A Zero-Probability Competing Cause

Why a Zero-Probability Cause Matters

Associated single-decrement probabilities answer a counterfactual question: what would the probability of decrement from a particular cause be if competing causes were removed? Usually, removing a competitor changes the probability of the cause of interest. If a competing cause has probability zero in the model, however, it contributes no exits to remove. Recognizing that simplification can avoid unnecessary calculations—but it does not make other nonzero causes disappear.

The key distinction is between a model probability of zero and zero exits observed in a finite sample. The conversions below concern exact probabilities under stated assumptions. A sample with no exits from a cause does not, by itself, establish that the underlying probability is zero.

Definition: Let \(q_j\) be the probability of decrement from cause \(j\) during an interval when the causes compete, and let \(q'_j\) be the associated single-decrement probability for cause \(j\) when the other causes are removed. A zero-probability cause has \(q_j=0\) in the multiple-decrement model. Under the UDD and constant-force models described here, it also has \(q'_j=0\) and does not affect conversions among the remaining causes.

Zero Causes Under UDD

For a precise UDD comparison, consider two causes that can have nonzero associated probabilities. Write their associated probabilities as \(a\) and \(b\). Assume each cause has a latent decrement time: conditional on that cause decrementing within the year, its time is uniform over the year, and the latent times are independent. The observed decrement is from whichever cause occurs first. If another cause has probability zero, its latent time cannot produce an exit and it is omitted from the calculation.

Under this two-active-cause model, the observed probabilities satisfy

q_1=a\left(1-\frac{b}{2}\right),\qquad q_2=b\left(1-\frac{a}{2}\right).

If cause 2 has zero probability, then \(b=0\), so \(q_1=a\) and \(q_2=0\). Thus, when all competing causes of cause 1 have zero probability, its observed and associated annual probabilities are equal. If cause 2 is still active, removing a different, zero-probability cause does not eliminate the effect of cause 2.

UDD check for two active causes: If all but one cause have zero probability, the remaining cause has \(q'_j=q_j\). If exactly one other cause remains active, omit the zero causes and use the two-cause UDD equations; do not set the associated probability equal to the observed probability just because one of the original causes is zero.

Worked Example: UDD Conversion With One Zero Cause

A hypothetical annual decrement model has three causes. The observed probabilities are \(q_1=0.19\), \(q_2=0.09\), and \(q_3=0\). Assume the two nonzero causes satisfy the UDD model above. Find the associated probabilities for causes 1 and 2, and check the result.

Method. Remove cause 3 from the conversion because its model probability is zero. Let \(a=q'_1\) and \(b=q'_2\). The two-cause equations are

0.19=a\left(1-\frac{b}{2}\right),\qquad 0.09=b\left(1-\frac{a}{2}\right).

Subtract the second observed probability from the first. The cross-products cancel:

$$ q_1-q_2 =a-b=0.19-0.09=0.10. $$

Therefore \(b=a-0.10\). Substitute this into the first equation:

0.19=a\left(1-\frac{a-0.10}{2}\right) =a(1.05-0.5a).

Rearranging gives \(0.5a^2-1.05a+0.19=0\). Its roots are

a=\frac{1.05\pm\sqrt{1.05^2-4(0.5)(0.19)}}{1} =\frac{1.05\pm0.85}{1},

so \(a=0.20\) or \(a=1.90\). The second value is not a probability, leaving \(a=0.20\). Then \(b=a-0.10=0.10\), and \(q'_3=0\).

Check. The UDD equations give \(0.20(1-0.10/2)=0.19\) and \(0.10(1-0.20/2)=0.09\), as required. The zero third cause contributes nothing, but cause 2 still reduces the observed probability for cause 1. The common mistake is to read \(q_3=0\) as meaning there are no competitors at all.

Zero Causes Under Constant Force

Under the constant-force model, each cause \(j\) has a constant force \(\mu_j\) through the interval. The forces are nonnegative, and the total force is \(\mu=\sum_j\mu_j\). For \(\mu>0\), the observed probability for cause \(j\) and its associated single-decrement probability are

q_j=\frac{\mu_j}{\mu}\left(1-e^{-\mu}\right), \qquad q'_j=1-e^{-\mu_j}.

If \(q_j=0\) while the total decrement probability is positive, the cause share \(q_j/\sum_k q_k\) is zero. Hence \(\mu_j=0\), and \(q'_j=1-e^0=0\). If the zero cause is omitted, the total observed probability and the sum of the other forces are unchanged. If only one cause remains active, its cause share is one and its observed probability equals its associated probability.

Constant-force check: Set \(\mu=-\ln(1-\sum_j q_j)\), and, when \(\sum_j q_j>0\), set \(\mu_j=\mu q_j/\sum_k q_k\). A zero observed cause has \(\mu_j=q'_j=0\). Causes with positive observed probabilities still compete with one another.

Worked Example: Constant-Force Conversion With One Zero Cause

A hypothetical model has observed annual probabilities \(q_1=0.12\), \(q_2=0.08\), and \(q_3=0\). Assume constant cause-specific forces. Calculate all three associated probabilities to four decimal places and verify the observed probabilities.

Method. The total observed decrement probability is \(Q=0.12+0.08+0=0.20\). The total force is

\mu=-\ln(1-Q)=-\ln(0.80)=0.2231436.

The cause shares among observed exits are \(0.12/0.20=0.60\) for cause 1 and \(0.08/0.20=0.40\) for cause 2. Cause 3 has share zero. Thus

\mu_1=0.60(0.2231436)=0.1338862,\quad \mu_2=0.40(0.2231436)=0.0892574,\quad \mu_3=0.

Convert each force to its associated probability:

q'_1=1-e^{-0.1338862}=0.1253,\quad q'_2=1-e^{-0.0892574}=0.0854,\quad q'_3=1-e^0=0.0000.

Check. The total force is \(0.2231436\), so the total observed exit probability is \(1-e^{-0.2231436}=0.20\). The shares are \(0.60\) and \(0.40\), giving observed probabilities \(0.60(0.20)=0.12\) and \(0.40(0.20)=0.08\), with zero for cause 3. The associated probabilities are not the observed probabilities for causes 1 and 2 because each still competes with the other. Rounding is to four decimal places, as requested.

Why the UDD and Constant-Force Checks Hold

The simplification follows from what it means for a cause to have zero probability under each model. Under UDD, the probability \(b\) of a latent decrement from that cause is zero. At each time \(t\) in the year, the probability that this cause has already occurred is \(tb=0\). It therefore cannot prevent an exit from another cause.

For example, in the two-cause UDD model, the probability that cause 1 produces the observed decrement is found by integrating over its possible time \(t\). Its unconditional event-time density is \(a\), and cause 2 has not occurred by \(t\) with probability \(1-bt\). Therefore

q_1=\int_0^1 a(1-bt)\,dt =a\left[t-\frac{bt^2}{2}\right]_0^1 =a\left(1-\frac{b}{2}\right).

Setting \(b=0\) gives \(q_1=a\). Under constant force, a zero probability corresponds to zero force: the expression \(1-e^{-\mu_j}\) is zero exactly when \(\mu_j=0\). The zero force contributes neither to the total force nor to any other cause's share.

These checks depend on the model probabilities being zero. If a data set records no exits for a cause, but the underlying probability is not specified to be zero, it is not valid to automatically remove that cause. A zero count could arise from sampling variation.

When the Simplification Is Useful—and the Main Trap

A zero cause can make a conversion shorter and provides a useful check on a calculation. Under either model, if it is the only possible cause of competition with the cause of interest, then a zero competing probability leaves the cause of interest's annual associated probability equal to its observed probability. But a zero cause does not remove any other active competitor.

A related distinction appears if probabilities are needed for part of a year. Once the only active cause has annual probability \(q\), UDD gives probability \(tq\) over the first fraction \(t\) of the year. Constant force gives \(1-(1-q)^t\). These fractional probabilities are generally different even though both models give the same associated annual probability when all other causes have probability zero.

For instance, let the sole active cause have annual probability \(0.31\). At \(t=0.5\), UDD gives \(0.5(0.31)=0.1550\). Under constant force the half-year probability is \(1-\sqrt{0.69}=0.1693\) to four decimal places. The zero competing causes make the annual association immediate; they do not make the two fractional-year assumptions interchangeable.

Do not discard active causes. In the UDD example, cause 3 was zero, but cause 2 still affected cause 1.
Do not treat a zero count as a zero probability. The conversion requires a model probability of zero, not merely an unobserved event in a sample.
Check the model before using equality. Equality of observed and associated probabilities for a cause follows when its competitors are all zero, not merely when one competitor is zero.

Key takeaway: A zero-probability cause has zero associated probability and can be omitted under both UDD and constant force. After omitting it, retain every nonzero competitor and apply the appropriate model to the remaining causes.

Check Your Understanding

Use the stated assumptions for each problem. Probabilities in the answers are rounded to four decimal places where needed.

Under UDD, two causes have observed probabilities \(q_1=0.14\) and \(q_2=0\). Find both associated probabilities. (Answer: \(q'_1=0.1400,\ q'_2=0.0000\))
Under constant force, the sole active cause has observed annual probability \(q_1=0.25\); two other causes have probability zero. Find its associated probability and total force to four significant figures. (Answer: \(q'_1=0.2500,\ \mu=-\ln(0.75)=0.2877\))
Under the two-active-cause UDD model, associated probabilities are \(q'_1=0.30\), \(q'_2=0.20\), and a third cause has associated probability zero. Find the observed probabilities and check that the zero cause has no effect. (Answer: \(q_1=0.2700,\ q_2=0.1700,\ q_3=0.0000\))
Under constant force, observed probabilities are \(q_1=0.15\), \(q_2=0.05\), and \(q_3=0\). Find \(q'_1\), \(q'_2\), and \(q'_3\). (Answer: \(q'_1=0.1541,\ q'_2=0.0543,\ q'_3=0.0000\))
A finite study records no exits from cause 4, while several other causes have exits. Is it justified to set the model probability for cause 4 to zero and omit it from an associated-table conversion? (Answer: No; zero observed exits alone do not establish a zero model probability.)

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