-year term policy pays $\$20{,}000$ at the end of the year of death and pays nothing on survival to the end of year 2. The effective annual interest rate is $i=5\%$. Assume $q_x=0.10$ and $q_{x+1}=0.20$, with one-year mortality probabilities applying at the stated ages. Find its value at age $x$.

Method. Start at the end of the coverage period, where the continuation value is zero, and work backward. The survival probabilities are $p_x=0.90$ and $p_{x+1}=0.80$, and $v=1/1.05$.

At age $x+1$, one year of term coverage remains. Its value is

A_{x+1:\overline{1}|}^{(B)} =v(q_{x+1}B+p_{x+1}\cdot 0) =\frac{0.20(\$20{,}000)}{1.05} =\frac{\$4{,}000}{1.05} =\$3{,}809.5238\ldots.

At age $x$, the first-year expected death payment is $0.10(\$20{,}000)=\$2{,}000$. The survival-weighted continuation value is $0.90(\$3{,}809.5238\ldots)=\$3{,}428.5714\ldots$. Therefore,

A_{x:\overline{2}|}^{(B)} =\frac{\$2{,}000+\$3{,}428.5714\ldots}{1.05} =\$5{,}170.0680\ldots.

The value is $\$5{,}170.07$ to the nearest cent. Check by enumerating the death years: death in year 1 has probability $0.10$; death in year 2 has probability $0.90(0.20)=0.18$. Thus the present value is $0.10(\$20{,}000)/1.05+0.18(\$20{,}000)/(1.05)^2=\$1{,}904.7619\ldots+\$3{,}265.3061\ldots=\$5{,}170.0680\ldots$, which agrees.

Rearranging the Recursion

A one-year recursion can also be used backward to recover the continuation value from a current value. Rearrangement is appropriate when $p_x>0$:

A_{x+1}^{(B)} =\frac{(1+i)A_x^{(B)}-q_xB}{p_x}.

The same algebra applies to a term value when the remaining term is positive, replacing the continuation with the value for one fewer year. This is useful when a current value, mortality probability, interest rate, and benefit are known, but the next-age value is missing. The result should be checked by substituting it back into the original recursion.

Worked Example: Recovering a Whole-Life Continuation Value

A hypothetical whole-life policy pays a level death benefit of $\$10{,}000$. At age $x$, its value is $A_x^{(B)}=\$6{,}000$, the effective annual interest rate is $i=5\%$, and $q_x=0.20$. Find the value conditional on surviving to age $x+1$.

Method. Rearrange the whole-life recursion. Here $p_x=1-0.20=0.80$, and $1+i=1.05$.

A_{x+1}^{(B)} =\frac{1.05(\$6{,}000)-0.20(\$10{,}000)}{0.80} =\frac{\$6{,}300-\$2{,}000}{0.80} =\frac{\$4{,}300}{0.80} =\$5{,}375.

Substitute to check: $v(q_xB+p_xA_{x+1}^{(B)})=(\$2{,}000+0.80(\$5{,}375))/1.05=(\$2{,}000+\$4{,}300)/1.05=\$6{,}000$. The recovered continuation value satisfies the original recursion exactly.

Common Errors and Checks

A reliable calculation starts by writing down what each value represents: benefit amount, duration, and valuation time. Then identify the first-year death payment and the conditional continuation value separately. The following checks catch frequent errors:

Using a unit death term with a nonunit continuation. In the first worked example, the incorrect calculation $v(q_x+p_xA_{x+1}^{(B)})=(0.10+0.90(\$4{,}000))/1.04=\$3{,}461.63$ treats the first-year death benefit as one dollar rather than $\$10{,}000$. The corrected death term is $q_xB=0.10(\$10{,}000)=\$1{,}000$, giving $(\$1{,}000+\$3{,}600)/1.04=\$4{,}423.08$. The incorrect sum also mixes a probability with dollar amounts.
Scaling the continuation value twice. If $A_{x+1}^{(B)}$ is already stated for benefit $B$, the survival term is $p_xA_{x+1}^{(B)}$, not $p_xB A_{x+1}^{(B)}$. Multiplication by $B$ is used when scaling a unit-benefit value, such as $BA_{x+1}$.
Dropping the one-year discount. Both the first-year death payment and the value conditional on survival are valued at time 1 inside the recursion. Their sum must be multiplied by $v$.
Giving a term policy an unintended maturity payment. For the term insurance defined here, survival to the end of the term produces no payment, so the remaining-coverage value at $n=0$ is zero.

A further check is to compare the recursion with a direct sum over possible death years when the term is short. Each year’s benefit probability must include survival through all earlier years, and the payment must be discounted to the end of its year. This independent calculation is especially useful for detecting a misplaced mortality probability or discount factor.

Key takeaway: For a constant death benefit $B$, use $v(q_xB+p_xA_{x+1}^{(B)})$ for whole life and $v(q_xB+p_xA_{x+1:\overline{n-1}|}^{(B)})$ for term insurance. The first-year death term includes $B$; the continuation value already includes the benefit and is carried forward once.

Check Your Understanding

For each problem, state which recursion or boundary applies before calculating.

A hypothetical whole-life policy pays $\$8{,}000$ at the end of the year of death. The effective annual interest rate is $4\%$, $q_x=0.25$, and the value conditional on survival to age $x+1$ is $\$3{,}000$. Find the value at age $x$, to the nearest cent. (Answer: $\$4{,}086.54$)
A hypothetical two-year term policy pays $\$12{,}000$ for death during the term and nothing on survival to the end. The effective annual interest rate is $0\%$, $q_x=0.10$, and $q_{x+1}=0.25$. Find its value at age $x$. (Answer: $\$3{,}900$)
A whole-life policy has a level death benefit of $\$9{,}000$. Its value at age $x$ is $\$5{,}000$, the effective annual interest rate is $4\%$, and $q_x=0.20$. Find its value conditional on survival to age $x+1$, and verify by substitution. (Answer: $\$4{,}250$)
A hypothetical three-year term policy pays $\$10{,}000$ at the end of the year of death and nothing at maturity. The effective annual interest rate is $5\%$, and the one-year death probabilities at ages $x$, $x+1$, and $x+2$ are $0.10$, $0.20$, and $0.25$, respectively. Work backward from the zero boundary and find the value at age $x$, to the nearest cent. (Answer: $\$4{,}139.94$)

What You'll Learn