Recovering Multiple-Decrement Probabilities From Associated Tables

Why Associated Tables Must Be Combined Carefully

An associated single-decrement table describes what would happen if one cause operated alone. A multiple-decrement table describes what happens when several causes can remove a life from the group. These are different probability models: in the multiple-decrement setting, an individual may exit from one cause before another cause has the opportunity to operate.

This tutorial recovers one-year multiple-decrement survival and cause-specific exit probabilities from associated single-decrement survival probabilities. The conversion uses a common assumption: each cause has a constant force throughout the year. The associated probabilities must all be consistent with that same assumption and time interval.

Definition: Suppose there are \(m\) competing causes during a one-year interval. Let \(\mu_j\) be the constant force for cause \(j\), for \(j=1,\ldots,m\). In the associated single-decrement table for cause \(j\), that cause acts alone, so its one-year survival probability is \(p'_j=e^{-\mu_j}\), and its one-year exit probability is \(q'_j=1-p'_j\). In the multiple-decrement model, write \(p\) for the probability of surviving the year without any exit and \(q_j\) for the probability of exiting from cause \(j\) during the year.

The prime marks an associated single-decrement probability. It does not mean that \(q'_j\) is the probability of a cause-\(j\) exit when all causes operate together. To combine the tables, first recover each force from its associated survival probability, then calculate the competing-risk probabilities under the common constant-force assumption.

The Recovery Formulas

For each cause, the associated survival probability gives the force directly: \( \mu_j=-\ln(p'_j) \). The multiple-decrement survival probability is the probability that none of the causes produces an exit during the year. Under constant forces, this is the exponential survival probability for the sum of the forces.

p=\exp\left(-\sum_{j=1}^{m}\mu_j\right) =\prod_{j=1}^{m}p'_j, \qquad q=1-p.

The total exit probability \(q\) is allocated across causes in proportion to their forces. When the total force is positive, the cause-specific probability is \(q_j=(\mu_j/\sum_k\mu_k)q\). Equivalently, the associated table values can be used directly:

q_j= \frac{-\ln(p'_j)} {\sum_{k=1}^{m}-\ln(p'_k)} \left(1-\prod_{k=1}^{m}p'_k\right), \qquad j=1,\ldots,m.

These formulas apply to a single interval when each cause has a constant force on that interval. They require \(0<p'_j\leq 1\) for every cause, so that the forces are finite and nonnegative. If every associated survival probability is 1, all forces are zero, and the correct result is \(p=1\) and \(q_j=0\) for every cause; in that special case the ratio formula has a zero denominator and should not be used. Probabilities from associated tables built on incompatible assumptions or intervals cannot be combined by these formulas without first reconciling the bases.

Why the Cause-Specific Formula Holds

The multiple-decrement exit probability for cause \(j\) is obtained by considering an exit at time \(t\) within the year. For that exit to occur, the individual must first have survived all causes to time \(t\), and then cause \(j\) must produce the exit. With constant forces, survival to time \(t\) is \(e^{-t\sum_k\mu_k}\), and the instantaneous exit rate from cause \(j\) is \(\mu_j\).

q_j =\int_0^1 e^{-t\sum_{k=1}^{m}\mu_k}\mu_j\,dt.

If \(\sum_k\mu_k>0\), integrating gives

q_j =\mu_j\left[ \frac{1-e^{-\sum_k\mu_k}}{\sum_k\mu_k} \right] =\frac{\mu_j}{\sum_k\mu_k} \left(1-e^{-\sum_k\mu_k}\right).

The term in parentheses is the probability of any exit during the year. Thus, each cause receives the share of total exits equal to its proportion of the total force. Summing over causes also verifies closure: \( \sum_j q_j=q \), and therefore \(p+\sum_jq_j=1\).

Convert associated survival to force.
For each cause, calculate \(\mu_j=-\ln(p'_j)\). If the input is an associated exit probability, first use \(p'_j=1-q'_j\).

Combine the forces.
Set \(\mu=\sum_j\mu_j\). The probability of surviving all causes is \(p=e^{-\mu}=\prod_jp'_j\), and the probability of any exit is \(q=1-p\).

Allocate exits by cause.
For positive \(\mu\), calculate \(q_j=(\mu_j/\mu)q\) for each cause.

Check closure.
Confirm that the cause-specific probabilities sum to \(q\), and that \(p+q=1\).

Worked Applications

Worked Example: Two Associated Survival Probabilities

A hypothetical one-year decrement model has two causes. The associated single-decrement survival probabilities are \(p'_1=e^{-0.12}\) and \(p'_2=e^{-0.08}\). Assuming both causes have constant forces during the year, find the multiple-decrement survival probability and each cause-specific exit probability. Give probabilities to four decimal places.

Method. Recover the two forces, combine them to obtain total survival and exits, and allocate exits in proportion to force. This avoids mistaking an associated exit probability for a competing-risk exit probability.

The forces are \( \mu_1=-\ln(e^{-0.12})=0.12 \) and \( \mu_2=-\ln(e^{-0.08})=0.08 \). Thus the total force is \( \mu=0.12+0.08=0.20 \).

The probability of surviving both causes is \( p=e^{-0.20}=0.8187308 \), so the probability of any exit is \( q=1-0.8187308=0.1812692 \).

Cause 1 receives \(0.12/0.20=0.60\) of total exits, giving \( q_1=0.60(0.1812692)=0.1087615 \). Cause 2 receives \(0.08/0.20=0.40\), giving \( q_2=0.40(0.1812692)=0.0725077 \).

Answer and check. To four decimal places, \(p=0.8187\), \(q_1=0.1088\), and \(q_2=0.0725\). Before rounding, \(q_1+q_2=0.1812692=q\), and \(p+q=1\).

Worked Example: Starting From Associated Exit Probabilities

A hypothetical policy group has three competing causes. Its associated one-year exit probabilities are \(q'_1=0.10\), \(q'_2=0.20\), and \(q'_3=0.05\). Convert these to multiple-decrement probabilities under constant forces over the year. Give probabilities to four decimal places.

Method. First convert each associated exit probability to associated survival, then recover the forces. Do not add the associated exit probabilities to obtain the total multiple-decrement exit probability.

The associated survival probabilities are \(p'_1=1-0.10=0.90\), \(p'_2=1-0.20=0.80\), and \(p'_3=1-0.05=0.95\). The corresponding forces are

\begin{aligned} \mu_1&=-\ln(0.90)=0.1053605,\\ \mu_2&=-\ln(0.80)=0.2231436,\\ \mu_3&=-\ln(0.95)=0.0512933. \end{aligned}

Their sum is \( \mu=0.1053605+0.2231436+0.0512933=0.3797974 \). The total survival probability can also be found by multiplying the associated survivals: \(p=0.90(0.80)(0.95)=0.6840\). Therefore \(q=1-0.6840=0.3160\).

The force proportions are approximately \( \mu_1/\mu=0.2774 \), \( \mu_2/\mu=0.5875 \), and \( \mu_3/\mu=0.1351 \). Applying these proportions to total exits gives

\begin{aligned} q_1&=\frac{0.1053605}{0.3797974}(0.3160)\approx0.0877,\\ q_2&=\frac{0.2231436}{0.3797974}(0.3160)\approx0.1857,\\ q_3&=\frac{0.0512933}{0.3797974}(0.3160)\approx0.0427. \end{aligned}

Answer and check. The multiple-decrement survival probability is \(0.6840\), and the cause-specific exit probabilities are \(0.0877\), \(0.1857\), and \(0.0427\). Using unrounded values, the three exits sum to \(0.3160\), so survival plus exits is 1. The displayed four-decimal exit probabilities sum to \(0.3161\) because of rounding.

Common mistake avoided. Adding the associated exit probabilities gives \(0.10+0.20+0.05=0.35\), which is not the multiple-decrement exit probability. Under the stated model, the correct total is \(1-(0.90)(0.80)(0.95)=0.3160\).

Worked Example: Recovering Expected Counts for a Cohort

A hypothetical cohort contains 1,000 individuals at the start of a year. The associated survival probabilities for two causes are \(p'_1=e^{-0.10}\) and \(p'_2=e^{-0.20}\). Under the common constant-force assumption, find the expected number who survive the year and the expected number who exit from each cause. Round expected counts to one decimal place.

Method. Recover the probabilities first, then multiply each by the initial cohort size. Expected counts need not be whole numbers.

The forces are \(\mu_1=0.10\) and \(\mu_2=0.20\), so \(\mu=0.30\). Overall survival is \(p=e^{-0.30}=0.7408182\), and total exit probability is \(q=1-0.7408182=0.2591818\).

The cause proportions are \(0.10/0.30=1/3\) and \(0.20/0.30=2/3\). Therefore \(q_1=(1/3)(0.2591818)=0.0863939\) and \(q_2=(2/3)(0.2591818)=0.1727879\).

Multiplying by 1,000 gives an expected \(740.8182\) survivors, \(86.3939\) exits from cause 1, and \(172.7879\) exits from cause 2. Rounded to one decimal place, these are 740.8, 86.4, and 172.8.

Answer and check. The unrounded expected counts sum to \(740.8182+86.3939+172.7879=1{,}000.0000\). The rounded counts also sum to 1,000.0, a useful check in this example.

Interpretation and Exam Traps

The associated tables describe hypothetical single-cause worlds; the multiple-decrement probabilities describe the joint competing-risk model. Under constant forces, multiplying associated survival probabilities works because each associated survival is an exponential of its cause-specific force. Taking logarithms recovers those forces, which then determine how total exits are divided among causes.

A frequent error is to use \(q'_j\) directly as \(q_j\), or to add the \(q'_j\) values to obtain total exits. The associated events are defined with the other causes removed, so their probabilities are not mutually exclusive probabilities from the multiple-decrement model. Another error is to calculate the correct total exit probability but divide it equally among causes; the allocation must instead follow the force proportions, which are generally unequal.

Key takeaway: Under a common constant-force assumption, multiply the associated survival probabilities to obtain multiple-decrement survival. Then allocate the total exit probability using the shares of the recovered forces. Always check \(p+\sum_jq_j=1\), allowing for rounding.

Check Your Understanding

For each problem, assume all associated tables use the same one-year interval and constant-force basis. Give probabilities to four decimal places unless expected counts are requested.

Two causes have associated survival probabilities \(e^{-0.06}\) and \(e^{-0.14}\). Find multiple-decrement survival and both cause-specific exit probabilities. (Answer: \(p=0.8187,\ q_1=0.0544,\ q_2=0.1269\))
Two causes have associated survival probabilities \(0.95\) and \(0.85\). Find the total exit probability and the two cause-specific probabilities. (Answer: \(q=0.1925,\ q_1=0.0462,\ q_2=0.1463\))
Three causes have associated exit probabilities \(0.04\), \(0.08\), and \(0.12\). Find the multiple-decrement survival probability and the exit probability for each cause. (Answer: \(p=0.7772,\ q_1\approx0.0361,\ q_2\approx0.0737,\ q_3\approx0.1130\))
A cohort of 1,000 has two causes with associated survival probabilities \(e^{-0.10}\) and \(e^{-0.20}\). Find the expected survivors and expected exits from each cause, to one decimal place. (Answer: 740.8 survivors, 86.4 exits from cause 1, and 172.8 exits from cause 2)
Three causes have the same associated survival probability, \(0.90\), for each cause. Find the multiple-decrement survival probability and the exit probability for each cause. (Answer: \(p=0.7290\), and each \(q_j\approx0.0903\))

What You'll Learn