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Constant-Force Cause Shares

Learn to allocate observed exits among competing causes under constant forces and to recover cause-specific forces from exit data.

Intermediate 11 min read

What You'll Learn

  • State the assumptions behind constant cause-specific forces.
  • Derive the observed exit probability for each cause.
  • Prove that each cause’s share of exits equals its share of the aggregate force.
  • Calculate cause-specific probabilities and expected exit counts.
  • Recover aggregate and cause-specific forces from observed exits.
  • Recognize when a cause-share ratio is undefined and avoid treating competing risks as isolated.

Why Cause Shares Matter

In a multiple-decrement model, a person can leave an active group for one of several mutually exclusive reasons, such as disability, withdrawal, or death. The probability of an observed exit from a particular cause depends on both that cause’s force and the possibility of an earlier exit from another cause. Under constant cause-specific forces, a useful simplification holds: each cause’s share of all observed exits is the same as its share of the aggregate force.

This result helps allocate total exits by cause and, in the reverse direction, recover force proportions from observed exit data. The key is to distinguish an observed exit probability in the presence of competing causes from the probability that would apply if only one cause operated.

Definition: Consider an active group exposed to \(r\) mutually exclusive causes of exit over an interval of length \(t\). The cause-specific force \(\mu_j\) is the instantaneous rate of exit from cause \(j\), conditional on still being active. Under the constant-force assumption, each \(\mu_j\) is constant throughout the interval. The aggregate force is \(\mu=\sum_{j=1}^{r}\mu_j\). Write \(q_j(t)\) for the probability that a person active at the start exits from cause \(j\) during the interval, and \(q(t)=\sum_{j=1}^{r}q_j(t)\) for the probability of any exit.

The model assumes that the causes are exhaustive for the exits being counted, that only one cause can be the first exit for a person, and that the stated forces apply while the person remains active. Forces are nonnegative. They are measured per unit of the same time scale used for \(t\); for example, an annual force must be paired with an interval measured in years.

Deriving the Cause-Share Result

Under constant forces, the probability of remaining active for \(s\) time units is \(e^{-\mu s}\). To exit from cause \(j\) in a short interval of length \(ds\) around time \(s\), a person must first remain active until \(s\), then exit from cause \(j\) in that short interval. The probability contribution is therefore approximately \(e^{-\mu s}\mu_j\,ds\). Integrating these contributions over the interval gives the observed probability for cause \(j\).

$$ q_j(t)=\int_0^t e^{-\mu s}\mu_j\,ds =\frac{\mu_j}{\mu}\left(1-e^{-\mu t}\right),\qquad \mu>0. $$

Summing over causes gives the total exit probability. Since \(\sum_j\mu_j=\mu\), the cause-specific probabilities add to the probability of any exit.

$$ q(t)=\sum_{j=1}^{r}q_j(t) =\left(\sum_{j=1}^{r}\frac{\mu_j}{\mu}\right)(1-e^{-\mu t}) =1-e^{-\mu t},\qquad \mu>0. $$

If \(\mu>0\) and \(t>0\), then \(q(t)>0\). Dividing the cause-specific exit probability by the total exit probability proves the cause-share result.

$$ \frac{q_j(t)}{q(t)} =\frac{(\mu_j/\mu)(1-e^{-\mu t})}{1-e^{-\mu t}} =\frac{\mu_j}{\mu}. $$
Theorem: If the cause-specific forces are constant over an interval of positive length and the aggregate force satisfies \(\mu>0\), then the share of observed exits attributable to cause \(j\) is \(q_j(t)/q(t)=\mu_j/\mu\). If \(\mu=0\), every \(\mu_j=0\), so \(q_j(t)=0\) for every cause and there are no exits. The cause-share ratio is defined only when \(q(t)>0\).

The derivation also explains why the result concerns shares, not the individual probabilities themselves. The probability of cause \(j\) includes the chance of surviving all causes long enough to experience cause \(j\). The common factor \(1-e^{-\mu t}\) accounts for the chance of any exit; the force ratio allocates that total among causes.

Worked Applications

Worked Example: Allocating Exits Among Three Causes

A hypothetical group of 10,000 people is active at the start of a two-year interval. The constant annual forces of recovery, mortality, and withdrawal are \(0.02\), \(0.03\), and \(0.01\), respectively. Find each cause’s observed exit probability and expected number of exits. Assume these are the only causes of exit and that a person exits at most once.

Method. Add the forces to get the aggregate force, calculate the probability of any exit over two years, and allocate that probability in proportion to the cause-specific forces. This method avoids treating each cause as if it operated alone.

The aggregate force is \(0.02+0.03+0.01=0.06\) per year. The interval is \(t=2\) years, so the total exit probability is

$$ q(2)=1-e^{-0.06(2)}=1-e^{-0.12}=0.1130796. $$

The force shares are \(0.02/0.06=1/3\) for recovery, \(0.03/0.06=1/2\) for mortality, and \(0.01/0.06=1/6\) for withdrawal. Thus the cause-specific probabilities are

$$ q_{\text{recovery}}(2)=\frac{1}{3}(0.1130796)=0.0376932, \qquad q_{\text{mortality}}(2)=\frac{1}{2}(0.1130796)=0.0565398, $$ $$ q_{\text{withdrawal}}(2)=\frac{1}{6}(0.1130796)=0.0188466. $$

Multiplying each probability by 10,000 gives expected exits of \(376.93\) from recovery, \(565.40\) from mortality, and \(188.47\) from withdrawal, rounded to two decimals. Their rounded sum is \(1{,}130.80\). The total expected exits calculated directly are \(10{,}000(0.1130796)=1{,}130.80\), confirming the allocation.

Worked Example: Comparing Cause Probabilities Over a Longer Interval

A hypothetical long-term care model has constant annual forces of claim termination due to recovery, \(0.03\), and death, \(0.05\). For a person active at the start, calculate the two-year observed exit probability for each cause and the expected cause-specific exits among 6,000 such people.

Method. Use the cause-share theorem with the aggregate two-year exit probability. A common mistake is to calculate \(1-e^{-0.03(2)}\) and \(1-e^{-0.05(2)}\) separately; those are not observed probabilities when both causes operate.

The aggregate force is \(0.03+0.05=0.08\) per year. Therefore,

$$ q(2)=1-e^{-0.08(2)}=1-e^{-0.16}=0.1478562. $$

Recovery’s share of the force is \(0.03/0.08=3/8\), and death’s share is \(0.05/0.08=5/8\). The observed probabilities are

$$ q_{\text{recovery}}(2)=\frac{3}{8}(0.1478562)=0.0554461, \qquad q_{\text{death}}(2)=\frac{5}{8}(0.1478562)=0.0924101. $$

The expected exits are \(6{,}000(0.0554461)=332.68\) from recovery and \(6{,}000(0.0924101)=554.46\) from death, rounded to two decimals. They sum to \(887.14\), which agrees with \(6{,}000(0.1478562)=887.14\). As a check on the competing-risk adjustment, the two cause-specific probabilities sum to the total exit probability, not to the sum of the two isolated-cause probabilities.

Worked Example: Recovering Cause-Specific Forces From Exit Counts

In a hypothetical one-year study, 1,000 people are active at the start. There are 100 observed exits from cause A and 200 each from causes B and C; the remaining people have no exit during the year. Assume constant forces, and that these three causes are exhaustive. Find the aggregate force and each cause-specific force.

Method. First calculate total survival and hence the aggregate force. Then use the observed exit shares as the force shares. This reverse calculation is valid under the constant-force assumption.

There are \(100+200+200=500\) observed exits, so the total exit probability is \(500/1{,}000=0.5\). The probability of remaining active for one year is also \(0.5\). Under a constant aggregate force, one-year survival is \(e^{-\mu}\), so

$$ e^{-\mu}=0.5 \quad\Longrightarrow\quad \mu=-\ln(0.5)=\ln 2=0.6931472 $$

The observed exit shares are \(100/500=0.2\) for A and \(200/500=0.4\) for each of B and C. By the theorem, those are also the force shares. Hence

$$ \mu_A=0.2(0.6931472)=0.1386294, \qquad \mu_B=\mu_C=0.4(0.6931472)=0.2772589. $$

The recovered forces sum to \(0.1386294+0.2772589+0.2772589=0.6931472\), matching the aggregate force. They are annual forces because the observed interval was one year. This check confirms both the force total and the allocation.

Common Mistakes and Interpretation

The formula \(q_j(t)=(\mu_j/\mu)(1-e^{-\mu t})\) applies to the observed probability of cause \(j\) in the multiple-decrement model. It is not the isolated single-cause probability \(1-e^{-\mu_jt}\). For example, with forces \(0.03\) and \(0.05\) over two years, the first cause’s observed probability is \(0.0554461\), not \(1-e^{-0.06}=0.0582355\). The latter treats the first cause as if the second cause could not remove anyone from the active group.

  • Using the wrong denominator. A cause’s exit share is \(q_j(t)/q(t)\), not \(q_j(t)\) itself. The force share is \(\mu_j/\mu\).
  • Ignoring time units. Multiply a force by the interval in the same units before evaluating the exponential.
  • Dividing by zero. If every cause force is zero, then \(\mu=0\), no exits occur, and a cause-share ratio cannot be formed. If \(t=0\), there are likewise no exits during the interval.
  • Confusing expected counts with realized counts. Multiplying a probability by the initial group size gives an expected number of exits, which need not be an integer.

The cause-share result is useful when a total exit probability is known but a cause-by-cause allocation is needed, or when cause-specific observed exits are available and force proportions must be inferred. In either direction, it depends on the constant-force model: the same aggregate force governs survival through the interval, and the relative cause forces remain fixed.

Check Your Understanding

For each problem, assume mutually exclusive and exhaustive causes with constant forces over the stated interval. Probabilities are rounded to four decimal places and expected counts to two decimal places.

  1. Two causes have annual forces \(0.01\) and \(0.04\). Find their observed probabilities over three years. (Answer: \(0.0279\) and \(0.1114\), respectively.)
  2. Three causes have annual forces \(0.015\), \(0.025\), and \(0.060\). Find the observed probability for each cause over 1.5 years. (Answer: \(0.0209\), \(0.0348\), and \(0.0836\), respectively.)
  3. A cohort of 2,400 has two causes with annual forces \(0.06\) and \(0.03\), and a third cause with force \(0.03\). Find the expected exits by cause over two years. (Answer: \(256.05\), \(128.02\), and \(128.02\), respectively.)
  4. Over 1.5 years, a cohort of 5,000 has 150 exits from cause A and 250 from cause B. The rest have no exit. Under constant forces, find the aggregate force and the two cause-specific forces. (Answer: aggregate \(0.05559\) per year; A \(0.02085\) per year; B \(0.03474\) per year.)
  5. Three causes have annual forces in the ratio \(1:2:2\), and the total exit probability over one year is \(0.5\). Find the observed probability for each cause and state whether the cause-share ratio is defined. (Answer: \(0.1000\), \(0.2000\), and \(0.2000\); yes, because the total exit probability is positive.)

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