Conditional Exit Probabilities Within an Interval

Why the Conditioning Event Matters

In a competing decrement model, an individual may leave a group through several causes. For a specified interval, two related questions are possible: what is the probability of exiting by a particular cause, and, given that an exit occurred, what is the probability that it was due to that cause? These probabilities have different denominators and answer different questions.

For example, an insurer may want to know the chance that a covered person withdraws during a year. That is an unconditional cause-specific probability: it includes everyone who was in the group at the start of the year. The insurer may instead ask what proportion of all exits during the year were withdrawals. That is conditional on an exit having occurred. People who remain in the group are excluded from this second calculation.

Definition: Let $T$ be the length of an interval, and let $E_j$ be the event that the individual exits through cause $j$ during that interval. Let $E$ be the event that the individual exits through any of the modeled causes during the interval. The conditional exit probability for cause $j$ is $P(E_j\mid E)$. The unconditional cause-specific exit probability is $P(E_j)$. When $P(E)>0$, $$ P(E_j\mid E)=\frac{P(E_j)}{P(E)}. $$

The denominator is the probability of any exit, not the probability of remaining and not the probability of exit by cause $j$. This conditional measure describes the cause mix among exits in the interval. It is not a probability for the original starting group.

Cause-Specific Exit Probabilities

Suppose an individual begins in one state at time 0, and can exit through one of $m$ causes. Write $\lambda_j(t)$ for the intensity of cause $j$ at time $t$, and let the total intensity be $\lambda(t)=\sum_{j=1}^{m}\lambda_j(t)$. Assume the intensities are nonnegative and sufficiently well behaved for the integrals below to exist. Also assume exits through the modeled causes are the only way to leave the state, and simultaneous exits have probability zero.

The probability of remaining in the state through time $t$ is

p_0(t)=\exp\left(-\int_0^t\lambda(s)\,ds\right).

The unconditional probability of exiting through cause $j$ by time $T$ is the integral of the cause-specific exit density:

P(E_j)=\int_0^T p_0(t)\lambda_j(t)\,dt.

Here $p_0(t)$ is the probability of still being exposed to the competing causes at time $t$, and $\lambda_j(t)$ is the instantaneous rate of exit by cause $j$ for an individual still in the state. The conditional probability is obtained by dividing this unconditional probability by the probability of any exit.

Key result: Under the assumptions above, the probability of any exit during $[0,T]$ is $1-p_0(T)$. Therefore, $$ P(E_j\mid E)= \frac{\displaystyle\int_0^T p_0(t)\lambda_j(t)\,dt} {1-p_0(T)}. $$ The same denominator can be calculated as $\sum_{k=1}^{m}P(E_k)$, since an exit during the interval must occur through exactly one modeled cause.

Why the Formula Holds

The cause-specific exit density follows from considering a short interval $[t,t+\Delta t)$. For an individual still in the state at time $t$, the probability of a cause-$j$ exit during that short interval is $\lambda_j(t)\Delta t+o(\Delta t)$, where $o(\Delta t)$ is a term that becomes negligible relative to $\Delta t$ as $\Delta t$ approaches zero. The probability of still being in the state at time $t$ is $p_0(t)$. Thus, the probability of reaching time $t$ in the state and then exiting by cause $j$ in the short interval is

p_0(t)\left(\lambda_j(t)\Delta t+o(\Delta t)\right).

Adding these probabilities over a partition of $[0,T]$ and taking the limit as the partition becomes arbitrarily fine gives the integral $\int_0^T p_0(t)\lambda_j(t)\,dt$. This establishes the unconditional cause-specific probability. Conditioning on the event of any exit then gives the ratio in the key result.

This argument also explains why the numerator is not generally just the integral of the intensity. An individual can exit through another cause before reaching time $t$, so the cause-$j$ intensity is weighted by the probability $p_0(t)$ of still being exposed to it.

Constant Intensities: A Useful Shortcut

If all cause intensities are constant over the interval, write $\lambda_j(t)=\lambda_j$ and $\lambda=\sum_j\lambda_j$, with $\lambda>0$. Then $p_0(t)=e^{-\lambda t}$, and

P(E_j)=\int_0^T e^{-\lambda t}\lambda_j\,dt =\frac{\lambda_j}{\lambda}\left(1-e^{-\lambda T}\right).

The probability of any exit is $1-e^{-\lambda T}$. Dividing the cause-$j$ probability by this denominator gives

P(E_j\mid E)=\frac{\lambda_j}{\lambda}.

The shortcut applies when the cause intensities are constant throughout the interval. A longer or shorter interval changes the probability of any exit, but under constant intensities it does not change the conditional cause proportions.

Worked Example: Constant Mortality and Withdrawal Intensities

In a hypothetical one-year model, an active policyholder can exit through mortality or withdrawal. Assume constant intensities of $0.06$ per year for mortality and $0.14$ per year for withdrawal. Calculate (a) the unconditional probability of mortality exit and (b) the probability that an exit is due to mortality, conditional on an exit during the year. Use the exponential survival model and give probabilities to four decimal places.

Method and assumptions. The total intensity is constant, the interval is $T=1$ year, and the only exits are the two specified causes. There is no interest calculation because this is an interval probability, not a present value.

Step 1: Find the total intensity and probability of any exit.

\lambda=0.06+0.14=0.20,\qquad P(E)=1-e^{-0.20(1)}=1-e^{-0.20}=0.1812692.

Step 2: Calculate the unconditional mortality exit probability.

P(E_{\mathrm{mortality}}) =\frac{0.06}{0.20}\left(1-e^{-0.20}\right) =0.30(0.1812692) =0.0543808.

Step 3: Condition on an exit.

P(E_{\mathrm{mortality}}\mid E) =\frac{0.0543808}{0.1812692} =0.3000.

The unconditional probability of mortality exit is $0.0544$, while mortality accounts for $0.3000$ of exits. As a check, the withdrawal probability is $0.70(0.1812692)=0.1268885$, and the two unconditional exit probabilities sum to $0.1812693$, subject to rounding. The method avoids treating the cause’s unconditional probability as the cause proportion among exits.

Time-Varying Intensities

When intensities vary during the interval, the constant-intensity shortcut generally does not apply. The cause-specific probability must account for both the changing cause intensity and the changing number of individuals still exposed to exit. In particular, the conditional probability is not generally the ratio of the cause intensity at the start of the interval to the total intensity at the start.

Worked Example: Two Time-Varying Cause Intensities

Over one year in a hypothetical model, the two cause intensities, measured per year, are $\lambda_1(t)=0.04+0.04t$ and $\lambda_2(t)=0.16-0.04t$, for $0\leq t\leq1$. Calculate the probability that an exit is due to cause 1, conditional on an exit during the year. Use the exponential remaining probability and give the result to four decimal places.

Method and assumptions. Time $t$ is measured in years. Both intensities are nonnegative on the interval, and their sum is constant. The integral of the cause-specific exit density is used for the numerator; the probability of any exit is the denominator.

Step 1: Find the total intensity and remaining probability.

\lambda_1(t)+\lambda_2(t)=0.20,\qquad p_0(t)=e^{-0.20t}.

Step 2: Evaluate the integrals needed for cause 1. Define $I_0=\int_0^1e^{-0.20t}\,dt$ and $I_1=\int_0^1te^{-0.20t}\,dt$. Direct integration gives

I_0=\frac{1-e^{-0.20}}{0.20}=0.9063462, \qquad I_1=\frac{1-(1+0.20)e^{-0.20}}{(0.20)^2}=0.4380774.

Step 3: Calculate the unconditional probability of cause 1.

\begin{aligned} P(E_1) &=\int_0^1 e^{-0.20t}(0.04+0.04t)\,dt\\ &=0.04I_0+0.04I_1\\ &=0.04(0.9063462)+0.04(0.4380774)\\ &=0.036253848+0.017523096\\ &=0.053776944. \end{aligned}

Step 4: Divide by the probability of any exit.

P(E_1\mid E) =\frac{0.053776944}{1-e^{-0.20}} =\frac{0.053776944}{0.181269247} =0.2966689 \approx 0.2967.

The final answer is $0.2967$. The conditional probability for cause 2 is the remaining share, $1-0.2966689=0.7033311$, so the two conditional shares sum to one.

Common mistake and correction. At $t=0$, the cause-1 intensity is $0.04$ and the total intensity is $0.20$. Using $0.04/0.20=0.20$ as the interval conditional probability is incorrect: it uses only the initial intensity ratio and ignores how the cause-1 intensity changes during the year. Integrating the cause-specific exit density and conditioning on any exit gives $0.2967$.

Several Causes and Interpretation

With more than two causes, the same calculation applies separately to each cause. For each $j$, integrate $p_0(t)\lambda_j(t)$ to get the unconditional cause-specific probability, then divide by the probability of any exit. No cause receives a separate conditioning event: all conditional shares use the same denominator.

Worked Example: Three Causes With Changing Intensities

A hypothetical group has three possible exits over one year. Their intensities per year are $\lambda_1(t)=0.02+0.01t$, $\lambda_2(t)=0.08$, and $\lambda_3(t)=0.10-0.01t$, for $0\leq t\leq1$. Find the conditional probability for each cause, given an exit during the year. Use the exponential remaining probability and round the final conditional probabilities to four decimal places.

Method and assumptions. The intensities are nonnegative over the year, and the three causes exhaust the possible exits. Their sum is $0.20$, so $p_0(t)=e^{-0.20t}$. Use $I_0=0.9063462$ and $I_1=0.4380774$, as evaluated by integrating $e^{-0.20t}$ and $te^{-0.20t}$ over $[0,1]$.

Step 1: Calculate each unconditional cause probability.

\begin{aligned} P(E_1) &=0.02I_0+0.01I_1\\ &=0.02(0.9063462)+0.01(0.4380774)\\ &=0.018126924+0.004380774\\ &=0.022507698,\\[4pt] P(E_2) &=0.08I_0 =0.08(0.9063462) =0.072507696,\\[4pt] P(E_3) &=0.10I_0-0.01I_1\\ &=0.10(0.9063462)-0.01(0.4380774)\\ &=0.090634620-0.004380774\\ &=0.086253846. \end{aligned}

Step 2: Find the common denominator. The probability of any exit is

P(E)=1-e^{-0.20}=0.181269247.

Step 3: Divide each cause probability by the probability of any exit.

\begin{aligned} P(E_1\mid E)&=\frac{0.022507698}{0.181269247}=0.1242,\\ P(E_2\mid E)&=\frac{0.072507696}{0.181269247}=0.4000,\\ P(E_3\mid E)&=\frac{0.086253846}{0.181269247}=0.4758. \end{aligned}

The rounded conditional shares sum to $1.0000$. This is a useful arithmetic check; small discrepancies can occur if intermediate values are rounded too early. The important distinction is that the unconditional probabilities describe exits per individual in the starting group, while the conditional probabilities describe the distribution of causes among those who exited.

Common Exam Traps

Using the wrong denominator. For a conditional cause probability, divide by the probability of any exit during the interval, $1-p_0(T)$, not by 1 and not by the probability of that same cause.
Confusing cause probability with cause share. $P(E_j)$ includes those who remain in the starting population in its denominator; $P(E_j\mid E)$ is calculated only among exits.
Using the constant-intensity shortcut when rates vary. The ratio $\lambda_j/\lambda$ gives the interval conditional probability when the cause intensities are constant. With time-varying intensities, use the integral formula unless a justified simplification is available.
Omitting survival weighting. The cause-specific exit density is $p_0(t)\lambda_j(t)$, not just $\lambda_j(t)$. The weighting accounts for competing exits before time $t$.
Mixing time units. If intensities are stated per year, the integration variable and interval length must be in years. Changing the time unit requires changing the numerical intensity as well.

For an actuarial calculation, it is useful to write down the target event before substituting numbers: first specify the interval, then identify the unconditional cause-specific probability, and finally state the conditioning event as “any modeled exit during the interval.” This keeps the numerator and denominator aligned.

Check Your Understanding

For each problem, distinguish the unconditional cause-specific probability from the conditional cause share when both are requested. Intensities are measured per year unless stated otherwise.

A hypothetical model has two constant intensities over a half-year interval: cause 1 has intensity $0.03$, and cause 2 has intensity $0.07$. Find the unconditional probability of cause-1 exit and the probability of cause 1 given any exit. (Answer: unconditional $0.0146$; conditional $0.3000$)
Three constant cause intensities are $0.02$, $0.03$, and $0.05$ over a two-year interval. Find the conditional probability of cause 2 and its unconditional exit probability. (Answer: conditional $0.3000$; unconditional $0.0544$)
Over one year, $\lambda_1(t)=0.02+0.02t$ and $\lambda_2(t)=0.08-0.02t$. Their total intensity is $0.10$. Using $I_0=0.9516258$ and $I_1=0.4678840$, calculate the probability of cause 1 given any exit. (Answer: $0.2983$)
Over one year, three intensities are $\lambda_1(t)=0.05+0.02t$, $\lambda_2(t)=0.10$, and $\lambda_3(t)=0.15-0.02t$. Their total is $0.30$. Find the conditional probability for each cause, using $I_0=0.8639393$, $I_1=0.4104035$, and $P(E)=0.2591818$. (Answer: cause 1 $0.1983$; cause 2 $0.3333$; cause 3 $0.4683$)
Over two years, a hypothetical two-cause model has $\lambda_1(t)=0.03+0.02t$ and $\lambda_2(t)=0.07-0.02t$. Their total intensity is $0.10$. Using $I_0=1.8126925$, $I_1=1.7523096$, and $P(E)=0.1812692$, calculate the conditional probability of cause 1. (Answer: $0.4933$)

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