Why Payment Dates and Exit Probabilities Must Be Tracked Together
An active-member pension valuation can involve several competing ways to leave active service. A member may retire, withdraw, or die; each outcome can trigger a different benefit, at a different time. A reliable calculation therefore keeps two questions separate: what is the probability of each exit, and when is its benefit paid? The probability of an exit in a later year must reflect the chance that the member remained active through all earlier years. The discount factor must then reflect the benefit’s payment dates, not simply the date of exit.
The model here is a hypothetical annual model. Exit probabilities apply to a member who is active at the start of a year, and each exit occurs at that year’s end. The probabilities for the competing exits in a year sum to at most one; the remainder is the probability of staying active for the next year. Interest is effective annual interest, and benefits are valued at time 0. This setup does not apply unchanged if exits occur continuously during the year or if the stated probabilities are not conditional on being active.
For example, if the year-1 probabilities of retirement, withdrawal, and death are 0.10, 0.10, and 0.10, the probability of staying active through year 1 is 0.70. If the conditional probability of death in year 2 is 0.04, the unconditional probability of death in year 2 is \(0.70(0.04)=0.028\), not 0.04. The member must first reach year 2 as an active member.
A General Valuation Method
For an exit in year \(j\), first find its unconditional probability. Next, write the benefit cash flows at their actual payment dates and discount each one to time 0. The expected present value (EPV) of the benefit is the exit probability multiplied by the present value conditional on that exit. For multiple distinct exit events and benefits, add their expected present values.
Multiply the conditional stay-active probabilities for all completed years before the year being valued.
Multiply the probability of being active at the start of the year by the conditional probability of the relevant exit during that year.
Use \(v=1/(1+i)\); a payment of amount \(B\) at time \(t\) has present value \(Bv^t\).
Multiply each event’s conditional benefit present value by its unconditional event probability, then sum across events.
The weighting rule follows from conditioning on the exit event. If event \(E\) occurs with probability \(P(E)\), and the present value of the benefit given \(E\) is \(Z\), then its expected present value is \(P(E)Z\) when \(Z\) is fixed given the event. For cash flows from mutually exclusive exit events, expectation is additive, so the total EPV is the sum of those event-weighted values. This is why each benefit must be paired with its own event probability and payment schedule.
Worked Example: A Survivor Benefit After a Later-Year Death
A hypothetical plan uses an effective annual interest rate of 5%. In year 1, the conditional probabilities of retirement, withdrawal, and death are 0.10, 0.10, and 0.05. A member who dies at the end of year 2 receives a survivor benefit of $180 at the time of death and at each of the next two year-ends. The conditional probability of death in year 2, given active status at its start, is 0.06. Find the EPV of this year-2 death benefit.
Method. Find the probability of reaching year 2 active, multiply by the year-2 conditional death probability, and discount the three payments at times 2, 3, and 4.
The year-1 probability of remaining active is \(1-0.10-0.10-0.05=0.75\). Therefore, the unconditional probability of death in year 2 is \(0.75(0.06)=0.045\). The interest rate gives \(v=1/1.05=0.95238095\), so \(v^2=0.90702948\), \(v^3=0.86383760\), and \(v^4=0.82270247\).
Conditional on death at the end of year 2, the benefit’s present value is \(180(v^2+v^3+v^4)=180(0.90702948+0.86383760+0.82270247)=180(2.59356955)=\$466.84252\).
Weighting by the unconditional death probability gives \(0.045(\$466.84252)=\$21.00791\), or $21.01 to the nearest cent. The common mistake avoided is using 0.06 as the unconditional death probability; that would ignore the 0.75 probability of reaching year 2 active.
Retirement Benefits Begin After Retirement
A retirement pension does not necessarily start at the retirement date. Its first payment date must be read from the plan terms. In the next example, the pension begins one year after retirement and is paid at two consecutive year-ends. Retirement in year 1 therefore produces payments at times 2 and 3; retirement in year 2 produces payments at times 3 and 4. Both the exit probability and the payment dates change with the retirement year.
Worked Example: Retirement in Either of Two Years
A hypothetical plan has an effective annual interest rate of 4%. In year 1, conditional on being active, the probabilities of retirement, withdrawal, and death are 0.12, 0.08, and 0.04. In year 2, conditional on being active at its start, the retirement probability is 0.25. A member retiring at a year-end receives a pension of $600 at each of the next two year-ends. Find the EPV of the retirement pension if retirement occurs in year 1 or year 2.
Method. Value each possible retirement separately, using its unconditional probability and the pension payment dates that follow that retirement date.
The probability of staying active through year 1 is \(1-0.12-0.08-0.04=0.76\). The unconditional probability of retirement in year 1 is 0.12. The unconditional probability of retirement in year 2 is \(0.76(0.25)=0.19\). The conditional year-2 probabilities of withdrawal and death are not needed to value year-2 retirement, because the year-2 retirement probability is already conditional on reaching that year active.
At 4%, \(v=1/1.04=0.96153846\), \(v^2=0.92455621\), \(v^3=0.88899636\), and \(v^4=0.85480419\). For year-1 retirement, the two payments occur at times 2 and 3: \(0.12\{600(v^2+v^3)\}=0.12\{600(0.92455621+0.88899636)\}=0.12(\$1,088.13154)=\$130.57579\).
For year-2 retirement, they occur at times 3 and 4: \(0.19\{600(v^3+v^4)\}=0.19\{600(0.88899636+0.85480419)\}=0.19(\$1,046.28033)=\$198.79326\).
The total EPV is \(\$130.57579+\$198.79326=\$329.36905\), or $329.37 to the nearest cent. The common mistake avoided is discounting both retirement cases from the same date: the year-2 pension is one year later than the year-1 pension.
Worked Pension Case: Retirement, Withdrawal, and Death
The final case combines all three active-member exits and their separate benefits. It is hypothetical and has a two-year period of active-member exposure. In each year, the stated exit probabilities are conditional on being active at the start of that year. Retirement triggers a three-payment pension beginning one year after retirement; withdrawal triggers an immediate refund; death triggers a survivor benefit paid immediately and at the next two year-ends. No benefit is paid for an exit type other than the benefit specified for that exit.
Worked Example: Combining the Three Exit Benefits
A hypothetical pension plan uses an effective annual interest rate of 5%. In year 1, the conditional probabilities of retirement, withdrawal, and death are 0.10, 0.10, and 0.10. In year 2, conditional on being active at its start, the corresponding probabilities are 0.20, 0.10, and 0.04. Retirement at a year-end pays $300 at each of the next three year-ends. Withdrawal pays a $1,000 refund immediately at the year-end of withdrawal. Death pays $250 immediately at the time of death and at each of the next two year-ends. Find the EPV of all benefits arising from exits in years 1 and 2.
Method. Calculate the probability of each exit event without conditioning on active status, value its benefits at the stated payment dates, and sum the weighted values.
The year-1 probability of remaining active is \(1-0.10-0.10-0.10=0.70\). The unconditional year-2 exit probabilities are therefore retirement \(0.70(0.20)=0.14\), withdrawal \(0.70(0.10)=0.07\), and death \(0.70(0.04)=0.028\). In particular, the year-2 death probability is 0.028, not 0.04. The probabilities in year 2 sum to 0.34 conditional on active status, leaving \(1-0.20-0.10-0.04=0.66\) active through that year.
At 5%, \(v=1/1.05=0.95238095\). The powers needed are \(v^2=0.90702948\), \(v^3=0.86383760\), \(v^4=0.82270247\), and \(v^5=0.78352617\).
Year-1 retirement. The unconditional event probability is 0.10. Payments occur at times 2, 3, and 4: \(0.10\{300(v^2+v^3+v^4)\}=0.10\{300(0.90702948+0.86383760+0.82270247)\}=\$77.81\).
Year-2 retirement. Its unconditional event probability is 0.14. Payments occur at times 3, 4, and 5: \(0.14\{300(v^3+v^4+v^5)\}=0.14\{300(0.86383760+0.82270247+0.78352617)\}=\$103.74\).
Withdrawals. The year-1 refund is paid at time 1, giving \(0.10(1{,}000v)=\$95.24\). The year-2 refund is paid at time 2, giving \(0.07(1{,}000v^2)=\$63.49\).
Deaths. For year-1 death, the unconditional probability is 0.10 and payments occur at times 1, 2, and 3: \(0.10\{250(v+v^2+v^3)\}=0.10\{250(0.95238095+0.90702948+0.86383760)\}=\$68.08\). For year-2 death, the unconditional probability is 0.028 and payments occur at times 2, 3, and 4: \(0.028\{250(v^2+v^3+v^4)\}=0.028\{250(0.90702948+0.86383760+0.82270247)\}=\$18.15\).
The total is \(\$77.81+\$103.74+\$95.24+\$63.49+\$68.08+\$18.15=\$426.51\) using the displayed rounded components. Summing the unrounded values gives \(\$426.51621\), so the correctly rounded total is $426.52 to the nearest cent. This final check illustrates why totals should be rounded only after adding unrounded components. The common mistakes avoided are using conditional exit probabilities as unconditional ones, treating the year-2 death probability as 0.04, and discounting benefits from the exit date when their specified payment dates differ.
Checks That Prevent Timing and Probability Errors
The event probabilities in a year are competing probabilities: a member can experience only one of the listed exits during that year. Do not multiply the retirement and withdrawal probabilities together to form an exit probability. Instead, multiply the probability of reaching the year active by the conditional probability of the particular exit. For an exit in a later year, reaching that year active requires staying active through every preceding year.
Payment timing is a separate check. A refund paid at withdrawal is discounted to the withdrawal date. A pension beginning one year after retirement has no payment at the retirement date. A survivor benefit described as payable immediately at death includes a payment at the death time; if death occurs at the end of year 2, that payment is at time 2, not time 1. The benefit wording determines the dates.
| Benefit event | Probability to use | Payment timing to check |
|---|---|---|
| Retirement in year \(j\) | Probability active at the start of year \(j\), multiplied by \(q_j^R\) | Use the pension commencement date and payment frequency |
| Withdrawal in year \(j\) | Probability active at the start of year \(j\), multiplied by \(q_j^W\) | Discount a refund to its stated payment date |
| Death in year \(j\) | Probability active at the start of year \(j\), multiplied by \(q_j^D\) | Place each survivor payment relative to the death date |
Check Your Understanding
For each problem, identify the unconditional event probability before valuing the associated cash flows.
- At 4% effective annual interest, the probability of staying active through year 1 is 0.80. The conditional probability of death in year 2 is 0.05. Death at the end of year 2 pays $100 at times 2 and 3. Find the EPV. (Answer: $7.25)
- At 5%, year-1 conditional exit probabilities are retirement 0.20, withdrawal 0.15, and death 0.10. In year 2, the conditional retirement probability is 0.30. Retirement pays $500 at each of the next two year-ends. Find the EPV from retirement in years 1 and 2. (Answer: $316.23)
- At 3%, year-1 conditional exit probabilities are withdrawal 0.12, retirement 0.08, and death 0.05. The probability of remaining active through year 1 is therefore 0.75. The conditional year-2 withdrawal probability is 0.10. A withdrawal refund of $800 is paid at the withdrawal year-end. Find the EPV of refunds from years 1 and 2. (Answer: $149.76)
- At 5%, the probabilities of remaining active through years 1 and 2, conditional on being active at their respective starts, are 0.80 and 0.70. The conditional probability of death in year 3 is 0.04. A benefit of $1,000 is paid at the end of the year of death. Find its EPV. (Answer: $19.35)
- At 4%, year-1 conditional exit probabilities are retirement 0.10, withdrawal 0.20, and death 0.05. In year 2, conditional probabilities of retirement, withdrawal, and death are 0.20, 0.10, and 0.10. A year-2 withdrawal pays $400 at time 2; a year-2 death pays $300 at times 2 and 3. Find the combined EPV of these two benefits. (Answer: $59.40)