Why the Time Path Matters
In a competing decrement model, the intensity for a cause can change while a person remains in the group. It may depend on attained age, on elapsed duration since a specified event, or on both. To calculate probabilities over an interval, first express each of those quantities as a function of time since the start of the interval. A formula in age and duration is not yet a formula in time until the model’s time origins are identified.
For example, attained age generally increases at one year per year of elapsed time. Duration also increases at that rate, but its starting value depends on what duration measures: time since policy issue, time since entering a state, or another explicitly defined event. Those starting values need not be the same.
From Intensities to Interval Probabilities
Assume a continuous-time competing decrement model with \(m\) mutually exclusive causes. While an individual remains in the group, cause \(j\) has nonnegative intensity \(\mu_j(t)\) per year along the individual’s path. Define the total intensity by \(\mu_{\mathrm{tot}}(t)=\sum_{j=1}^{m}\mu_j(t)\). The probability of remaining to time \(t\), conditional on being in the group at time zero, is the exponential of the negative cumulative total intensity. The probability of exit by cause \(j\) by time \(T\) is the integral of the probability of still being present just before \(t\), multiplied by the cause-\(j\) intensity at \(t\).
Here \(q_j(T)\) is conditional on membership in the group at the beginning of the interval. These formulas apply when the specified intensities describe the competing exits for an individual who remains in the group; they do not treat an intensity as an interval probability. If cause intensities are functions of age and duration, substitute the corresponding paths before integrating.
Deriving the Cause-Specific Probability Formula
The cause-specific formula follows by considering a short interval of length \(dt\). Conditional on still being in the group at time \(t\), the probability of an exit from cause \(j\) during that short interval is \(\mu_j(t)\,dt+o(dt)\). The probability of reaching time \(t\) is \({}_tp_{\mathrm{remain}}\). Thus the unconditional probability of a cause-\(j\) exit in that short interval is \({}_tp_{\mathrm{remain}}\mu_j(t)\,dt+o(dt)\). Integrating this exit density from zero to \(T\) gives \(q_j(T)\).
The remaining probability itself follows from the survival differential equation. Its derivative is the negative of the total intensity times the probability of still remaining:
Separating variables and integrating gives \(\ln({}_tp_{\mathrm{remain}})=-\int_0^t\mu_{\mathrm{tot}}(u)\,du\), and exponentiating yields the stated remaining-probability formula. Since every exit is assigned to one of the modeled causes, the probabilities satisfy \({}_Tp_{\mathrm{remain}}+\sum_jq_j(T)=1\), provided the causes cover all exits from the group.
Worked Calculations Along Age and Duration Paths
Worked Example: A Shared Age-Dependent Multiplier
A hypothetical group is observed for two years. At the start, a member is age 60. The cause-specific intensities per year depend on attained age as follows: \(\mu_1(a)=0.020[1+0.5(a-60)]\) and \(\mu_2(a)=0.010[1+0.5(a-60)]\). Find the probability of remaining and the probability of exit from each cause by the end of the interval. Assume age increases continuously in years and these are the only exit causes.
Method. Set \(a(t)=60+t\), for \(0\le t\le2\). Substitution gives \(\mu_1(t)=0.020(1+0.5t)\) and \(\mu_2(t)=0.010(1+0.5t)\). Therefore the total intensity is \(\mu_{\mathrm{tot}}(t)=0.030(1+0.5t)\).
The cumulative total intensity is \(\int_0^2 0.030(1+0.5t)\,dt=0.030[t+0.25t^2]_0^2=0.030(2+1)=0.0900\). The probability of remaining is \(e^{-0.0900}=0.9139\), to four decimal places. The probability of any exit is \(1-0.913931\ldots=0.0861\).
The two causes have a constant intensity ratio: \(\mu_1(t)/\mu_{\mathrm{tot}}(t)=2/3\), with the remaining \(1/3\) assigned to cause 2. Because that ratio is constant throughout the interval, the exit probabilities split the total exit probability in those proportions: \(q_1(2)=(2/3)(1-e^{-0.09})=0.0574\) and \(q_2(2)=(1/3)(1-e^{-0.09})=0.0287\), to four decimal places.
Check. \(0.9139+0.0574+0.0287=1.0000\) after rounding. The shared age multiplier makes the cause proportions constant; that shortcut would not apply if the causes had different age patterns. A common mistake is to integrate the multiplier as \(2+0.5(2)^2=4\); the correct integral of \(1+0.5t\) is \(2+0.25(2)^2=3\).
Worked Example: Duration Since Entry and Attained Age Both Advance
A hypothetical member enters a covered state at age 60 after five years of policy duration. Over the next two years, the two cause-specific intensities per year are \(\mu_1(a,s)=0.010[1+0.1(a-60)+0.1(s-5)]\) and \(\mu_2(a,s)=0.020[1+0.1(a-60)+0.1(s-5)]\). Here \(s\) is policy duration, not time since entering the covered state. Calculate the probability of remaining and both cause-specific exit probabilities.
Method. At the interval start, \(a_0=60\) and \(s_0=5\). Therefore \(a(t)=60+t\) and \(s(t)=5+t\), so the shared factor in both intensities is \(1+0.2t\). The total intensity along the path is \(0.030(1+0.2t)\).
The cumulative total intensity through two years is \(\int_0^2 0.030(1+0.2t)\,dt=0.030[t+0.1t^2]_0^2=0.030(2+0.4)=0.072\). Thus the probability of remaining is \(e^{-0.072}=0.9305\), to four decimal places, and the total probability of exit is \(1-e^{-0.072}=0.0695\).
The cause intensities remain in the ratio \(1:2\) at every time. Hence \(q_1(2)=(1/3)(1-e^{-0.072})=0.0232\) and \(q_2(2)=(2/3)(1-e^{-0.072})=0.0463\), to four decimal places.
Check. The rounded probabilities sum to \(0.9305+0.0232+0.0463=1.0000\). A common time-origin error is to set \(s(0)=0\) because the member has just entered the covered state. That would be wrong here: policy duration is already five years. If duration instead meant time since state entry, its starting value would be zero, but that is not the definition in this problem.
Worked Example: Different Age and Duration Effects by Cause
A hypothetical member is age 50 and at policy duration four at the beginning of a two-year interval. The cause intensities per year are \(\mu_1(a)=0.010+0.002(a-50)\) and \(\mu_2(s)=0.030-0.002(s-4)\). Age and policy duration both advance continuously. Find the remaining probability and each cause-specific exit probability.
Method. Along the path \(a(t)=50+t\) and \(s(t)=4+t\), the intensities are \(\mu_1(t)=0.010+0.002t\) and \(\mu_2(t)=0.030-0.002t\). Both are nonnegative throughout \(0\le t\le2\). Their sum is the constant \(0.040\), so the probability of remaining to time \(t\) is \(e^{-0.040t}\).
The probability of remaining to year two is \(e^{-0.040(2)}=e^{-0.08}=0.9231\). For cause 1, \(q_1(2)=\int_0^2e^{-0.04t}(0.010+0.002t)\,dt\). To evaluate it, use \(\int_0^T e^{-kt}\,dt=(1-e^{-kT})/k\) and \(\int_0^T te^{-kt}\,dt=[1-e^{-kT}(1+kT)]/k^2\), the latter obtained by integration by parts. With \(k=0.04\) and \(T=2\), these integrals are \(1.922091\) and \(1.896466\), respectively. Therefore \(q_1(2)=0.010(1.922091)+0.002(1.896466)=0.0192209+0.0037929=0.0230138\), or \(0.0230\) to four decimal places.
The total exit probability is \(1-e^{-0.08}=0.0768837\). Thus \(q_2(2)=0.0768837-0.0230138=0.0538699\), or \(0.0539\) to four decimal places. This subtraction is valid because the two causes are exhaustive and mutually exclusive.
Check. The rounded values sum to \(0.9231+0.0230+0.0539=1.0000\). The causes do not have a constant ratio, so splitting total exits in proportion to their initial intensities would be incorrect. The correction is to integrate each cause’s intensity against the probability of remaining at each time.
Common Time-Origin and Integration Traps
Age and duration often advance at the same rate, but that does not make them interchangeable. Their values at the interval start can differ, and their intensity functions can use different origins. Before integrating, write down the time path for every variable that appears in a cause intensity.
State the attained age and the value of each duration measure at time zero. Read the definition of duration rather than inferring it from a state transition.
For age and duration measured in years, use \(a(t)=a_0+t\) and \(s(t)=s_0+t\), unless the model specifies a different clock or the quantity resets at an event.
Add the cause-specific intensities along the path, then integrate that total to obtain the remaining probability.
Integrate the cause’s intensity multiplied by the probability of remaining up to that time. Use a fixed share of total exits only when the cause share is constant over the interval.
A second frequent error is to integrate each cause intensity by itself and treat the result as its exit probability. That ignores competition: an individual can exit from a cause at time \(t\) only if still present then. The factor \({}_tp_{\mathrm{remain}}\) in the cause-specific exit density accounts for that condition.
Check Your Understanding
For each problem, assume the stated causes are exhaustive and intensities are measured per year. Give probabilities to four decimal places.
- A hypothetical member begins at age 70. Over two years, \(\mu_1(a)=0.012[1+0.05(a-70)]\) and \(\mu_2(a)=0.018[1+0.05(a-70)]\). Find the remaining probability and both cause-specific exit probabilities. (Answer: remaining \(0.9389\); cause 1 \(0.0244\); cause 2 \(0.0366\).)
- A member begins at policy duration three. Over two years, \(\mu_1(s)=0.005[1+0.1(s-3)]\) and \(\mu_2(s)=0.015[1+0.1(s-3)]\). Find the remaining probability and both cause-specific exit probabilities. (Answer: remaining \(0.9570\); cause 1 \(0.0108\); cause 2 \(0.0323\).)
- Over one year, a hypothetical member follows \(\mu_1(t)=0.020+0.010t\) and \(\mu_2(t)=0.030-0.010t\). Find the remaining probability and both cause-specific exit probabilities. (Answer: remaining \(0.9512\); cause 1 \(0.0243\); cause 2 \(0.0244\).)
- A member has attained age 40 and policy duration two at the start of a one-year interval. The cause intensities are \(\mu_1(s)=0.010[1+0.1s]\) and \(\mu_2(s)=0.020[1+0.1s]\). Find the remaining probability and both exit probabilities, using policy duration rather than time since interval start in the intensity formula. (Answer: remaining \(0.9632\); cause 1 \(0.0123\); cause 2 \(0.0245\).)
- A member starts at age 65 and policy duration seven. For three years, \(\mu_1(a,s)=0.006[1+0.02(a-65)+0.03(s-7)]\) and \(\mu_2(a,s)=0.012[1+0.02(a-65)+0.03(s-7)]\). Find the remaining probability and each cause-specific exit probability. (Answer: remaining \(0.9436\); cause 1 \(0.0188\); cause 2 \(0.0376\).)