Deriving the One-Year Equation by Conditioning on Death or Survival

Why Condition on Death or Survival?

A one-year reserve recursion links the reserve at the start of a policy year to the reserve one year later. Its key step is to split the year into two mutually exclusive outcomes: the insured dies during the year, or the insured survives to its end. Each outcome has its own end-of-year cash requirement. Weighting those requirements by their conditional probabilities gives the expected amount that must be available at the end of the year.

The timing convention matters. In this tutorial, the reserve ${}_tV_x$ is measured immediately before the premium due at duration $t$. The premium $P_{x+t}$ is paid at the start of that year. A death benefit $B_{t+1}$ is paid at the end of the year of death, and if the insured survives, the reserve at the next anniversary is ${}_{t+1}V_x$, measured before the next premium. We assume an annual discrete model, a deterministic effective annual interest rate $i$, and no expenses. The equation below uses these conventions; changing payment timing can change the equation.

Definition: Conditional on the insured being alive at age $x+t$, let ${}_1q_{x+t}$ be the probability of death before the end of the next year and ${}_1p_{x+t}$ the probability of survival to that anniversary. These probabilities satisfy ${}_1p_{x+t}+{}_1q_{x+t}=1$. The one-year reserve recursion relates the fund available after the start-of-year premium, ${}_tV_x+P_{x+t}$, to the expected end-of-year amount required for the death or survival outcome.

Deriving the One-Year Equation

Let $D$ denote death during the policy year and $S$ denote survival to its end. Conditional on being alive at the beginning of the year, $D$ and $S$ are exhaustive and mutually exclusive. Under the stated timing convention, the amount required at year-end is $B_{t+1}$ if $D$ occurs and ${}_{t+1}V_x$ if $S$ occurs. The premium and opening reserve accumulate at rate $i$, so their year-end value is $({}_tV_x+P_{x+t})(1+i)$.

The recursion equates this accumulated amount to the probability-weighted end-of-year requirement. In particular, the death benefit is weighted by the probability of death, not by the probability of survival; the next reserve is weighted by the probability of survival.

\begin{aligned} ({}_tV_x+P_{x+t})(1+i) &=E[\text{end-of-year amount required}\mid\text{alive at age }x+t]\\ &={}_1q_{x+t}B_{t+1}+{}_1p_{x+t}{}_ {t+1}V_x. \end{aligned}

In the display, the spaced subscript in the last term denotes the next-duration reserve ${}_{t+1}V_x$. Equivalently, the standard compact form is

({}_tV_x+P_{x+t})(1+i) ={}_1q_{x+t}B_{t+1}+{}_1p_{x+t}{}_{t+1}V_x.

The discount-factor form follows by dividing both sides by $1+i$, or multiplying by $v=1/(1+i)$:

{}_tV_x+P_{x+t} =v\left({}_1q_{x+t}B_{t+1}+{}_1p_{x+t}{}_{t+1}V_x\right).

This is a conditional expected-value equation: the probabilities are conditional on survival to the start of the year. It does not say that each policyholder’s individual reserve grows into the exact claim or reserve that will be needed for that person. Rather, it balances the opening fund and premium against the expected year-end requirement under the model.

Why the Probability-Weighted Amount Is Correct

The justification is the expectation rule for a quantity with two possible outcomes. Define $Z$ as the end-of-year amount required. It equals $B_{t+1}$ on $D$ and ${}_{t+1}V_x$ on $S$. Since $D$ and $S$ partition the possible outcomes, the definition of conditional expectation gives

\begin{aligned} E[Z\mid\text{alive at age }x+t] &=B_{t+1}P(D\mid\text{alive at age }x+t)\\ &\quad+{}_{t+1}V_xP(S\mid\text{alive at age }x+t)\\ &={}_1q_{x+t}B_{t+1}+{}_1p_{x+t}{}_{t+1}V_x. \end{aligned}

The opening reserve and premium are both present at the beginning of the year, so under a deterministic effective annual rate their accumulated value is $({}_tV_x+P_{x+t})(1+i)$. Equating that value to $E[Z\mid\text{alive at age }x+t]$ yields the recursion. This argument depends on the stated timing: if a death benefit is paid immediately at death rather than at the end of the year, it does not generally accumulate for a full year, and this end-of-year formula cannot be used unchanged.

Set the year’s timing.
Identify the opening reserve, premium timing, death-benefit timing, and reserve carried forward on survival.

Write the two outcomes.
Death requires the death benefit at year-end; survival requires the next-anniversary reserve at year-end.

Weight by conditional probabilities.
Use ${}_1q_{x+t}$ for death and ${}_1p_{x+t}$ for survival, given alive at the start of the year.

Equate the accumulated opening fund to the expected requirement.
Apply the annual interest factor to the opening reserve plus the premium, then solve for the unknown.

Worked Applications

Worked Example: Finding the Opening Reserve

A hypothetical one-year segment of a policy has an effective annual interest rate of $4\%$. The insured is age $x+t$, with ${}_1q_{x+t}=0.02$ and ${}_1p_{x+t}=0.98$. The premium of $\$1{,}200$ is paid at the start of the year, the end-of-year death benefit is $\$100{,}000$, and the next-anniversary reserve is $\$5{,}000$. Find ${}_tV_x$. Assume no expenses.

Method. Apply the one-year equation with death and survival treated as separate end-of-year outcomes:

({}_tV_x+1{,}200)(1.04) =0.02(100{,}000)+0.98(5{,}000).

The death-outcome contribution is $0.02(100{,}000)=2{,}000$. The survival-outcome contribution is $0.98(5{,}000)=4{,}900$. Thus, the expected end-of-year requirement is $2{,}000+4{,}900=6{,}900$. Divide by $1.04$ to find the opening reserve plus premium:

{}_tV_x+1{,}200=\frac{6{,}900}{1.04}=6{,}634.6153846\ldots

Subtract the premium: ${}_tV_x=6{,}634.6153846\ldots-1{,}200=5{,}434.6153846\ldots$. Rounded to the nearest cent, the reserve is $\$5{,}434.62$.

Check. The accumulated opening amount is $(5{,}434.6153846\ldots+1{,}200)(1.04)=6{,}900$, equal to the probability-weighted end-of-year requirement. The common mistake avoided here is adding the two conditional outcome amounts without weighting them by their probabilities.

Worked Example: Solving for an Unknown Death Benefit

A hypothetical policy has an effective annual interest rate of $5\%$. At the start of a year, the insured is alive at age $x+t$, with ${}_1q_{x+t}=0.03$ and ${}_1p_{x+t}=0.97$. The reserve immediately before the premium is $\$4{,}000$, the premium is $\$900$, and the next-anniversary reserve is $\$4{,}000$. The death benefit is payable at year-end. Find the death benefit $B_{t+1}$, assuming no expenses.

Method. Substitute the known values into the accumulated-fund equation:

(4{,}000+900)(1.05)=0.03B_{t+1}+0.97(4{,}000).

The left side is $4{,}900(1.05)=5{,}145$. The survival contribution on the right is $0.97(4{,}000)=3{,}880$. Therefore,

0.03B_{t+1}=5{,}145-3{,}880=1{,}265, \qquad B_{t+1}=\frac{1{,}265}{0.03}=42{,}166.6666\ldots

Rounded to the nearest cent, the death benefit is $\$42{,}166.67$.

Check. The death contribution is $0.03(42{,}166.6666\ldots)=1{,}265$; adding the survival contribution gives $1{,}265+3{,}880=5{,}145$, matching the accumulated opening amount. The common mistake avoided is subtracting the full next reserve instead of its probability-weighted value.

Worked Example: Including a Survival Cash Benefit

A hypothetical contract pays $\$50{,}000$ at the end of the year if the insured dies during that year. If the insured survives, the contract pays a $\$1{,}000$ survival benefit at the anniversary and continues with a reserve of $\$6{,}000$, measured before the next premium. The effective annual interest rate is $3\%$, ${}_1q_{x+t}=0.04$, ${}_1p_{x+t}=0.96$, and the start-of-year premium is $\$800$. Find the opening reserve, assuming no expenses.

Method. On survival, the year-end requirement includes both the survival cash benefit and the continuing reserve. Therefore, the survival outcome requires $\$1{,}000+\$6{,}000=\$7{,}000$. The recursion is

({}_tV_x+800)(1.03) =0.04(50{,}000)+0.96(1{,}000+6{,}000).

The death contribution is $0.04(50{,}000)=2{,}000$. The survival contribution is $0.96(7{,}000)=6{,}720$. The expected end-of-year requirement is $2{,}000+6{,}720=8{,}720$. Hence,

{}_tV_x+800=\frac{8{,}720}{1.03}=8{,}466.0194174\ldots, \qquad {}_tV_x=7{,}666.0194174\ldots.

Rounded to the nearest cent, the opening reserve is $\$7{,}666.02$.

Check. $(7{,}666.0194174\ldots+800)(1.03)=8{,}720$, as required. The common mistake avoided is including the survival benefit in the death outcome or omitting it from the survival outcome.

Common Errors and Interpretation

The recursion is simple once the outcomes and their timing are fixed, but several errors can produce plausible-looking answers.

Reversing the probabilities. Death benefit cash flow is weighted by ${}_1q_{x+t}$, while the next reserve is weighted by ${}_1p_{x+t}$. Both probabilities are conditional on the insured being alive at the start of the year.
Using an unconditional probability. The recursion is for a policy in force at the start of the year. Use the one-year probabilities conditional on that starting survival, not a probability of survival from issue unless the duration and conditioning are adjusted.
Applying interest to only part of the opening amount. Under this convention, the opening reserve and the premium are both available at the start of the year, so both accumulate by $1+i$.
Discounting in the wrong direction. The accumulated-fund equation multiplies the opening amount by $1+i$. To express the equation at the beginning of the year, divide the expected end-of-year requirement by $1+i$; do not multiply that requirement by $1+i$.
Ignoring payment timing. This version assumes a death benefit paid at the end of the year of death. A benefit paid at the moment of death does not generally have the same year-end value.
Forgetting a survival payment. If survival produces both a cash benefit and a continuing policy, both the survival payment and the next reserve belong in the survival outcome.

A reliable check is to write the two end-of-year requirements in words before substituting numbers. Then verify that the death probability and survival probability sum to one, and that the calculated accumulated fund equals the weighted requirement. This is also a useful way to detect a timing error before doing the algebra.

Key takeaway: Under the stated annual timing convention, the one-year equation is $({}_tV_x+P_{x+t})(1+i)={}_1q_{x+t}B_{t+1}+{}_1p_{x+t}{}_{t+1}V_x$. It follows by conditioning on death or survival, with each end-of-year amount weighted by its corresponding probability. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the timing convention stated above and assume no expenses unless the problem says otherwise.

A hypothetical policy has $i=0.02$, ${}_1q_{x+t}=0.01$, a death benefit of $\$80{,}000$, next reserve $\$2{,}000$, and start-of-year premium $\$500$. Find the opening reserve. (Answer: $\$2{,}225.49$)
A hypothetical policy has $i=0.04$, ${}_1q_{x+t}=0.05$, ${}_1p_{x+t}=0.95$, opening reserve $\$3{,}000$, premium $\$700$, and next reserve $\$2{,}500$. Find the end-of-year death benefit. (Answer: $\$29{,}460.00$)
A hypothetical policy pays $\$100{,}000$ on death during the year, and on survival pays $\$500$ while continuing with a next reserve of $\$9{,}000$. Given $i=0.05$, ${}_1q_{x+t}=0.02$, ${}_1p_{x+t}=0.98$, and premium $\$1{,}000$, find the opening reserve. (Answer: $\$9{,}771.43$)
A hypothetical policy has $i=0.03$, death benefit $\$50{,}000$, opening reserve $\$2{,}500$, premium $\$600$, and next reserve $\$3{,}000$. Find the one-year conditional probability of death. (Answer: $0.0041$)
A hypothetical policy has $i=0.04$, ${}_1q_{x+t}=0.02$, death benefit $\$60{,}000$, opening reserve $\$5{,}000$, and premium $\$1{,}000$. Find the next reserve. (Answer: $\$5{,}142.86$)

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