Checking Probability Conservation

Why Probability Conservation Is a Useful Check

In a competing-decrement model, an individual who is in the group at the start of an interval either remains in the group through the end of the interval or exits by one of the modeled causes. These outcomes must account for the entire starting probability. Probability conservation turns that basic fact into a practical check: if the probabilities do not add to one, at least one calculation, conditioning base, or reported value needs attention.

This check is especially useful when several cause-specific probabilities have been calculated separately. A cause probability can look plausible on its own yet still be inconsistent with the remaining probability or with the probabilities for the other causes. The identity does not tell us which entry is wrong, but it quickly reveals that the set cannot all be correct.

Definition: Consider an individual who is in the group at time $s$, and a finite set of mutually exclusive exit causes $j=1,\ldots,m$. Let $R(s,t)$ be the conditional probability of remaining in the group through time $t$, and let $Q_j(s,t)$ be the conditional probability of exiting by cause $j$ during $(s,t]$. Under the competing-decrement model, the outcomes are exhaustive and mutually exclusive, so $$ R(s,t)+\sum_{j=1}^{m}Q_j(s,t)=1. $$

Model Assumptions and the Conservation Identity

Assume the individual begins the interval in the group and that the model assigns a cause-specific intensity $\mu_j(u)$ to each possible exit cause at time $u$. The total intensity is $\mu(u)=\sum_{j=1}^{m}\mu_j(u)$. Assume these intensities are nonnegative and integrable over the interval. An exit removes the individual from the group, and the model treats the causes as a set of competing first exits. In a continuous-time model with finite intensities, simultaneous exits have probability zero.

The remaining probability from $s$ to $t$ and the probability of exit by cause $j$ can be written as

R(s,t)=\exp\left(-\int_s^t\mu(u)\,du\right), \qquad Q_j(s,t)=\int_s^t R(s,u)\mu_j(u)\,du.

The second expression accumulates the probability of still being in the group just before time $u$, multiplied by the cause-specific exit rate at that time. These formulas use probabilities conditional on being in the group at $s$. They do not use the original cohort size or an unconditional probability from an earlier starting time.

Theorem: For the competing-decrement model just described, $$ R(s,t)+\sum_{j=1}^{m}Q_j(s,t)=1. $$ The identity applies to probabilities on the same interval and with the same conditioning base. It is not valid to add a conditional interval exit probability to an unconditional probability measured from an earlier time.

A short derivation shows why the identity must hold. Fix $s$, and let $R(s,u)$ denote the probability of remaining through time $u$. Its derivative is $-R(s,u)\mu(u)$. Define the total probability accounted for by time $t$ as the remaining probability plus all cumulative cause-specific exit probabilities.

\begin{aligned} \frac{d}{dt}\left(R(s,t)+\sum_{j=1}^{m}Q_j(s,t)\right) &=-R(s,t)\mu(t)+\sum_{j=1}^{m}R(s,t)\mu_j(t)\\ &=-R(s,t)\left(\sum_{j=1}^{m}\mu_j(t)\right) +R(s,t)\sum_{j=1}^{m}\mu_j(t)\\ &=0. \end{aligned}

The expression is therefore constant over the interval. At $t=s$, no time has elapsed: $R(s,s)=1$ and $Q_j(s,s)=0$ for every cause. The constant is one, which proves the identity. This argument requires the total intensity to equal the sum of the cause-specific intensities, and it assumes the modeled causes exhaust the possible exits being considered.

Checking an Interval Calculation Step by Step

For constant cause-specific intensities over an interval of length $h$, write the rates as $\mu_1,\ldots,\mu_m$, and let $\mu=\sum_j\mu_j$. Then the remaining probability is $e^{-\mu h}$. If $\mu>0$, integrating the cause-specific exit density gives

Q_j=\frac{\mu_j}{\mu}\left(1-e^{-\mu h}\right).

The cause-specific probabilities sum to $1-e^{-\mu h}$, the probability of any exit. Adding the remaining probability gives one. The calculation is a useful check, but it depends on constant intensities over the interval; if intensities vary, the integral form of $Q_j(s,t)$ must be used instead.

Worked Example: Three Constant Causes Over Two Years

A hypothetical insured group is subject to three mutually exclusive exits, with constant cause-specific intensities $0.06$, $0.09$, and $0.15$ per year for two years. Calculate the probability of remaining and each cause-specific exit probability, then check conservation. Use four decimal places for the final probabilities.

Method. Add the intensities to obtain the total intensity, calculate the remaining probability, and allocate the total exit probability in proportion to the cause-specific intensities. This avoids treating an intensity itself as an interval probability.

The total intensity is

\mu=0.06+0.09+0.15=0.30\text{ per year}.

Over two years, the cumulative total intensity is $0.30(2)=0.60$. Thus the remaining probability is

R=e^{-0.60}=0.5488116361\ldots

The probability of any exit is $1-R=1-0.5488116361\ldots=0.4511883639\ldots$. The cause shares are $0.06/0.30=0.20$, $0.09/0.30=0.30$, and $0.15/0.30=0.50$. Therefore,

\begin{aligned} Q_1&=0.20(0.4511883639\ldots)=0.0902376728\ldots,\\ Q_2&=0.30(0.4511883639\ldots)=0.1353565092\ldots,\\ Q_3&=0.50(0.4511883639\ldots)=0.2255941820\ldots. \end{aligned}

Rounded to four decimal places, the remaining probability is $0.5488$, and the exit probabilities are $0.0902$, $0.1354$, and $0.2256$. The rounded check is

$$ 0.5488+0.0902+0.1354+0.2256=1.0000. $$

Using the unrounded values, the sum is also one: the cause-specific probabilities add to $0.4511883639\ldots$, which complements the remaining probability. A rounding check should use the displayed rounded entries, while the underlying identity holds for the unrounded values.

Worked Example: From Conditional Probabilities to Expected Counts

A hypothetical cohort of 2,500 people is active at the start of a one-year interval. During that interval, the two competing exit intensities are constant at $0.12$ and $0.08$ per year. Find the conditional probability of remaining and the two cause-specific exit probabilities. Then calculate the expected number remaining and the expected exits by each cause.

Method. First calculate probabilities conditional on being active at the interval’s start. Multiply each probability by 2,500 only after checking that the conditional probabilities sum to one.

The total intensity is $0.12+0.08=0.20$ per year. For a one-year interval, the remaining probability is

R=e^{-0.20}=0.8187307531\ldots

The probability of any exit is $1-0.8187307531\ldots=0.1812692469\ldots$. The cause shares are $0.12/0.20=0.60$ and $0.08/0.20=0.40$, giving

\begin{aligned} Q_1&=0.60(0.1812692469\ldots)=0.1087615481\ldots,\\ Q_2&=0.40(0.1812692469\ldots)=0.0725076988\ldots. \end{aligned}

The conditional probability check is $0.8187307531\ldots+0.1087615481\ldots+0.0725076988\ldots=1$, allowing for the displayed digits. Multiplying by the 2,500 people gives expected counts:

\begin{aligned} \text{remaining}&=2500(0.8187307531\ldots)=2046.8268828\ldots,\\ \text{cause 1 exits}&=2500(0.1087615481\ldots)=271.9038703\ldots,\\ \text{cause 2 exits}&=2500(0.0725076988\ldots)=181.2692470\ldots. \end{aligned}

To two decimal places, the expected counts are $2{,}046.83$, $271.90$, and $181.27$. They sum to $2{,}500.00$. These are expected counts, so decimals are appropriate even though actual observed counts are whole numbers.

A common mistake is to multiply the first cause’s conditional exit probability by the original cohort size but apply the second cause’s probability to the number remaining after cause 1. That would apply a sequential process rather than competing first exits and would change the model. Both conditional cause probabilities here use the same starting cohort of 2,500.

Using Conservation to Audit a Probability Table

A conservation check is an audit, not a method for silently adjusting every entry. If a table fails the check, first confirm that all values refer to the same interval, the same starting group, and mutually exclusive outcomes. Then identify which entries are independently known or supported by the model. Only when the problem supplies a basis for treating some entries as correct should the remaining entry be found by subtraction.

Worked Example: Finding an Inconsistent Cause Probability

A hypothetical interval report gives a conditional remaining probability of $0.86$, exit probability $0.07$ for cause A, $0.04$ for cause B, and $0.02$ for cause C. A separate check has confirmed the remaining probability and the cause A and B values. Diagnose the report and find the corrected cause C probability. Then give expected counts for 5,000 people active at the interval’s start.

Method. Add the entries using the common conditional starting base. Since the remaining probability and two cause probabilities are confirmed, use conservation to recover the only unconfirmed entry.

As reported, the sum is

$$ 0.86+0.07+0.04+0.02=0.99. $$

The reported cause C value is therefore inconsistent with the confirmed entries. Its corrected value must satisfy

$$ Q_C=1-0.86-0.07-0.04=0.03. $$

The corrected conditional probabilities check to $0.86+0.07+0.04+0.03=1.00$. For 5,000 people, the expected counts are $5{,}000(0.86)=4{,}300$ remaining, $5{,}000(0.07)=350$ exits by A, $5{,}000(0.04)=200$ exits by B, and $5{,}000(0.03)=150$ exits by C. The counts sum to $4{,}300+350+200+150=5{,}000$.

If none of the entries had been independently validated, the sum of $0.99$ would establish an inconsistency but would not identify which value was wrong. Renormalizing all four values by dividing by $0.99$ would impose an adjustment not justified by the given information.

Common Exam Traps and Practical Checks

Omitting the remaining probability. Exit probabilities alone generally sum to less than one. Their sum is the probability of any exit, not the probability of all outcomes.
Mixing conditioning bases. Conditional probabilities for an interval beginning at $s$ must be added to the conditional remaining probability for that same interval. Do not combine them with an unconditional cohort probability from an earlier starting time.
Using overlapping outcomes. The cause categories must describe mutually exclusive first exits. If outcomes overlap or a modeled exit category is omitted, the stated conservation identity cannot be applied as written.
Overinterpreting a failed check. A sum different from one detects a problem; it does not determine which component is wrong unless additional information identifies trusted entries.
Rounding too early. Carry unrounded values through the calculation, then round the final probabilities. Small differences in a rounded sum can arise from rounding, so compare both the displayed total and the unrounded calculation.

For an interval calculation, a reliable final audit is to write the remaining probability and every cause-specific exit probability with the same conditioning statement, add them, and compare the total with one. For cohort counts, multiply all of those conditional probabilities by the same number active at the interval’s start; the resulting expected counts must add back to that starting cohort.

Key takeaway: In a competing-decrement model, remaining in the group and exiting by each modeled cause are exhaustive, mutually exclusive outcomes. Their probabilities must sum to one when they refer to the same interval and conditioning base.

Check Your Understanding

For each question, use a common conditioning base for the remaining probability and all cause-specific exits. Round probabilities to four decimal places where needed.

A hypothetical group has two constant exit intensities, $0.04$ and $0.06$ per year, over one year. Find the remaining probability and each cause-specific exit probability. Check the rounded sum. (Answer: $0.9048$ remain; $0.0381$ and $0.0571$ exit; rounded total $1.0000$.)
Two constant cause-specific intensities are $0.08$ and $0.12$ per year over half a year. Find the remaining probability and the two exit probabilities. (Answer: $0.9048$ remain; $0.0381$ and $0.0571$ exit; total $1.0000$.)
A conditional interval table reports remaining probability $0.72$, and exit probabilities $0.11$, $0.09$, and an unknown value for the third cause. Find the missing value. (Answer: $0.08$.)
A cohort of 1,200 is active at an interval’s start. Its conditional probabilities are $0.85$ remaining, $0.09$ exiting by A, $0.04$ by B, and $0.02$ by C. Calculate the expected counts and verify their total. (Answer: 1,020 remain; 108 exit by A; 48 by B; 24 by C; total $1{,}200$.)
For two constant causes with intensities $0.05$ and $0.10$ per year over two years, calculate the remaining probability and each cause-specific exit probability, then check conservation. (Answer: $0.7408$ remain; $0.0864$ and $0.1728$ exit; rounded total $1.0000$.)

What You'll Learn