Conditional Exit Probabilities After Survival

Introduction

In a multiple-decrement model, a member of a group may leave through one of several competing causes. A probability for a later interval must be read relative to the lives still at risk at the start of that interval. If the question conditions on survival through earlier intervals, those earlier exits determine who remains in the relevant risk set; they do not get added to the later interval’s cause-specific probability.

This distinction is useful whenever rates change over time or a problem provides both interval-specific exit probabilities and counts from an original cohort. The essential question is: what population forms the denominator after the stated survival condition? The calculations below are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Suppose a member is in the decrement group at the start of interval \(k\), and let \(q_k^{(j)}\) be the probability of exit from cause \(j\) during that interval. Let \(p_k\) be the probability of remaining in the group through the interval. With mutually exclusive causes that exhaust all exits, \(p_k=1-\sum_j q_k^{(j)}\). The probability \(q_k^{(j)}\) is conditional on being in the group at the start of interval \(k\).

Conditional and Joint Probabilities

Index the intervals from 1 onward. Let \(S_m\) denote the event that a member survives in the decrement group through the first \(m\) intervals. Let \(E_{m+1}^{(j)}\) denote exit from cause \(j\) during interval \(m+1\). The event \(E_{m+1}^{(j)}\) already requires the member to be at risk at the start of interval \(m+1\). Therefore, conditioning on \(S_m\) gives the interval probability:

P\bigl(E_{m+1}^{(j)}\mid S_m\bigr)=q_{m+1}^{(j)}.

By contrast, the probability of surviving the first \(m\) intervals and then exiting from cause \(j\) in the next interval is a joint probability measured from the original starting population. Under the interval model,

P\bigl(S_m\cap E_{m+1}^{(j)}\bigr) =\left(\prod_{r=1}^{m}p_r\right)q_{m+1}^{(j)}.

The product appears because a member must survive each earlier interval to enter the next one. No independence assumption between different members is needed for this individual-level probability calculation. The model does require that the interval probabilities describe the applicable risk set at each interval’s start and that the listed causes are mutually exclusive.

Why the Risk-Set Denominator Is Correct

A short derivation makes the denominator rule explicit. The conditional probability definition gives

P\bigl(E_{m+1}^{(j)}\mid S_m\bigr) =\frac{P\bigl(S_m\cap E_{m+1}^{(j)}\bigr)}{P(S_m)}.

The probability of reaching the start of interval \(m+1\) is \(P(S_m)=\prod_{r=1}^{m}p_r\). Among those who reach that interval, the fraction exiting from cause \(j\) in it is \(q_{m+1}^{(j)}\). Hence the joint probability in the numerator is \(\left(\prod_{r=1}^{m}p_r\right)q_{m+1}^{(j)}\). Substitution and cancellation give

\frac{\left(\prod_{r=1}^{m}p_r\right)q_{m+1}^{(j)}}{\prod_{r=1}^{m}p_r} =q_{m+1}^{(j)}.

This cancellation is why one should not multiply by the earlier survival probability when the question asks for a probability conditional on that survival. Multiplication is appropriate only when the question asks for the joint probability from the original starting point.

Locate the conditioned-on time.
Translate “survived the first \(m\) intervals” into the start of interval \(m+1\).

Identify the target cause and interval.
Use the cause-specific probability for that interval, not a probability from an earlier interval.

Check the requested probability type.
For a conditional probability, use the interval probability. For a joint probability from the original cohort, multiply it by survival through the earlier intervals.

Worked Calculations

Worked Example: Later-Year Cause Probability From Cohort Counts

A hypothetical group begins year 1 with 1,000 members. During year 1, 100 exit through cause A and 80 through cause B. During year 2, among those at risk at the start of that year, 82 exit through A and 123 through B. Find the probability of a year-2 exit through B conditional on surviving year 1. Also find the probability, from the original group, of surviving year 1 and then exiting through B in year 2.

Method. First find the year-2 risk-set size, then divide the year-2 cause-B exits by that risk set. To obtain the joint probability from the original group, instead divide those exits by the original 1,000 members.

At the start of year 2, the number still in the group is \(1{,}000-100-80=820\). Thus the conditional probability of a year-2 exit through B is \(123/820=0.15\).

The joint probability from the original starting group is \(123/1{,}000=0.123\). As a check, year-1 survival is \(820/1{,}000=0.82\), and \(0.82(0.15)=0.123\).

Answer. Conditional on surviving year 1, the probability is \(0.15\). The joint probability from the original group is \(0.123\). The common mistake avoided is using the original 1,000-member denominator for the conditional probability.

Worked Example: Changing Probabilities Across Three Years

For a hypothetical insured group, the multiple-decrement probabilities are as follows. In year 1, \(q_1^{(A)}=0.10\) and \(q_1^{(B)}=0.05\). In year 2, \(q_2^{(A)}=0.08\) and \(q_2^{(B)}=0.12\). In year 3, \(q_3^{(A)}=0.14\) and \(q_3^{(B)}=0.06\). Find the probability of a year-3 exit through A conditional on surviving the first two years. Then find the probability of surviving both years and exiting through A in year 3, measured from the original group.

Method. Calculate the all-cause survival probability for each of the first two intervals by subtracting both cause probabilities from 1. The conditional year-3 probability is supplied directly by the year-3 cause-A probability.

Year-1 survival is \(p_1=1-0.10-0.05=0.85\). Year-2 survival is \(p_2=1-0.08-0.12=0.80\). Therefore, survival through the first two years has probability \(0.85(0.80)=0.68\).

Conditional on surviving those two years, the probability of exiting through A during year 3 is \(q_3^{(A)}=0.14\). The corresponding joint probability from the original group is \(0.68(0.14)=0.0952\).

Answer. The conditional probability is \(0.14\), and the joint probability from the original group is \(0.0952\). The common mistake avoided is carrying year-1 or year-2 cause probabilities into year 3 as though the rates were constant; each interval has its own probabilities.

Worked Example: Recovering a Conditional Probability From Original-Cohort Shares

A hypothetical cohort has 2,000 members at time 0. At the end of year 1, 1,600 remain; at the end of year 2, 1,200 remain. During year 3, 96 members exit through a specified cause C. Find the probability of exit through C during year 3 conditional on surviving the first two years. Verify the result using the original cohort.

Method. The year-3 risk set consists of the 1,200 members who survived through year 2. Divide the year-3 cause-C exits by this risk-set count.

The conditional probability is \(96/1{,}200=0.08\). As a second calculation, survival through the first two years from the original cohort is \(1{,}200/2{,}000=0.60\), and the joint probability of reaching year 3 and then exiting through C is \(96/2{,}000=0.048\). Dividing the joint probability by the survival probability gives \(0.048/0.60=0.08\), as required.

Answer. The conditional probability is \(0.08\). The common mistake avoided is treating \(96/2{,}000=0.048\), which is a joint probability, as the conditional probability.

More Than One Later Interval

Sometimes the question asks for a cause-specific exit over several later intervals, conditional on surviving to the start of that period. If the target is cause \(j\) in one particular interval \(k\), after survival through interval \(m\), the conditional probability is the product of survival probabilities from \(m+1\) through \(k-1\) and the cause-\(j\) probability in interval \(k\):

P\bigl(\text{survive intervals }m+1,\ldots,k-1 \text{ and exit from }j\text{ in }k\mid S_m\bigr) =\left(\prod_{r=m+1}^{k-1}p_r\right)q_k^{(j)}.

When \(k=m+1\), the product has no factors and is defined to equal 1, leaving \(q_{m+1}^{(j)}\). If the question asks for cause \(j\) over a range of later intervals, add the mutually exclusive possibilities for each interval, with the appropriate intervening survival probabilities. For example, from the start of interval \(m+1\), the probability of cause \(j\) in any of intervals \(m+1\) through \(n\) is

\sum_{k=m+1}^{n} \left(\prod_{r=m+1}^{k-1}p_r\right)q_k^{(j)}.

These expressions use the multiple-decrement probabilities in each interval. They do not treat cause \(j\) as the only possible decrement: survival to a later interval requires avoiding all causes in each intervening interval.

Common Exam Trap

Common mistake: A group has 1,000 members initially; 200 leave in year 1, and 80 of the 800 who remain leave through cause A in year 2. The wrong conditional calculation is \(80/1{,}000=0.08\). That calculation answers the joint probability from the original group. The correct conditional probability, given survival through year 1, is \(80/800=0.10\). The check is \(0.80(0.10)=0.08\), which recovers the joint probability.

A related error is to add the earlier cause-specific exit probability to the later one. Earlier exits are not additional ways to satisfy an event explicitly restricted to the later interval. First apply the stated survival condition; then calculate the probability of the specified cause in the target interval among the survivors.

Key takeaway: A cause-specific interval probability is conditional on being at risk at that interval’s start. If a question conditions on survival to that start, use the interval probability itself. Multiply by prior survival only when the requested answer is a joint probability measured from an earlier starting population.

Check Your Understanding

For each problem, identify whether the requested value is conditional on a later risk set or joint from the original group before choosing the denominator.

In year 1, a hypothetical group has cause-A probability \(0.07\) and cause-B probability \(0.11\). Among those surviving to year 2, the cause-B probability is \(0.09\). Find the probability of a year-2 B exit conditional on surviving year 1. (Answer: \(0.09\))
A cohort starts with 1,500 members. In year 1, 120 exit through cause A and 180 through cause B. In year 2, 96 of those at risk at its start exit through A. Find the conditional year-2 A probability and the joint probability from the original cohort. (Answer: conditional \(0.0800\); joint \(0.0640\))
Year 1 has cause probabilities \(0.06\) for A and \(0.14\) for B; year 2 has \(0.12\) for A and \(0.05\) for B; year 3 has \(0.09\) for B. Find the year-3 B probability conditional on surviving the first two years, and the joint probability from the original group. (Answer: conditional \(0.0900\); joint \(0.0598\))
A hypothetical cohort starts with 1,000 members and has 300 members remaining at the end of year 2. In year 3, 42 members exit through cause C. Find the year-3 conditional probability of cause C and the joint probability of surviving the first two years and exiting through C. (Answer: conditional \(0.1400\); joint \(0.0420\))
From an original cohort, the cumulative proportions exiting through A by the ends of years 1 and 2 are \(0.12\) and \(0.28\), respectively. The year-1 proportion exiting through B is \(0.08\), and there are no other causes. Find the probability of a year-2 A exit conditional on surviving year 1. (Answer: \(0.2000\))

What You'll Learn