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Why Interval Cause Shares Are Not Instantaneous Shares

See how changing cause intensities and survival weighting determine the proportion of interval exits attributed to each cause.

Intermediate 10 min read

What You'll Learn

  • Distinguish an instantaneous cause share from a cause’s share of exits over an interval
  • Derive the cause-specific exit probability from the remaining probability and cause intensity
  • Interpret the interval share as a survival-weighted average
  • Calculate interval shares for changing, piecewise-constant, and continuously varying intensities
  • Identify why initial shares and ordinary time averages can misstate interval results

Why an Interval Share Can Differ

At a particular instant, the relative sizes of competing decrement intensities describe the relative rates of exit at that instant. An interval’s cause-specific exit proportion answers a different question: among all exits that actually occur during the interval, what proportion are due to each cause? When cause intensities change over time, these quantities need not agree. Even if a cause has a high instantaneous share early in the interval, relatively few exits may occur later when its share is lower; conversely, the cohort available to exit also becomes smaller as time passes.

We will work with a cohort initially at risk and a finite set of mutually exclusive causes. Assume cause \(j\) has a nonnegative intensity \(\mu_j(t)\) at time \(t\), measured per year when \(t\) is measured in years. Define the total intensity as the sum of the cause intensities. Intensities are instantaneous rates conditional on still being at risk; they are not interval probabilities.

Definition: At a time \(t\) when the total intensity is positive, the instantaneous share for cause \(j\) is \(\rho_j(t)=\mu_j(t)/\mu(t)\). The interval cause share over \([0,T]\), when the probability of an exit in the interval is positive, is the probability of an exit from cause \(j\) during that interval divided by the probability of any exit during the interval. These are different quantities: the first is defined at a time, while the second summarizes exits over an interval.

Let \(R(t)\) denote the probability that an initially at-risk member of the cohort remains at risk through time \(t\). For a continuous-time competing decrement model, the total intensity is \(\mu(t)=\sum_j\mu_j(t)\), and the remaining probability satisfies \(R'(t)=-R(t)\mu(t)\), with \(R(0)=1\). Thus \(R(t)=\exp(-\int_0^t\mu(s)\,ds)\). This result assumes the specified intensities describe the cohort throughout the interval.

Conditional on remaining at risk at time \(t\), the probability of an exit from cause \(j\) during a short interval of length \(dt\) is approximately \(\mu_j(t)dt\). Therefore the unconditional probability density of a cause-\(j\) exit at time \(t\) is \(R(t)\mu_j(t)\). Integrating this density gives both the cause-specific probability and the interval share.

$$ q_j(0,T)=\int_0^T R(t)\mu_j(t)\,dt, \qquad q(0,T)=\int_0^T R(t)\mu(t)\,dt=1-R(T), \qquad \operatorname{Share}_j(0,T)=\frac{q_j(0,T)}{q(0,T)}. $$

Why the Formula Is a Weighted Average

Where \(\mu(t)>0\), write \(\mu_j(t)=\mu(t)\rho_j(t)\). Substituting this into the cause-specific probability shows that the interval share is an average of instantaneous shares, but not usually an ordinary time average.

$$ \operatorname{Share}_j(0,T) = \frac{\int_0^T R(t)\mu(t)\rho_j(t)\,dt} {\int_0^T R(t)\mu(t)\,dt}. $$

To see why, consider a small time slice of width \(dt\). Its contribution to all exits is approximately \(R(t)\mu(t)dt\), and its contribution to cause-\(j\) exits is approximately \(R(t)\mu(t)\rho_j(t)dt\). Dividing the total cause-\(j\) contribution by the total exit contribution gives the displayed ratio. Thus times with more exits receive more weight. Both the total intensity and the size of the remaining cohort affect that weight. If \(\mu(t)=0\), there are no exits at that instant, so its instantaneous share need not be defined; it makes no contribution to either integral.

A useful check follows directly from the same argument: the cause-specific probabilities sum to the total exit probability. Indeed, summing \(R(t)\mu_j(t)\) over causes gives \(R(t)\mu(t)\), whose integral is \(1-R(T)\). Therefore, when the total exit probability is positive, the interval cause shares sum to 1.

Worked Example: A Share That Falls Halfway Through the Interval

A hypothetical cohort is followed for one year. The total intensity is \(0.50\) per year throughout. Cause 1 has \(80\%\) of the total intensity during the first half-year and \(20\%\) during the second half-year; cause 2 has the remaining share. Find the probability of an exit from each cause and each cause’s share of all exits. Compare the interval share for cause 1 with its initial instantaneous share.

Method. Split the interval at the change in intensities. In each half-year, the total integrated intensity is \(0.50(0.5)=0.25\), so the probability of remaining after the first half-year is \(e^{-0.25}=0.7788008\). The probability of an exit in either half-year, conditional on being at risk at its start, is \(1-e^{-0.25}=0.2211992\).

In the first half-year, the cause intensities are \(0.40\) and \(0.10\) per year. The exit probabilities by cause in that segment, measured from the original cohort, are therefore

$$ q_{1,\text{first}}=0.80(1-e^{-0.25})=0.1769594, \qquad q_{2,\text{first}}=0.20(1-e^{-0.25})=0.0442398. $$

At the start of the second half-year, only \(e^{-0.25}\) of the original cohort remains at risk. The cause shares then switch to \(0.20\) and \(0.80\). Thus

$$ q_{1,\text{second}} =e^{-0.25}(0.20)(1-e^{-0.25}) =0.0344540, $$
$$ q_{2,\text{second}} =e^{-0.25}(0.80)(1-e^{-0.25}) =0.1378161. $$

Adding each cause’s segment probabilities gives \(q_1=0.1769594+0.0344540=0.2114134\) and \(q_2=0.0442398+0.1378161=0.1820559\). The total exit probability is \(q_1+q_2=0.3934693\). Independently, the total integrated intensity over the year is \(0.50\), so \(1-e^{-0.50}=0.3934693\), confirming probability conservation.

The interval shares, rounded to four decimal places, are \(0.2114134/0.3934693=0.5373\) for cause 1 and \(0.1820559/0.3934693=0.4627\) for cause 2. Cause 1’s initial instantaneous share was \(0.8000\), not \(0.5373\). Common mistake avoided: applying the initial \(80\%\) share to every exit in the year ignores the change halfway through and the survivors available in the second half.

Changing Shares Throughout the Interval

When shares vary continuously, the same integral formula applies. In some cases the integrals can be evaluated directly. It is important to use the remaining probability at each time: multiplying the instantaneous cause share by time alone does not account for the fact that exits are more likely in some parts of the interval than others.

Worked Example: A Linearly Increasing Cause Share

In a hypothetical one-year model, the total intensity is constant at \(0.60\) per year. The instantaneous share for cause 1 is \(\rho_1(t)=0.20+0.60t\), where \(t\) is measured in years for \(0\leq t\leq1\); cause 2 has the remaining share. Find the cause-1 exit probability and interval share. Compare the result with the initial share and with the ordinary time-average share.

The total intensity being constant gives \(R(t)=e^{-0.60t}\), and the cause-1 intensity is \(0.60(0.20+0.60t)\). Hence

$$ q_1(0,1) = \int_0^1 e^{-0.60t}\,0.60(0.20+0.60t)\,dt. $$

Separate the constant and linear terms. The two needed integrals are

$$ \int_0^1 0.60e^{-0.60t}\,dt=1-e^{-0.60}=0.4511884, $$
$$ \int_0^1 0.60t e^{-0.60t}\,dt = \frac{1-(1+0.60)e^{-0.60}}{0.60} =0.2031690. $$

The coefficient of the linear term in the cause-1 intensity is \(0.60\), so

$$ q_1(0,1) =0.20(0.4511884)+0.60(0.2031690) =0.0902377+0.1219014 =0.2121391. $$

The total exit probability is \(1-e^{-0.60}=0.4511884\). Therefore the interval share for cause 1 is \(0.2121391/0.4511884=0.4702\), to four decimal places. The initial instantaneous share is \(0.2000\), while the ordinary time average of \(0.20+0.60t\) over the year is \(0.50\). Neither is the interval share. The exits are weighted toward earlier times, when more members remain at risk and cause 1’s share is lower.

Common mistake avoided: reporting \(0.50\), the simple average of the endpoint shares, treats each time equally. The correct weighting is by \(R(t)\mu(t)\), as seen in the integral.

Worked Example: Changing Both Total Intensity and Cause Share

A hypothetical cohort is observed for one year. During the first half-year, the cause intensities are \(0.40\) and \(0.10\) per year. During the second half-year, they are \(0.20\) and \(0.60\) per year. Calculate the probability of exit from each cause and the interval share for cause 1. Compare it with cause 1’s instantaneous share at the start and end of the year.

Method. Calculate the probability of remaining through each segment using its total intensity, then multiply by the segment’s cause-specific exit probability. In the first half-year, the total intensity is \(0.50\), the integrated intensity is \(0.25\), and the probability of remaining to the midpoint is \(e^{-0.25}=0.7788008\). The cause probabilities in this segment are

$$ q_{1,\text{first}}=\frac{0.40}{0.50}(1-e^{-0.25})=0.1769594, \qquad q_{2,\text{first}}=\frac{0.10}{0.50}(1-e^{-0.25})=0.0442398. $$

In the second half-year, the total intensity is \(0.80\), so the conditional probability of an exit during that segment is \(1-e^{-0.80(0.5)}=1-e^{-0.40}=0.3296800\). The cohort’s probability of remaining at the start of the segment is \(e^{-0.25}\). The cause probabilities, again measured from the original cohort, are

$$ q_{1,\text{second}} =e^{-0.25}\frac{0.20}{0.80}(1-e^{-0.40}) =0.0641888, $$
$$ q_{2,\text{second}} =e^{-0.25}\frac{0.60}{0.80}(1-e^{-0.40}) =0.1925663. $$

Thus \(q_1=0.1769594+0.0641888=0.2411481\), and \(q_2=0.0442398+0.1925663=0.2368061\). Their sum is \(0.4779542\). The total probability of remaining at the end is \(e^{-0.25}e^{-0.40}=e^{-0.65}=0.5220458\), so the total exit probability is \(1-e^{-0.65}=0.4779542\), as required.

The cause-1 interval share is \(0.2411481/0.4779542=0.5045\), to four decimal places. Its instantaneous share is \(0.40/0.50=0.8000\) during the first segment and \(0.20/0.80=0.2500\) during the second. The interval share lies between them, but is not their simple average. The total intensity also changes, so the second segment has a different exit probability conditional on being at risk. Common mistake avoided: using the end-of-year share \(0.2500\) for the entire interval discards the substantial cause-1 exits in the first segment.

Interpretation and Exam Traps

An interval share is a cohort-level summary of exits over a stated period. It depends on the entire paths of the cause intensities, not just their values at the interval’s beginning or end. The formula also explains why changing the total intensity matters: it changes the weights \(R(t)\mu(t)\) assigned to different times, even if the pattern of instantaneous shares is unchanged.

When evaluating a problem, keep the following distinctions clear:

  • Instantaneous share: \(\mu_j(t)/\mu(t)\), a ratio of rates at a specified time.
  • Cause-specific interval probability: \(\int_0^T R(t)\mu_j(t)\,dt\), measured from the original cohort.
  • Interval cause share: the cause-specific interval probability divided by \(1-R(T)\), provided \(1-R(T)>0\).

A frequent exam error is to take an unweighted average of the instantaneous shares and label it an interval share. That is correct only under special circumstances, such as a constant instantaneous share over the interval. Another error is to confuse a conditional segment probability with a probability measured from the original cohort. For a later segment, multiply its conditional exit probability by the probability of still being at risk at that segment’s start.

Key takeaway: Interval cause shares are weighted by when exits occur. Calculate each cause-specific exit probability using the remaining probability and its cause intensity, sum those probabilities to check the total exit probability, and divide by the total only at the end.

Check Your Understanding

For each hypothetical model, calculate the requested cause-specific probability or interval share. Probabilities and shares are rounded to four decimal places.

  1. Over one year, the total intensity is \(0.40\) per year and the instantaneous share for cause 1 is constant at \(0.25\). Find the cause-1 exit probability and its interval share. (Answer: probability \(0.0824\); share \(0.2500\))
  2. Over one year, the total intensity is \(0.50\) per year. Cause 1’s instantaneous share is \(0.80\) in the first half-year and \(0.20\) in the second half-year. Find its interval share. (Answer: \(0.5373\))
  3. Over one year, the total intensity is \(0.80\) per year and cause 1’s share is \(0.25+0.50t\), with \(t\) in years. Find the cause-1 exit probability and its interval share. (Answer: probability \(0.2572\); share \(0.4670\))
  4. During the first half-year, cause intensities are \(0.30\) and \(0.30\) per year; during the second half-year, they are \(0.50\) and \(0.10\) per year. Find cause 1’s interval share over the full year. (Answer: \(0.6419\))

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