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How an Added Cause Changes Existing Exit Probabilities

Learn to recalculate remaining probabilities and original-cause exits after adding a competing decrement, without changing the original intensities.

Intermediate 10 min read

What You'll Learn

  • Define remaining and cumulative cause-specific probabilities from cause-specific intensities
  • Recalculate probabilities when an additional cause is introduced
  • Compare constant-intensity and piecewise-constant models
  • Explain why original-cause exits cannot increase when a competing intensity is added
  • Avoid confusing fixed cause-specific intensities with fixed cumulative probabilities
  • Check that all cause probabilities and the remaining probability sum to one

Why an Added Cause Changes More Than the Total Exit Probability

In a multiple decrement model, each cause has its own cause-specific intensity. Holding an original cause’s intensity fixed does not hold its cumulative exit probability fixed. An added cause removes some members from the group that remains exposed to the original causes, so fewer can eventually exit from those original causes. This distinction matters whenever probabilities, rather than just instantaneous rates, are being compared.

The calculations below assume a single cohort that begins in the active state. Exit causes are mutually exclusive, and after an exit the individual is no longer at risk of a later exit in this model. Time is measured in years. Intensities may be constant or time-varying, but the relevant rates and assumptions are stated in each example. The results are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: In a continuous-time competing decrement model, \(\mu_j(t)\) is the cause-specific intensity for cause \(j\) at time \(t\), conditional on being in the active state immediately before \(t\). The remaining probability at time \(t\) is the probability that no cause has yet occurred. The cumulative probability of exit from cause \(j\) by time \(t\) is the probability that the first exit has occurred by \(t\) and was due to cause \(j\).

The intensity is conditional on still being at risk; it is not itself a probability over an interval. To obtain an interval exit probability, the model must account for the chance of remaining exposed until each possible exit time. In particular, if an additional cause is added, it contributes to the total intensity governing how quickly the active cohort is depleted.

Remaining Probability and Cause-Specific Exits

Suppose the cause-specific intensities are \(\mu_1(t),\ldots,\mu_r(t)\). The total intensity of exit at time \(t\) is their sum. Write \(S(t)\) for the probability that an individual remains active through time \(t\), and \(q_j(t)\) for the cumulative probability of exit due to cause \(j\) by time \(t\). For deterministic intensities and no other transitions, the model gives:

$$ S(t)=\exp\left(-\int_0^t\sum_{k=1}^{r}\mu_k(s)\,ds\right), \qquad q_j(t)=\int_0^t S(s)\mu_j(s)\,ds. $$

The first expression uses the total intensity because remaining active requires avoiding every cause. The second expression accumulates the probability of being active at time \(s\), multiplied by the instantaneous intensity of cause \(j\), over all possible exit times up to \(t\). Probability is conserved:

$$ S(t)+\sum_{j=1}^{r}q_j(t)=1. $$

These formulas describe continuous-time exits with the stated intensities. They do not apply unchanged if exits can occur simultaneously with positive probability, if the intensities are conditional on something other than still being active, or if the model includes additional transitions that alter who is exposed. In the standard competing decrement setup, simultaneous exits have probability zero.

Theorem: Suppose the original cause-specific intensities \(\mu_j(t)\) remain unchanged and an additional cause with intensity \(\nu(t)\geq 0\) is introduced. Then the new remaining probability is no greater than the original remaining probability at every time. The cumulative exit probability for each original cause is also no greater than before.

Why Original-Cause Probabilities Decrease

Let \(H(t)=\int_0^t\sum_j\mu_j(s)\,ds\) be the cumulative total intensity from the original causes, and let \(G(t)=\int_0^t\nu(s)\,ds\) be the cumulative intensity of the added cause. Before the additional cause is introduced, the remaining probability is \(\exp(-H(t))\). After it is introduced, the remaining probability is \(\exp(-H(t)-G(t))\). Since \(G(t)\geq 0\), the new remaining probability is no greater.

For any original cause \(j\), its old and new cumulative probabilities are respectively:

$$ q_j(t)=\int_0^t e^{-H(s)}\mu_j(s)\,ds, \qquad q_j^{\text{new}}(t)=\int_0^t e^{-H(s)-G(s)}\mu_j(s)\,ds. $$

Because \(\mu_j(s)\geq 0\) and \(e^{-G(s)}\leq 1\), the new integrand is no greater at every time \(s\). Integrating over the same interval proves \(q_j^{\text{new}}(t)\leq q_j(t)\). The inequality is strict if the added cause has positive cumulative intensity over times when the original cause has positive intensity. This comparison does not say that every original exit probability falls by the same percentage.

Constant-intensity result: If the original causes have constant intensities \(\lambda_1,\ldots,\lambda_r\), and an added cause has constant intensity \(\gamma\), over an interval of length \(t\), then the new remaining probability and the cumulative probability for original cause \(j\) are
$$ S_{\text{new}}(t)=e^{-(\lambda_1+\cdots+\lambda_r+\gamma)t}, \qquad q_j^{\text{new}}(t)= \frac{\lambda_j}{\lambda_1+\cdots+\lambda_r+\gamma} \left(1-e^{-(\lambda_1+\cdots+\lambda_r+\gamma)t}\right). $$

This follows by substituting the constant total intensity into the remaining-probability formula, then integrating the cause-specific exit density:

$$ q_j^{\text{new}}(t) =\int_0^t e^{-(\lambda_1+\cdots+\lambda_r+\gamma)s}\lambda_j\,ds =\frac{\lambda_j}{\lambda_1+\cdots+\lambda_r+\gamma} \left(1-e^{-(\lambda_1+\cdots+\lambda_r+\gamma)t}\right). $$

Worked Calculations

Worked Example: Adding a Third Constant Cause

A hypothetical one-year contract has two competing exits, lapse \(A\) and disability \(B\), with constant annual intensities \(\lambda_A=0.10\) and \(\lambda_B=0.20\). A revised version adds a third cause \(C\) with constant annual intensity \(\gamma=0.30\). All rates are annual, the interval is one year, and there are no other exits. Find the remaining probability and the three cumulative exit probabilities before and after the change.

Method. Use the constant-intensity formula, first with total intensity \(0.10+0.20=0.30\), then with total intensity \(0.10+0.20+0.30=0.60\).

Before adding \(C\), the remaining probability is \(e^{-0.30}=0.740818\), so the total probability of exit is \(1-0.740818=0.259182\). The cause probabilities are:

$$ q_A=\frac{0.10}{0.30}(0.259182)=0.086394, \qquad q_B=\frac{0.20}{0.30}(0.259182)=0.172788. $$

After adding \(C\), the total intensity is \(0.60\), and the remaining probability is \(e^{-0.60}=0.548812\). Thus the total exit probability is \(1-0.548812=0.451188\). The cause probabilities are:

$$ q_A^{\text{new}}=\frac{0.10}{0.60}(0.451188)=0.075198, \qquad q_B^{\text{new}}=\frac{0.20}{0.60}(0.451188)=0.150396, $$ $$ q_C^{\text{new}}=\frac{0.30}{0.60}(0.451188)=0.225594. $$

Check. The new probabilities sum to \(0.075198+0.150396+0.225594+0.548812=1.000000\). Both original cause probabilities decreased, even though neither original intensity changed. Values are rounded to six decimals.

Common mistake avoided: Keeping the old \(q_A\) and \(q_B\) because their intensities were unchanged confuses an intensity conditional on remaining active with a cumulative probability for the starting cohort.

Worked Example: Adding a Cause Partway Through the Year

A hypothetical cohort has two original causes, \(A\) and \(B\), each with constant annual intensity \(0.10\) throughout a one-year period. A new cause \(C\) has intensity zero during the first half-year and \(0.20\) during the second half-year. Find the one-year remaining probability and cumulative exit probabilities after adding \(C\), then compare the original-cause probabilities with the model without \(C\).

Method. Apply the constant-intensity formulas separately over the two half-year intervals, carrying the remaining probability at the end of the first interval into the second. In the first half-year the total intensity is \(0.20\), so remaining probability is \(e^{-0.20(0.5)}=e^{-0.10}=0.904838\). The first-half exits are:

$$ q_{A,1}=q_{B,1} =\frac{0.10}{0.20}(1-e^{-0.10}) =0.047581. $$

In the second half-year the total intensity is \(0.10+0.10+0.20=0.40\). Conditional on reaching the start of that interval, the remaining probability through its end is \(e^{-0.40(0.5)}=e^{-0.20}=0.818731\). The interval exit probabilities, measured from the original cohort, are:

$$ q_{A,2}=0.904838\frac{0.10}{0.40}(1-e^{-0.20})=0.041005, $$ $$ q_{B,2}=0.041005, \qquad q_{C,2}=0.904838\frac{0.20}{0.40}(1-e^{-0.20})=0.082010. $$

Therefore \(q_A=0.047581+0.041005=0.088586\), \(q_B=0.088586\), and \(q_C=0.082010\). The final remaining probability is \(0.904838e^{-0.20}=e^{-0.30}=0.740818\).

Without \(C\), the total intensity is \(0.20\) for the full year. The remaining probability would be \(e^{-0.20}=0.818731\), and each original cause probability would be \(\frac{0.10}{0.20}(1-e^{-0.20})=0.090635\). Thus each original cause probability falls from \(0.090635\) to \(0.088586\).

Check. With \(C\), \(0.088586+0.088586+0.082010+0.740818=1.000000\) after rounding. Values are rounded to six decimals.

Common mistake avoided: Applying the second-half rates to the entire year would incorrectly make the added cause active before it begins. The probability entering each interval must also be carried forward.

Worked Example: Comparing a Two-Year Model

For a hypothetical two-year coverage period, cause \(A\) has constant annual intensity \(0.12\) and cause \(B\) has constant annual intensity \(0.08\). A proposed change adds cause \(C\) with constant annual intensity \(0.10\). Find the remaining and cumulative cause probabilities with and without \(C\), and quantify the reduction in \(A\)'s probability.

Method. Use the constant-intensity formula with \(t=2\). Before the change, total intensity is \(0.20\), so \(S=e^{-0.40}=0.670320\), and total exit probability is \(0.329680\). Hence:

$$ q_A=\frac{0.12}{0.20}(0.329680)=0.197808, \qquad q_B=\frac{0.08}{0.20}(0.329680)=0.131872. $$

After the change, total intensity is \(0.30\), so \(S_{\text{new}}=e^{-0.60}=0.548812\), and total exit probability is \(0.451188\). The cause probabilities are:

$$ q_A^{\text{new}}=\frac{0.12}{0.30}(0.451188)=0.180475, \qquad q_B^{\text{new}}=\frac{0.08}{0.30}(0.451188)=0.120317, $$ $$ q_C^{\text{new}}=\frac{0.10}{0.30}(0.451188)=0.150396. $$

The reduction in \(A\)'s cumulative probability is \(0.197808-0.180475=0.017333\), to six decimals. The model with \(C\) conserves probability: \(0.180475+0.120317+0.150396+0.548812=1.000000\).

Common mistake avoided: Dividing by the original total intensity \(0.20\) after adding \(C\) would give incorrect cause shares. Both the new total intensity and the new remaining probability determine the result.

Interpreting the Change and Checking Work

When all intensities are constant, the ratio of the cumulative probabilities for two original causes is the ratio of their intensities: for example, \(q_A/q_B=\lambda_A/\lambda_B\). Adding a constant cause reduces both original probabilities but leaves that ratio unchanged. With time-varying intensities, however, original causes may be relatively more or less active at different times. The added cause changes how much of the cohort survives to each time, so even the ratio of their cumulative probabilities need not remain the same.

A reliable calculation separates three quantities: the cause-specific intensities, the remaining probability, and the cumulative cause probabilities. Check that the total intensity includes every cause active at the relevant time, and use the remaining probability that corresponds to that same set of causes. Finally, verify probability conservation by adding all cumulative exits to the probability still remaining.

  • An added nonnegative intensity cannot increase remaining probability.
  • With original intensities fixed, it cannot increase an original cause’s cumulative exit probability.
  • The total probability of exit generally increases, since more causes can remove members from the active group.
  • A cause introduced only partway through an interval must not be applied before its start time.

Check Your Understanding

For each problem, assume a continuous-time competing decrement model with mutually exclusive exits and no other transitions. Rates are annual, and round probabilities to four decimals.

  1. Over one year, cause \(A\) has constant intensity \(0.05\), cause \(B\) has constant intensity \(0.15\), and a new cause \(C\) is added at intensity \(0.10\). Find the remaining probability and all three cumulative exit probabilities. (Answer: remaining \(0.7408\); \(q_A=0.0432,\ q_B=0.1296,\ q_C=0.0864\).)
  2. Over one year, two original causes have constant intensities \(0.08\) and \(0.12\). A new cause with constant intensity \(0.20\) is added. Find the remaining probability and the cumulative probability for each cause. (Answer: remaining \(0.6703\); original causes \(0.0659\) and \(0.0989\); added cause \(0.1648\).)
  3. Over two years, cause \(A\) has constant intensity \(0.10\), cause \(B\) has constant intensity \(0.10\), and cause \(C\) is added with constant intensity \(0.10\). Find the remaining probability and each cause probability; compare \(q_A\) with its value before \(C\) was added. (Answer: remaining \(0.5488\); each cause probability \(0.1504\); without \(C\), \(q_A=0.1648\).)
  4. Over one year, \(A\) and \(B\) each have constant intensity \(0.10\). Cause \(C\) has intensity zero in the first half-year and \(0.20\) in the second half-year. Find the one-year remaining probability and cumulative cause probabilities. (Answer: remaining \(0.7408\); \(q_A=0.0886,\ q_B=0.0886,\ q_C=0.0820\).)
  5. Suppose original intensities \(\mu_A(t)\) and \(\mu_B(t)\) are unchanged, and an added cause has intensity \(\nu(t)\geq0\). State the direction of change in remaining probability and in each original cumulative exit probability, and give the reason. (Answer: remaining probability and both original cumulative probabilities cannot increase; the added nonnegative intensity reduces survival, so each original cause’s exit-density integrand is no greater.)

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