Why Benefit Timing Changes the Recursion
In a level-benefit policy, the same death benefit appears in each annual reserve equation. If the benefit changes by policy year, the equation itself does not change, but the benefit amount must be indexed to the year in which death occurs. Confusing that benefit with the reserve at the end of the year is a common source of errors.
This tutorial uses a discrete annual model. Premiums are paid at the beginning of each policy year while the policy is in force. A death benefit for death during policy year \(k+1\) is paid at the end of that year. Reserves are measured immediately before the premium due at time \(k\); at the end of the final year, the reserve is zero. These timing conventions make the premium and reserve cash flows explicit.
The index on \(B_{k+1}\) matters: it identifies the benefit associated with death in the year ahead. The reserve \({}_{k+1}V_x\), by contrast, is the value at the end of that year for a policy that survives. The two amounts are different cash flows on different contingencies.
The Annual Recursion with a Changing Benefit
Suppose the effective annual interest rate is \(i\), and the one-year mortality probabilities for the policy year under consideration are \({}_1q_{x+k}\) and \({}_1p_{x+k}\). At time \(k\), the reserve and the premium are available to fund the next year’s obligations. At time \(k+1\), the insurer pays \(B_{k+1}\) if death occurred, or holds the reserve \({}_{k+1}V_x\) if the insured survived.
The justification is a one-year cash-flow balance. The amount \({}_kV_x+P\) accumulates for one year to \(({}_kV_x+P)(1+i)\). Its expected amount needed at the year-end is the death-contingent payment plus the survival-contingent reserve, weighted by their respective probabilities. Equating these amounts gives the recursion. The benefit is not multiplied by the survival probability: it is paid on the death branch, whose probability is \({}_1q_{x+k}\).
If the current reserve is known and \({}_1p_{x+k}>0\), the same equation can be rearranged to calculate the next reserve:
A forward calculation therefore needs the current reserve. A backward calculation instead needs the next reserve. For an \(n\)-year term policy with no maturity benefit, the terminal condition is \({}_nV_x=0\). A net level premium can be found by equivalence: the expected present value of premiums equals the expected present value of benefits. Once that premium is set, the recursion should reproduce the reserve boundary.
Worked Reserve Sequences
Worked Example: Increasing Benefits in a Two-Year Term Policy
A hypothetical two-year term policy pays \$100 if death occurs in year 1 and \$200 if death occurs in year 2. Assume an effective annual interest rate of \(i=0.10\), a one-year death probability of \(0.20\) in each year, and survival probability \(0.80\) in each year. Find the level net premium and the reserve at time 1.
Method. First use equivalence to determine the premium, then use the backward recursion from the zero reserve at expiry. The key is to weight the year-2 benefit by survival through year 1 when valuing it at issue.
The present value of expected benefits at issue is
Premiums are due at time 0 and, if the insured survives year 1, at time 1. Their expected present-value factor is
Thus the level net premium is \(P=44.6281/1.7273=\$25.84\), rounded to the nearest cent. To find the reserve at time 1, use \({}_2V_x=0\):
So \({}_1V_x=\$10.53\). As a check, the issue-year recursion gives
The small discrepancy in the displayed check is due to rounding intermediate values; using the unrounded premium and reserve gives \({}_0V_x=0\). This verifies the equivalence premium and the boundary condition.
Worked Example: Decreasing Benefits and a Negative Reserve
A hypothetical two-year term policy pays \$300 for death in year 1 and \$150 for death in year 2. Assume \(i=0.05\), with \({}_1q_{x}=0.10\), \({}_1p_x=0.90\), \({}_1q_{x+1}=0.10\), and \({}_1p_{x+1}=0.90\). Find the level net premium and the reserve at time 1.
Method. Use equivalence for the premium and work backward from \({}_2V_x=0\). This example also shows why an intermediate reserve need not be positive: the first-year benefit is substantially larger than the later benefit.
The expected present value of benefits at issue is
The expected present-value factor for premiums is
Therefore \(P=40.8163/1.8571=\$21.98\), rounded to the nearest cent. The reserve at time 1 is
Thus \({}_1V_x=-\$7.69\). Check the issue-time recursion using the unrounded values:
A negative reserve here is consistent with the specified net-premium model: the future level premiums, together with the reserve, balance expected future benefits. It does not mean the policyholder owes a negative benefit. The calculation should not be “corrected” by replacing the reserve with zero.
Worked Example: Changing Mortality and an Increasing Benefit
A hypothetical two-year term policy pays \$100 in year 1 and \$150 in year 2. The effective annual interest rate is \(i=0.25\). In year 1, the death and survival probabilities are \(0.10\) and \(0.90\); in year 2, they are \(0.20\) and \(0.80\). Find the level net premium and the time-1 reserve.
Method. Calculate expected present values using the mortality probability for each policy year, then apply the backward recursion. The mortality probabilities are duration-specific, so using the year-1 probability in year 2 would be incorrect.
The issue-time present value of expected benefits is
The expected present-value factor for premiums is
Hence \(P=25.28/1.72=\$14.70\), rounded to the nearest cent. Starting from \({}_2V_x=0\), the time-1 reserve is
Thus \({}_1V_x=\$9.30\). The issue-time check is
Using unrounded values gives exactly zero at issue, as required by equivalence. The benefit in the second-year death branch is \$150, while the reserve in the survival branch is \({}_2V_x=0\); they must remain separate in the calculation.
Common Traps and Practical Checks
The recursion is structurally the same for increasing and decreasing benefits. What changes is the indexed benefit \(B_{k+1}\), and, where mortality varies by duration, the relevant one-year probabilities. Before calculating, write the policy-year benefit schedule beside the reserve sequence. Then label each recursion with the year’s death probability and survival probability.
- Using the wrong year’s benefit. In the equation for \({}_kV_x\), use \(B_{k+1}\), the benefit for death during the year ahead—not \(B_k\) or the next reserve.
- Applying the wrong probability. The death benefit is weighted by \({}_1q_{x+k}\); the next reserve is weighted by \({}_1p_{x+k}\).
- Reversing the direction of calculation. The forward formula solves for \({}_{k+1}V_x\) from a known \({}_kV_x\). The backward formula solves for \({}_kV_x\) from a known \({}_{k+1}V_x\).
- Ignoring premium timing. This recursion assumes a premium at the beginning of each year while the policy is in force. A different premium timing requires a different cash-flow equation.
- Assuming reserves must be positive. Decreasing benefits can produce a negative intermediate net reserve. Follow the stated model and the boundary condition rather than imposing a positivity condition.
The most reliable check is to substitute each calculated reserve back into its one-year equation. For a level net premium on a term policy, also check that \({}_0V_x=0\) under equivalence and that the terminal reserve is zero when no maturity benefit is payable. The method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use the stated annual timing and keep the death benefit for the year ahead distinct from the next reserve.
- A one-year policy has \(i=0.04\), \({}_1q_x=0.15\), a death benefit of \$80,000, current reserve \({}_0V_x=0\), and premium \(P=\$15{,}000\). Find \({}_1V_x\) by the forward recursion. (Answer: \(\$4{,}235.29\))
- At time \(k\), a policy has \(i=0.05\), \({}_1q_{x+k}=0.20\), next reserve \({}_{k+1}V_x=\$700\), premium \(P=\$900\), and death benefit \(B_{k+1}=\$5{,}000\). Find \({}_kV_x\). (Answer: \(\$585.71\))
- A two-year term policy pays \$1,000 in year 1 and \$2,000 in year 2. Assume \(i=0.10\), with death probability \(0.10\) and survival probability \(0.90\) in each year. Find the level net premium and \({}_1V_x\). (Answer: \(P=\$131.82\), \({}_1V_x=\$50.00\))
- A three-year term policy pays \$300, \$200, and \$100 for death in years 1, 2, and 3, respectively. Let \(i=0\); the successive death probabilities are \(0.20\), \(0.25\), and \(0.20\), with survival probabilities \(0.80\), \(0.75\), and \(0.80\). Find the level net premium and \({}_1V_x\). (Answer: \(P=\$46.67\), \({}_1V_x=-\$16.67\))
- At time \(k\), \(i=0.10\), \({}_1q_{x+k}=0.10\), \(B_{k+1}=\$500\), \(P=\$100\), and \({}_kV_x=\$200\). Find \({}_{k+1}V_x\). (Answer: \(\$311.11\))