Introduction
With multiple causes of decrement, an observed annual probability reflects both the cause itself and competition from the other causes. Under a constant-force assumption, the conversion between observed probabilities and cause-specific forces has a precise form. Once those forces are found, they also determine the associated single-decrement probabilities: the probabilities that would apply if each cause acted alone.
This tutorial works through the full three-cause calculation, including how to reconstruct the observed exits from the forces. The key is to distinguish a cause’s observed probability in the multiple-decrement setting from its associated probability in a hypothetical single-decrement setting. The methods are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Model and Notation
Consider a group exposed to three mutually exclusive causes of decrement during a one-year interval. Let \(q_1,q_2,q_3\) be the observed probabilities of exit from each cause, measured from the start of the year. Assume the causes are exhaustive, so that anyone who exits during the year exits from exactly one of these causes. The probability of remaining in the group for the full year is \(1-Q\), where \(Q=q_1+q_2+q_3\).
The constant-force assumption is a model for the rate of decrement over the year. It does not say that the observed annual probability for a cause equals its associated probability. Nor does it imply that the causes are independent events with separate annual probabilities: the observed exits compete for the same group of lives.
From Observed Probabilities to Forces
Under the model, the probability of no exit from any cause during one year is \(e^{-\mu}\). Therefore \(1-Q=e^{-\mu}\), and the total force is determined by the total observed exit probability. Conditional on an exit during the year, cause \(j\) accounts for the fraction \(\mu_j/\mu\) of exits. Thus the observed probability for cause \(j\) is that fraction times the total exit probability.
These formulas require \(0<Q<1\) and the stated constant-force model. If \(Q=0\), all three observed cause probabilities are zero and the cause shares \(q_j/Q\) are undefined; the forces are then all zero under this model. If \(Q=1\), the implied total force is infinite, so finite constant forces cannot reproduce the data exactly.
After finding each force, calculate the associated probability separately:
The associated probabilities generally do not sum to the observed total probability. Each describes a different hypothetical setting in which only one cause is present. The observed cause probabilities, by contrast, are generated jointly by all competing forces.
Why the Conversion Formula Holds
For a small interval of length \(dt\), the probability of an exit from cause \(j\) is approximately \(\mu_j\,dt\), while the probability of remaining exposed to all causes up to time \(t\) is \(e^{-\mu t}\). Consequently, the probability of exiting from cause \(j\) between times \(t\) and \(t+dt\) is approximately \(e^{-\mu t}\mu_j\,dt\). Integrating over the year gives the observed probability for cause \(j\):
The middle equality follows by integrating the exponential, and \(1-e^{-\mu}=Q\) because the probability of no decrement is \(e^{-\mu}=1-Q\). Rearranging yields the force-conversion formula. The same calculation also explains the cause shares: under constant forces, each cause’s share of the total force equals its share of observed exits.
Worked Examples
Worked Example: Three Observed Causes and Their Associated Probabilities
A hypothetical membership group has observed one-year decrement probabilities \(q_1=0.08\), \(q_2=0.12\), and \(q_3=0.20\). Assume the three causes are exhaustive and mutually exclusive, and that each cause has a constant force per year. Find all three forces and associated probabilities, then check that the forces reproduce the observed probabilities.
Method. First find the total exit probability and total force. Then allocate the force in proportion to each cause’s observed share of exits. Finally, convert each force to its associated probability.
Step 1: Total probability and force. The total observed exit probability is \(Q=0.08+0.12+0.20=0.40\). Hence the probability of no exit is \(1-Q=0.60\), and
Step 2: Cause-specific forces. The cause shares are \(0.08/0.40=0.20\), \(0.12/0.40=0.30\), and \(0.20/0.40=0.50\). Therefore
Step 3: Associated probabilities. Apply \(q'_j=1-e^{-\mu_j}\) to each cause:
The associated probabilities are reported to four decimal places. They differ from the observed probabilities \(0.08,0.12,0.20\), because each associated probability removes competition from the other causes.
Check. The forces sum to \(0.5108256\). The total exit probability implied by that sum is \(1-e^{-0.5108256}=0.40\). The force shares are \(0.20,0.30,0.50\), so the reconstructed observed probabilities are \(0.20(0.40)=0.08\), \(0.30(0.40)=0.12\), and \(0.50(0.40)=0.20\), as required. The common mistake avoided here is treating an observed cause probability as if that cause acted alone.
Worked Example: Finding One Associated Probability and Reconstructing All Exits
For a hypothetical service plan, the observed annual exit probabilities are \(0.05\), \(0.08\), and \(0.12\) for causes 1, 2, and 3. Assume mutually exclusive, exhaustive causes and constant forces per year. Find the associated probability for cause 2, then verify the complete set of observed probabilities.
Method. Use the total observed probability to find the total force; use cause 2’s share of observed exits to obtain its force; then apply the associated-probability formula. To reconstruct the data, use the force shares and the total exit probability.
The total probability is \(Q=0.05+0.08+0.12=0.25\), so the total force is
Cause 2’s observed share is \(0.08/0.25=0.32\). Its force is therefore
Thus its associated probability is
The associated probability is \(0.0879\) to four decimal places. For the reconstruction, the three force shares are \(0.05/0.25=0.20\), \(0.08/0.25=0.32\), and \(0.12/0.25=0.48\). Multiplying each share by the total exit probability \(0.25\) gives \(0.0500\), \(0.0800\), and \(0.1200\). Their sum is \(0.2500\), confirming the original total. A common mistake is to use \(0.08\) as cause 2’s associated probability; that is the observed probability under competition, not the isolated-cause probability.
Worked Example: Reconstructing Observed Exits From Associated Bases
A hypothetical three-cause model is specified by constant annual forces \(\mu_1=0.10\), \(\mu_2=0.20\), and \(\mu_3=0.30\). Find the associated annual probabilities and the observed annual exit probability for each cause. Assume the causes are mutually exclusive and exhaustive, with no re-entry during the year.
Method. Calculate each associated probability using its own force. For observed exits, first add the forces, then calculate the total exit probability and allocate it in proportion to the forces.
The associated probabilities are
The total force is \(0.10+0.20+0.30=0.60\). The probability of no exit from any cause is \(e^{-0.60}=0.548812\), so the total observed exit probability is \(Q=1-e^{-0.60}=0.451188\). The force shares are \(0.10/0.60=1/6\), \(0.20/0.60=1/3\), and \(0.30/0.60=1/2\). Therefore,
To six decimal places, the observed probabilities are \(0.075198\), \(0.150396\), and \(0.225594\). Their sum is \(0.451188\), matching the total. The check also shows why it is incorrect to add the associated probabilities to obtain the observed total: those probabilities apply to separate single-cause models, not to the joint competing-risk model.
Common Exam Trap
A tempting but incorrect shortcut is to calculate each cause’s force as \(-\ln(1-q_j)\), as if its observed probability came from a single-decrement model. For the first cause in Worked Example 1, that shortcut gives \(-\ln(0.92)\approx0.08338\). The correct force is \(0.1021651\), because \(q_1=0.08\) is an observed competing-risk probability and the total probability of no decrement is \(0.60\), not \(0.92\).
Check Your Understanding
Assume constant forces per year, mutually exclusive causes, and no re-entry for each problem. Give probabilities to four decimal places where requested.
- Observed annual probabilities for three causes are \(0.04\), \(0.06\), and \(0.10\). Find the associated probability for each cause. (Answer: \(0.0436,\ 0.0648,\ 0.1056\))
- Three associated single-cause probabilities correspond to forces \(0.05\), \(0.10\), and \(0.15\). Find the observed annual probability for each cause. (Answer: \(0.0432,\ 0.0864,\ 0.1296\))
- Observed annual probabilities are \(0.08\), \(0.10\), and \(0.14\). Find the associated probability for cause 3. (Answer: \(0.1553\))
- Observed annual probabilities are \(0.10\), \(0.12\), and \(0.18\). Find the associated probability for cause 2 and verify the total observed probability. (Answer: associated probability \(0.1421\); total observed probability \(0.4000\))
- Cause-specific forces are \(0.04\), \(0.08\), and \(0.12\). Find the total observed exit probability and the three observed cause probabilities. (Answer: total \(0.2134\); causes \(0.0356,\ 0.0711,\ 0.1067\))