Discrete EPV of Retirement Lump Sums

Why retirement timing matters

A retirement lump sum is paid only if a member is still active when an eligible retirement occurs. Its value therefore depends on more than the stated benefit amount: the member must remain active through earlier years, retire in an eligible year, and receive the payment at the specified time. Competing exits such as withdrawal or death can prevent a retirement payment from being made.

This tutorial develops the discrete expected present value (EPV) for a lump sum paid at the end of the year of retirement. The calculation uses annual probabilities conditional on active membership at the beginning of each year. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Suppose a member is active at time 0 and time is measured in years. Let $q_k^{(r)}$ be the probability of retirement during year $k+1$, conditional on being active at the start of that year. Let $q_k^{(j)}$ denote the corresponding conditional probability of exit from any other competing cause $j$. The probability of remaining active throughout year $k+1$, conditional on being active at its start, is $$ p_k^{(a)}=1-q_k^{(r)}-\sum_{j\ne r}q_k^{(j)}. $$ The probability of being active at time $k$ is $$ {}_kp^{(a)}=\prod_{h=0}^{k-1}p_h^{(a)}, \qquad {}_0p^{(a)}=1. $$

The cause probabilities for a given year are mutually exclusive: a member cannot retire and withdraw during the same year in this model. Their sum cannot exceed 1. The annual probabilities may vary by duration, age, or other factors represented in the valuation basis. The formulas below require that the probabilities used for each year are conditional on being active at that year's start.

Discrete EPV for an eligible retirement lump sum

Index years by $k=0,1,2,\ldots$. Year $k+1$ runs from time $k$ to time $k+1$. Let $B_{k+1}$ be the lump sum payable at the end of year $k+1$ if retirement occurs in that year, and let $e_{k+1}$ equal 1 if retirement in that year satisfies all benefit eligibility conditions and 0 otherwise. With effective annual interest rate $i$, the discount factor is $v=1/(1+i)$.

Theorem: If the lump sum is paid at the end of the year of eligible retirement, its EPV at time 0 is $$ \operatorname{EPV} =\sum_{k=0}^{n-1} v^{k+1}\,{}_kp^{(a)}\,q_k^{(r)}\,e_{k+1}B_{k+1}. $$ A term contributes only when retirement is eligible in that year. If the benefit amount itself is defined as zero in an ineligible year, the indicator $e_{k+1}$ may instead be absorbed into $B_{k+1}$.

The expression ${}_kp^{(a)}q_k^{(r)}$ is the probability that the member is active at the start of year $k+1$ and retires during that year. It is an unconditional probability measured from time 0. Multiplying it by the applicable benefit gives the expected payment at time $k+1$; multiplying by $v^{k+1}$ discounts that expected payment to time 0.

Why the probability weight has this form

The calculation follows from conditioning on whether the member is active at the beginning of the retirement year. Retirement during year $k+1$ requires both that the member has remained active for the first $k$ years and that retirement is then the cause of exit in the next year. The second probability is conditional on the first event, so the multiplication rule gives

P(\text{retirement in year } k+1) =P(\text{active at time }k) P(\text{retirement in year } k+1\mid \text{active at time }k) ={}_kp^{(a)}q_k^{(r)}.

For each eligible year, the present-value random variable equals $v^{k+1}B_{k+1}$ if retirement occurs in that year and equals zero otherwise. Its expectation is therefore $v^{k+1}B_{k+1}{}_kp^{(a)}q_k^{(r)}$. The events of retirement in different years are mutually exclusive, so adding their expected present values gives the stated sum. This reasoning relies on using the probability of retirement conditional on being active at the beginning of the year; it does not require the annual cause probabilities to be constant across years.

Mark the eligibility years.
Apply every eligibility condition to the retirement date, not merely to the date at which the valuation is made.

Find active-member survival.
For each year before a possible retirement, multiply the probabilities of remaining active through the preceding years.

Find each retirement probability.
Multiply the probability of being active at the year’s start by the retirement probability conditional on being active then.

Attach the benefit and discount.
Use the lump sum applicable to that retirement year and discount to the actual payment date.

Add the eligible terms.
Exclude ineligible retirement years and sum the present values of the remaining expected payments.

Worked calculations

Worked Example: A Lump Sum in the Final Two Eligible Years

A hypothetical pension plan pays a lump sum at the end of the year of retirement. Retirement is eligible only in years 2 and 3. The benefits are $12,000 for retirement in year 2 and $14,000 for retirement in year 3. The effective annual interest rate is $i=5\%$. The annual probabilities, conditional on being active at the start of the year, are:

Year 1: retirement $0.10$, withdrawal $0.08$, death $0.02$.
Year 2: retirement $0.15$, withdrawal $0.05$, death $0.02$.
Year 3: retirement $0.20$, withdrawal $0.04$, death $0.02$.

Method. Calculate active survival to the start of each eligible year, multiply by that year's conditional retirement probability, and discount each eligible payment to time 0.

In year 1, the probability of remaining active is $p_0^{(a)}=1-0.10-0.08-0.02=0.80$. Thus the probability of being active at the start of year 2 is ${}_1p^{(a)}=0.80$.

In year 2, the probability of remaining active is $p_1^{(a)}=1-0.15-0.05-0.02=0.78$. Thus the probability of being active at the start of year 3 is ${}_2p^{(a)}=0.80(0.78)=0.624$. The probabilities of eligible retirement are consequently $0.80(0.15)=0.12$ in year 2 and $0.624(0.20)=0.1248$ in year 3.

Here $v=1/1.05$. The year 2 contribution is $12{,}000(0.12)v^2=1{,}440/1.1025=\$1{,}306.12$. The year 3 contribution is $14{,}000(0.1248)v^3=1{,}747.20/1.157625=\$1{,}509.30$. Adding the contributions gives $\operatorname{EPV}=\$1{,}306.12+\$1{,}509.30=\$2{,}815.42$, to the nearest cent.

Check. The probability of retirement in either eligible year is $0.12+0.1248=0.2448$, which is less than 1. Both payment amounts are discounted for their actual payment years, and no year 1 retirement benefit is included because that year is ineligible.

Common mistake avoided. The year 3 retirement probability is not just $0.20$. That is conditional on being active at the start of year 3; from time 0, the probability is $0.624(0.20)=0.1248$.

Worked Example: Eligibility and a Benefit Based on Service

A hypothetical plan pays a retirement lump sum at the end of the retirement year. A member becomes eligible after completing one full year of active service, so only retirements in years 2 and 3 qualify. The benefit is $18,000 for retirement in year 2 and $21,000 for retirement in year 3. The effective annual interest rate is $4\%$. Conditional annual exit probabilities are:

Year 1: retirement $0.05$, withdrawal $0.10$, death $0.01$.
Year 2: retirement $0.12$, withdrawal $0.08$, death $0.02$.
Year 3: retirement $0.10$, withdrawal $0.06$, death $0.02$.

Method. First translate the service condition into eligible retirement years. Then calculate active survival and value only the eligible retirement payments.

The year 1 probability of remaining active is $1-0.05-0.10-0.01=0.84$, so ${}_1p^{(a)}=0.84$. The year 2 probability of remaining active is $1-0.12-0.08-0.02=0.78$, so ${}_2p^{(a)}=0.84(0.78)=0.6552$. The probabilities of eligible retirement are $0.84(0.12)=0.1008$ in year 2 and $0.6552(0.10)=0.06552$ in year 3.

For year 2, the present value of the expected payment is $18{,}000(0.1008)/(1.04)^2=1{,}814.40/1.0816=\$1{,}677.51$. For year 3, it is $21{,}000(0.06552)/(1.04)^3=1{,}375.92/1.124864=\$1{,}223.19$. Therefore, $\operatorname{EPV}=\$1{,}677.51+\$1{,}223.19=\$2{,}900.70$, to the nearest cent.

Check. The member must survive as an active member through year 1 to qualify for a year 2 retirement benefit. The year 3 term includes the additional requirement of remaining active through year 2.

Common mistake avoided. Eligibility is determined at the retirement date. Treating a year 1 retirement as eligible simply because the member is projected to complete a year of service by the end of that year would contradict the stated rule that eligibility begins after completing one full year.

Worked Example: An Escalating Lump Sum Over an Eligibility Window

A hypothetical plan permits retirement with a lump sum in years 2 through 4, inclusive. The lump sum for retirement in year $k$ is $B_k=\$10{,}000(1.03)^{k-1}$, paid at the end of year $k$. The effective annual interest rate is $3\%$. Conditional probabilities of retirement, withdrawal, and death are:

Year 1: $0.08,\ 0.12,\ 0.02$, respectively.
Year 2: $0.10,\ 0.08,\ 0.02$, respectively.
Year 3: $0.15,\ 0.05,\ 0.02$, respectively.
Year 4: $0.20,\ 0.04,\ 0.01$, respectively.

Method. Calculate the active-member probability at the start of each eligible year and weight the year-specific retirement benefit by the corresponding conditional retirement probability.

The probabilities of remaining active are $0.78$ in year 1, $0.80$ in year 2, and $0.78$ in year 3. Therefore, ${}_1p^{(a)}=0.78$, ${}_2p^{(a)}=0.78(0.80)=0.624$, and ${}_3p^{(a)}=0.624(0.78)=0.48672$. The probabilities of eligible retirement are $0.78(0.10)=0.078$ in year 2, $0.624(0.15)=0.0936$ in year 3, and $0.48672(0.20)=0.097344$ in year 4.

Each discounted benefit has the same value before its retirement probability is applied: $$ v^k B_k =\frac{10{,}000(1.03)^{k-1}}{(1.03)^k} =\frac{10{,}000}{1.03} =\$9{,}708.7379\ldots. $$ Thus the total probability weight is $0.078+0.0936+0.097344=0.268944$, and $\operatorname{EPV}=(10{,}000/1.03)(0.268944)=\$2{,}611.11$, to the nearest cent.

Check. The escalation rate equals the interest rate, so the discounted amount for each eligible year is constant. The year 4 active probability includes survival through years 1, 2, and 3, but not through year 4; retirement itself occurs during year 4.

Common mistake avoided. Do not multiply the year 4 retirement probability by the probability of remaining active through year 4. The member needs to be active at the start of year 4 and then retire during that year.

Timing and conditioning traps

The most frequent errors are about what a probability is conditional on and when the payment is made. The following incorrect calculation illustrates both the danger of using a conditional probability as if it were unconditional and the importance of matching the discount exponent to the payment date.

Suppose a member has probability $0.80$ of remaining active in year 1, and the year 2 retirement probability is $0.15$ conditional on being active at the start of year 2. A $12,000 benefit is paid at the end of year 2, and $i=5\%$. The incorrect calculation $12{,}000(0.15)v$ treats the conditional year 2 retirement probability as the probability from time 0 and discounts only one year. It gives $12{,}000(0.15)/1.05=\$1{,}714.29$.

The correct probability of retirement in year 2 from time 0 is $0.80(0.15)=0.12$. The payment is at time 2, so the correct EPV is $12{,}000(0.12)v^2=1{,}440/1.1025=\$1{,}306.12$. The incorrect calculation overstates the value because it omits both the chance of an earlier competing exit and one year of discounting.

Before finalizing a calculation, check that the probabilities used in each term answer the same question: “What is the probability, from time 0, that this eligible retirement happens in this year?” Also confirm whether the benefit is paid at retirement, at year-end, or at another stated date. If payment occurs at a time other than the end of the retirement year, the discount factor must reflect that payment date; the displayed theorem does not apply unchanged.

Check Your Understanding

For each problem, use the stated payment timing and annual conditional probabilities. Amounts are in dollars, and final answers are rounded to the nearest cent.

A hypothetical plan pays $15,000 at the end of year 2 if retirement occurs in that year. The effective annual interest rate is $5\%$. In year 1, conditional on active membership, retirement is $0.10$ and other exits total $0.10$. In year 2, retirement is $0.20$. Calculate the EPV. (Answer: $2,176.87)
Retirement is eligible in years 1 and 2, and the benefit is $10,000 in either year, payable at year-end. The effective annual interest rate is $0\%$. In year 1, retirement has probability $0.10$ and other exits total $0.20$; in year 2, retirement has probability $0.25$. Calculate the EPV. (Answer: $2,750.00)
A $20,000 lump sum is payable at the end of year 3 if retirement occurs then. It is the only eligible year, and $i=4\%$. The probability of remaining active is $0.80$ in year 1 and $0.75$ in year 2. The year 3 retirement probability, conditional on being active at the start of that year, is $0.10$. Calculate the EPV. (Answer: $1,066.79)
A plan allows retirement in years 2 through 4 and pays $B_k=\$10{,}000(1.03)^{k-1}$ at the end of year $k$. The effective annual interest rate is $3\%$. Conditional probabilities of remaining active in years 1, 2, and 3 are $0.78,\ 0.80,\ 0.78$, respectively; retirement probabilities in years 2, 3, and 4 are $0.10,\ 0.15,\ 0.20$, respectively. Calculate the EPV. (Answer: $2,611.11)
A hypothetical plan pays $8,000 for retirement in year 2 and $12,000 for retirement in year 3; both are paid at year-end. Both years are eligible, and $i=0\%$. In year 1, retirement is $0.10$ and other exits total $0.20$. In year 2, retirement is $0.20$ and other exits total $0.10$. In year 3, conditional retirement probability is $0.10$. Calculate the EPV. (Answer: $1,708.00)

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