Introduction
A life insurance value can be divided according to what happens during the first policy year. If death occurs, the contract pays a benefit at the end of that year. If the insured survives, no death benefit is paid in that year, and the remaining contract is valued from the next policy anniversary. This simple partition is a useful way to calculate an insurance value and to check whether its timing has been handled correctly.
The focus here is the one-year recursion for a unit whole-life benefit. It is a local, year-by-year valuation identity: it separates the first year from the future, rather than converting an insurance value into an annuity value.
What the Partition Separates
The first term, \(vq_x\), represents a benefit of 1 paid at the end of the first year if death occurs in that year. Its probability is \(q_x\), and its payment time is one year, so it is discounted once. The second term, \(vp_xA_{x+1}\), represents survival through the first year, followed by the value of the remaining whole-life insurance. Survival has probability \(p_x\). The future insurance value \(A_{x+1}\) is measured at the next policy anniversary, so it is discounted back one year using \(v\).
This setup assumes an annual effective interest rate, a unit benefit payable at the end of the year of death, and no benefit at the end of the first year if the insured is alive. After survival to age \(x+1\), the future contract follows the same mortality and interest assumptions used to define \(A_{x+1}\). The identity is not the right model without modification for a benefit paid immediately on death, a benefit paid at a different time, or a contract with different benefits on death and survival.
Deriving the One-Year Identity
Condition on the two mutually exclusive outcomes for the first year: death or survival. The expected present value is the probability-weighted value across those outcomes. On death during the year, the present value at age \(x\) is \(v\), because the unit benefit is paid at time 1. On survival, the first-year death benefit is not paid; the remaining insurance has expected value \(A_{x+1}\) at time 1, or \(vA_{x+1}\) at age \(x\).
The partition follows directly from conditional expectation: multiply the value in each first-year outcome by the probability of that outcome, then add the results. Since \(p_x+q_x=1\), the two cases cover all possibilities. The formula does not assume that the insured dies at the middle of the year; it uses the specified end-of-year-of-death payment convention.
Use \(q_x\) for death within one year and \(p_x\) for survival through one year.
The death benefit has present value \(v\). The surviving policy has present value \(vA_{x+1}\).
Multiply each outcome value by its probability: \(A_x=q_xv+p_xvA_{x+1}\).
Worked Applications
Worked Example: Calculate the Current-Age Insurance Value
A hypothetical whole-life policy pays 1 at the end of the year of death. The effective annual interest rate is 4%. At age \(x\), the probability of death within one year is \(q_x=0.08\); the probability of survival is therefore \(p_x=0.92\). The insurance value at age \(x+1\), under the same assumptions, is \(A_{x+1}=0.25\). Calculate \(A_x\). State the method and a timing error to avoid.
Method. Apply the one-year partition, first checking that the probabilities sum to 1: \(p_x+q_x=0.92+0.08=1.00\). The discount factor is \(v=1/(1.04)=0.96153846\).
The probability-weighted value before discounting is \(q_x+p_xA_{x+1}=0.08+(0.92)(0.25)=0.08+0.23=0.31\). Thus, \(A_x=v(0.31)=0.96153846(0.31)=0.29807692\). Rounded to four decimal places, \(A_x=0.2981\).
Check. The first-year death contribution is \(v q_x=0.96153846(0.08)=0.07692308\). The survival contribution is \(v p_xA_{x+1}=0.96153846(0.92)(0.25)=0.22115385\). Their sum is \(0.29807693\), matching the result apart from rounding. The common timing error is to discount the first-year death benefit more than once; it is paid at time 1, so one factor of \(v\) is correct.
Worked Example: Recover an Unknown One-Year Death Probability
A hypothetical unit whole-life policy has an effective annual interest rate of 5%. Its value at age \(x\) is \(A_x=0.18\), and its value at age \(x+1\) is \(A_{x+1}=0.12\). Using the one-year partition, find \(q_x\) and check that the resulting survival probability is valid.
Method. Rearrange the partition to solve for \(q_x\), using \(p_x=1-q_x\). Since \(v=1/1.05\), multiplying the identity by \(1.05\) gives \(1.05A_x=q_x+(1-q_x)A_{x+1}\).
Substitute the given values: \(1.05(0.18)=q_x+(1-q_x)(0.12)\), so \(0.189=0.12+0.88q_x\). Therefore \(0.069=0.88q_x\), and \(q_x=0.069/0.88=0.07840909\). To four decimal places, \(q_x=0.0784\), and \(p_x=1-q_x=0.92159091\), or \(0.9216\) to four decimal places.
Check. The probabilities are nonnegative and sum to 1. Substitution gives \(q_x+p_xA_{x+1}=0.07840909+(0.92159091)(0.12)=0.189\), and \(0.189/1.05=0.18\). A common mistake is to solve \(1.05A_x=q_x+A_{x+1}\), omitting the probability \(p_x\) attached to the future value. The future contract contributes only if the insured survives.
Worked Example: Apply the Partition to a Dollar Benefit
A hypothetical whole-life policy pays $80,000 at the end of the year of death. The effective annual interest rate is 4%. At age \(x\), \(q_x=0.05\) and \(p_x=0.95\). At age \(x+1\), the expected present value of the remaining benefit, measured per dollar of benefit, is \(A_{x+1}=0.10\). Calculate the expected present value at age \(x\) in dollars.
Method. First calculate the value per dollar using the partition. Here \(v=1/1.04=0.96153846\), and the probabilities check because \(0.05+0.95=1\).
The probability-weighted value per dollar before discounting is \(q_x+p_xA_{x+1}=0.05+(0.95)(0.10)=0.05+0.095=0.145\). Thus the value per dollar at age \(x\) is \(v(0.145)=0.96153846(0.145)=0.13942308\). Multiplying by the $80,000 benefit gives \(\$80{,}000(0.13942308)=\$11{,}153.8464\), or **$11,153.85** to the nearest cent.
Check. The first-year death contribution in dollars is \(\$80{,}000(0.05)(0.96153846)=\$3{,}846.15\). The survival contribution is \(\$80{,}000(0.95)(0.10)(0.96153846)=\$7{,}307.69\). The contributions total $11,153.84 using the displayed rounded amounts; using unrounded values gives $11,153.85. The common mistake is to treat \(A_{x+1}\) as if it were already measured at age \(x\), and then omit the additional one-year discount factor.
Interpretation and Limits
The recursion expresses a general valuation habit: split at a meaningful time point, value each possible outcome at that time, and discount values measured in the future back to the valuation date. It is particularly useful when the value at age \(x+1\) is known or can be calculated separately. It also gives a direct consistency check: the current value is the discounted, probability-weighted combination of first-year death and survival outcomes.
The benefit timing and the survival condition must match the contract. If a benefit were paid at the beginning of the year of death rather than at its end, its value on the death branch would not be \(v\). If survival itself produced a payment, that payment would need its own term. Similarly, if the post-survival contract had a different benefit or different assumptions, its continuation value could not automatically be represented by the stated \(A_{x+1}\).
| First-year outcome | Probability | Value at age \(x\) | Contribution to \(A_x\) |
|---|---|---|---|
| Death within one year | \(q_x\) | \(v\) | \(vq_x\) |
| Survival for one year | \(p_x\) | \(vA_{x+1}\) | \(vp_xA_{x+1}\) |
For exam-style calculations aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus, keep three checks visible: the probabilities sum to 1, each cash flow is discounted from its actual payment time, and a continuation value is weighted by the probability that the contract reaches that future time. These checks catch many errors before any more extensive calculation is needed.
Check Your Understanding
For each problem, use the stated annual interest and mortality assumptions and the one-year partition. Probabilities in the answers are rounded to four decimal places.
- A hypothetical unit whole-life policy has \(i=3\%\), \(q_x=0.06\), and \(A_{x+1}=0.20\). Find \(p_x\) and \(A_x\). (Answer: \(p_x=0.94,\ A_x=0.2408\))
- For a unit whole-life policy, \(i=5\%\), \(A_x=0.25\), and \(A_{x+1}=0.15\). Find \(q_x\), then verify that \(p_x+q_x=1\). (Answer: \(q_x=0.1324,\ p_x=0.8676\))
- A unit whole-life policy has \(q_x=0.04\), \(A_{x+1}=0.18\), and \(A_x=0.16\). Use the recursion to find the effective annual interest rate \(i\). (Answer: \(i=33\%\))
- A hypothetical policy pays $120,000 at the end of the year of death. At age \(x\), \(i=4\%\), \(q_x=0.02\), and \(p_x=0.98\); at age \(x+1\), the value per dollar of benefit is \(A_{x+1}=0.08\). Find the expected present value at age \(x\), to the nearest cent. (Answer: $11,353.85)