Introduction
Competing-decrement probabilities can be calculated in several ways, but the appropriate method depends on how the cause-specific intensities are specified. Constant or specially related intensities may allow a direct formula. Intensities that change by interval can often be handled exactly interval by interval. Smooth functions without a convenient antiderivative usually call for numerical integration.
The central calculation is the same in all three cases: at each time, an exit from one cause can occur only among those who have not already exited from any cause. The method changes how the resulting integral is evaluated, not what the integral represents.
This definition uses elapsed time from the start of the interval. If the problem gives age or policy duration instead, use that variable consistently in the intensity functions and integration limits. Rates stated per year require time measured in years; a monthly interval must either be converted to years or use rates expressed per month.
How to Choose a Calculation Method
First inspect the total intensity \(\lambda(t)=\sum_j\lambda_j(t)\), because it determines survival. Then inspect each cause-specific intensity in the integrand \(S(t)\lambda_j(t)\). A simple total intensity does not necessarily make every cause integral simple, and a simple cause share does not by itself determine the survival factor.
| Form of the intensities | Usual method | Key check |
|---|---|---|
| Constant rates or functions with manageable antiderivatives | Closed-form integration | Use the cumulative total intensity in survival |
| Rates constant on specified successive intervals | Exact piecewise-constant calculation | Carry forward the cohort remaining at each boundary |
| Smooth functions without a convenient antiderivative | Numerical integration | Check approximation accuracy and probability totals |
These are choices of computational method, not different decrement models. For exact piecewise-constant calculations, each interval has its own total rate and cause shares. For numerical integration, a quadrature rule approximates the integral of the cause-specific exit density. In either case, the probability of surviving all causes to \(T\) is determined by the same cumulative total intensity.
Worked Example: Closed Form With Proportional Intensities
A hypothetical benefit program tracks active members for one year. For elapsed time \(0\le t\le1\), the cause-specific intensities per year are \(\lambda_1(t)=0.2(1+t)\) and \(\lambda_2(t)=0.1(1+t)\). Find each cause probability and the probability of remaining active at year 1.
Method. The intensities have a common time-varying factor. Their ratio is constant, so the cause-specific probabilities can be found from the total exit probability and the fixed cause shares.
The total intensity is $$ \lambda(t)=0.2(1+t)+0.1(1+t)=0.3(1+t). $$ Its cumulative value to time \(t\) is $$ \int_0^t 0.3(1+s)\,ds=0.3\left(t+\frac{t^2}{2}\right). $$ Thus the probability of remaining active to year 1 is $$ S(1)=\exp\left[-0.3\left(1+\frac{1}{2}\right)\right] =e^{-0.45}=0.6376282. $$ The total probability of exit by either cause is \(1-S(1)=0.3623718\). The cause shares are \(0.2/0.3=2/3\) and \(0.1/0.3=1/3\), constant at every time. Therefore $$ q_1(1)=\frac{2}{3}(0.3623718)=0.2415812,\qquad q_2(1)=\frac{1}{3}(0.3623718)=0.1207906. $$ To four decimal places, the answers are \(q_1=0.2416\), \(q_2=0.1208\), and \(S(1)=0.6376\). The check is \(0.2416+0.1208+0.6376=1.0000\). The common mistake avoided here is treating the intensities as constant just because their ratio is constant; the cumulative total intensity still has to be integrated over time.
Why the Cause-Specific Integral Works
Over a short interval of width \(\Delta t\) beginning at time \(t\), an individual who is still active has probability approximately \(\lambda_j(t)\Delta t\) of exiting from cause \(j\), while the probability of having remained active up to that point is \(S(t)\). Thus the unconditional probability of a cause-\(j\) exit in that short interval is approximately \(S(t)\lambda_j(t)\Delta t\). Adding these contributions over the interval and taking the limit gives the integral defining \(q_j(T)\).
This argument also explains an important check. Differentiating survival gives \(S'(t)=-S(t)\sum_j\lambda_j(t)\). Integrating from 0 to \(T\), with \(S(0)=1\), shows that the probabilities of all cause-specific exits and survival add to one:
This identity assumes the listed causes account for all exits from the active state over the interval. If the model includes another exit cause, it must be included in the total intensity and in the probability check.
Exact Calculation Across Piecewise-Constant Intervals
Suppose the intensities are constant within each specified interval. On an interval of length \(h\), let the cause-\(j\) intensity be \(\lambda_j\), and let \(\lambda=\sum_j\lambda_j\). Conditional on being active at the start of that interval, the probability of an exit from cause \(j\) during it is $$ \frac{\lambda_j}{\lambda}\left(1-e^{-\lambda h}\right), $$ when \(\lambda>0\). This formula is exact. The term in parentheses is the probability of any exit during the interval; the fraction is the cause’s share of the total intensity, which is constant throughout that interval. If \(\lambda=0\), all cause exit probabilities on that interval are zero.
Worked Example: Two Successive Constant-Intensity Intervals
For a hypothetical coverage group observed over two years, the cause intensities per year are \((0.2,0.1)\) during year 1 and \((0.1,0.3)\) during year 2. Calculate the two cause probabilities over the full period and the probability of remaining active.
Method. Calculate conditional probabilities within each interval, then weight year 2 exits by the probability of still being active at the start of year 2.
In year 1 the total intensity is \(0.2+0.1=0.3\). The probability of any exit is \(1-e^{-0.3}=0.2591818\). Hence the cause probabilities conditional on being active at time 0 are $$ q_{1,1}=\frac{0.2}{0.3}(1-e^{-0.3})=0.1727879,\qquad q_{2,1}=\frac{0.1}{0.3}(1-e^{-0.3})=0.0863939. $$ The probability still active at year 1 is \(e^{-0.3}=0.7408182\).
In year 2 the total intensity is \(0.1+0.3=0.4\), and the conditional probability of any exit during that year is \(1-e^{-0.4}=0.3296799\). The conditional cause shares are \(0.1/0.4=0.25\) and \(0.3/0.4=0.75\). Thus the unconditional year 2 cause probabilities, measured from time 0, are $$ q_{1,2}=e^{-0.3}(0.25)(1-e^{-0.4}) =0.25(e^{-0.3}-e^{-0.7})=0.0610582, $$ $$ q_{2,2}=e^{-0.3}(0.75)(1-e^{-0.4}) =0.75(e^{-0.3}-e^{-0.7})=0.1831747. $$ The full-period probabilities are \(q_1=0.1727879+0.0610582=0.2338461\) and \(q_2=0.0863939+0.1831747=0.2695686\). The probability of remaining active is \(e^{-0.3}e^{-0.4}=e^{-0.7}=0.4965853\). To four decimal places, these are \(0.2338\), \(0.2696\), and \(0.4966\); their rounded sum is 1.0000. The common mistake avoided is adding year 2’s conditional cause probabilities without multiplying by the probability \(0.7408182\) of reaching that year still active.
Numerical Integration for Smooth Intensities
When intensities vary smoothly but the integrals are inconvenient to evaluate exactly, apply a numerical integration rule to \(S(t)\lambda_j(t)\). Simpson’s rule is one option when the integrand is sufficiently smooth. With an even number \(n\) of equal subintervals of width \(h=T/n\), it approximates an integral using endpoint values, alternating weights of 4 and 2 at interior points, and the multiplier \(h/3\). More subintervals generally improve the approximation for a smooth integrand, although a numerical result should still be checked against probability identities and, where needed, a finer calculation.
Worked Example: Simpson’s Rule for Time-Varying Causes
A hypothetical group has cause intensities \(\lambda_1(t)=0.1+0.1t\) and \(\lambda_2(t)=0.2\) per year over \(0\le t\le1\). Use Simpson’s rule with \(n=4\) equal subintervals to approximate each exit probability and find the survival probability.
Method. First calculate survival from the integrated total intensity, then apply Simpson’s rule to each cause-specific exit density \(S(t)\lambda_j(t)\). The total intensity is \(0.3+0.1t\), so $$ S(t)=\exp\left(-0.3t-0.05t^2\right). $$ At \(t=0,0.25,0.5,0.75,1\), the survival values are approximately \(1.000000, 0.924847, 0.850017, 0.776372, 0.704688\). The cause 1 intensities at those points are \(0.100, 0.125, 0.150, 0.175, 0.200\). Therefore the cause 1 integrand \(S(t)\lambda_1(t)\) has values approximately \(0.100000, 0.115606, 0.127503, 0.135865, 0.140938\).
Here \(h=0.25\). Simpson’s rule gives $$ q_1(1)\approx \frac{0.25}{3}\left[0.100000+0.140938 +4(0.115606+0.135865)+2(0.127503)\right] =0.1252. $$ For cause 2, multiply each survival value by \(0.2\). The cause 2 integrand values are approximately \(0.200000, 0.184969, 0.170003, 0.155274, 0.140938\). Hence $$ q_2(1)\approx\frac{0.25}{3}\left[0.200000+0.140938 +4(0.184969+0.155274)+2(0.170003)\right] =0.1702. $$ The exact survival probability at year 1 is \(S(1)=e^{-0.35}=0.7047\). The approximate exit probabilities sum to \(0.1252+0.1702=0.2954\), compared with the total exit probability \(1-0.704688=0.295312\). The small difference is due to numerical integration and rounding; to four decimals the sum of the separately rounded probabilities is \(1.0001\). A finer quadrature can be used if greater accuracy is required. The common mistake avoided is integrating the cause intensity alone: the integrand must also include the probability of still being exposed at time \(t\).
A Common Trap and a Practical Selection Routine
In the numerical example, cause 1’s instantaneous share of total intensity at time 0 is \(0.1/0.3=1/3\). Multiplying that initial share by the total exit probability would give approximately \((1/3)(0.295312)=0.0984\), not the correct \(0.1252\). The share changes over time because cause 1’s intensity increases while cause 2’s is constant. A constant cause share can be used across an interval only when the cause intensities maintain that share throughout the interval.
Confirm that intensity units and time limits match, and identify all causes included in the model.
Add the cause-specific intensities and integrate the total to obtain survival.
Use closed-form integration when the relevant integrals are manageable, exact interval calculations when rates are piecewise constant, and numerical quadrature for smooth functions without a convenient antiderivative.
Verify that each cause probability is nonnegative and that cause probabilities plus survival equal one, allowing for numerical approximation and rounding.
Method selection matters because a correct-looking formula can still produce an incorrect answer if it ignores time variation, loses the surviving cohort at an interval boundary, or uses inconsistent time units. These checks are useful whenever competing decrements are calculated, whether the final output is a probability, an expected number of exits, or an input to a later actuarial valuation.
Check Your Understanding
For each problem, identify or apply the appropriate calculation method. Probabilities requested as decimals should be reported to four decimal places.
- A hypothetical model has constant annual intensities \(0.12\) for cause 1 and \(0.08\) for cause 2 over three years. Find both cause probabilities and the probability of remaining active. (Answer: \(q_1=0.2707,\ q_2=0.1805,\ S(3)=0.5488\).)
- For the first half-year, a group has constant annual intensities \(0.2\) and \(0.1\); for the next half-year, the intensities are \(0.1\) and \(0.3\). Find the full-year probabilities for both causes and for remaining active. (Answer: \(q_1=0.1319,\ q_2=0.1634,\ S(1)=0.7047\).)
- Over two years, intensities are \(\lambda_1(t)=0.3(1+t)\) and \(\lambda_2(t)=0.2(1+t)\). Use the proportional-intensity closed-form approach to find the cause probabilities and survival. (Answer: \(q_1=0.5188,\ q_2=0.3459,\ S(2)=0.1353\).)
- A cause-specific intensity is \(0.05+0.04t\) and a second cause has constant intensity \(0.12\), both per year on \(0\le t\le1\). There is no convenient elementary antiderivative for both cause integrals. State a suitable method and the integrand to approximate for cause 1. (Answer: numerical integration of \((0.05+0.04t)\exp[-0.17t-0.02t^2]\) over \(0\le t\le1\).)