Why Conditional Exit Probabilities Are Not Cohort Probabilities
In a pension plan with competing exits, a probability such as “a member exits through withdrawal during the second year, given that the member is active at the start of that year” is conditional on being there to experience that year. It is not yet the probability, measured from the original valuation date, that a randomly selected member exits through withdrawal during the second year. The latter must also allow for the possibility that the member exited earlier.
The calculation has two parts: find the probability that the member is still active at the start of the year, then multiply by the conditional probability of the specified exit during that year. This distinction is essential when probabilities by cause and duration are used to describe an original cohort.
From Active-Member Survival to Cause-Specific Exits
The conditional probabilities \(q_k^{(j)}\) describe what happens to members who are active at the beginning of year \(k+1\). To obtain probabilities for the original cohort, first determine the proportion that reaches that year active. Since remaining active is required in every earlier interval,
The unconditional probability of exiting from cause \(j\) during year \(k+1\), measured from time 0, is then
“Unconditional” here means unconditional on being active at the start of that year. The probability is still conditional on the assumed starting status of the cohort: every member is active at time 0. These are discrete annual probabilities; they are not continuous-time exit rates.
Why the Multiplication Formula Holds
An exit from cause \(j\) during year \(k+1\) can occur only if the member is active at time \(k\). Thus the event of interest is the intersection of “active at time \(k\)” and “exit through cause \(j\) during \((k,k+1]\).” By the definition of conditional probability,
The multiplication is therefore not an independence assumption. The second factor is specifically the probability of cause \(j\) conditional on being active at the start of the interval.
The same reasoning gives a useful probability check. In year \(k+1\), the sum of all unconditional cause-specific exit probabilities is \({}_kp^{(A)}\sum_jq_k^{(j)}\). The probability of remaining active through that year is \({}_{k+1}p^{(A)}={}_kp^{(A)}p_k\). Because \(p_k=1-\sum_jq_k^{(j)}\), the exits in that year plus the members who remain active equal the number who began it active.
Over several years, the same check telescopes: the cumulative probability of an exit by any cause by time \(n\), plus the probability of still being active at time \(n\), is 1. The cumulative probability for cause \(j\) is the sum of that cause’s unconditional annual probabilities; it is not a product of annual cause probabilities.
Worked Example: Two Years with Three Exit Causes
Problem. A hypothetical pension plan has annual intervals. For members active at the start of year 1, the conditional probabilities of retirement, death, and withdrawal are 0.0200, 0.0800, and 0.1000. For members active at the start of year 2, they are 0.0300, 0.1000, and 0.0500. Find the unconditional probability of each cause-specific exit in each year, the cumulative probability by cause through year 2, and the probability still active at the end of year 2.
Method. First find the conditional probability of remaining active in each year. Then multiply each year’s conditional cause probabilities by the probability of reaching that year active. This avoids treating year 2’s conditional probabilities as probabilities for the original cohort.
In year 1, the total conditional exit probability is \(0.0200+0.0800+0.1000=0.2000\), so the conditional active probability is \(p_0=1-0.2000=0.8000\). Everyone starts year 1 active, so the unconditional year 1 exit probabilities are 0.0200 for retirement, 0.0800 for death, and 0.1000 for withdrawal.
The probability of reaching year 2 active is \({}_1p^{(A)}=0.8000\). In year 2, the conditional total exit probability is \(0.0300+0.1000+0.0500=0.1800\), leaving \(p_1=0.8200\). Therefore, the unconditional year 2 exit probabilities are:
Adding each cause across the two years gives cumulative probabilities of \(0.0200+0.0240=0.0440\) for retirement, \(0.0800+0.0800=0.1600\) for death, and \(0.1000+0.0400=0.1400\) for withdrawal. The probability still active after year 2 is \(0.8000(0.8200)=0.6560\).
Check. The four mutually exclusive outcomes through year 2 sum to \(0.0440+0.1600+0.1400+0.6560=1.0000\). The probabilities are consistent with the cohort accounting.
Common mistake avoided. Reporting 0.1000 as the unconditional year 2 withdrawal probability would ignore the 0.8000 probability of reaching year 2 active. The cohort probability is 0.8000 times 0.1000, or 0.0800.
Worked Example: Cumulative Probabilities Over Three Years
Problem. Consider a hypothetical plan with two possible exit causes, A and B. For a member active at the start of year 1, the conditional probabilities are \(q_0^{(A)}=0.0500\) and \(q_0^{(B)}=0.1500\). At the start of year 2 they are 0.1000 and 0.1000; at the start of year 3 they are 0.1500 and 0.0500. Find the cumulative probabilities of exit through A and B by the end of year 3 and the probability still active then.
Method. Compute the probability of reaching each year active using the prior years’ remaining-active probabilities. Then weight each year’s cause probabilities by the active probability at that year’s start.
In each of the three years, the conditional probability of any exit is 0.2000. Hence the conditional probability of remaining active is 0.8000 each year. The probabilities of reaching the starts of years 1, 2, and 3 active are, respectively, \(1.0000\), \(0.8000\), and \(0.8000(0.8000)=0.6400\).
For cause A, the unconditional exit probabilities in years 1, 2, and 3 are \(1.0000(0.0500)=0.0500\), \(0.8000(0.1000)=0.0800\), and \(0.6400(0.1500)=0.0960\). The cumulative probability for A is \(0.0500+0.0800+0.0960=0.2260\).
For cause B, the corresponding unconditional exit probabilities are \(1.0000(0.1500)=0.1500\), \(0.8000(0.1000)=0.0800\), and \(0.6400(0.0500)=0.0320\). The cumulative probability for B is \(0.1500+0.0800+0.0320=0.2620\).
The probability of remaining active through year 3 is \(0.8000(0.8000)(0.8000)=0.5120\). Checking the total, \(0.2260+0.2620+0.5120=1.0000\).
Common mistake avoided. The year 3 factor is 0.6400 because the member must survive actively through years 1 and 2. Do not multiply it by 0.8000 again: the year 3 conditional cause probabilities already describe exits during year 3 among members active at its start.
Worked Example: Duration-Varying Probabilities and a Full Reconciliation
Problem. A hypothetical pension plan has three mutually exclusive exit causes: retirement, disability, and death. Conditional on being active at the start of each year, the probabilities are as follows. Find the unconditional exit probabilities by cause and year, the cumulative probabilities by cause through year 4, and the probability still active at the end of year 4.
| Year | Retirement | Disability | Death | Remain active |
|---|---|---|---|---|
| 1 | 0.1000 | 0.0500 | 0.0500 | 0.8000 |
| 2 | 0.2000 | 0.1000 | 0.0500 | 0.6500 |
| 3 | 0.2500 | 0.0500 | 0.1000 | 0.6000 |
| 4 | 0.3000 | 0.1000 | 0.1000 | 0.5000 |
Method. Work across the years in order. For each year, multiply each conditional cause probability by the probability of reaching that year active. The probability of reaching year 1 is 1.0000. After year 1 it is \(1.0000(0.8000)=0.8000\); after year 2 it is \(0.8000(0.6500)=0.5200\); and after year 3 it is \(0.5200(0.6000)=0.3120\).
In year 1, the unconditional exit probabilities are the conditional probabilities themselves: retirement 0.1000, disability 0.0500, and death 0.0500.
In year 2, 0.8000 of the original cohort reaches the year active. The unconditional probabilities are retirement \(0.8000(0.2000)=0.1600\), disability \(0.8000(0.1000)=0.0800\), and death \(0.8000(0.0500)=0.0400\).
In year 3, 0.5200 reaches the year active. The unconditional probabilities are retirement \(0.5200(0.2500)=0.1300\), disability \(0.5200(0.0500)=0.0260\), and death \(0.5200(0.1000)=0.0520\).
In year 4, 0.3120 reaches the year active. The unconditional probabilities are retirement \(0.3120(0.3000)=0.0936\), disability \(0.3120(0.1000)=0.0312\), and death \(0.3120(0.1000)=0.0312\).
Adding across years, the cumulative retirement probability is \(0.1000+0.1600+0.1300+0.0936=0.4836\). The cumulative disability probability is \(0.0500+0.0800+0.0260+0.0312=0.1872\). The cumulative death probability is \(0.0500+0.0400+0.0520+0.0312=0.1732\). The probability still active after year 4 is \(0.3120(0.5000)=0.1560\).
Check. The final total is \(0.4836+0.1872+0.1732+0.1560=1.0000\). In each year, the unconditional exits plus the probability reaching the next year active also equal the probability starting that year active.
Common mistake avoided. The conditional cause probabilities change with duration, so one must use the values for the actual year. Applying year 1’s cause probabilities to every year would not represent the stated assumptions.
Common Interpretation Traps
- Leaving out prior active survival. A conditional probability for a later year applies only to members active at that year’s start. Multiply it by the probability of reaching that start active.
- Applying survival twice. Once the probability of reaching the year active has been included, multiply by the conditional cause probability once. Do not add another active-survival factor for the same year.
- Treating causes as independent exits. In this setup, the \(q_k^{(j)}\) values are mutually exclusive probabilities of first exit conditional on active status. The probability of remaining active is one minus their sum, not a product of separate cause-specific survival probabilities.
- Confusing annual and cumulative probabilities. A cause-specific probability for one year is not the cumulative probability through that year. Add the unconditional annual probabilities for that cause to obtain the cumulative value.
- Failing to reconcile the cohort. At any horizon, the cumulative probabilities of all exit causes plus the probability still active must equal 1 when all possible outcomes are included.
These probabilities describe the timing and cause of exits from the active population. When a later calculation assigns benefits to those exits, it must use the cause- and duration-specific probabilities on the appropriate cohort basis. The probability calculation itself does not determine benefit amounts or their payment dates.
Check Your Understanding
For each problem, interpret the stated cause probabilities as conditional on being active at the beginning of the stated year.
- A plan has two causes, retirement and death. In year 1 their conditional probabilities are 0.0400 and 0.0600; in year 2 they are 0.1000 and 0.0500. Find the unconditional year 2 probabilities by cause. (Answer: Retirement 0.0900; death 0.0450.)
- In a hypothetical three-year plan, the conditional probabilities of retirement and death are, respectively, \((0.1200,0.0300)\) in year 1, \((0.0800,0.0700)\) in year 2, and \((0.2000,0.0500)\) in year 3. Find the cumulative probability of each exit cause through year 3 and the probability still active. (Answer: Retirement 0.3325; death 0.1256; still active 0.5419.)
- A plan has three causes, A, B, and C. Their conditional probabilities in year 1 are 0.0200, 0.0800, and 0.1000. In year 2 they are 0.0300, 0.0700, and 0.1000. Find each cause’s unconditional year 2 probability, each cumulative probability through year 2, and the probability still active. (Answer: Year 2 A 0.0240, B 0.0560, C 0.0800; cumulative A 0.0440, B 0.1360, C 0.1800; still active 0.6400.)
- At the start of year 1 a cohort is entirely active. The conditional probabilities of any exit in years 1 and 2 are 0.1000 and 0.2000, respectively. In year 3, the conditional probabilities for causes A and B are 0.2500 and 0.1000. Find the unconditional year 3 probabilities for A and B and the probability still active after year 3. (Answer: A 0.1800; B 0.0720; still active 0.4680.)
- A cohort has probability 0.5600 of reaching year 3 active. Its unconditional year 3 probabilities of exit through A and B are 0.0840 and 0.1120. Find the conditional probabilities of A and B among members active at the start of year 3, the conditional probability of remaining active in year 3, and the cohort probability that remains active after year 3. (Answer: Conditional A 0.1500; conditional B 0.2000; conditional remain active 0.6500; cohort still active after year 3 0.3640.)