Recovering One Cause Probability From the Others

Why a Missing Cause Probability Is a Residual

In a competing decrement model, an individual who is active at the start of an interval can remain active through the end of the interval or leave through one of several causes. If those outcomes are defined consistently and cover every possibility, their probabilities must add to one. That conservation rule makes it possible to recover one missing cause-specific exit probability from the remaining probability and the probabilities for the other causes.

The subtraction is simple; deciding which quantities may be added is the important part. The remaining probability and each cause-specific probability must refer to the same starting group and the same interval. In addition, each cause-specific probability must be an unconditional probability from that starting group, not a share calculated only among those who exit.

Definition: Consider a fixed interval and a group of individuals all active at its start. Let \(p\) be the probability that an individual remains active at the interval’s end. Let \(q_j\) be the probability that the individual exits during the interval through cause \(j\), for \(j=1,\ldots,m\). If remaining active and the \(m\) exit causes are mutually exclusive and exhaustive outcomes, then \(p+\sum_{j=1}^{m}q_j=1\).

Here, a cause-specific probability \(q_j\) is measured relative to the group active at the start, not just relative to the individuals who exit. For example, \(q_j=0.03\) means that 3% of the starting group exits through cause \(j\) in the interval. It does not mean that 3% of all exits are due to cause \(j\).

The Residual Formula and Its Conditions

If cause \(r\) is the only missing cause probability, rearrange the conservation equation to obtain:

q_r=1-p-\sum_{\substack{j=1\\j\ne r}}^{m}q_j

In words, first calculate the total probability of exiting, \(1-p\), and then subtract the probabilities assigned to all the other exit causes. The residual is the probability for the missing cause.

The reason the formula holds is that the outcomes form a partition: exactly one outcome occurs for each individual in the starting group—remaining active, exiting through cause 1, or exiting through one of the other causes. For mutually exclusive outcomes, the probability of their union is the sum of their probabilities. Since that union is the whole set of possible outcomes, its probability is one. Rearranging that sum gives the residual formula.

Conditions for use: The outcomes must be mutually exclusive and exhaustive over the stated interval. All probabilities must be conditional on the same status at the interval’s start and refer to the same interval. If any cause is omitted from the listed outcomes, or if categories overlap, the residual does not necessarily represent the intended cause.

A valid residual must satisfy \(0\le q_r\le 1-p\). The upper bound follows because the missing cause is only one part of all exits. Equivalently, the known cause probabilities must be nonnegative and their sum cannot exceed \(1-p\). A negative residual is not a probability: it signals inconsistent inputs, incompatible definitions, or a calculation error. A residual above \(1-p\) also signals a problem.

Worked Examples

Worked Example: Completing a Four-Cause Probability Set

Method: Calculate the total exit probability from the remaining probability, then subtract the known cause probabilities. A hypothetical block of policies is active at the start of a one-year interval. The one-year probability of remaining active is 0.9100. The probabilities of exit through causes A, B, and C are 0.0325, 0.0180, and 0.0115, respectively. Find the one-year probability of exit through cause D, assuming these four causes and remaining active are mutually exclusive and exhaustive.

First calculate the probability of exiting through any cause:

$$ 1-p=1-0.9100=0.0900 $$

The known exit probabilities sum to:

$$ q_A+q_B+q_C=0.0325+0.0180+0.0115=0.0620 $$

Therefore the missing probability is:

$$ q_D=0.0900-0.0620=0.0280 $$

Check. Add the remaining probability and all four exit probabilities: \(0.9100+0.0325+0.0180+0.0115+0.0280=1.0000\). Also, \(0\le0.0280\le0.0900\), so the residual is within the possible range for an exit cause.

The probability of exit through D is 0.0280, or 2.80% of the policies active at the start. The common mistake to avoid is subtracting the known causes from one without first accounting for those who remain active.

Worked Example: Using a Cause Share Among Exits

Method: Convert the conditional cause share into an interval probability before applying conservation. In a hypothetical one-year model, 0.8600 of the starting group remains active. Cause A has interval exit probability 0.0700. Among all individuals who exit during the year, 40% exit through cause B. The only other exit cause is C. Find the interval probability of exit through C.

The total probability of exit is:

P(\text{exit})=1-0.8600=0.1400

The 40% figure is conditional on exit. Thus the probability of both exiting and exiting through B, measured from the starting group, is:

$$ q_B=0.4000(0.1400)=0.0560 $$

Subtract the known causes from the total exit probability:

$$ q_C=0.1400-0.0700-0.0560=0.0140 $$

Check. The outcomes sum to \(0.8600+0.0700+0.0560+0.0140=1.0000\). The exit probabilities also sum to \(0.1400\), as required. The answer is 0.0140.

The common mistake is to use 0.4000 as \(q_B\). That treats a share among exits as a probability from the original starting group. In fact, only 0.1400 of the group exits, so cause B’s interval probability is \(0.4000\times0.1400=0.0560\).

Worked Example: Recovering Retirement Exit and Its Share

Method: Find the missing cause as a residual, then divide by the total exit probability if a share among exits is requested. In a hypothetical one-year disability-income model, the probability of remaining in the model’s active state is 0.925. The probabilities of exit through death, recovery, and lapse are 0.024, 0.018, and 0.009. Retirement is the only other possible exit. Find both the retirement exit probability and the proportion of all exits attributable to retirement.

The total probability of exiting is:

$$ 1-p=1-0.925=0.075 $$

The known exit probabilities sum to:

$$ 0.024+0.018+0.009=0.051 $$

The retirement exit probability is therefore:

q_{\text{retirement}}=0.075-0.051=0.024

To find the proportion of all exits attributable to retirement, divide its interval probability by the total exit probability:

P(\text{retirement}\mid\text{exit})=\frac{0.024}{0.075}=0.32

Check. The remaining probability and exit probabilities sum to \(0.925+0.024+0.018+0.009+0.024=1.000\). The retirement probability is no greater than the total exit probability, and its share among exits is between zero and one. The answers are a retirement exit probability of 0.024 and a retirement share of 32% of exits.

The common mistake is to report 32% as the retirement probability. It is a conditional share among those who exit. The interval probability from the original group is 0.024.

Validation and Common Exam Traps

A residual calculation should include a check, not just a final subtraction. The following sequence helps keep the meaning of each quantity clear.

Fix the reference group and interval.
Confirm that the remaining probability and every cause probability are conditional on the same group being active at the interval’s start, and cover the same time span.

Identify the outcome types.
Check that “remaining active” and the listed causes are mutually exclusive and include every possible outcome for the model being used.

Separate interval probabilities from exit shares.
If a cause value is stated as a proportion of exits, multiply that share by the total exit probability \(1-p\) to express it from the starting group.

Calculate and test the residual.
Subtract the known interval exit probabilities from \(1-p\). Confirm that the result is nonnegative and that all outcomes sum to one.

For example, suppose \(p=0.86\), \(q_A=0.07\), and cause B accounts for 40% of all exits. The total exit probability is 0.14, so \(q_B=0.40(0.14)=0.056\). Treating 0.40 as the unconditional probability for B would already give \(0.86+0.07+0.40=1.33\), before even including any other cause. That sum exceeds one and exposes the error. The corrected calculation uses B’s interval probability, not its share among exits.

Another trap is mixing quantities from different intervals. A probability of remaining active for one year cannot be combined directly with a cause probability for a different year. Similarly, probabilities conditional on being active at the start of a subinterval cannot be added to probabilities conditional on being active at the start of a larger interval without first putting them on a common basis. The residual rule itself does not perform such conversions; its inputs must already describe the same partition.

Finally, a small negative residual can arise when supplied values have been rounded. If a calculation gives a negative value only by an amount consistent with the stated rounding precision, report the issue and explain that the unrounded inputs may differ. Do not silently replace a materially negative result with zero: first check the definitions, reference group, interval, and arithmetic.

Key takeaway: For a complete set of mutually exclusive interval outcomes, a missing cause probability equals the total exit probability \(1-p\) minus the known cause-specific interval probabilities. A conditional share among exits must first be converted to an interval probability.

Check Your Understanding

For each problem, treat the stated outcomes as mutually exclusive and exhaustive unless the problem says otherwise.

A hypothetical one-year model has a remaining probability of 0.923. Two cause-specific exit probabilities are 0.025 and 0.017. Find the probability for the only other cause. (Answer: 0.035)
In a hypothetical interval, the remaining probability is 0.820. Three causes have probabilities 0.045, 0.031, and an unknown value. Find the unknown probability and check it against the total exit probability. (Answer: 0.104; total exit probability 0.180)
A model has a remaining probability of 0.900. Cause A has interval probability 0.040, and cause B accounts for 35% of all exits. Cause C is the only other exit cause. Find the interval probability for C. (Answer: 0.025)
The remaining probability is 0.880, cause A has probability 0.050, and cause B accounts for one-third of all exits. Find the interval probability for the only other cause. (Answer: 0.030)
A hypothetical model has remaining probability 0.940 and known cause probabilities 0.020, 0.015, and 0.010. Find the probability of the one missing cause and its share among all exits. (Answer: missing probability 0.015; share among exits 0.2500, or 25%)

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