Expected Cause-Specific Exits From a Cohort

From Probabilities to Expected Cohort Counts

Competing-decrement calculations often produce probabilities: the probability a member remains in a group, or exits from it by a particular cause during an interval. A cohort calculation translates those probabilities into expected numbers. This distinction matters in applications such as planning hypothetical benefit payments or summarizing the expected outcomes for a block of policies.

The essential operation is simple, but the population to which a probability applies must be identified correctly. A probability specified for an interval among members still present at its start is applied to the expected number reaching that interval—not automatically to the original cohort. Similarly, when members fall into groups with different probabilities, the expected counts must be calculated group by group or with properly weighted probabilities.

Definition: For a cohort of \(N\) members, let \(p_0\) be the probability that a member remains through a specified interval, and let \(p_j\) be the probability that the member exits by cause \(j\) during that interval. If the causes are mutually exclusive and the outcomes are exhaustive, then \(p_0+\sum_j p_j=1\). The expected number remaining is \(Np_0\), and the expected number exiting by cause \(j\) is \(Np_j\).

These are expected counts, not promises about the realized counts. Actual counts are whole numbers, while an expected count can be fractional. For example, an expected 62.5 exits does not mean that half a person exits; it is the average count implied by the probability model over repeated cohorts under the same conditions.

Why Multiplication Gives the Expected Count

The count formula follows directly from the definition of expectation. For each member \(k\), define an indicator \(I_{kj}\) that is 1 if that member exits by cause \(j\) during the interval and 0 otherwise. The total number \(C_j\) exiting by cause \(j\) is the sum of these indicators. If every member has the same cause-\(j\) probability \(p_j\), then each indicator has expectation \(p_j\).

C_j=\sum_{k=1}^{N} I_{kj}, \qquad E[C_j]=E\!\left[\sum_{k=1}^{N} I_{kj}\right] =\sum_{k=1}^{N}E[I_{kj}] =\sum_{k=1}^{N}p_j =Np_j.

This argument uses linearity of expectation; it does not require the members’ outcomes to be independent. Independence would matter for other questions, such as calculating the variance of the count, but not for this expected-count formula. The same reasoning gives \(Np_0\) for the expected number remaining.

Key result: For a cohort with common outcome probabilities, multiply the cohort size by the probability of each mutually exclusive outcome. If the outcomes are exhaustive, the expected counts conserve the cohort: \(Np_0+\sum_j Np_j=N\).

One Interval With Cause-Specific Probabilities

The probabilities used in a cohort calculation must refer to the same starting population and the same interval. In a single interval, if the remaining probability and all cause-specific exit probabilities sum to 1, each expected count is obtained directly from the initial cohort size.

Worked Example: Expected Exits in One Interval

A hypothetical cohort has 2,500 members at the start of a one-year interval. For each member, the probability of remaining throughout the year is 0.9200, the probability of death during the year is 0.0250, and the probability of withdrawal during the year is 0.0550. Assume these outcomes are mutually exclusive and exhaustive, and that the probabilities apply equally to every member. Find the expected number remaining and the expected exits by each cause.

Method. Check the probability total, then multiply each outcome probability by the initial cohort size. The probability check is \(0.9200+0.0250+0.0550=1.0000\), so the stated outcomes account for the full cohort.

The expected number remaining is \(2{,}500(0.9200)=2{,}300\).

The expected number of deaths is \(2{,}500(0.0250)=62.5\).

The expected number of withdrawals is \(2{,}500(0.0550)=137.5\).

As a check, \(2{,}300+62.5+137.5=2{,}500\), equal to the initial cohort. The fractional results are valid expected counts. The common mistake this calculation avoids is rounding the cause-specific expected counts separately to whole people and then treating those rounded values as the exact expected counts.

Updating the Cohort Across Successive Intervals

When an interval’s probabilities are conditional on a member being present at the interval’s start, first determine the expected number reaching that start. Apply the next interval’s probabilities to that number. This is the cohort-count counterpart of updating remaining probabilities across subintervals.

Start with the appropriate population.
Use the initial cohort for the first interval and the expected number remaining from the previous interval for the next one.

Calculate each interval’s outcomes.
Multiply the population entering that interval by its conditional remaining and cause-specific exit probabilities.

Aggregate cause totals if needed.
Add expected counts for the same cause across intervals, while keeping interval-specific counts distinguishable when the question requires them.

Reconcile the cohort.
The total expected exits over all causes and intervals, plus the expected number remaining at the end, must equal the initial cohort.

Worked Example: Updating Expected Counts Over Two Intervals

A hypothetical cohort begins with 1,200 members. In the first interval, the probabilities conditional on being present at its start are 0.9400 of remaining, 0.0200 of death, and 0.0400 of withdrawal. In the second interval, conditional on being present at that interval’s start, the probabilities are 0.9000 of remaining, 0.0400 of death, and 0.0600 of withdrawal. Assume the probabilities apply equally to all members who enter the relevant interval. Find the expected deaths and withdrawals over both intervals and the expected number remaining at the end.

Method. Update the expected cohort remaining after the first interval, then apply the second interval’s conditional probabilities to that updated number. Each interval’s probabilities sum to 1: \(0.9400+0.0200+0.0400=1.0000\) and \(0.9000+0.0400+0.0600=1.0000\).

In the first interval, expected deaths are \(1{,}200(0.0200)=24\), and expected withdrawals are \(1{,}200(0.0400)=48\). The expected number entering the second interval is \(1{,}200(0.9400)=1{,}128\). The first-interval check is \(24+48+1{,}128=1{,}200\).

Apply the second-interval probabilities to 1,128, not to 1,200. Expected deaths in the second interval are \(1{,}128(0.0400)=45.12\). Expected withdrawals in that interval are \(1{,}128(0.0600)=67.68\). The expected number remaining after the second interval is \(1{,}128(0.9000)=1{,}015.2\). The second-interval check is \(45.12+67.68+1{,}015.2=1{,}128\).

Across both intervals, expected deaths are \(24+45.12=69.12\), and expected withdrawals are \(48+67.68=115.68\). Adding these to the final expected number remaining gives \(69.12+115.68+1{,}015.2=1{,}200\), as required.

Common mistake. Applying the second interval’s death probability to all 1,200 members gives \(1{,}200(0.0400)=48\), rather than the correct 45.12. The second-interval probability is conditional on being present at that interval’s start; only an expected 1,128 members reach that point. The same incorrect base would overstate second-interval withdrawals as 72 instead of 67.68.

Combining Groups With Different Probabilities

A single probability cannot be applied to an entire cohort if its members are partitioned into groups with different outcome probabilities. For disjoint groups indexed by \(g\), let \(N_g\) be the size of group \(g\), and let \(p_{gj}\) be that group’s probability of exit by cause \(j\). The expected total for cause \(j\) is the sum of the group-specific expected counts:

E[C_j]=\sum_g N_g p_{gj}.

Equivalently, an overall probability may be used if it is weighted by group size: \(p_j^{\mathrm{overall}}=\sum_g (N_g/N)p_{gj}\), where \(N=\sum_g N_g\). An unweighted average of group probabilities is generally wrong when group sizes differ.

Worked Example: Aggregating Two Groups

A hypothetical one-year cohort contains 400 members in Group A and 200 in Group B. In Group A, the probabilities of remaining, exiting by Cause 1, and exiting by Cause 2 are 0.9500, 0.0300, and 0.0200. In Group B, the corresponding probabilities are 0.8000, 0.1200, and 0.0800. Within each group, outcomes are mutually exclusive and exhaustive. Find the expected exits by each cause and the expected number remaining.

Method. Calculate each group’s expected counts separately, then add corresponding outcomes. In Group A, the expected Cause 1 exits are \(400(0.0300)=12\); the expected Cause 2 exits are \(400(0.0200)=8\); and the expected number remaining is \(400(0.9500)=380\). These total \(12+8+380=400\).

In Group B, the expected Cause 1 exits are \(200(0.1200)=24\); the expected Cause 2 exits are \(200(0.0800)=16\); and the expected number remaining is \(200(0.8000)=160\). These total \(24+16+160=200\).

Combining the groups gives \(12+24=36\) expected Cause 1 exits, \(8+16=24\) expected Cause 2 exits, and \(380+160=540\) expected members remaining. The total is \(36+24+540=600\), equal to the initial cohort.

As a check using weighted probabilities, the overall Cause 1 probability is \((400(0.0300)+200(0.1200))/600=36/600=0.0600\). The overall Cause 2 probability is \(24/600=0.0400\), and the overall remaining probability is \(540/600=0.9000\). These produce the same expected counts when multiplied by 600. An unweighted average of the two Cause 1 probabilities would be \((0.0300+0.1200)/2=0.0750\), giving 45 exits, which is incorrect because the groups are unequal in size.

Exam Checks and Common Traps

Expected-count questions are usually straightforward once the probability’s reference population is clear. Before multiplying, identify the interval, cause, and population to which each probability applies. Then check that the outcome probabilities are consistent and that the resulting expected counts reconcile.

Confusing conditional and unconditional probabilities. An interval probability conditional on reaching the interval is multiplied by the expected count at that interval’s start.
Using the wrong cohort size. In a sequence of intervals, do not apply a later interval’s conditional probabilities to the original cohort unless every original member reaches that interval.
Averaging group probabilities without weights. When group sizes differ, calculate expected counts by group or use a group-size-weighted overall probability.
Rounding too early. Preserve fractional expected counts through intermediate calculations; round only if the question requests a rounded presentation.
Failing the conservation check. For exhaustive outcomes, the expected counts over all causes and the remaining group must sum to the relevant starting cohort.

Key takeaway: Expected counts are cohort size times the corresponding probability, but the cohort size must match the probability’s conditions. Update the population between intervals, weight heterogeneous groups by their sizes, and verify that expected outcomes reconcile to the initial cohort.

Check Your Understanding

For each problem, identify the population to which the probabilities apply before calculating expected counts.

A hypothetical cohort of 800 members has one-year probabilities 0.9000 of remaining, 0.0600 of Cause A exit, and 0.0400 of Cause B exit. Find the expected count for each outcome. (Answer: 720 remaining, 48 Cause A exits, 32 Cause B exits.)
A cohort of 500 has first-interval probabilities 0.8000 remaining, 0.1000 Cause A, and 0.1000 Cause B. Among those reaching a second interval, the probabilities are 0.9000 remaining, 0.0600 Cause A, and 0.0400 Cause B. Find the expected total exits by each cause and the final expected count remaining. (Answer: 74 Cause A exits, 66 Cause B exits, 360 remaining.)
A hypothetical block has 300 members in Group X and 100 in Group Y. Their Cause A probabilities for one interval are 0.0200 and 0.0800, respectively. Find the expected Cause A exits and the size-weighted overall Cause A probability. (Answer: 14 expected exits; overall probability 0.0350.)
A cohort of 1,000 has 0.9500 probability of remaining through the first interval. In the next interval, conditional on reaching it, the Cause A probability is 0.0300. What is the expected Cause A count in the second interval? (Answer: 28.5.)
A cohort of 600 is split into two groups of 500 and 100. Their probabilities of a specified exit are 0.0400 and 0.1000. Find the expected exits and explain why the unweighted mean probability is not suitable. (Answer: 30 expected exits; the size-weighted probability is 0.0500, whereas the unweighted mean 0.0700 ignores the unequal group sizes.)

What You'll Learn