Common Errors in Competing-Risk Pension Valuation

Why Competing-Risk Pension Values Are Easy to Misstate

A pension value under competing decrements can be wrong even when every rate and benefit amount has been copied correctly. The usual causes are errors of conditioning and timing: a decrement probability conditional on being active is used as though it were measured from the original population; survival to a later year is left out; active-member mortality is continued after retirement; or a cash flow is discounted to the wrong date.

The basic discipline is to keep the exposure state and the timeline visible at each step. A decrement probability for a year applies to members active at that year's start. A pension paid after retirement is instead subject to the retired member's survival from the retirement date to the payment date. These probabilities describe different groups and different periods.

Definition: In a discrete-time multiple-decrement model, $q_k^{(j)}$ is the probability that an active member at the start of year $k$ exits during that year from cause $j$. The probability of remaining active for the year is $p_k=1-\sum_j q_k^{(j)}$, assuming the listed causes are mutually exclusive and exhaustive for exits during that year. The probability of exiting from cause $j$ in year $k$, measured from the initial active status, is prior active survival multiplied by $q_k^{(j)}$.

The annual probabilities are conditional on being exposed at the start of the year. They are not automatically the probabilities that a member active at the valuation date exits from that cause in a particular later year. This distinction is the source of the first two errors below.

From Conditional Decrements to Unconditional Exit Probabilities

Suppose the member starts active at time 0. For a cause $j$, if decrement probabilities can vary by year, the probability of remaining active through the first $k-1$ years is

{}_{{k-1}}p^{\text{active}}=\prod_{r=1}^{k-1}p_r.

Consequently, the probability of exit from cause $j$ in year $k$, measured from time 0, is

P(\text{exit from cause }j\text{ in year }k) = \left(\prod_{r=1}^{k-1}p_r\right)q_k^{(j)}.

For year 1, the empty product is 1, so the unconditional probability equals $q_1^{(j)}$. For later years, the member must first remain active through every preceding year. The formula follows by the multiplication rule: the event of a year-$k$ exit from cause $j$ requires both survival in the active state up to year $k$ and then that cause-specific exit during year $k$.

Worked Example: A Later Retirement Probability

A hypothetical pension plan has annual active-member probabilities that are constant in each of the first two years: retirement $q^{(r)}=0.15$, death $q^{(d)}=0.05$, and withdrawal $q^{(w)}=0.10$. The causes are mutually exclusive and exhaustive. A benefit of $20,000 is payable at the end of the year in which retirement occurs. Find the expected present value of this benefit for a member active at time 0, using an effective annual interest rate of $i=5\%$. Assume no further payment condition after retirement is relevant to this exit benefit.

Method. Calculate the probability of reaching each year active, multiply by that year's conditional retirement probability, and discount the payment from its stated year-end date. This avoids treating a conditional decrement probability as an unconditional weight.

First calculate the probability of remaining active for one year: $p=1-0.15-0.05-0.10=0.70$. The probability of retirement in year 1, measured from time 0, is $0.15$. The probability of retirement in year 2 is $0.70(0.15)=0.105$.

With $v=1/1.05=0.9523809524$, the expected present value is

20{,}000\left(0.15v+0.105v^2\right) = 20{,}000\left(0.1428571429+0.0952380952\right) = \$4{,}761.90.

As a check, the probabilities of these two mutually exclusive retirement events sum to $0.15+0.105=0.255$, and each is weighted at its own payment date. The common error is to use $0.15$ again for year 2, producing $20{,}000(0.15v+0.15v^2)=\$5{,}578.23$. That overstates the value because it assigns year-2 retirement probability to members who already exited in year 1.

Keep Active and Retired Mortality Separate

Active-member decrement probabilities describe exits while the person is still an active member. Once retirement occurs, a pension payable while the retiree is alive depends on retired-member survival. The relevant mortality basis may differ from active-member mortality, and the periods to which the two assumptions apply must not be mixed.

For an annual pension payable at the ends of the first $n$ years after retirement, let $q_R$ be a constant annual retired-member death probability, with death during a year preventing that year's year-end payment. Then the survival probabilities from retirement to the payment dates are $(1-q_R)^1,\ldots,(1-q_R)^n$. At effective annual interest rate $i$, the value at the retirement date is

B\sum_{j=1}^{n}(1-q_R)^jv^j, \qquad v=\frac{1}{1+i},

where $B$ is the annual pension amount. The first survival factor is $1-q_R$, not 1: the first payment is at the end of the first year after retirement, and the member must survive that year.

Worked Example: Mortality After Retirement

A hypothetical member retires at time 0 and is entitled to a pension of $10,000 at the end of each of the next three years, provided the member is alive on the payment date. The effective annual interest rate is $i=5\%$. The retired-member annual death probability is $0.20$, independently from year to year. The active-member annual death probability before retirement would have been $0.05$. Find the pension value at retirement.

Method. Use retired survival from the retirement date to each payment date, then discount each payment to retirement. The active-member mortality assumption no longer applies after retirement.

The retired-member survival probabilities to the three payment dates are $0.80$, $0.80^2=0.64$, and $0.80^3=0.512$. With $v=1/1.05=0.9523809524$, the three discounted expected payments are:

\begin{aligned} \text{Year 1: }&10{,}000(0.80)v=\$7{,}619.05,\\ \text{Year 2: }&10{,}000(0.64)v^2=\$5{,}804.99,\\ \text{Year 3: }&10{,}000(0.512)v^3=\$4{,}422.85. \end{aligned}

Therefore the value is $7{,}619.05+5{,}804.99+4{,}422.85=\$17{,}846.89$ when the displayed components are rounded to cents. Equivalently, evaluating the unrounded expression $10{,}000(0.80v+0.64v^2+0.512v^3)$ and rounding only at the end gives $17,846.88. The one-cent difference in the component sum is due to rounding each component separately; the final value is $17,846.88.

The common mistake is to substitute active-member mortality of $0.05$ for retired mortality. That gives $10{,}000(0.95v+0.95^2v^2+0.95^3v^3)=\$24{,}639.89$, an overstated value because it uses the wrong population's mortality assumption.

Include Prior Survival and Discount to the Payment Date

When an exit benefit is payable in a later year, two separate adjustments are required. First, include the probability that the member is still active at the start of the exit year. Second, discount the benefit from its actual payment date to the valuation date. If a benefit is payable at the end of year $k$, the discount factor is $v^k$; a year-2 payment is not discounted for only one year.

For a benefit amount $C_k^{(j)}$ paid at the end of year $k$ on an exit from cause $j$, the expected present value contribution is

v^k\left(\prod_{r=1}^{k-1}p_r\right)q_k^{(j)}C_k^{(j)}.

This is an expected present value because it multiplies the cash flow, conditional on the specified exit, by the probability of that exit from the valuation date and discounts it to that date.

Worked Example: Prior Survival and the Year-End Payment Date

A hypothetical active member faces the following constant annual probabilities in each of two years: retirement $0.20$, death $0.10$, withdrawal $0.10$, and remaining active $0.60$. A $15,000 retirement benefit is paid at the end of the year of retirement. Find the present value contribution from retirement in year 2 for a member active at time 0, using $i=4\%$.

Method. Multiply the year-2 conditional retirement probability by the probability of remaining active through year 1, then discount the payment for two years. This avoids both omitting prior survival and discounting to the wrong date.

The member must remain active through year 1, an event with probability $0.60$. Conditional on being active at the start of year 2, the retirement probability is $0.20$. Thus the unconditional probability of retirement in year 2 is $0.60(0.20)=0.12$. The payment is at time 2, so $v^2=1/(1.04)^2=1/1.0816$. The contribution is

15{,}000(0.12)v^2 = \frac{1{,}800}{1.0816} = \$1{,}664.20.

The result is consistent with the event probability: only 12% of the initial members retire in year 2 under these assumptions. Omitting prior survival gives $15{,}000(0.20)v^2=\$2{,}773.67$. Using $v$ instead of $v^2$ gives $1{,}800/1.04=\$1{,}730.77$. Both calculations value the wrong event or the wrong payment date.

A Practical Audit Sequence

Before accepting a competing-risk pension value, write down the state occupied at each point in the cash-flow timeline. The following sequence makes the conditioning and timing explicit.

Identify the exposure state.
Confirm whether the member is active or retired at the start of each period. Apply active-member decrements only during active exposure.

Check the probability basis.
Determine whether each stated rate is conditional on being active at the period's start or is already unconditional from the valuation date. Do not multiply by prior survival twice, or omit it when it is needed.

Trace each benefit trigger.
Identify the cause that generates the benefit, and include only the probability of that trigger event. For a later-year trigger, include survival in the relevant state through all preceding periods.

Set the payment date.
Distinguish the exit date from a later year-end payment date. Use the discount factor corresponding to the time from valuation to payment.

Apply post-retirement survival if needed.
For a pension contingent on being alive after retirement, use retired-member survival from retirement to each payment date.

A useful check is to compare each probability with the group to which it refers. A conditional year-2 retirement probability can be as large as the year-1 probability, but its unconditional probability from time 0 cannot exceed it when survival to year 2 is less than 1 and the conditional rates are equal. Similarly, a pension's first payment at the end of the first year after retirement must have a survival factor below 1 if there is positive mortality during that year.

These checks are especially important in pension valuations that combine an active-member exit benefit with a post-retirement pension. The active exit probability determines whether and when the member retires; retired-member survival then determines whether later pension payments occur. Keeping those stages separate is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Key takeaway: For each payment, identify the triggering event, multiply all required conditional probabilities along the path to that event, apply survival after the trigger only when the benefit depends on it, and discount from the actual payment date. Never substitute a conditional decrement probability for an unconditional weight without accounting for prior survival.

Check Your Understanding

For each problem, identify the relevant exposure and payment date before calculating.

An active member faces annual retirement, death, and withdrawal probabilities of $0.12$, $0.08$, and $0.10$, respectively, in each of the first three years. What is the unconditional probability of retirement in year 3 for a member active at time 0? (Answer: $0.70^2(0.12)=0.0588$.)
A $25,000 death benefit is paid at the end of year 2 if an active member dies in that year. The first-year active probability is $0.85$, the year-2 conditional death probability is $0.04$, and $i=5\%$. Find the present value contribution from year-2 deaths. (Answer: $25{,}000(0.85)(0.04)v^2=\$770.98$.)
A retiree receives $6,000 at the ends of years 1 and 2 after retirement, if alive. Retired annual death probability is $0.10$, independently by year, and $i=5\%$. Find the value at retirement. (Answer: $6{,}000(0.90v+0.90^2v^2)=\$9{,}551.02$.)
An active member has a first-year probability of remaining active of $0.80$, followed by a year-2 retirement probability of $0.25$ conditional on being active at the start of year 2. A $18,000 benefit is paid at the end of year 2 on retirement, with $i=4\%$. Find its present value contribution from year-2 retirements. (Answer: $18{,}000(0.80)(0.25)v^2=\$3{,}328.40$.)
At time 0, an active member has probability $0.20$ of retiring at the end of year 1. The pension is $1,000 at each of the first three year-ends after retirement, conditional on survival; retired annual death probability is $0.20$, and $i=0\%$. Find the expected present value at time 0 of these three possible payments. (Answer: $0.20(1{,}000)(0.80+0.64+0.512)=\$390.40$.)

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