Why Comparing Decrement Models Requires More Than One Measure
Suppose two hypothetical groups are exposed to the same types of exit, such as death and withdrawal, but their rates differ. A comparison based only on the death rate can answer which group has the higher instantaneous death intensity among people still active. It does not, by itself, answer which group has more deaths over a year, which has more exits from all causes, or which has a greater probability of remaining active.
The distinction matters because a person can experience only the first exit in a multiple-decrement model. A competing exit removes that person from the group still exposed to the other causes. Thus both the rates and the time spent remaining exposed contribute to the number of exits attributed to each cause.
The cause-specific intensity is an instantaneous rate among those still active; it is not an interval probability. The remaining probability reflects the effect of every cause. Cumulative incidence combines the cause-specific intensity with the probability of still being exposed to it.
Core Relationships for Comparing Models
For continuous, nonnegative cause-specific intensities, the remaining probability satisfies a simple differential equation. In a short interval of length \(dt\), a person still active at \(t\) exits with approximate probability \(\lambda_{\text{tot}}(t)dt\). Therefore the instantaneous decrease in the probability of remaining active is \(R(t)\lambda_{\text{tot}}(t)dt\).
Among the group initially active, the probability of a cause-\(j\) exit in a short interval around \(t\) is approximately \(R(t)\lambda_j(t)dt\). Integrating that exit density gives the cumulative incidence. Summing the cause-specific incidences gives the probability of exiting from any cause by time \(t\).
For constant intensities over an interval, the expressions simplify. If the cause-specific intensities are \(\lambda_1,\ldots,\lambda_m\) throughout an interval of length \(t\), let \(\lambda_{\text{tot}}=\sum_j\lambda_j\). Then \(R(t)=e^{-\lambda_{\text{tot}}t}\). For \(\lambda_{\text{tot}}>0\), the cumulative incidence for cause \(j\) is \(\lambda_j/\lambda_{\text{tot}}\) times the all-cause exit probability. This result follows because \(R(s)=e^{-\lambda_{\text{tot}}s}\) and the cause-\(j\) exit density is \(\lambda_j e^{-\lambda_{\text{tot}}s}\); integrating from 0 to \(t\) gives the stated formula. If all intensities are zero, there are no exits and every cause-specific cumulative incidence is zero.
Worked Comparisons
Worked Example: A Higher Cause-Specific Rate Does Not Mean More All-Cause Exits
Problem. Compare two hypothetical groups over one year. In Model A, the constant death intensity is \(0.20\) per year and the constant withdrawal intensity is \(0.10\) per year. In Model B, the corresponding intensities are \(0.15\) and \(0.45\) per year. Assume these are the only first-exit causes, there is no re-entry, and intensities are constant within each group. There is no interest or discounting because this is a probability comparison. Find each model’s one-year remaining probability, all-cause exit probability, and death cumulative incidence.
Method. Add the cause-specific intensities to obtain the total intensity, then calculate \(R(1)\), \(1-R(1)\), and the death cumulative incidence. The constant-intensity formula applies because both cause-specific intensities are constant for the full year.
Model A. The total intensity is \(0.20+0.10=0.30\) per year. Thus \(R_A(1)=e^{-0.30}=0.740818\), and the probability of all-cause exit is \(1-0.740818=0.259182\). The death share of the total intensity is \(0.20/0.30=2/3\), so the death cumulative incidence is \((2/3)(0.259182)=0.172788\). The withdrawal cumulative incidence is \((0.10/0.30)(0.259182)=0.086394\).
Model B. The total intensity is \(0.15+0.45=0.60\) per year. Thus \(R_B(1)=e^{-0.60}=0.548812\), and the probability of all-cause exit is \(1-0.548812=0.451188\). The death share of the total intensity is \(0.15/0.60=0.25\), so the death cumulative incidence is \(0.25(0.451188)=0.112797\). The withdrawal cumulative incidence is \(0.75(0.451188)=0.338391\).
Check and conclusion. In Model A, the cause-specific death intensity is higher, and its death cumulative incidence is also higher: \(0.1728\) versus \(0.1128\). However, Model B has the higher all-cause exit probability: \(0.4512\) versus \(0.2592\). In each model, the two cause-specific incidences sum to the all-cause exit probability. This example avoids the mistake of treating “higher death intensity” and “more exits overall” as the same comparison.
Worked Example: The Same Cause Intensity Can Produce Different Cumulative Incidence
Problem. Compare two hypothetical one-year models. In both, the constant disability intensity is \(0.20\) per year. In Model A, disability is the only exit cause. In Model B, there is also a constant withdrawal intensity of \(0.80\) per year. Assume no re-entry, no other causes, and constant intensities. There is no interest or discounting. Calculate the disability cumulative incidence and all-cause exit probability in each model.
Method. Use the total intensity to calculate remaining probability, then integrate the disability exit density. The assumed constant intensities imply an exponential remaining probability with parameter equal to the sum of the intensities.
Model A. The total intensity is \(0.20\). Therefore \(R_A(1)=e^{-0.20}=0.818731\), and all-cause exit probability is \(1-0.818731=0.181269\). Since there is no competing exit, all exits are from disability: \(F_{\text{disability},A}(1)=0.181269\).
Model B. The total intensity is \(0.20+0.80=1.00\). Therefore \(R_B(1)=e^{-1}=0.367879\), and all-cause exit probability is \(1-0.367879=0.632121\). The disability share of the total intensity is \(0.20/1.00=0.20\), giving \(F_{\text{disability},B}(1)=0.20(0.632121)=0.126424\). The withdrawal cumulative incidence is \(0.80(0.632121)=0.505697\).
Check and conclusion. The disability intensity is \(0.20\) per year in both models, but the disability cumulative incidence is lower in Model B, \(0.1264\) instead of \(0.1813\). Its competing withdrawal intensity removes people from the active group before they can exit through disability. The incidences in Model B sum to \(0.126424+0.505697=0.632121\), matching the all-cause exit probability. This avoids the mistake of assuming that equal cause-specific intensities must imply equal cause-specific probabilities.
Worked Example: Timing Can Change Cause Incidence Even When Total Exits Match
Problem. Compare two hypothetical models over two years. Each model has total intensity \(0.50\) per year in both years, but the cause-specific intensities switch between the years. In Model A, the death and withdrawal intensities are \(0.40\) and \(0.10\) in year 1, then \(0.10\) and \(0.40\) in year 2. Model B has \(0.10\) and \(0.40\) in year 1, then \(0.40\) and \(0.10\) in year 2. Assume piecewise-constant intensities, no re-entry, and no other causes. There is no interest or discounting. Find the two-year remaining probability, all-cause exit probability, and death cumulative incidence for each model.
Method. Apply the constant-intensity calculation separately in each year. For year 2, multiply its conditional death exit probability by the probability of remaining active through year 1. This accounts for exposure to year 2.
In both models, \(R(1)=e^{-0.50}=0.606531\), and \(R(2)=e^{-0.50}e^{-0.50}=e^{-1}=0.367879\). Thus the two-year all-cause exit probability is \(1-0.367879=0.632121\) for each model.
Model A death incidence. During year 1, the death share of the total intensity is \(0.40/0.50=0.80\). The year-1 death incidence is \(0.80(1-e^{-0.50})=0.80(0.393469)=0.314775\). During year 2, the conditional death incidence among those active at year 1 is \((0.10/0.50)(1-e^{-0.50})=0.20(0.393469)=0.078694\). Its contribution from the original group is \(0.606531(0.078694)=0.047730\). Therefore \(F_{\text{death},A}(2)=0.314775+0.047730=0.362505\), or \(0.3625\) to four decimals.
Model B death incidence. During year 1, the death share is \(0.10/0.50=0.20\), giving \(0.20(0.393469)=0.078694\). During year 2, the death share is \(0.40/0.50=0.80\), giving conditional incidence \(0.80(0.393469)=0.314775\). The contribution from the original group is \(0.606531(0.314775)=0.190921\). Therefore \(F_{\text{death},B}(2)=0.078694+0.190921=0.269615\), or \(0.2696\) to four decimals.
Check and conclusion. Both models have the same remaining probability and all-cause exit probability after two years, but Model A has the higher death cumulative incidence. The death incidence plus the withdrawal incidence equals \(0.632121\) in each model; specifically, the withdrawal incidences are \(0.269615\) in A and \(0.362505\) in B. This avoids the mistake of ignoring the order of the cause-specific intensities when exposure changes over time.
What the Different Rankings Mean
A useful comparison states which quantity is being ranked and over what interval. A cause-specific intensity compares instantaneous rates among those still active at a given time. Remaining probability compares the chance of avoiding every exit through a stated time. All-cause exit probability compares the chance of having left by any cause. Cause-specific cumulative incidence compares the chance of a first exit from a particular cause by that time.
| Measure | Question answered | What it depends on |
|---|---|---|
| Cause-specific intensity \(\lambda_j(t)\) | How quickly does cause \(j\) produce exits among those still active at time \(t\)? | The cause-\(j\) rate at that time |
| Remaining probability \(R(t)\) | What is the probability of still being active at time \(t\)? | The combined intensities of all causes over time |
| All-cause exit probability \(1-R(t)\) | What is the probability of any exit by time \(t\)? | The combined intensities of all causes over time |
| Cause-specific cumulative incidence \(F_j(t)\) | What is the probability of a first exit from cause \(j\) by time \(t\)? | Both \(\lambda_j(s)\) and the remaining probability \(R(s)\) throughout the interval |
For piecewise-constant intensities, a reliable comparison process is to calculate remaining probability at each interval boundary, calculate each cause’s conditional exit probability within each interval, and weight that interval probability by the probability of having reached the interval still active. This last weighting is essential: an interval’s conditional probability among survivors is not yet its contribution to the original group’s cumulative incidence.
Name the cause, time horizon, and measure being compared. Do not use “risk” without clarifying whether it means intensity, cumulative incidence, or all-cause exit probability.
Sum the intensities to find the total intensity and use it to calculate the probability of remaining active.
Integrate \(R(s)\lambda_j(s)\) over the interval, or add appropriately exposure-weighted interval contributions when intensities are piecewise constant.
Confirm that all cause-specific cumulative incidences sum to \(1-R(t)\), allowing for rounding only after the calculation.
Common Exam Trap
A frequent error is to calculate a cause-\(j\) probability as \(1-e^{-\lambda_jt}\) while other competing causes are present. That expression treats cause \(j\) as if it were the only exit cause. In Model B of the disability example, it would give \(1-e^{-0.20}=0.1813\). The correct disability cumulative incidence is \(0.1264\), because the withdrawal intensity of \(0.80\) per year reduces the probability of remaining exposed to disability. The appropriate density is \(R(s)\lambda_j(s)\), not just a function of \(\lambda_j(s)\) in isolation.
A second trap is to compare cause-specific incidence without checking the time horizon or the competing causes. A higher intensity can coexist with a lower cumulative incidence if the population has less remaining exposure, and equal all-cause exit probabilities can coexist with different cause-specific incidences. State the measure, apply the model over the stated period, and use the probability-conservation check.
Check Your Understanding
For each hypothetical model, use the stated horizon and assume no re-entry or unlisted exit causes. Give probabilities to four decimals.
- A model has constant cause-1 and cause-2 intensities of \(0.25\) and \(0.25\) per year for two years. Find the remaining probability, all-cause exit probability, and cause-1 cumulative incidence. (Answer: \(R(2)=0.3679\), all-cause exit \(=0.6321\), \(F_1(2)=0.3161\).)
- A hypothetical mortality-and-withdrawal model has constant death intensity \(0.10\) and withdrawal intensity \(0.20\) per year for one year. Find the death cumulative incidence. (Answer: \(0.0864\).)
- Both models have a constant cause-1 intensity of \(0.30\) per year. Model A has cause-2 intensity \(0.10\); Model B has cause-2 intensity \(0.70\). Over one year, find each model’s cause-1 cumulative incidence and identify which has the higher all-cause exit probability. (Answer: \(F_{1,A}(1)=0.2473\), \(F_{1,B}(1)=0.1896\); Model B has the higher all-cause exit probability.)
- In year 1, a model has cause-1 and cause-2 intensities \(0.30\) and \(0.10\) per year; in year 2, they are \(0.10\) and \(0.30\). Find the two-year remaining probability and cause-1 cumulative incidence. (Answer: \(R(2)=0.4493\), \(F_1(2)=0.3025\).)
- Model A has constant cause-1 and cause-2 intensities \(0.18\) and \(0.02\) per year. Model B has \(0.16\) and \(0.24\) per year. Over one year, compare cause-1 cumulative incidence and all-cause exit probability. (Answer: \(F_{1,A}(1)=0.1631\), \(F_{1,B}(1)=0.1319\); all-cause exit probabilities are \(0.1813\) and \(0.3297\), respectively, so Model B has more all-cause exits.)