Why the Recursion Changes When Premiums Cease
A limited-payment life insurance policy requires premiums for only a specified number of years, but its insurance protection can continue after those payments end. The reserve recursion must reflect both parts of that contract: future premiums become zero, while the promised death benefit remains payable. Confusing the premium-paying period with the coverage period can produce a reserve that is materially wrong.
This tutorial uses an annual model with benefits payable at the end of the year of death. Let the policy provide a level death benefit \(B\), and let premiums be due annually for \(m\) years, at times \(0,1,\ldots,m-1\). The policy remains in force after time \(m\). The mortality probability for a life aged \(x+k\) is \(q_{x+k}\), and \(p_{x+k}=1-q_{x+k}\). The effective annual interest rate is \(i\).
This timing convention matters at the boundary. The last premium is paid at time \(m-1\), not at time \(m\). The reserve \(V_m\) therefore supports the policy after the final premium has been collected. Once \(k\ge m\), the insurer receives no further premium for the contract.
The Post-Premium Recursion
Consider a policy year beginning at time \(k\ge m\). The insurer starts with reserve \(V_k\), which accumulates with interest for one year. If death occurs during the year, the insurer pays \(B\) at year-end. If the insured survives, the insurer holds reserve \(V_{k+1}\) at time \(k+1\). There is no premium in either case.
This form calculates \(V_k\) from the reserve one year later. Solving for the later reserve instead gives the forward form:
The forward form is useful when the reserve at the end of the premium-paying period is known and the question asks for subsequent reserves. The backward form is useful when a later reserve is given and an earlier post-premium reserve is required. These are two rearrangements of the same annual balance, not different reserve definitions.
Why the Formula Holds
The recursion follows by conditioning on the two possible outcomes over the year. The probability of death is \(q_{x+k}\), and the probability of survival is \(p_{x+k}\). A compact derivation is:
With no premium received during the year, the reserve available at year-end before paying claims is \((1+i)V_k\).
With probability \(q_{x+k}\), the insurer pays \(B\) at year-end. The expected year-end payment for this outcome is \(q_{x+k}B\).
With probability \(p_{x+k}\), the contract continues and the required year-end reserve is \(V_{k+1}\). Its probability-weighted amount is \(p_{x+k}V_{k+1}\).
Equating the accumulated starting reserve to the expected death payment plus the expected surviving reserve gives \((1+i)V_k=q_{x+k}B+p_{x+k}V_{k+1}\).
The argument depends on the benefit being payable at the end of the year of death. A benefit payable at the beginning of the year, or continuously at the moment of death, would require a different cash-flow timing and therefore a different recursion. Likewise, if the policy expires at the end of the year, the surviving reserve may be zero rather than the reserve for continuing coverage.
Worked Applications
In each example, first confirm that the year being considered is after the last premium due. That determines whether a premium belongs in the cash-flow equation.
Worked Example: Calculate a Post-Premium Reserve Backward
A hypothetical limited-payment whole-life policy has a level death benefit of \(B=\$100{,}000\). All premiums have ceased. During a particular policy year, the effective annual interest rate is \(i=5\%\), the probability of death is \(q_{x+k}=0.05\), and the reserve one year later is \(V_{k+1}=\$40{,}000\). Calculate \(V_k\), to the nearest cent, and verify the result.
Method. Since premiums have ceased, use the zero-premium backward recursion. The survival probability is \(p_{x+k}=1-0.05=0.95\).
The probability-weighted death benefit is \(q_{x+k}B=0.05(\$100{,}000)=\$5{,}000\). The probability-weighted reserve on survival is \(p_{x+k}V_{k+1}=0.95(\$40{,}000)=\$38{,}000\).
The total expected year-end amount is therefore \(\$5{,}000+\$38{,}000=\$43{,}000\). Discounting one year at \(5\%\) gives \(V_k=\$43{,}000/1.05=\$40{,}952.38095\ldots\), or \(\$40{,}952.38\).
Check. Accumulating the rounded reserve gives approximately \(\$40{,}952.38(1.05)=\$43{,}000.00\), which matches the death payment plus the survival reserve. The common mistake avoided is adding a premium to the starting reserve after the premium-paying period has ended.
Worked Example: Project Reserves Forward After the Final Premium
A hypothetical policy has completed its premium-paying period at time \(m\). Its reserve at that time is \(V_m=\$12{,}000\), and its continuing level death benefit is \(B=\$60{,}000\). The effective annual interest rate is \(i=4\%\). The death probabilities in the next two policy years are \(0.02\) and \(0.03\), respectively. Find \(V_{m+1}\) and \(V_{m+2}\), to the nearest cent.
Method. Both years are after premium cessation, so use the forward zero-premium recursion in each year. In the first year, \(q_{x+m}=0.02\) and \(p_{x+m}=0.98\). The accumulated starting reserve is \(\$12{,}000(1.04)=\$12{,}480\), and the probability-weighted death benefit is \(0.02(\$60{,}000)=\$1{,}200\). Hence
Thus \(V_{m+1}=\$11{,}510.20\) to the nearest cent. For the second year, \(q_{x+m+1}=0.03\) and \(p_{x+m+1}=0.97\). Using the unrounded first-year result, the accumulated reserve is \(\$11{,}510.2040816\ldots(1.04)=\$11{,}970.6122449\ldots\). The probability-weighted death benefit is \(0.03(\$60{,}000)=\$1{,}800\). Therefore
Check. For the second year, \(0.97(\$10{,}485.1672638\ldots)+\$1{,}800=\$11{,}970.6122449\ldots\), the accumulated starting reserve. The common mistake avoided is rounding the first reserve before using it in the next recursion; retaining full precision until the final answers avoids carrying a rounding difference forward.
Worked Example: Do Not Carry the Premium Into a Later Year
A hypothetical policy requires premiums for four years, with the last premium due at time \(3\). Coverage continues for life with a level death benefit of \(B=\$100{,}000\). At time \(4\), the reserve is \(V_4=\$20{,}000\). For the year from time \(4\) to time \(5\), the effective annual interest rate is \(i=5\%\) and the death probability is \(q_{x+4}=0.02\). Find \(V_5\). A candidate incorrectly includes the annual premium \(P=\$1{,}000\) in the calculation.
Method. The final premium was due at time \(3\), so no premium is due in the year beginning at time \(4\). The survival probability is \(p_{x+4}=1-0.02=0.98\). The accumulated reserve is \(\$20{,}000(1.05)=\$21{,}000\), and the probability-weighted death benefit is \(0.02(\$100{,}000)=\$2{,}000\). Thus
The incorrect calculation adds a premium even though none is due: \((\$20{,}000+\$1{,}000)(1.05)=\$22{,}050\), then \((\$22{,}050-\$2{,}000)/0.98=\$20{,}459.18\). That result is too high because it includes a cash inflow the contract does not provide.
Check. Using the correct result, \(0.98(\$19{,}387.755102\ldots)+\$2{,}000=\$21{,}000\). This equals the accumulated reserve at time \(5\), as required.
Boundary and Exam Traps
The most important boundary check is to count premium due dates, not simply to count years of coverage. If a policy is described as “four-pay,” premiums are typically due at times \(0,1,2,3\); the recursion for the year beginning at time \(4\) has no premium. The death benefit can still be payable in that year and in later years.
A useful comparison is the ordinary premium-paying recursion under the same timing convention. For \(k<m\), the premium is added at time \(k\), so the balance is \((1+i)(V_k+P)=q_{x+k}B+p_{x+k}V_{k+1}\). For \(k\ge m\), set the premium cash flow to zero; the balance becomes \((1+i)V_k=q_{x+k}B+p_{x+k}V_{k+1}\). The change is the premium term, not the death-benefit term.
| Year being valued | Premium due at its start? | Death benefit if death occurs? | Starting amount accumulated |
|---|---|---|---|
| \(k<m\) | Yes, \(P\) | \(B\) | \((1+i)(V_k+P)\) |
| \(k\ge m\), coverage continues | No | \(B\) | \((1+i)V_k\) |
| After coverage has expired | No | No, under an ordinary term contract | Depends on the contract’s terminal reserve condition |
Do not assume that a reserve must rise or fall just because premiums have stopped. Its change depends on the interest rate, mortality probability, death benefit, and future reserve. The recursion gives the relevant balance; the sign of the change should be determined from the supplied values rather than guessed from the phrase “paid-up.”
Check Your Understanding
For each problem, assume an annual effective interest rate and an end-of-year death benefit. Round dollar answers to the nearest cent.
- A limited-payment policy’s premiums have ceased. In a year with \(i=4\%\), \(q_{x+k}=0.02\), \(B=\$50{,}000\), and \(V_{k+1}=\$10{,}000\), calculate \(V_k\). (Answer: \(\$10{,}384.62\))
- After the final premium, \(V_k=\$15{,}000\), \(i=5\%\), \(q_{x+k}=0.03\), and \(B=\$40{,}000\). Calculate \(V_{k+1}\). (Answer: \(\$15{,}000.00\))
- With no future premiums, \(i=2\%\), \(q_{x+k}=0.04\), \(B=\$75{,}000\), and \(V_{k+1}=\$25{,}000\), calculate \(V_k\). (Answer: \(\$26{,}470.59\))
- A policy’s premiums have ended, and \(V_m=\$20{,}000\). The benefit is \(\$100{,}000\), \(i=5\%\), and the next two death probabilities are \(0.02\) and \(0.04\). Calculate \(V_{m+2}\), retaining full precision between years. (Answer: \(\$17{,}038.69\))
- A four-pay policy has its final premium due at time \(3\), and coverage continues. At time \(4\), \(V_4=\$12{,}000\); for the next year \(i=4\%\), \(q_{x+4}=0.025\), and \(B=\$60{,}000\). If the former annual premium was \(\$900\), calculate \(V_5\) and state whether that premium enters the recursion. (Answer: \(V_5=\$11{,}261.54\); the premium does not enter.)