Build Retrospective Reserves from Past Cash Flows

Why Build a Reserve from Past Cash Flows?

A retrospective reserve looks backward from a valuation date. It accumulates premiums received and benefits paid in the past, then determines the resulting amount per policy still in force. The calculation combines three ingredients: when each cash flow occurred, how likely it was to occur, and how much interest it earns after it occurs.

The reserve is not simply the accumulated premiums on one surviving policy minus that policy’s past benefits. A policy that survives to the valuation date did not receive a death benefit on its own contract. Instead, the retrospective method uses expected cash flows across the original group of policies, including claims paid on policies that ended before the valuation date. It then divides the group’s accumulated net amount by the probability of survival, to express the result per policy still in force.

Definition: A retrospective reserve at time $t$ is the accumulated value at $t$ of expected past premiums and other inflows, less the accumulated value of expected past benefits and other outflows, divided by the probability that a policy issued at age $x$ remains in force to time $t$. This definition assumes the cash-flow model, interest basis, and in-force conditions are specified consistently.

The Calculation Framework

Consider a policy issued at age $x$, with annual premiums collected at the beginnings of policy years and death benefits paid at the ends of policy years. Let $P_k$ be the premium due at time $k$, provided the policyholder is alive then, and let $B_{k+1}$ be the benefit paid at the end of year $k+1$ if death occurs during that year. Let $i$ be the effective annual interest rate. For a valuation at time $t$, immediately after the premium due at time $t-1$ has been collected, the retrospective reserve is

{}_tV_x^{R} = \frac{ \displaystyle\sum_{k=0}^{t-1} {}_kp_x P_k(1+i)^{t-k} - \displaystyle\sum_{k=0}^{t-1} {}_kp_x q_{x+k} B_{k+1}(1+i)^{t-k-1} }{ {}_tp_x }.

The first sum includes premiums due at times $0,1,\ldots,t-1$. A premium due at time $k$ is received only if the insured survives to that time, which has probability ${}_kp_x$. It accumulates for $t-k$ years. The second sum includes death benefits for deaths in years $1,\ldots,t$. The probability of death in year $k+1$, measured from issue, is ${}_kp_xq_{x+k}$; a benefit paid at time $k+1$ accumulates for $t-k-1$ years. Finally, ${}_tp_x$ is the probability of being in force at time $t$.

This version assumes level annual effective interest, premiums at the start of each covered year, and death benefits at the end of the year of death. If premiums are paid at different times, benefits are paid immediately on death, or a cash flow occurs exactly at the valuation boundary, its accumulation exponent must change to match its actual timing. For a benefit that depends on survival to a future maturity date, do not include it among cash flows paid before $t$; include it only if the valuation is after that payment.

Key rule: Accumulate each expected cash flow from its payment date to the valuation date, using the probability that it is paid. Subtract accumulated outflows from accumulated inflows, then divide by the probability of survival to the valuation date.

Why the Survival Division Appears

The numerator represents an expected accumulated amount for the original group of policies, not for just the survivors. If there are $N$ identical policies at issue, the expected number surviving to time $t$ is $N{}_tp_x$. The group’s expected accumulated net cash flows are $N$ times the numerator in the reserve formula. Dividing the group amount by the expected number of survivors cancels $N$ and leaves the amount per surviving policy.

For example, if the expected accumulated net amount for an original group is $N C$, and the expected number of policies still in force is $N{}_tp_x$, then the amount per survivor is $NC/(N{}_tp_x)=C/{}_tp_x$. This explains the denominator. Omitting it would leave the calculation on the scale of the original cohort rather than the scale of the policies still in force.

Worked Calculations

Worked Example: Level Premiums and a Two-Year Term Benefit

A hypothetical two-year term policy has a $1,000 death benefit paid at the end of the year of death. A $200 premium is collected at the beginning of each year if the policyholder is alive. The annual effective interest rate is $5\%$, and the mortality assumption is $q_{x}=q_{x+1}=0.10$. Calculate the retrospective reserve at time 2, after the second premium has been collected.

Method. Accumulate each expected premium and claim to time 2, then divide the net accumulated amount by the probability of survival to time 2.

Step 1: Survival and death probabilities. Survival through both years is $ {}_2p_x=0.90(0.90)=0.81$. The probability of death in year 1 is $0.10$; the probability of death in year 2 is $0.90(0.10)=0.09$.

Step 2: Accumulated premiums. The time-0 premium accumulates for two years: $200(1.05)^2=200(1.1025)=\$220.50$. The time-1 premium is received with probability $0.90$ and accumulates for one year: $0.90(200)(1.05)=\$189.00$. Total expected accumulated premiums are $\$220.50+\$189.00=\$409.50$.

Step 3: Accumulated benefits. A year-1 death benefit accumulates one year: $0.10(1{,}000)(1.05)=\$105.00$. A year-2 death benefit is paid at time 2, so it has no accumulation after payment: $0.09(1{,}000)=\$90.00$. Total expected accumulated benefits are $\$195.00$.

Step 4: Divide by survival probability. The accumulated net amount is $\$409.50-\$195.00=\$214.50$. Thus ${}_2V_x^R=\$214.50/0.81=\$264.8148\ldots$, or $264.81 to the nearest cent.

Check. Multiplying the reserve by the survival probability gives $0.81(\$264.8148\ldots)=\$214.50$, the accumulated net amount for the original cohort. The common error of dividing by $0.90$, the one-year survival probability, would use the wrong valuation duration.

Worked Example: Unequal Premiums and Changing Mortality

A hypothetical policy pays $5,000 at the end of the first policy year if death occurs in that year, and $6,000 at the end of the second year if death occurs in that year. Premiums of $1,200 at time 0 and $1,500 at time 1 are due if the policy is then in force. Assume $i=4\%$, $q_x=0.10$, and $q_{x+1}=0.20$. Find the reserve at time 2, immediately after any time-1 premium.

Method. Apply the retrospective cash-flow sum separately to each premium and each year’s death benefit. Use the mortality probability appropriate to that policy year.

Step 1: Survival and death probabilities. The probability of surviving to time 1 is $0.90$. The probability of surviving to time 2 is $0.90(0.80)=0.72$. Death probabilities in years 1 and 2 are $0.10$ and $0.90(0.20)=0.18$, respectively.

Step 2: Accumulated premiums. The time-0 premium accumulates for two years: $1{,}200(1.04)^2=1{,}200(1.0816)=\$1{,}297.92$. The time-1 premium is received only if the policyholder survives the first year, so its expected accumulated value is $0.90(1{,}500)(1.04)=\$1{,}404.00$. Total: $\$2{,}701.92$.

Step 3: Accumulated benefits. The year-1 claim is paid at time 1 and accumulates for one year: $0.10(5{,}000)(1.04)=\$520.00$. The year-2 claim is paid at time 2: $0.18(6{,}000)=\$1{,}080.00$. Total: $\$1{,}600.00$.

Step 4: Calculate the reserve. The accumulated net amount is $\$2{,}701.92-\$1{,}600.00=\$1{,}101.92$. Divide by the two-year survival probability: ${}_2V_x^R=\$1{,}101.92/0.72=\$1{,}529.0555\ldots$, or $1,529.06 to the nearest cent.

Check. The premiums and benefits are each weighted by the probability of being in the state that triggers that cash flow. Using $0.20$ alone as the probability of a second-year death would omit the required survival through year 1.

Worked Example: Endowment Timing at the Valuation Boundary

A hypothetical three-year endowment policy collects a $4,000 premium at the beginning of each policy year while in force. It pays $8,000 at the end of the year of death, or $7,000 at time 3 to a policyholder who survives to maturity. Assume $i=5\%$, with $q_x=0.10$, $q_{x+1}=0.20$, and $q_{x+2}=0.25$. Calculate the retrospective reserve at time 3 immediately before the maturity benefit is paid.

Method. Accumulate all premiums and death benefits paid before time 3. Because this valuation is just before maturity payment, the $7,000 survival benefit is not yet a past cash flow.

Step 1: Survival and death probabilities. The probabilities of being alive at times 1, 2, and 3 are $0.90$, $0.90(0.80)=0.72$, and $0.72(0.75)=0.54$. The probabilities of death in years 1, 2, and 3 are $0.10$, $0.90(0.20)=0.18$, and $0.72(0.25)=0.18$.

Step 2: Accumulated premiums. The time-0 premium accumulates three years: $4{,}000(1.05)^3=\$4{,}630.50$. The time-1 premium contributes $0.90(4{,}000)(1.05)^2=\$3{,}969.00$. The time-2 premium contributes $0.72(4{,}000)(1.05)=\$3{,}024.00$. Total: $\$11{,}623.50$.

Step 3: Accumulated death benefits. The year-1 claim contributes $0.10(8{,}000)(1.05)^2=\$882.00$. The year-2 claim contributes $0.18(8{,}000)(1.05)=\$1{,}512.00$. The year-3 claim is paid at time 3 and contributes $0.18(8{,}000)=\$1{,}440.00$. Total: $\$3{,}834.00$.

Step 4: Divide by survival to maturity. The accumulated net amount is $\$11{,}623.50-\$3{,}834.00=\$7{,}789.50$. Therefore, ${}_3V_x^R=\$7{,}789.50/0.54=\$14{,}425.00$.

Check. The terminal benefit is excluded because the reserve is measured immediately before it is paid. Including $0.54(\$7{,}000)$ as a past outflow would instead value the cash flows immediately after that payment. The valuation boundary must match the cash flows included.

Common Errors and Practical Checks

Retrospective calculations are especially sensitive to timing. A reliable solution identifies the valuation instant first, then lists every cash flow before that instant. The accumulation factor follows from the time between the cash flow and valuation; it is not determined by whether the cash flow is a premium or a benefit.

Forgetting survival weights. A premium due at time $k$ is received only if the policy is in force then. Weight it by ${}_kp_x$; a claim in year $k+1$ is weighted by ${}_kp_xq_{x+k}$.
Using the wrong accumulation period. A cash flow paid at the valuation time accumulates for zero years. A premium paid one year earlier accumulates for one year.
Using the wrong denominator. Divide by ${}_tp_x$, the probability of survival to the stated duration, not by a one-year survival probability unless $t=1$.
Counting a boundary payment on the wrong side. State whether the reserve is immediately before or after a premium or benefit due at time $t$. Include the payment only if it has occurred by the valuation instant.
Mixing mortality assumptions. For changing mortality, use the $q$ for the age and policy year in which the benefit is triggered. Survival to that year must also be included.

The same structure applies to a broader set of contractual cash flows: add each expected premium or other inflow, subtract each expected benefit or expense, accumulate each to the same date, and divide by the probability of remaining in force if the reserve is stated per surviving policy. The formula must be adjusted whenever payment timing or in-force conditions differ from the annual assumptions used above.

Key takeaway: A retrospective reserve is a cohort cash-flow calculation expressed per survivor. Correct probability weights, interest accumulation, and valuation-date timing are all essential; a correct arithmetic total with an incorrect time boundary is still the wrong reserve.

Check Your Understanding

For each problem, identify the relevant survival and claim probabilities before accumulating cash flows. Round money to the nearest cent.

A one-year term policy has a $2,000 death benefit paid at year-end, a $300 premium at time 0, $i=4\%$, and $q_x=0.10$. Find the retrospective reserve at time 1. (Answer: $124.44)
A two-year term policy pays $3,000 at the end of the year of death. It collects $500 at times 0 and 1 while in force. Given $i=5\%$, $q_x=0.20$, and $q_{x+1}=0.10$, find the reserve at time 2. (Answer: $140.63)
A two-year policy collects $1,000 at time 0 and $1,500 at time 1, if in force. It pays $6,000 at the end of the year of death. Given $i=4\%$, $q_x=0.05$, and $q_{x+1}=0.15$, calculate the reserve at time 2. (Answer: $1,729.54)
A two-year endowment collects $1,000 at times 0 and 1 while in force, pays $4,000 at the end of the year of death, and pays $5,000 at time 2 on survival. Assume $i=0$, $q_x=0.10$, and $q_{x+1}=0.20$. Find the reserve immediately before the maturity payment. (Answer: $1,083.33)
A two-year term policy pays $10,000 at the end of the year of death. It collects an unknown level premium $P$ at times 0 and 1 while in force. Given $i=5\%$, $q_x=q_{x+1}=0.10$, and a reserve of $1,000 at time 2, determine $P$. (Answer: $1,347.99)

What You'll Learn