Why Active-Member Survival Requires Successive Conditioning
A pension valuation may depend on how many members remain in active service at each future year. During any year, an active member might retire, withdraw, become disabled, or die. These are competing exits: a member who exits through one cause is no longer exposed to the other causes as an active member during later years. The probability of remaining active therefore depends on the chance of avoiding every exit in each year, conditional on still being active when that year begins.
The key calculation is a product across years, not a sum of annual decrement probabilities. The annual probabilities can vary with duration, age, service, or other rating factors. The product rule still applies when each year’s probabilities are correctly defined conditional on active membership at the start of that year.
Here “exhaustive among exits” means that every exit from active membership is assigned to one of the listed causes. If there are additional causes, such as dismissal, their probabilities must also be included. In every year, the cause probabilities must be nonnegative and their sum cannot exceed 1. The remainder \(p_k^{(A)}\) is the probability of staying active for that year.
Let \(A_0\) denote the event that a member is active at the initial valuation date, and let \(A_k\) denote the event that the member is still active at the end of year \(k\). Thus, remaining active through year \(n\) means that no exit has occurred in any of the first \(n\) years. Write \({}_np_0^{(A)}=P(A_n\mid A_0)\) for this probability. Each annual transition probability is applied to the members who have survived in active status to the start of that year.
Deriving the Product Formula
The derivation follows from conditional probability. Staying active through year \(n\) requires staying active through year \(n-1\), followed by staying active during year \(n\). Given that the member has stayed active through year \(n-1\), the member is active at the start of year \(n\), so the conditional probability of surviving that year in active status is \(p_n^{(A)}\).
The conditional probability in the product can be written as \(P(A_k\mid A_{k-1})\) because \(A_k\) implies that the member was active at the start of year \(k\). Equivalently, one can apply the chain rule to the nested events \(A_1,A_2,\ldots,A_n\), conditional on \(A_0\). This does not require the member’s annual outcomes to be independent. The conditional probabilities already describe the transition for a member who is active at the relevant year’s start.
If the four cause probabilities are constant across all years, define \(q^{(j)}\) as the constant annual conditional probability for cause \(j\), and let \(p^{(A)}=1-\sum_j q^{(j)}\). The formula reduces to \({}_np_0^{(A)}=(p^{(A)})^n\). This simplified expression is not appropriate when the annual probabilities change with duration or member characteristics; in that case, use the year-specific product.
Confirm that each annual cause probability is conditional on being active at the start of that year, not on the original cohort.
Add the mutually exclusive exit probabilities for the year, then subtract their sum from 1.
Multiply the annual active-retention probabilities for all years in the period of interest.
For a homogeneous cohort of \(N_0\) initially active members, the expected active count after \(n\) years is \(N_0{}_np_0^{(A)}\).
Worked Applications
Worked Example: Changing Annual Probabilities
A hypothetical pension plan has 2,500 active members at the valuation date. For each year, the probabilities below are conditional on being active at the start of that year. The four listed exits are mutually exclusive and are the only exits from active status. Find the probability a member remains active through three years and the expected number of active members then.
Method. Calculate the retention probability separately for each year, then multiply. The annual probabilities are not constant, so a constant-retention power is not suitable.
In year 1, the total probability of exit is \(0.08+0.05+0.02+0.01=0.16\). Therefore, \(p_1^{(A)}=1-0.16=0.84\).
In year 2, the total probability of exit is \(0.10+0.04+0.03+0.01=0.18\). Therefore, \(p_2^{(A)}=1-0.18=0.82\).
In year 3, the total probability of exit is \(0.12+0.03+0.02+0.01=0.18\). Therefore, \(p_3^{(A)}=1-0.18=0.82\).
The probability of remaining active through all three years is
The expected number of active members is \(2{,}500(0.564816)=1{,}412.04\). Thus the probability, rounded to four decimal places, is 0.5648, and the expected count, rounded to two decimal places, is 1,412.04 members.
Check. Each annual retention lies between 0 and 1, and the three-year retention \(0.564816\) is lower than each individual annual retention, as required. The calculation avoids treating the later-year exit probabilities as proportions of the original 2,500 members.
Worked Example: Constant Decrement Probabilities
For a hypothetical group of 1,200 active members, assume that in every year the conditional probabilities of retirement, withdrawal, disability, and death are \(0.06\), \(0.03\), \(0.01\), and \(0.02\), respectively. Assume these causes are mutually exclusive and exhaustive among exits, and that the same probabilities apply in each of the next five years. Find the probability of remaining active for five years and the expected number of active members after five years.
Method. First calculate the total annual exit probability: \(0.06+0.03+0.01+0.02=0.12\). The annual active-retention probability is therefore \(1-0.12=0.88\).
Because the annual probabilities are constant, the five-year probability is the annual retention raised to the fifth power:
The expected number remaining active is \(1{,}200(0.5277319168)=633.27830016\). Rounded only at the final step, the probability is 0.5277 to four decimal places, and the expected count is 633.3 members to one decimal place.
Check. The expected count is below the initial 1,200, and multiplying it by the annual retention gives the expected count after one further year. The power is valid here because the stated conditional probabilities are the same at each annual duration.
Worked Example: Separate Member Groups
A hypothetical plan has 600 active members in group A and 400 in group B. Each group has its own annual conditional decrement probabilities; within each group these probabilities are the same for the next two years. For group A, the probabilities of retirement, withdrawal, disability, and death are \(0.04, 0.03, 0.02, 0.01\) each year. For group B, they are \(0.02, 0.01, 0.01, 0.01\) each year. Causes are mutually exclusive and exhaustive among exits. Find the total expected active membership after two years.
Method. Calculate each group’s retention separately. Combining the groups’ cause probabilities before allowing for their different initial sizes would use the wrong weighting.
For group A, the annual exit probability is \(0.04+0.03+0.02+0.01=0.10\), so the annual retention is \(0.90\). Its two-year retention is \((0.90)^2=0.81\), giving an expected active count of \(600(0.81)=486\).
For group B, the annual exit probability is \(0.02+0.01+0.01+0.01=0.05\), so the annual retention is \(0.95\). Its two-year retention is \((0.95)^2=0.9025\), giving an expected active count of \(400(0.9025)=361\).
The total expected active count is \(486+361=\mathbf{847}\). As a check using year-by-year counts, after year 1 the expected counts are \(600(0.90)=540\) in group A and \(400(0.95)=380\) in group B. After year 2 they are \(540(0.90)=486\) and \(380(0.95)=361\), again totaling 847.
Check. The total is also \(1{,}000(0.847)=847\), so the aggregate two-year survival probability is \(0.847\). The group calculations are essential: the groups do not share the same annual retention probability.
Common Mistake: Adding Decrements Across Years
A common error is to add the annual exit probabilities and subtract that sum from 1. That operation is valid for finding the total exit probability within one year when the causes are mutually exclusive. It is not valid across years when the annual probabilities are conditional on still being active at the beginning of each year.
For the three-year example above, the annual exit probabilities were \(0.16\), \(0.18\), and \(0.18\). The incorrect calculation would be
This incorrectly treats the year 2 and year 3 conditional probabilities as though they applied directly to the original cohort and as though year-by-year exits could simply be added. The correct calculation first allows for survival in each year, then applies the next year’s conditional probability to those still active:
Another related error is to treat each cause probability as a fraction of the initial membership when it is specified conditional on being active at the start of a later year. If counts by cause are needed, first determine the expected active count at that year’s start; the conditional cause probabilities apply to that count, not automatically to the original cohort.
When the Product Needs Refinement
The product formula is valid when the annual probabilities are appropriate for the member group being followed and each \(p_k^{(A)}\) describes the transition from active status during year \(k\). In practice, probabilities may vary with attained age, service, duration, or plan rules. The index \(k\) then identifies the relevant year-specific probabilities, and the product must retain those changes rather than replacing them with an unexamined average.
If members have different decrement assumptions, calculate their survival probabilities by group or member type and combine the resulting expected counts using the appropriate initial membership. A single product for the whole plan is justified only when a common sequence of conditional annual retentions applies to the cohort being valued. If the probabilities are defined on a different conditioning basis, first translate them to the conditional probabilities required for this calculation.
This survival calculation is a building block for pension valuation under competing exits: it identifies the proportion of an initially active group expected to reach each future year still active. A complete benefit valuation may also require the timing and amount of payments triggered by exits, but those amounts are separate from the probability of remaining in active status.
Check Your Understanding
For each question, treat the stated annual exit causes as mutually exclusive and exhaustive, and the probabilities as conditional on active membership at the start of the stated year.
- In year 1, retirement, withdrawal, disability, and death probabilities are \(0.04, 0.02, 0.01, 0.02\). In year 2 they are \(0.05, 0.03, 0.01, 0.01\). Find the probability of remaining active through both years. (Answer: 0.8190)
- Each year, the four cause probabilities are \(0.03, 0.04, 0.02, 0.01\), and they remain constant for four years. Find the four-year active-member survival probability. (Answer: 0.6561)
- The annual cause probabilities are \(0.07, 0.02, 0.02, 0.01\) in year 1; \(0.08, 0.03, 0.02, 0.02\) in year 2; and \(0.05, 0.04, 0.01, 0.02\) in year 3. Find the probability of remaining active through all three years. (Answer: 0.6582)
- A group of 8,000 active members has annual retention probabilities \(0.94, 0.90,\) and \(0.96\) over the next three years. Find the expected number still active at the end of year 3. (Answer: 6,497.28 members)
- The probability of remaining active through two years is \(0.75\). The year 1 retention probability is \(0.90\). In year 2, the retirement, withdrawal, and disability probabilities are \(0.04, 0.03,\) and \(0.01\). What death probability in year 2 is consistent with the stated two-year survival? (Answer: 0.0867)