Multi-Year Reserve Recursion Problem Workshop

Why Multi-Year Reserve Problems Need a Consistent Timeline

A linked reserve problem is not just a sequence of substitutions. Each year can have a different benefit, mortality rate, expense, or premium condition, and a single timing error can affect every reserve that follows. A reliable solution begins by stating exactly when each cash flow occurs and what the reserve represents.

This workshop uses a gross reserve recursion: premiums and expenses are included, and reserves are valued conditionally on the insured being alive at the valuation time. The problems are hypothetical and illustrate techniques aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $V_k$ be the reserve at time $k$, immediately before any premium or expense due at that time, conditional on the insured being alive and the policy in force. During year $k$, a premium $P_k$ is collected and an expense $e_k$ is paid at time $k$. If death occurs during that year, benefit $B_{k+1}$ is paid at the end of the year. If the insured survives, the reserve at time $k+1$ is $V_{k+1}$. With effective annual interest rate $i$, one-year death probability $q_k$, and survival probability $p_k=1-q_k$, the recursion is $V_k=v(q_kB_{k+1}+p_kV_{k+1})-P_k+e_k$, where $v=1/(1+i)$.

The timing convention matters. The premium and expense in this definition occur at the beginning of the year, before mortality during the year. The benefit is paid at the end of the year of death. The expense is not conditional on surviving the year; it is incurred at the start while the policy is in force. A problem with end-of-year expenses, monthly premiums, or a different benefit timing requires a different cash-flow equation.

Deriving the One-Year Recursion

The recursion follows by comparing the funds available at the start of the year with expected obligations at the end. Immediately after the beginning-of-year premium and expense, the funds are $V_k+P_k-e_k$. These funds accumulate for one year at rate $i$. At year-end, death occurs with probability $q_k$, producing payment $B_{k+1}$; survival occurs with probability $p_k$, leaving reserve $V_{k+1}$.

(V_k+P_k-e_k)(1+i)=q_kB_{k+1}+p_kV_{k+1}.

Solving this equation for $V_k$ and multiplying the expected end-of-year amount by $v$ gives the stated recursion. This is an expected-value balance conditional on being in force at time $k$; it does not mean that a particular policy pays the expected claim amount.

V_k=v(q_kB_{k+1}+p_kV_{k+1})-P_k+e_k.

A pure term policy with no benefit at expiry has boundary reserve $V_n=0$ at time $n$, provided there are no expenses or other payments after expiry. A maturity benefit changes that boundary: its value must be included in the survival branch for the final year. Boundaries should be stated before applying the recursion.

A Reliable Workflow for Linked Reserves

For a multi-year problem, write down a separate row for each policy year. Record the mortality probability, death benefit, premium, and expense for that year, together with the reserve at the end of the year. Then identify whether the question asks for a backward calculation, a forward calculation, or a premium that must be solved from a boundary condition.

Fix the time convention.
State when the reserve is measured and when each premium, expense, and benefit is paid.

Set the boundary.
For a term policy without a maturity payment, use the specified expiry reserve, commonly $V_n=0$.

Write one equation per year.
Use the year-specific values $q_k$, $B_{k+1}$, $P_k$, and $e_k$, rather than assuming they are constant.

Solve in the useful direction.
Work backward from a known terminal reserve, forward from a known initial reserve, or solve the linked equations for an unknown premium.

Check before rounding.
Substitute unrounded reserves into the recursion and confirm that the boundary and any requested initial condition hold.

Worked Problems

Worked Example: Backward Recursion with Increasing Benefits

A hypothetical two-year term policy has an effective annual interest rate of $i=10\%$. The death probabilities are $q_0=0.10$ and $q_1=0.20$, so $p_0=0.90$ and $p_1=0.80$. The death benefits are $B_1=\$80$ and $B_2=\$100$. A premium of $\$8$ and an expense of $\$1$ are paid at the beginning of each year while the policy is in force. There is no maturity benefit, so $V_2=0$. Find $V_1$ and $V_0$.

Method. Work backward from the expiry boundary. Here $v=1/1.10=10/11$. In the second year, the expected end-of-year amount is $0.20(100)+0.80(0)=20$. Therefore,

V_1=\frac{20}{1.10}-8+1 =\frac{200}{11}-7 =\frac{123}{11} \approx \$11.18.

For year 0, the expected end-of-year amount uses the first-year benefit and the unrounded $V_1$:

\begin{aligned} V_0 &=\frac{0.10(80)+0.90(123/11)}{1.10}-8+1\\ &=\frac{8+1107/110}{1.10}-7\\ &=\frac{1987}{121}-7 =\frac{1140}{121} \approx \$9.42. \end{aligned}

The exact values are $V_1=123/11$ and $V_0=1140/121$; dollar amounts are rounded to the nearest cent only at the end. The boundary is satisfied because the last-year calculation used $V_2=0$, as required for term coverage with no maturity payment.

Common mistake. Omitting the death probability and using the full second-year benefit in the expected amount would give $V_1=100/1.10-8+1$, about $\$83.91$. That incorrectly treats the death benefit as payable to every policy in force. The correct expected death-benefit payment is $q_1B_2=0.20(100)=\$20$.

Worked Example: Solving a Level Gross Premium from the Issue Boundary

A hypothetical two-year term policy has $i=5\%$, $q_0=0.02$, and $q_1=0.03$. Thus $p_0=0.98$ and $p_1=0.97$. The benefits are $B_1=\$10{,}000$ and $B_2=\$15{,}000$. A level premium $P$ and a beginning-of-year expense of $\$100$ apply in both years. There is no maturity benefit, so $V_2=0$. Find $P$ such that the initial reserve is $V_0=0$.

Method. First write the second-year recursion, then substitute it into the first-year recursion. The discount factor is $v=1/1.05=20/21$. For year 1,

V_1=\frac{0.03(15{,}000)+0.97(0)}{1.05}-P+100 =\frac{3000}{7}-P+100 =\frac{3700}{7}-P.

For year 0, $V_0=0$, so the first-year recursion becomes

0=\frac{200+0.98V_1}{1.05}-P+100, \qquad P=100+\frac{20}{21}(200+0.98V_1).

Substitute $V_1=3700/7-P$. Since $0.98(3700/7)=518$, the premium equation is

\begin{aligned} P &=100+\frac{20}{21}(718-0.98P)\\ &=100+\frac{14360}{21}-\frac{14}{15}P,\\ \frac{29}{15}P&=\frac{16460}{21},\\ P&=\frac{82300}{203} \approx \$405.42. \end{aligned}

Using the exact premium, the year-1 reserve is

V_1=\frac{3700}{7}-\frac{82300}{203} =\frac{25000}{203} \approx \$123.15.

Check the initial recursion without rounding: $0.98V_1=24500/203$, so $200+0.98V_1=65100/203$. Dividing by $1.05$ gives $305.418719\ldots$. Then $305.418719\ldots-405.418719\ldots+100=0$, verifying the required $V_0$.

Common mistake. The expense is paid at the beginning, so it enters the reserve recursion as $+100$, while the premium enters as $-P$. Reversing either sign changes the premium equation and no longer satisfies the stated cash-flow balance.

Worked Example: Premium-Paying Period Followed by Paid-Up Coverage

A hypothetical three-year policy has $i=5\%$, with $q_k=0.10$ and $p_k=0.90$ in each year. Its death benefits are $B_1=\$100$, $B_2=\$120$, and $B_3=\$150$. A premium of $\$15$ is paid at times 0 and 1, but no premium is due at time 2: coverage is paid-up for the final year. A beginning-of-year expense of $\$2$ applies in all three years. There is no maturity benefit, so $V_3=0$. Find the reserves $V_2,V_1,V_0$.

Method. Work backward, changing $P_k$ at the paid-up boundary. With $v=20/21$, the final year has $P_2=0$, giving

V_2=\frac{0.10(150)+0.90(0)}{1.05}+2 =\frac{100}{7}+2 =\frac{114}{7} \approx \$16.29.

In year 1, the premium is still due. Using the exact $V_2$,

\begin{aligned} V_1 &=\frac{12+0.90(114/7)}{1.05}-15+2\\ &=\frac{1244}{49}-13 =\frac{607}{49} \approx \$12.39. \end{aligned}

At time 0, the premium is also due. Keep $V_1$ unrounded:

\begin{aligned} V_0 &=\frac{10+0.90(607/49)}{1.05}-15+2\\ &=\frac{20726}{1029}-13 =\frac{7349}{1029} \approx \$7.14. \end{aligned}

The result has the required paid-up structure: no premium was subtracted in the calculation of $V_2$, while the $\$15$ premium was subtracted in the calculations of $V_1$ and $V_0$. To check the first-year balance, $V_0+15-2=V_0+13$; multiplying this amount by $1.05$ gives $10+0.90V_1$, using the exact fractions above.

Common mistake. Continuing to subtract $\$15$ in year 2 would treat the contract as premium-paying after the paid-up boundary. The premium schedule is part of the model, not a value to carry forward automatically.

Checks That Catch Multi-Year Errors

A reserve can be numerically plausible and still be wrong. Before rounding, substitute the computed values into each year’s equation. The check should use the same year-specific probabilities and cash flows as the original recursion. Also verify that a terminal or issue condition is met exactly, rather than approximately because of rounded intermediate values.

Keep exact fractions or full calculator precision through the linked calculations, and round only the requested final result. If an answer is required to the nearest cent, retain the unrounded reserve when calculating the next year; rounding a reserve to cents first can change later results.

The sign convention offers another quick check. Under the timing used here, a premium reduces the reserve required at the valuation time, so it appears as $-P_k$. A beginning expense increases the required reserve, so it appears as $+e_k$. Mortality weights the death benefit by $q_k$, and survival weights the next reserve by $p_k$; the probabilities should sum to one.

Key takeaway: A multi-year reserve solution is a chain of year-specific expected-value balances. State the timing, impose the correct boundary, apply each year’s premium and expense schedule, and verify the full chain using unrounded values.

Check Your Understanding

For each problem, use the beginning-of-year premium and expense, end-of-year death benefit, and reserve timing defined above. Round money to the nearest cent where needed.

A two-year term policy has $i=0$, $q_0=0.10$, $q_1=0.20$, benefits $B_1=\$50$, $B_2=\$80$, a premium of $\$6$ at each year’s start, no expenses, and $V_2=0$. Find $V_1$ and $V_0$. (Answer: $V_1=\$10.00,\;V_0=\$8.00$)
A one-year term policy has $i=5\%$, $q_0=0.10$, $B_1=\$100$, a beginning premium of $\$8$, a beginning expense of $\$1$, and $V_1=0$. Find $V_0$. (Answer: $\$2.52$)
A two-year term policy has $i=0$, $q_0=q_1=0.10$, benefits $B_1=\$100$ and $B_2=\$200$, no expenses, and the same premium $P$ is paid at times 0 and 1. If $V_2=0$, find $P$ that makes $V_0=0$. (Answer: $P=280/19\approx \$14.74$)
A two-year policy has $i=0$, $q_0=q_1=0.20$, benefits $B_1=\$40$ and $B_2=\$70$, and a $\$2$ expense at the start of each year. A single premium $P$ is paid at time 0 only. There is no maturity benefit, so $V_2=0$. Find $P$ that makes $V_0=0$. (Answer: $P=\$22.80$)
A two-year policy has $i=0$, $q_0=0.10$, $q_1=0.25$, benefits $B_1=\$60$ and $B_2=\$100$, and no premiums or expenses. With $V_2=0$, find $V_1$ and $V_0$. (Answer: $V_1=\$25.00,\;V_0=\$28.50$)

What You'll Learn