Recursions for Time-Varying Death Benefits

Why the Benefit Schedule Matters

A term insurance contract may pay different amounts depending on when death occurs. For example, a hypothetical policy might provide a smaller benefit during an initial period and a larger benefit later. Its value cannot be calculated correctly by treating every year’s death benefit as the same amount. At each duration, the recursion must use the benefit for death in the coming year and, separately, the value of coverage that remains if the insured survives.

The key is to label benefits by the year of death they apply to. The recursion then has two mutually exclusive outcomes over the next policy year: death, which triggers that year’s benefit, or survival, which leads to the next duration’s policy value. This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Consider an $n$-year term insurance issued at age $x$. Let $B_j$ be the death benefit payable at the end of policy year $j$, if death occurs during that year. Let ${}_kV_x$ be the expected present value, at duration $k$, of the remaining death benefits, conditional on the insured being alive then. Assume annual effective interest rate $i$, $v=1/(1+i)$, and one-year death and survival probabilities ${}_1q_{x+k}$ and ${}_1p_{x+k}=1-{}_1q_{x+k}$. Benefits are paid at the end of the year of death, there is no maturity benefit or premium cash flow in this value, and mortality probabilities apply conditionally on being alive at the start of each year.

The One-Year Recursion

At duration $k$, the coming year is policy year $k+1$. If death occurs during that year, probability ${}_1q_{x+k}$, the benefit is $B_{k+1}$, payable one year from the valuation date. If the insured survives, probability ${}_1p_{x+k}$, the remaining contract has value ${}_{k+1}V_x$ at the end of the year. Discount either end-of-year amount by $v$ to obtain its value at duration $k$.

{}_kV_x = v\left[{}_1q_{x+k}B_{k+1} +{}_1p_{x+k}\,{}_{k+1}V_x\right], \qquad k=0,1,\ldots,n-1.

The terminal condition is ${}_nV_x=0$: after the $n$-year term has expired, there are no future death benefits under this contract. The zero is a boundary value, not an assumption that the contract has no value during the final year. In that year, a death benefit can still be paid.

Indexing rule: $B_{k+1}$ is the benefit for death in the year immediately ahead of duration $k$. The continuation value is ${}_{k+1}V_x$, which applies only on survival. Do not replace the year-specific benefit with a single level benefit unless the contract actually specifies one.

Why the Recursion Holds

The formula follows by conditioning on the insured’s status at the end of the next year. The death and survival events are disjoint and together exhaust the possible outcomes. Their conditional probabilities are ${}_1q_{x+k}$ and ${}_1p_{x+k}$, respectively. The corresponding payments or values are measured one year later, so each is discounted by $v$.

Repeated substitution also gives a direct expression in terms of the probability of death in each policy year. To die in year $j$, the insured must survive the first $j-1$ years and then die in year $j$. The probability is ${}_{j-1}p_x\,{}_1q_{x+j-1}$, and the benefit is discounted $j$ years. Thus:

{}_0V_x = \sum_{j=1}^{n} v^j\,{}_{j-1}p_x\,{}_1q_{x+j-1}B_j.

This expression is a useful independent check, and it shows why each benefit must stay attached to its own year. It is also equivalent to evaluating the recursion backward from ${}_nV_x=0$. Backward evaluation is usually convenient when all the one-year mortality probabilities and benefits are given.

Start at the end of the term.
Set ${}_nV_x=0$, because no covered years remain after duration $n$.

Move back one year at a time.
For each $k=n-1,\ldots,0$, pair ${}_1q_{x+k}$ with $B_{k+1}$, and pair ${}_1p_{x+k}$ with ${}_{k+1}V_x$.

Discount the two outcomes.
Multiply the probability-weighted end-of-year amounts by $v$ to obtain ${}_kV_x$.

Check the result.
When practical, compare the initial value with the direct sum of the discounted benefits by year of death.

Worked Examples

Worked Example: Three Different Death Benefits

A hypothetical three-year term policy pays $500, $850, and $1,400 for death in years 1, 2, and 3, respectively. The annual effective interest rate is $i=1/0.94-1=0.06383$, so $v=0.94$. The one-year death probabilities at the start of the three years are $0.08$, $0.18$, and $0.35$. Calculate the initial expected present value.

Method. Use backward recursion. The survival probabilities are $0.92$, $0.82$, and $0.65$, and the terminal value is ${}_3V_x=0$.

At duration 2, only the year-3 death benefit remains:

{}_2V_x=0.94[0.35(1400)+0.65(0)] =0.94(490)=460.60.

At duration 1, use the year-2 benefit on death and the duration-2 value on survival:

{}_1V_x =0.94[0.18(850)+0.82(460.60)] =0.94[153+377.692] =498.85048.

At duration 0, the year-1 benefit is $500:

{}_0V_x =0.94[0.08(500)+0.92(498.85048)] =0.94[40+458.9424416] =469.005895104.

Rounded to the nearest cent, the initial value is $469.01. As a check, the probabilities of death in years 1, 2, and 3 are $0.08$, $0.92(0.18)=0.1656$, and $0.92(0.82)(0.35)=0.26404$. The direct present-value sum is $0.94(0.08)(500)+0.94^2(0.1656)(850)+0.94^3(0.26404)(1400)=469.005895104$, agreeing with the recursion. The method avoids using one benefit amount for all three years.

Worked Example: A Benefit That Increases Each Year

A hypothetical four-year term policy pays $400, $700, $1,000, and $1,300 for death in years 1 through 4. The annual effective interest rate is $i=1/0.96-1=0.04167$, so $v=0.96$. The successive one-year death probabilities are $0.05$, $0.10$, $0.20$, and $0.25$. Find the value at issue.

Method. Calculate backward from ${}_4V_x=0$. The corresponding survival probabilities are $0.95$, $0.90$, $0.80$, and $0.75$.

For the final covered year, followed by the preceding years:

\begin{aligned} {}_3V_x &=0.96[0.25(1300)+0.75(0)] =312,\\ {}_2V_x &=0.96[0.20(1000)+0.80(312)] =0.96(200+249.6) =431.616,\\ {}_1V_x &=0.96[0.10(700)+0.90(431.616)]\\ &=0.96(70+388.4544) =440.116224,\\ {}_0V_x &=0.96[0.05(400)+0.95(440.116224)]\\ &=0.96(20+418.1104128) =420.585996288. \end{aligned}

The initial value is $420.59, rounded to the nearest cent. Each step uses the benefit for death in the next policy year, not the benefit associated with the duration index itself. In particular, at duration 1 the coming death benefit is $700, even though the continuation value already reflects benefits in years 3 and 4.

Worked Example: Verify the Recursion by Direct Valuation

A hypothetical three-year contract pays $100, $200, and $300 for death in years 1, 2, and 3. Assume a constant one-year death probability of $0.20$ at each duration and annual effective interest $i=1/0.90-1=0.1111$, so $v=0.90$. Calculate its value by recursion and check it directly.

Method. Work backward, with survival probability $0.80$ and ${}_3V_x=0$:

\begin{aligned} {}_2V_x&=0.90[0.20(300)]=54,\\ {}_1V_x&=0.90[0.20(200)+0.80(54)] =0.90(40+43.2)=74.88,\\ {}_0V_x&=0.90[0.20(100)+0.80(74.88)]\\ &=0.90(20+59.904)=71.9136. \end{aligned}

The probabilities of death in the three years are $0.20$, $0.80(0.20)=0.16$, and $0.80^2(0.20)=0.128$. Direct valuation gives

$$ 0.90(0.20)(100)+0.90^2(0.16)(200)+0.90^3(0.128)(300) =18+25.92+27.9936 =71.9136. $$

Both methods give $71.91 to the nearest cent. This check confirms that the continuation value accounts for survival to later years, while the benefit in each term is paid only if death occurs in that particular year.

Common Mistakes and Scope

The most common error is to put the death benefit on the survival branch. For example, in the third example, an incorrect duration-1 calculation might be $0.90[200+0.80(54)]=218.88$. That calculation treats the $200 benefit as payable whether or not death occurs. The correct calculation weights it by the death probability: $0.90[0.20(200)+0.80(54)]=74.88$.

Misaligning the indices. At duration $k$, use $B_{k+1}$, the benefit for death in the coming year.
Forgetting the survival probability. The next-duration value contributes only if the insured survives, so it is multiplied by ${}_1p_{x+k}$.
Discounting incorrectly. Both the death benefit and the continuation value are amounts at the end of the coming year, so both are multiplied by one factor $v$.
Using a nonzero terminal value without a maturity benefit. For a pure term contract with no benefit at expiry, ${}_nV_x=0$. A maturity payment would require a different boundary condition.

These formulas value the remaining death benefits only, conditional on the insured being alive at the valuation duration. If a problem includes premiums or other cash flows, their values must also be included with the appropriate timing and sign. The recursion does not make those cash flows disappear; it isolates the death-benefit component.

Key takeaway: A time-varying term benefit is handled one year at a time: weight that year’s benefit by the probability of death, weight the next value by the probability of survival, and discount both. Preserve the benefit-year index throughout the backward calculation.

Check Your Understanding

For each problem, use the stated annual effective interest rate and assume death benefits are payable at the end of the year of death, with no maturity benefit.

A two-year policy pays $100 in year 1 and $400 in year 2. Let $v=0.90$, ${}_1q_x=0.20$, and ${}_1q_{x+1}=0.50$. Find the value at issue. (Answer: $147.60)
A three-year policy pays $250, $500, and $750 in years 1, 2, and 3. Let $v=0.95$ and the successive death probabilities be $0.10$, $0.20$, and $0.30$. Find the value at issue, rounded to the nearest cent. (Answer: $243.87)
A three-year policy pays $1,000, $600, and $200 in successive years. Let $v=0.96$ and the death probabilities be $0.05$, $0.25$, and $0.40$. Calculate the value at issue, rounded to the nearest cent. (Answer: $229.76)
A four-year policy pays $500 in years 1 and 2 and $1,000 in years 3 and 4. Let $v=0.97$, with one-year death probabilities $0.10$, $0.15$, $0.20$, and $0.25$. Find the value at issue, rounded to the nearest cent. (Answer: $387.10)

What You'll Learn