Eligibility Windows and Retirement Benefits

Why Eligibility Dates Matter

A retirement benefit may be available only at particular ages, after a required service period, or during a specified window. A member who leaves active service before such a date cannot receive that retirement benefit under the model. Even if retirement is permitted on a date, the member must first remain active until that date and then retire rather than experience another exit.

The valuation therefore has two distinct probability steps: the probability of reaching an eligible date in active service, and the probability of retiring at that date conditional on being active at its start. Multiplying these probabilities gives the unconditional probability of a retirement payment. The amount payable can differ across eligibility dates, and each amount must be discounted from its own payment date.

Definition: Consider discrete periods $t=1,2,\ldots,n$, with each period ending at an annual valuation date. Let $q_{j,t}$ be the probability that an active member exits during period $t$ for cause $j$, conditional on being active at the start of that period. The causes are mutually exclusive. Let $q_{R,t}$ be the conditional probability of retirement and $q_{A,t}$ the probability of any other exit. Retirement is prohibited in period $t$ when $q_{R,t}=0$. The probability of remaining active through period $t$, conditional on being active at its start, is $p_t^A=1-q_{R,t}-q_{A,t}$. If the retirement benefit $B_t$ is paid at the end of period $t$, and $v=1/(1+i)$, its expected present value is calculated using the probability of reaching that date active, the conditional retirement probability, and $v^tB_t$.

Here $q_{A,t}$ can be the sum of several mutually exclusive competing-exit probabilities, such as death and withdrawal. All probabilities for a period are conditional on being active at the start of that period and must sum to no more than one. The probability of remaining active is the remainder. Once a member exits, the member does not return to active service in this model.

Let $s_t^A$ denote the probability of being active at the start of period $t$, for a member active at time 0. Then $s_1^A=1$, and $s_{t+1}^A=s_t^A p_t^A$. If a retirement benefit is payable at the end of period $t$, the unconditional probability of receiving it is $s_t^Aq_{R,t}$. Consequently, for benefits paid at period-end, the expected present value is

\operatorname{EPV}=\sum_{t=1}^{n}v^t B_t s_t^A q_{R,t}, \qquad s_t^A=\prod_{k=1}^{t-1}p_k^A.

The sum includes only periods in which retirement benefits are available; equivalently, prohibited periods have $q_{R,t}=0$. A benefit payable immediately on retirement, at the start of a period, or on a different schedule would require a different discount exponent. The formula above assumes period-end payment.

How the Eligibility-Window Formula Arises

The valuation formula follows from conditioning on the date of retirement. For a payment at the end of period $t$, the member must remain active through every preceding period, then retire during period $t$. The probability of the first event is $s_t^A$. By the definition of the conditional decrement probability, the probability of retirement during period $t$, given active status at its start, is $q_{R,t}$. Thus the unconditional probability of retirement in period $t$ is $s_t^Aq_{R,t}$.

The present value of $B_t$ paid at the end of period $t$ is $v^tB_t$. Multiplying this amount by its probability gives the expected present value contribution for that date. Retirement at different dates consists of mutually exclusive outcomes, so adding the contributions gives the total expected present value. This argument uses conditional probabilities; it does not require the exits in different periods to be independent.

Mark the permitted dates.
Set $q_{R,t}=0$ in every period when retirement is not allowed. Record the retirement probability conditional on being active for each permitted period.

Track active status.
For each period, subtract all mutually exclusive exits, including retirement when permitted, to obtain $p_t^A$. Multiply the preceding active probability by $p_t^A$ to find the next period's active probability.

Value each retirement date.
Multiply the active probability at the start of the period by its conditional retirement probability and the benefit at that date. Discount to the valuation date using the stated payment timing.

Add the date-specific values.
Sum the contributions from all permitted dates. Check that the active probability never increases and that the cause probabilities in each period are valid.

Worked Valuations

Worked Example: Two Eligibility Dates with Earlier Competing Exits

A hypothetical plan pays a lump sum at the end of year 2 or year 4, but not at other dates. In years 1 and 3, retirement is prohibited; death and withdrawal probabilities are 0.03 and 0.07. In years 2 and 4, the conditional probabilities of retirement, death, and withdrawal are 0.20, 0.03, and 0.07. The benefit is $5,000 at year 2 and $9,000 at year 4. Assume $i=0$. Find the expected present value.

Method. First calculate the conditional probability of remaining active in each period, then multiply the active-at-start probability by the retirement probability at each permitted date. With $i=0$, $v=1$, so no discount adjustment is needed.

In year 1, retirement is prohibited and the other exits total $0.03+0.07=0.10$. Thus $p_1^A=1-0.10=0.90$, so $s_2^A=0.90$. The unconditional probability of retirement at the end of year 2 is $0.90(0.20)=0.18$. Its expected payment is $0.18(\$5{,}000)=\$900.00$.

In year 2, the probability of remaining active after the competing retirement, death, and withdrawal exits is $1-0.20-0.03-0.07=0.70$. Therefore $s_3^A=0.90(0.70)=0.63$. Retirement is prohibited in year 3, so $p_3^A=1-0.03-0.07=0.90$, giving $s_4^A=0.63(0.90)=0.567$. The unconditional probability of retirement at the end of year 4 is $0.567(0.20)=0.1134$. Its expected payment is $0.1134(\$9{,}000)=\$1{,}020.60$.

Adding the date-specific expected payments gives $\$900.00+\$1{,}020.60=\boxed{\$1{,}920.60}$. The result is checked by noting that the probability of reaching year 4 active, $0.567$, is below the probability reaching year 2 active, $0.90$, as required.

Common mistake avoided: Applying the year-4 retirement probability to the original member population would ignore the deaths, withdrawals, and year-2 retirement that can occur beforehand.

Worked Example: Retirement Available Only at the End of Year 3

A hypothetical pension permits retirement only at the end of year 3. In year 1, the conditional death and withdrawal probabilities are 0.05 and 0.15. In year 2, they are 0.04 and 0.16. At the end of year 3, the conditional probabilities of retirement, death, and withdrawal are 0.25, 0.05, and 0.10. The retirement benefit is $20,000, payable at year-end. Assume an effective annual interest rate $i=10\%$. Find its expected present value.

Method. Calculate the probability of remaining active through years 1 and 2. Multiply that probability by the conditional probability of retirement in year 3, then discount the year-3 payment by $v^3$.

Retirement is prohibited in years 1 and 2. The active probabilities for those periods are $p_1^A=1-0.05-0.15=0.80$ and $p_2^A=1-0.04-0.16=0.80$. Hence the probability of being active at the start of year 3 is $s_3^A=0.80(0.80)=0.64$. The unconditional probability of retirement in year 3 is $0.64(0.25)=0.16$.

The expected year-3 payment before discounting is $0.16(\$20{,}000)=\$3{,}200$. Since $v=1/1.10$, the present value is

\operatorname{EPV}=\frac{\$3{,}200}{(1.10)^3} =\frac{\$3{,}200}{1.331} =\$2{,}404.207\ldots

Rounded to the nearest cent, the expected present value is $\boxed{\$2{,}404.21}$. As a check, the probability of retirement is 0.16, not 0.25: the latter is conditional on reaching year 3 active.

Common mistake avoided: Discounting the payment for only one year because the retirement probability is specified in the year-3 decrement table. The payment is made at the end of year 3, so the correct factor is $v^3$.

Worked Example: A Window That Opens After a Service Requirement

A hypothetical plan allows retirement only at the ends of years 3 and 4. Retirement is prohibited in years 1 and 2. In each of those first two years, the death probability is 0.02 and the withdrawal probability is 0.08. At the end of year 3, the conditional probabilities of retirement, death, and withdrawal are 0.25, 0.03, and 0.07. At the end of year 4, they are 0.40, 0.02, and 0.08. The benefits are $7,000 at year 3 and $10,000 at year 4. Assume $i=0$. Find the expected present value.

Method. The two ineligible years determine the probability of reaching the start of the eligibility window. Then account for the year-3 retirement and other exits before calculating the year-4 contribution.

In years 1 and 2, the probability of remaining active is $1-0.02-0.08=0.90$ in each year. Thus the probability of being active at the start of year 3 is $0.90(0.90)=0.81$. The unconditional probability of retirement at the end of year 3 is $0.81(0.25)=0.2025$, giving an expected payment of $0.2025(\$7{,}000)=\$1{,}417.50$.

If active at the start of year 3, the probability of remaining active through that year is $1-0.25-0.03-0.07=0.65$. Therefore the probability of being active at the start of year 4 is $0.81(0.65)=0.5265$. The unconditional probability of retirement in year 4 is $0.5265(0.40)=0.2106$. The expected year-4 payment is $0.2106(\$10{,}000)=\$2{,}106.00$.

With $i=0$, both payments have discount factor 1. The total is $\$1{,}417.50+\$2{,}106.00=\boxed{\$3{,}523.50}$. The probabilities of retirement in years 3 and 4 sum to $0.4131$, which is less than the probability $0.81$ of reaching the eligibility window, providing a useful probability check.

Common mistake avoided: Using $0.65$ as the probability of reaching year 4 active. That is conditional on being active at the start of year 3; the unconditional probability is $0.81(0.65)=0.5265$.

Common Errors and Practical Checks

Eligibility rules should be translated into dates before doing any arithmetic. A service requirement may mean a benefit first becomes available at a particular year-end, while an age rule may allow retirement at several dates. In either case, distinguish the date on which the member is eligible from the period over which a probability is conditional. A conditional retirement probability applies only to members active at the start of that period.

A frequent exam error is to value each benefit as though every original member could reach every permitted date. For example, using $q_{R,t}B_t$ instead of $s_t^Aq_{R,t}B_t$ omits the probability of surviving all earlier competing exits in active service. Another error is to omit retirement itself when calculating the probability of remaining active after an eligible date. Members who retire at an earlier permitted date are no longer active and cannot receive a later retirement benefit in this model.

Before finalizing a valuation, check that all period-specific probabilities are conditional on active status at the period start, that prohibited dates have retirement probability zero, and that each active probability is updated using every applicable exit. Confirm that each payment is discounted to the valuation date according to its actual timing. These checks are especially important when benefit amounts vary across the eligibility window.

Key takeaway: For a retirement benefit payable at the end of an eligible period $t$, use the probability of being active at its start multiplied by the conditional probability of retirement in that period. Multiply by the date-specific benefit and its period-$t$ discount factor, then sum over eligible dates.

Check Your Understanding

For each problem, treat the stated decrement probabilities as conditional on being active at the start of the period, and assume benefits are paid at period-end.

A hypothetical plan permits retirement only at the end of year 2. The probability of remaining active in year 1 is 0.85; the conditional retirement probability in year 2 is 0.30. The benefit is $8,000 and $i=0$. Find the expected present value. (Answer: $2,040.00)
Retirement is permitted at the ends of years 1 and 3. In year 1, retirement is 0.10 and other exits total 0.20. Year 2 retirement is prohibited and other exits total 0.15. In year 3, the conditional retirement probability is 0.20. The benefits are $3,000 and $6,000 at years 1 and 3, respectively, and $i=5\%$. Find the expected present value. (Answer: $902.49)
A hypothetical member may retire at the ends of years 2 and 3 only. The probability of remaining active in year 1 is 0.90. In year 2, retirement is 0.15, death is 0.05, and withdrawal is 0.10. In year 3, retirement is 0.25. Benefits are $4,000 and $7,000 at years 2 and 3. Assume $i=0$. Find the expected present value. (Answer: $1,642.50)
Retirement is prohibited until the end of year 4. The probabilities of remaining active in years 1, 2, and 3 are 0.95, 0.90, and 0.80. At the end of year 4, the conditional retirement probability is 0.30 and the benefit is $15,000. Assume $i=5\%$. Find the expected present value. (Answer: $2,532.28)
A hypothetical plan allows retirement at the ends of years 2, 3, and 4. The active probability in year 1 is 0.90. In years 2, 3, and 4, respectively, the conditional retirement probabilities are 0.20, 0.25, and 0.30; the probabilities of other exits are 0.10, 0.15, and 0.20. Benefits at those dates are $4,000, $5,000, and $6,000. Assume $i=0$. Find the expected present value. (Answer: $2,187.90)

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