Reserve Recursion Across Policy Years

Why a Reserve Recursion Is Useful

A reserve at one policy anniversary is linked to the reserve at the next anniversary by what happens during the intervening year. The opening reserve is available to earn interest; premiums and expenses alter the amount invested; and benefits may be paid before the next anniversary. The reserve recursion records that balance in one equation.

This relationship is useful both for valuing reserves backward, from a known terminal reserve, and for projecting reserves forward, from a known opening reserve. The equation is only valid when the timing of premiums, expenses, benefits, and reserve measurements is stated consistently.

Definition: Let $V_t$ be the reserve immediately before the start-of-year cash flows at time $t$, conditional on the policy being in force at that time. Let $P_t$ be a premium received at time $t$, and $E_t$ an expense paid at time $t$. During the year, death occurs with probability $q_t$ or the policyholder survives with probability $p_t=1-q_t$. Assume a death benefit $D_{t+1}$ is paid at the end of the year if death occurs, and a reserve $V_{t+1}$ is held at the end of the year if the policyholder survives. With annual effective interest rate $i$, the recursion is $(V_t+P_t-E_t)(1+i)=q_tD_{t+1}+p_tV_{t+1}$.

This model treats death and survival as the only outcomes during the year. It assumes interest is earned at a deterministic effective annual rate, start-of-year premiums and expenses are invested for the full year, and death benefits and closing reserves are measured at year-end. There are no lapses or other payments in this basic version. If a contract has a different payment timing or additional possible outcomes, the balance must be adjusted to reflect them.

Deriving the Annual Balance

The recursion follows by balancing the assets available at the end of the year against the expected amounts needed then. Start with the opening reserve. Add the premium received at the start of the year and subtract the start-of-year expense. The net amount earns interest for one year. At year-end, a death benefit is due on the death outcome; on the survival outcome, the fund must support the next reserve.

\underbrace{(V_t+P_t-E_t)(1+i)}_{\text{fund accumulated to year-end}} = \underbrace{q_tD_{t+1}+p_tV_{t+1}}_{\text{expected year-end requirement}}

The expectation on the right is conditional on the policy being in force at time $t$. The death benefit is multiplied by the probability of death, while the next reserve is multiplied by the probability of survival. In particular, $V_{t+1}$ is not required on a policy that has ended through death under this model.

Rearranging the same balance gives two useful forms. The first solves for the opening reserve, which is convenient when valuing backward. The second solves for the closing reserve, which is convenient when projecting forward.

V_t=\frac{q_tD_{t+1}+p_tV_{t+1}}{1+i}-P_t+E_t

V_{t+1} = \frac{(V_t+P_t-E_t)(1+i)-q_tD_{t+1}}{p_t}, \qquad p_t>0

The forward form requires $p_t>0$, since it solves for a reserve conditional on survival. If $p_t=0$, there are no surviving policies for which $V_{t+1}$ needs to be calculated from this balance. The backward form remains usable without division by $p_t$.

Set the time points.
State when the opening reserve is measured and when each premium, expense, benefit, and closing reserve occurs.

Write the year-end balance.
Accumulate the opening reserve plus start-of-year premium less start-of-year expense at rate $i$.

Weight each outcome.
Multiply each outcome-specific year-end amount by its probability conditional on being in force at the start of the year.

Solve in the required direction.
Use the backward form when $V_{t+1}$ is known and the forward form when $V_t$ is known.

Worked Examples

Worked Example: Valuing a Two-Year Term Reserve Backward

A hypothetical two-year term policy has a $1,000 death benefit paid at the end of the year of death. A $50 premium is received at the start of each year, with no expenses. The effective annual interest rate is 5%. In each year, the conditional probability of death is 0.10 and the probability of survival is 0.90. The terminal reserve at time 2 is $V_2=0$. Calculate $V_1$ and $V_0$. Assume no other cash flows or decrements.

Method. Work backward from the terminal reserve using the opening-reserve form. For the year from time 1 to time 2:

V_1=\frac{(0.10)(1000)+(0.90)(0)}{1.05}-50 =\frac{100}{1.05}-50 =95.238095-50 =45.238095

For the year from time 0 to time 1, the same assumptions apply, and $V_1$ is now the closing reserve for a policyholder who survives that year:

\begin{aligned} V_0 &=\frac{(0.10)(1000)+(0.90)(45.238095)}{1.05}-50\\ &=\frac{100+40.714286}{1.05}-50\\ &=134.013605-50\\ &=84.013605. \end{aligned}

Rounded to the nearest cent, $V_1=\$45.24$ and $V_0=\$84.01$. As a check, the year-0 accumulated fund is $(84.013605+50)(1.05)=140.714285$, matching the expected year-end requirement $100+(0.90)(45.238095)=140.714286$, apart from rounding. The common error avoided is treating the full next reserve as payable on death as well as on survival.

Worked Example: Projecting Reserves Forward

A hypothetical policy has opening reserve $V_0=\$200$. At the start of the year it receives a premium of $80 and pays an expense of $10. The annual effective interest rate is 4%. The death probability is 0.05, the survival probability is 0.95, and the year-end death benefit is $500. Calculate the reserve for a survivor at time 1. For the next year, assume a premium of $80, expense of $10, interest rate of 4%, death probability of 0.10, survival probability of 0.90, and year-end death benefit of $500; project the reserve to time 2. Treat these as the stated assumptions for the two successive years.

Method. Accumulate the net opening amount, subtract the expected death-benefit payment, and divide the remainder by the survival probability to obtain the reserve conditional on survival. In the first year, the accumulated amount is:

$$ (200+80-10)(1.04)=270(1.04)=280.80. $$

The expected death-benefit payment is $(0.05)(500)=\$25$. The amount assigned to surviving policies is therefore $280.80-25=\$255.80$, so:

V_1=\frac{255.80}{0.95}=269.263158.

For the second year, the accumulated amount before the death benefit is:

$$ (269.263158+80-10)(1.04) =339.263158(1.04) =352.833684. $$

The expected death-benefit payment is $(0.10)(500)=\$50$. Thus:

V_2=\frac{352.833684-50}{0.90} =\frac{302.833684}{0.90} =336.481871.

Rounded to the nearest cent, $V_1=\$269.26$ and $V_2=\$336.48$. The first-year balance checks because $(0.95)(269.263158)=\$255.80$. The second checks because $(0.90)(336.481871)=\$302.833684$. The common error avoided is dividing the full accumulated fund by the survival probability before subtracting the expected death benefit.

Worked Example: Solving for a Premium Consistent With Two Reserves

A hypothetical one-year policy has an opening reserve of $30 and a closing reserve of $50 conditional on survival. A premium $P_0$ is received at the start of the year, and a $5 expense is paid then. The effective annual interest rate is 3%. The death probability is 0.02, survival probability is 0.98, and the death benefit of $1,000 is paid at year-end. Find the premium that makes the reserve balance hold.

Method. Use the year-end balance with the premium as the unknown. The expected year-end requirement is the death benefit weighted by death probability plus the closing reserve weighted by survival probability:

$$ (0.02)(1000)+(0.98)(50)=20+49=69. $$

The accumulated fund is $(30+P_0-5)(1.03)$. Equating the two amounts gives:

\begin{aligned} (25+P_0)(1.03)&=69,\\ 25+P_0&=\frac{69}{1.03}=66.990291,\\ P_0&=66.990291-25=41.990291. \end{aligned}

The required premium is $\$41.99$ to the nearest cent. Substitution checks the result: $(30+41.990291-5)(1.03)=69$, equal to the expected year-end requirement. The common error avoided is omitting the $5 expense from the amount available to accumulate.

Timing and Probability: Common Exam Traps

The recursion depends on both when a cash flow occurs and which outcome makes it payable. For the model defined here, the premium and expense occur at the start of the year, so they are included before multiplication by $1+i$. The death benefit and survivor reserve occur at year-end, so they are not accumulated for another year in this equation.

A frequent incorrect calculation is to use $q_tD_{t+1}+V_{t+1}$ on the right-hand side. That charges the next reserve with probability 1, as though it were required whether or not the policyholder survives. The correct expression is $q_tD_{t+1}+p_tV_{t+1}$. For example, with $q_t=0.10$, $p_t=0.90$, a $1,000 death benefit, and $V_{t+1}=\$50$, the correct expected requirement is $(0.10)(1000)+(0.90)(50)=\$145$, not $\$1,050$. The incorrect expression counts the full reserve for a policy that has ended through death.

Another error is to put a start-of-year expense outside the accumulation incorrectly, or to accumulate a year-end benefit for a further year. The safest approach is to draw the year’s timeline first. If a contract instead pays a premium at year-end, that premium cannot be included in the same way as a start-of-year premium. If a death benefit is paid immediately at death rather than at year-end, this annual end-of-year model does not represent its exact timing; an appropriate within-year model or a stated approximation is needed.

Key takeaway: For this annual model, the accumulated opening fund is $(V_t+P_t-E_t)(1+i)$. The expected year-end requirement is the outcome-weighted sum $q_tD_{t+1}+p_tV_{t+1}$. State the timing and probabilities before applying the recursion.

Check Your Understanding

For each problem, use the stated annual timing: premiums and expenses occur at the start of the year; death benefits and closing reserves are measured at year-end.

A hypothetical one-year policy has $V_0=\$100$, premium $40, expense $5, $i=2\%$, death probability 0.10, survival probability 0.90, and year-end death benefit $200. Find $V_1$. (Answer: $130.78)
A policy has $V_t=\$75$, $V_{t+1}=\$90$, $i=4\%$, death probability 0.03, survival probability 0.97, death benefit $400, and start-of-year expense $8. Find the start-of-year premium $P_t$. (Answer: $28.48)
A two-year hypothetical policy has a $70 premium and $10 expense at the start of each year, $i=5\%$, and $V_2=0$. In the year from time 1 to time 2, death probability is 0.10 and the death benefit is $600. In the year from time 0 to time 1, death probability is 0.05 and the death benefit is $800. Find $V_1$ and $V_0$. (Answer: $V_1=-\$2.86$, $V_0=-\$24.49$)
A hypothetical policy has $V_t=\$120$, premium $30, expense $4, $i=6\%$, death probability 0.08, survival probability 0.92, and death benefit $700. Project the reserve conditional on survival. (Answer: $107.35)
Explain why the closing reserve is weighted by the survival probability in the annual recursion, but the death benefit is weighted by the death probability. (Answer: The closing reserve is required only if the policyholder survives to the next anniversary; the death benefit is paid only on the death outcome.)

What You'll Learn