Why Different Benefits Change the Recursion
An endowment contract can pay one amount if the insured dies during the term and another if the insured survives to maturity. The two outcomes occur at the same year-end in a discrete model, but their benefit amounts are not interchangeable. The one-year recursion must weight each amount by the probability of its corresponding outcome.
This tutorial considers a contract issued at age \(x\) for \(n\) years. It assumes level death and maturity benefits, annual effective interest, and death benefits paid at the end of the policy year of death. These timing assumptions matter: a benefit paid immediately at death would require different discounting within the year.
For each policy year \(k+1\), let \({}_1q_{x+k}\) be the probability that a life aged \(x+k\) dies during that year, and let \({}_1p_{x+k}=1-{}_1q_{x+k}\) be the probability of surviving it. The recursion discounts one year and then averages the two possible outcomes: death produces \(B\), while survival leads to the next reserve.
The One-Year Recursion
The boundary convention is important. \({}_nV_x=C\) means the value for a survivor immediately before the maturity payment. Immediately after that payment, the contract has ended and its reserve is zero. Using the pre-payment value makes the same recursion valid in the final year as in every earlier year.
Why the Recursion Holds
Condition on what happens during the next policy year, given that the insured is alive at duration \(k\). If death occurs, the payment \(B\) is made at time \(k+1\), so its value at duration \(k\) is \(vB\). If the insured survives, the future benefits have value \({}_{k+1}V_x\) at duration \(k+1\), and their value one year earlier is \(v{}_{k+1}V_x\). Weight these values by the probabilities of the corresponding outcomes:
This is a conditional expected-present-value argument. It assumes the stated annual mortality probabilities and a common annual interest rate for discounting. It does not include premiums, expenses, lapses, or benefits that vary by duration.
Working Backward and Checking the Value
The terminal value is known, so the practical method is to calculate one year at a time in reverse order. The recursion also has a direct check: the issue value is the sum of the discounted death-benefit values for each year and the discounted maturity benefit for survival through the whole term. For a constant benefit \(B\) and maturity benefit \(C\), this is
where \({}_0p_x=1\). A death in year \(j+1\) requires survival through the first \(j\) years and death in the next year; survival to maturity requires survival through all \(n\) years. These mutually exclusive outcomes account for every possible benefit payment.
Use \({}_nV_x=C\), the value immediately before the maturity payment.
At each duration, multiply the death probability by \(B\), the survival probability by the next reserve, add, and discount by \(v\).
Compare the issue value with the present value of the mutually exclusive death and maturity outcomes.
Worked Example: A Two-Year Endowment with a Larger Maturity Benefit
A hypothetical two-year contract pays \(B=\$12{,}000\) at the end of the year of death and \(C=\$20{,}000\) at maturity if the insured survives. The annual effective interest rate is \(i=5\%\), with \({}_1q_x=0.10\) and \({}_1q_{x+1}=0.20\). Find the issue value and the reserve at duration 1. Assume the given probabilities apply to each policy year and the death benefit is paid at that year-end.
Method. Work backward from maturity. Here \(v=1/1.05=0.95238095\), \({}_1p_x=0.90\), and \({}_1p_{x+1}=0.80\).
At duration 2, immediately before maturity, \({}_2V_x=\$20{,}000\). Therefore, at duration 1,
At issue,
Rounded to the nearest cent, the duration-1 reserve is \(\$17{,}523.81\) and the issue value is \(\$16{,}163.27\). The method avoids treating the maturity amount as a death benefit or discounting it from the wrong date.
Check. The probabilities of death in years 1 and 2 are \(0.10\) and \(0.90(0.20)=0.18\); the probability of survival to maturity is \(0.90(0.80)=0.72\). Thus
The direct present-value check agrees with the recursive calculation.
Worked Example: A Three-Year Contract with a Smaller Death Benefit
A hypothetical three-year policy pays \(B=\$30{,}000\) on death during the term and \(C=\$50{,}000\) at maturity. Let \(i=4\%\), and suppose the successive one-year death probabilities are \(0.10\), \(0.20\), and \(0.25\). Calculate the issue value and the reserve at duration 1. Round money to the nearest cent.
Method. Set \({}_3V_x=\$50{,}000\) and work backward. The corresponding survival probabilities are \(0.90\), \(0.80\), and \(0.75\), and \(v=1/1.04\).
For the final year, at duration 2,
At duration 1,
At issue,
Rounded to the nearest cent, \({}_1V_x=\$39{,}053.25\) and \({}_0V_x=\$36{,}680.70\). The method distinguishes the benefit for death from the larger maturity benefit at every step; the maturity amount enters only through the final survival outcome and the continuation reserves.
Worked Example: Check the Recursion by Enumerating Outcomes
A hypothetical three-year contract has \(B=\$10{,}000\), \(C=\$30{,}000\), and \(i=0\%\). The successive annual death probabilities are \(0.10\), \(0.20\), and \(0.25\). Calculate the issue value by recursion and independently by listing the outcomes. State why the zero interest rate simplifies the calculation.
Method. With \(i=0\%\), \(v=1\), so no benefit is reduced by discounting. Start with \({}_3V_x=\$30{,}000\). At duration 2,
At duration 1 and then at issue,
For the independent check, the death probabilities in years 1, 2, and 3 are \(0.10\), \(0.90(0.20)=0.18\), and \(0.90(0.80)(0.25)=0.18\). The probability of reaching maturity is \(0.90(0.80)(0.75)=0.54\). These sum to \(0.10+0.18+0.18+0.54=1.00\). The expected benefit is
The check agrees. Zero interest removes discounting, but it does not remove the need to distinguish death from survival.
Common Mistake: Using the Death Benefit on Survival
A frequent error is to discount \(B\) regardless of whether the insured dies or survives. Consider a one-year contract with \(B=\$80{,}000\), \(C=\$200{,}000\), \(i=4\%\), and \({}_1q_x=0.25\). The incorrect calculation \(80{,}000/1.04=\$76{,}923.08\) assigns the death benefit to both outcomes.
The correct calculation first weights the death benefit by \(0.25\) and the maturity benefit by the survival probability \(0.75\):
On survival, the maturity benefit of \(\$200{,}000\) applies. The incorrect result is not a small rounding difference; it uses the wrong cash flow for the survival outcome.
Check Your Understanding
For each problem, use end-of-year death benefits and the stated annual effective interest rate. Round monetary answers to the nearest cent where needed.
- A one-year contract pays \(\$18{,}000\) on death and \(\$30{,}000\) at maturity. If \(i=5\%\) and \({}_1q_x=0.20\), find the issue value and identify the payment if the insured survives. (Answer: \(\$26{,}285.71\); the maturity benefit of \(\$30{,}000\) applies on survival.)
- A two-year contract pays \(B=\$10{,}000\) on death and \(C=\$25{,}000\) at maturity. Let \(i=0\%\), \({}_1q_x=0.10\), and \({}_1q_{x+1}=0.20\). Calculate \({}_1V_x\) and \({}_0V_x\). (Answer: \({}_1V_x=\$22{,}000\); \({}_0V_x=\$20{,}800\).)
- A two-year contract pays \(B=\$20{,}000\) on death and \(C=\$40{,}000\) at maturity. Let \(i=4\%\), \({}_1q_x=0.10\), and \({}_1q_{x+1}=0.25\). Calculate the issue value. (Answer: \(\$31{,}046.60\).)
- A three-year contract pays \(B=\$12{,}000\) on death and \(C=\$20{,}000\) at maturity. If \(i=0\%\) and the successive annual death probabilities are \(0.10\), \(0.20\), and \(0.25\), calculate \({}_1V_x\) and \({}_0V_x\). (Answer: \({}_1V_x=\$16{,}800\); \({}_0V_x=\$16{,}320\).)
- A two-year contract pays \(B=\$30{,}000\) on death and \(C=\$10{,}000\) at maturity. Let \(i=10\%\), with \({}_1q_x={}_1q_{x+1}=0.20\). Calculate \({}_1V_x\) and \({}_0V_x\), and state which benefit is larger. (Answer: \({}_1V_x=\$12{,}727.27\); \({}_0V_x=\$14{,}710.74\); the death benefit is larger.)