Retirement Pension Values with Competing Exits

Valuing a Pension Across Competing Exit Dates

An active member may retire at different eligible durations, or leave first through another cause such as withdrawal or death. The expected present value of a retirement pension must allow for both possibilities: the probability the member remains active until each eligible retirement date and the pension value if retirement occurs then. A retirement probability for a particular year is not, by itself, the probability that an active member at the valuation date retires in that year.

This tutorial uses a discrete annual model. Decrements occur at the ends of policy years, and the cause probabilities for a year are conditional on being active at its start. Retirement benefits are valued at the retirement date, then discounted to time 0. The pension value at each retirement date is treated as a supplied valuation input; its construction depends on the pension payment terms and the post-retirement valuation basis.

Definition: Let $A_k$ be the probability that a member is active immediately after the decrements at the end of year $k$, with $A_0=1$. In year $k$, let $r_k$, $w_k$, and $d_k$ be the conditional probabilities of retirement, withdrawal, and death, respectively, among members active at the start of that year. Assume these causes are mutually exclusive and exhaustive together with remaining active. Then $$ A_k=A_{k-1}(1-r_k-w_k-d_k), $$ and the unconditional probability of retirement at the end of year $k$ is $A_{k-1}r_k$. If retirement is not permitted in year $k$, set $r_k=0$.

Write $B_k$ for the annual pension amount established at retirement in year $k$, and $g_k$ for the value at that retirement date of a pension of 1 dollar per year, on the stated pension valuation basis. Thus $B_kg_k$ is the pension’s value at the retirement date. For example, if the pension is payable in advance, $g_k$ may be an annuity-due factor; the factor must match the stated payment and survival assumptions. It is an input here rather than a factor derived from active-member exit probabilities.

Theorem: If retirement at the end of year $k$ creates pension value $B_kg_k$, and the effective annual interest rate is $i$, then the expected present value of the retirement pension is $$ \operatorname{EPV}=\sum_{k=1}^{n}v^k A_{k-1}r_kB_kg_k, \qquad v=\frac{1}{1+i}. $$ The sum includes only the durations being considered. A term is zero when retirement is ineligible at that duration.

Why the Valuation Formula Works

For each eligible year $k$, define the event $R_k$ as retirement at the end of year $k$. To retire then, the member must be active at the start of year $k$, an event with probability $A_{k-1}$, and retirement must be the decrement during that year, with conditional probability $r_k$. Therefore $P(R_k)=A_{k-1}r_k$. On $R_k$, the pension value at time $k$ is $B_kg_k$, so its present value at time 0 is $v^kB_kg_k$. If $R_k$ does not occur, this term contributes zero. Its expected contribution is consequently $v^kB_kg_kP(R_k)$. The retirement events for distinct years are disjoint, so adding these expected contributions gives the formula.

This argument explains why the starting active probability matters. The conditional probability $r_k$ describes only members active at the beginning of year $k$. Multiplying it by $A_{k-1}$ converts it into the probability, from the original valuation date, of retirement in that year. The formula assumes annual decrements are represented by the stated mutually exclusive cause probabilities and that the pension value is measured at the same year-end as retirement. If benefits are paid at another time, the discount timing must be changed accordingly.

Set the eligibility window.
Use a zero retirement probability in any year when the member is not eligible, while retaining any withdrawal or death decrements that can occur.

Find active probabilities.
Starting from $A_0=1$, multiply each year’s starting active probability by its probability of remaining active through that year.

Find retirement probabilities from time 0.
For each eligible year, calculate $A_{k-1}r_k$, not just $r_k$.

Value and discount each pension.
Multiply the retirement probability by the pension value $B_kg_k$ at that date, then by $v^k$.

Add the contributions.
The sum is the expected present value of the retirement pension. Other exit benefits are separate cash flows and are not included in this sum.

Worked Applications

Worked Example: Two Eligible Retirement Dates

A hypothetical plan member is ineligible to retire at the end of year 1. In that year, the conditional withdrawal probability is 0.10 and the conditional death probability is 0.05. At the end of years 2 and 3, the conditional retirement probabilities are 0.20 and 0.30, respectively. In each of those years, withdrawal has probability 0.10 and death has probability 0.05. The annual pension is $12,000 whichever eligible date retirement occurs. The pension value factor at either retirement date is 8.5 per dollar of annual pension. The effective annual interest rate is 5%. Find the expected present value at time 0.

Method. Carry the active probability through each year, then weight each pension value by the unconditional probability of retirement at that year. This avoids treating the conditional retirement rates as probabilities from time 0.

In year 1, retirement is not eligible. The probability of remaining active is $1-0.10-0.05=0.85$, so $A_1=0.85$.

In year 2, the probability of retirement from time 0 is $A_1r_2=0.85(0.20)=0.17$. The pension value at retirement is $12{,}000(8.5)=\$102{,}000$. The member remains active after year 2 with probability $$ A_2=0.85(1-0.20-0.10-0.05)=0.85(0.65)=0.5525. $$

In year 3, the probability of retirement from time 0 is $A_2r_3=0.5525(0.30)=0.16575$. The pension value at retirement is again $102,000. Since $v=1/1.05$, the two present-value contributions are $$ v^2(0.17)(102{,}000)=\frac{17{,}340}{1.05^2}=\$15{,}727.89, $$ and $$ v^3(0.16575)(102{,}000)=\frac{16{,}906.50}{1.05^3}=\$14{,}604.47. $$

Adding the contributions gives an expected present value of $\$15{,}727.89+\$14{,}604.47=\boxed{\$30{,}332.36}$. Amounts are rounded to the nearest cent only at the final step. As a check, the probability of retirement in either eligible year is $0.17+0.16575=0.33575$, less than 1; some members leave through other causes or remain active past year 3.

Common mistake avoided: Multiplying the year 3 retirement probability 0.30 directly by the year 3 pension value ignores that only 0.5525 of the original members are active at the start of year 3.

Worked Example: Changing Pension Values by Retirement Date

A hypothetical member cannot retire in year 1. In that year, the conditional withdrawal probability is 0.08 and the death probability is 0.02. Retirement is permitted in years 2 and 3. In year 2, the conditional probabilities of retirement, withdrawal, and death are $2/9$, 0.30, and 0.20. In year 3, they are 0.40, 0.10, and 0.05. The pension values at the retirement dates, after applying the appropriate pension factors, are $100,000 in year 2 and $110,000 in year 3. At an effective annual interest rate of 10%, calculate the expected present value.

Method. First obtain the unconditional probabilities of retirement. Then apply the time-specific pension values and year-specific discount factors; the pension values need not be equal across retirement dates.

After year 1, the active probability is $A_1=1-0.08-0.02=0.90$. The year 2 retirement probability from time 0 is $$ A_1r_2=0.90\left(\frac{2}{9}\right)=0.20. $$ The probability of remaining active through year 2 is $1-2/9-0.30-0.20=5/18$, so $A_2=0.90(5/18)=0.25$. Therefore the year 3 retirement probability from time 0 is $A_2r_3=0.25(0.40)=0.10$.

The year 2 contribution is $$ \frac{0.20(100{,}000)}{1.10^2} =\frac{20{,}000}{1.21} =\$16{,}528.93. $$ The year 3 contribution is $$ \frac{0.10(110{,}000)}{1.10^3} =\frac{11{,}000}{1.331} =\$8{,}264.46. $$ Thus the expected present value is $\$16{,}528.93+\$8{,}264.46=\boxed{\$24{,}793.39}$, rounded to the nearest cent. The year 3 value is higher at retirement, but its contribution is smaller because fewer members reach that retirement date and the payment is discounted for longer.

Common mistake avoided: Applying the year 2 pension value to both terms, or discounting both terms by $v^2$, would ignore the stated change in pension value or the actual retirement date.

Worked Example: An Eligibility Window and Three Pension Factors

A hypothetical plan permits retirement only at the ends of years 2, 3, and 4. In year 1, conditional withdrawal and death probabilities are 0.10 and 0.05. For years 2, 3, and 4, respectively, the conditional retirement, withdrawal, and death probabilities are $(0.20,0.10,0.05)$, $(0.25,0.05,0.10)$, and $(0.40,0.10,0.05)$. The annual pensions and pension value factors at the three dates are, respectively, ($8,000, 9.5), ($9,000, 9.0), and ($10,000, 8.4). The effective annual interest rate is 5%. Find the expected present value.

Method. Use the eligibility window to identify the retirement terms, carry active probabilities year by year, and calculate a separate retirement-date value for each pension.

At the end of year 1, $A_1=1-0.10-0.05=0.85$. Thus the year 2 retirement probability is $0.85(0.20)=0.17$, and the active probability after year 2 is $A_2=0.85(0.65)=0.5525$. The year 3 retirement probability is $0.5525(0.25)=0.138125$, and the active probability after year 3 is $$ A_3=0.5525(1-0.25-0.05-0.10)=0.5525(0.60)=0.3315. $$ The year 4 retirement probability is $0.3315(0.40)=0.1326$.

The pension values at retirement are $8{,}000(9.5)=\$76{,}000$, $9{,}000(9.0)=\$81{,}000$, and $10{,}000(8.4)=\$84{,}000$. The discounted contributions are $$ \frac{0.17(76{,}000)}{1.05^2} =\frac{12{,}920}{1.1025} =\$11{,}718.82, $$ $$ \frac{0.138125(81{,}000)}{1.05^3} =\frac{11{,}188.125}{1.157625} =\$9{,}664.72, $$ and $$ \frac{0.1326(84{,}000)}{1.05^4} =\frac{11{,}138.40}{1.21550625} =\$9{,}163.59. $$ The expected present value is therefore $\boxed{\$30{,}547.13}$, rounded to the nearest cent. The three unconditional retirement probabilities sum to $0.440725$; the remaining probability accounts for earlier competing exits and members still active after year 4.

Common mistake avoided: Retirement is not eligible in year 1, but withdrawal and death still reduce the population that can retire later. Setting all year 1 decrements to zero would incorrectly overstate each later retirement probability.

Interpretation and Exam Traps

The result is an expected present value from the original valuation date. It is not the pension value conditional on retirement, nor is it the pension value for a member known to remain active to a particular duration. The distinction is visible in each term: $B_kg_k$ is the value conditional on retirement at year $k$, while $A_{k-1}r_k$ is the probability of reaching that event from time 0.

Use the correct starting population. Multiply each conditional retirement probability by the active probability at the start of that year.
Respect eligibility. Set retirement probability to zero before the eligibility window begins or after it ends, but continue to account for other decrements.
Discount from the valuation date to the benefit valuation date. If retirement is at the end of year $k$, use $v^k$ for a value measured at that date.
Match the pension factor to the stated pension. The factor must correspond to its payment pattern and valuation basis; do not substitute an active-member survival probability for a post-retirement pension factor.
Keep distinct cash flows distinct. This sum values retirement pensions only. A withdrawal refund, death benefit, or other exit payment requires its own cause-specific valuation.

These calculations are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In a longer valuation, the same structure applies when the retirement amount, eligibility, or pension factor varies by duration: specify the inputs at each date, calculate the active probabilities consistently, and sum the discounted date-specific contributions.

Check Your Understanding

For each problem, calculate an expected present value from time 0. Treat the stated pension values as values at the retirement dates, and round dollar answers to the nearest cent.

A hypothetical member may retire at the ends of years 1 and 2. In year 1, the conditional retirement, withdrawal, and death probabilities are 0.15, 0.05, and 0.02. In year 2, they are 0.20, 0.10, and 0.05. The pension values are $40,000 and $42,000, respectively, and $i=4\%$. (Answer: $11,826.92)
Retirement is not eligible in year 1, when withdrawal and death probabilities are 0.12 and 0.08. At the end of year 2, the conditional retirement probability is 0.30, and the pension value is $50,000. At $i=5\%$, find the retirement pension EPV. (Answer: $10,884.35)
A member can retire at the ends of years 1 and 2. In year 1, the conditional retirement, withdrawal, and death probabilities are 0.10, 0.10, and 0.05. In year 2, the conditional retirement probability is 0.20. Pension values are $20,000 and $24,000, respectively, and $i=10\%$. Find the EPV. (Answer: $4,793.39)
Retirement is permitted at the ends of years 2 and 3. In year 1, withdrawal and death probabilities are 0.06 and 0.04. In year 2, the retirement, withdrawal, and death probabilities are 0.25, 0.10, and 0.05; in year 3, the retirement probability is 0.20. The pension values are $60,000 and $66,000, respectively, and $i=5\%$. Find the EPV. (Answer: $18,402.33)

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