Why the Contract’s Cash Flows Determine the Recursion
A life insurance recursion is a one-year accounting identity: the value at a valuation date equals the discounted expected value of the benefits paid during the next year and the value remaining if the insured survives. The arithmetic is usually straightforward. The important decision is choosing the recursion whose benefit timing, coverage period, and endpoint match the contract.
This tutorial treats annual discrete contracts valued at integer durations. Unless stated otherwise, a death benefit is paid at the end of the year of death, and a maturity benefit is paid at the end of the stated term if the insured survives. These timing assumptions matter: a benefit paid immediately on death does not use the same one-year discounting formula. The distinctions and calculations here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
If a maturity benefit \(M\) is due to a survivor at time \(n\), the endpoint value is \(V_n=M\). If no benefit is due at the endpoint, the boundary value is \(V_n=0\). In either case, values are calculated backward from the endpoint. For a year in which a death benefit \(B_k\) is covered, the general recursion is:
The value \(B_k\) is the benefit for death in the next year; \(V_{k+1}\) is the value of benefits after survival to the next valuation time. In a year with no death coverage and no other payment during the year, the recursion reduces to \(V_k=vp_{x+k}V_{k+1}\).
Match the Contract to Its Boundary and Recursion
The same one-year identity can describe several products. What changes is whether a death benefit applies in each year and what benefit, if any, remains at the endpoint.
| Contract during a year | Recursion in a covered year | Endpoint or eligibility condition |
|---|---|---|
| Whole life, unit death benefit | \(A_{x+k}=v(q_{x+k}+p_{x+k}A_{x+k+1})\) | Coverage continues for life; use a later whole-life value if calculating backward from a known age. |
| Term, unit death benefit for \(n\) years | \(V_k=v(q_{x+k}+p_{x+k}V_{k+1})\) | \(V_n=0\): no benefit is payable after the term ends. |
| Endowment, death benefit \(B\) and maturity benefit \(M\) | \(V_k=v(q_{x+k}B+p_{x+k}V_{k+1})\) | \(V_n=M\): a survivor receives the maturity benefit. |
| Deferred insurance during the waiting period | \(V_k=vp_{x+k}V_{k+1}\) | Use the value of the insurance that becomes eligible at the end of the waiting period. |
For a whole-life policy with a unit benefit, \(A_{x+k}\) denotes the value of the remaining whole-life insurance at age \(x+k\). There is no fixed term endpoint at which the death benefit abruptly stops. For a term policy, by contrast, the zero boundary is essential: after the term, the death benefit is no longer payable. An endowment policy combines a death benefit during the term with a maturity payment to a survivor at its end.
A deferred policy has a period during which the insured must survive before death coverage begins. In that waiting period, a death benefit is not payable. Once eligibility begins, use the recursion for the actual coverage then in force: whole life if coverage continues for life, or term insurance if coverage lasts for a specified number of years.
Identify the valuation time, each year of death coverage, any waiting period, and the contract endpoint.
For each year, state the death benefit if death occurs, and note any maturity payment if the insured survives to the endpoint.
Use zero after a term with no maturity benefit, the maturity amount when it is payable, or the relevant insurance value when calculating back from an eligibility date or known age.
Use the covered-year formula only where the death benefit is eligible; use the survival-only formula during a waiting period with no benefit.
Why the One-Year Identity Holds
Condition on the insured’s status at the end of the next year. Death during that year has probability \(q_{x+k}\) and produces the death benefit \(B_k\) at time \(k+1\). Survival has probability \(p_{x+k}\) and leaves the contract with value \(V_{k+1}\) at time \(k+1\). Both possible values at that time are discounted one year to time \(k\).
This is an application of conditioning and the expected present value principle. It assumes the stated annual probabilities and annual effective interest rate apply over the interval, and that the death benefit is paid at the end of the year of death. If the benefit is paid at another time, its discounting must reflect that timing.
Worked Applications
Worked Example: Whole-Life Value Versus a Two-Year Term
A hypothetical insurer values a unit death benefit with \(i=0.05\), so \(v=1/1.05=20/21\). The annual death probabilities are \(q_x=0.10\) and \(q_{x+1}=0.20\). The whole-life value at age \(x+2\), based on mortality beyond that age, is \(A_{x+2}=0.50\). Calculate the whole-life value at age \(x\). Then calculate the value at age \(x\) of a two-year term policy with the same death probabilities and no maturity benefit.
Method. For whole life, coverage continues beyond time 2, so use the supplied whole-life value as the backward-calculation boundary. For the two-year term, no benefit is payable after time 2, so use a zero boundary.
Whole life, time 1. The one-year recursion gives:
Whole life, time 0. Substitute \(A_{x+1}=4/7\):
Two-year term, time 2. The policy has ended and has no maturity benefit, so \(V_2=0\). At time 1:
Two-year term, time 0.
The whole-life value is higher because the unit death benefit remains payable after the first two years. The calculation avoids the mistake of giving the term policy the continuing-coverage boundary \(A_{x+2}=0.50\); its correct boundary is zero.
Worked Example: Endowment With Unequal Death and Maturity Benefits
A hypothetical two-year endowment pays 2 at the end of the year of death during the term, or 3 at time 2 if the insured survives. The interest rate is \(i=0.05\), and the death probabilities are \(q_x=0.10\) and \(q_{x+1}=0.20\). Find the value at time 0.
Method. Start with the maturity boundary \(V_2=3\), then apply the covered-year recursion backward using death benefit 2 in both years. Here \(v=20/21\), \(p_x=0.90\), and \(p_{x+1}=0.80\).
At time 1, a death during the second year produces 2, while survival to time 2 produces 3:
At time 0:
A path-by-path check gives the same result. Death in year 1 contributes \(0.10(2)v=0.190476\). Death in year 2 contributes \(0.90(0.20)(2)v^2=0.326531\). Survival to maturity contributes \(0.90(0.80)(3)v^2=1.959184\). Their sum is \(0.190476+0.326531+1.959184=2.476191\), with the one-unit difference in the last decimal due to displayed rounding; the exact recursion value is \(52/21\). The boundary is 3, not zero, because a maturity payment is due to a survivor.
Worked Example: Deferred Two-Year Term Insurance
A hypothetical policy has a two-year waiting period followed by two years of term coverage. It pays 5 at the end of the year of death only during the covered years. The annual interest rate is \(i=0.25\), so \(v=0.80\). The death probabilities in years 1 through 4 are, respectively, \(0.10\), \(0.20\), \(0.25\), and \(0.20\). Find the value at time 0.
Method. First value the two covered years, working back from the zero boundary at time 4. Then move backward through the two waiting years using survival only. During the waiting period, death does not produce a payment.
At time 4, the term has ended with no maturity benefit, so \(V_4=0\). In the second covered year, the death probability is \(0.20\), giving:
At time 2, the first covered year has death probability \(0.25\):
There is no death benefit in the waiting years. At time 1, survival probability is \(0.80\):
At time 0, survival probability is \(0.90\):
The value is \(0.681984\), or \(0.68\) to two decimal places. A common error is to treat this as four years of immediate term coverage. That incorrect contract would include a first-year death payment with value \(0.80(0.10)(5)=0.40\), even though the actual policy pays nothing for death during the waiting period. Applying the immediate four-year term recursion to all four stated years gives \(1.657984\), not \(0.681984\). The error is not in the discount factor; it is in applying the death benefit before eligibility.
Changing Benefits and Common Selection Errors
A death benefit need not be level. If the benefit in a covered year is \(B_k\), use that year’s amount in the death term \(q_{x+k}B_k\). Keep the survival term \(p_{x+k}V_{k+1}\) unchanged: it represents the value of the contract that remains after survival. For an endowment, use the death benefit specified for that year and retain the stated maturity amount as the endpoint. These adjustments follow directly from the one-year conditioning argument.
- Using the wrong endpoint. A term policy with no maturity payment ends at zero; an endowment ends at its maturity benefit.
- Covering the waiting period by mistake. Apply the death-benefit term only after coverage becomes eligible.
- Confusing the term’s length with its start date. A two-year deferred, two-year term policy has four years between issue and expiry, but only two years of death coverage.
- Using the wrong year’s mortality or benefit. In each recursion, pair \(q_{x+k}\) with the benefit for death during that same year.
- Ignoring payment timing. The formulas here discount end-of-year payments by \(v\). A benefit paid at another time requires a timing-appropriate discount factor.
The main practical test is to describe the cash flow before writing an equation: “If death occurs in this year, what is paid?” and “If the insured survives to the next date, what contract remains?” Once those two answers are clear, the recursion and boundary follow.
Check Your Understanding
For each problem, identify the coverage years and endpoint before calculating.
- A one-year term policy pays 4 at the end of the year of death. If \(i=0.25\) and \(q_x=0.15\), find its value at time 0. (Answer: \(0.48\))
- A two-year term policy pays 1 at the end of the year of death. Let \(i=0.25\), \(q_x=0.10\), and \(q_{x+1}=0.20\). Find its value at time 0. (Answer: \(0.1952\))
- A two-year endowment pays 2 on death during the term and 4 at maturity. Let \(i=0.25\), \(q_x=0.25\), and \(q_{x+1}=0.20\). Find its value at time 0. (Answer: \(2.128\))
- A deferred whole-life policy has a two-year waiting period. Its whole-life value at the end of the waiting period is \(0.40\). If \(i=0.25\), \(q_x=0.10\), and \(q_{x+1}=0.20\), find its value at issue. (Answer: \(0.18432\))
- A two-year term policy pays 3 for death in the first year and 5 for death in the second year. There is no maturity benefit. If \(i=0.25\), \(q_x=0.10\), and \(q_{x+1}=0.20\), find its value at time 0. (Answer: \(0.816\))