Why Two Probabilities Describe One Cause
When several causes can end an exposure, the probability of exiting from one particular cause depends on whether the individual remains exposed long enough for that cause to occur. The observed exit probability includes this competition. The corresponding associated single-decrement probability asks a different question: what would the probability be if only the specified cause operated, while its conditional probabilities across the interval stayed the same?
This distinction matters when comparing decrement experience with an associated single-decrement table. The associated probability is generally larger because removing other causes allows more exposure to the cause being measured. It is not, however, a probability obtained by simply adding the other causes’ probabilities to the observed probability.
In this partitioned model, the probability of surviving a subinterval while all causes operate is \(1-p_{j,r}-p_{k,r}\). In the associated model for cause \(j\), survival through that subinterval is \(1-p_{j,r}\). Define \(R_r\) as the probability of reaching the start of subinterval \(r\) with all causes operating, and \(R'_r\) as the corresponding probability in the associated model. Both equal 1 for the first subinterval. The observed cause-\(j\) probability is the sum of the probabilities of reaching each subinterval and then exiting from \(j\):
These formulas require conditional probabilities defined for individuals exposed at the start of each subinterval, with no re-entry after an exit. The partition is a calculation device; no assumption about a continuous-time distribution of exit times is needed here. In particular, these formulas do not by themselves specify how to convert between an observed table and an associated table without the relevant conditional probabilities.
Worked Two-Cause Interval Calculation
The following example follows a hypothetical insured group over one year. The year is divided into three subintervals. Cause A and cause B are competing exits; all probabilities below are conditional on being exposed at the start of the stated subinterval.
Worked Example: Comparing Observed and Associated Probabilities
For a hypothetical one-year coverage review, the conditional probabilities of exit from causes A and B in the three successive subintervals are:
| Subinterval | Cause A | Cause B | Neither cause |
|---|---|---|---|
| 1 | 0.10 | 0.20 | 0.70 |
| 2 | 0.20 | 0.10 | 0.70 |
| 3 | 0.10 | 0.30 | 0.60 |
Find the observed probability of exit from A, the associated single-decrement probability for A, the observed probability of exit from B, and the probability of surviving the full interval. Then verify the total.
Method. First calculate how many individuals reach each subinterval when both causes operate. The probabilities of surviving subintervals 1 and 2 are \(0.70\) and \(0.70\), so the probabilities of reaching the three subintervals are
The observed probability of exit from A is therefore
For the associated table, remove cause B. The probabilities of reaching the starts of the subintervals with cause A alone are \(R'_1=1\), \(R'_2=1-0.10=0.90\), and \(R'_3=0.90(1-0.20)=0.72\). Thus
For cause B, the observed probability is
Survival through all three subintervals is
The total-probability check gives \(q_A+q_B+{}_1p_{\text{interval}}=0.289+0.417+0.294=1.000\). The calculations are consistent. In particular, the observed probability for A is \(0.063\) below its associated probability. The method avoids the mistake of calculating \(q'_A\) using the probability of reaching each subinterval under both causes; the associated model removes B when calculating exposure.
Why the Observed Probability Is No Greater
At every subinterval, survival with both causes operating is no greater than survival with cause \(j\) alone:
Multiplying the applicable survival probabilities through the earlier subintervals gives \(R_r\leq R'_r\) for every \(r\). Since every \(p_{j,r}\) is nonnegative, multiplying each exposure probability by \(p_{j,r}\) and summing yields the comparison.
To see the strictness condition, subtract the two probabilities:
Every term on the right is nonnegative. The sum is positive precisely when at least one term has both \(R'_r-R_r>0\) and \(p_{j,r}>0\). This qualification is important: a competing exit does not necessarily make the inequality strict merely because it has positive probability somewhere in the interval. It must reduce exposure to a subinterval where cause \(j\) could occur.
Worked Example: When Competition Does Not Make the Inequality Strict
In a hypothetical three-subinterval model, let the conditional probabilities for cause A be \(1,0,0\), and let the probabilities for cause B be \(0,0.50,0\). Find \(q_A\) and \(q'_A\), and explain why they are equal despite the possibility of cause B.
Observed model. Cause A occurs with probability 1 in the first subinterval, so nobody reaches the second subinterval. In detail, the reaching probabilities are \(R_1=1\), \(R_2=1(1-1-0)=0\), and \(R_3=0(1-0-0.50)=0\). Therefore
Associated model. With B removed, the cause-A conditional probabilities remain \(1,0,0\). Hence
Thus \(q_A=q'_A=1\). Cause B has a positive conditional probability in the second subinterval, but no one reaches it under either model. Competition does not reduce exposure to any subinterval with a positive cause-A probability: the only such subinterval is the first, which is reached with probability 1 in both models. The equality is therefore consistent with the strictness condition. The common error is to conclude that any positive competing probability forces \(q_A<q'_A\); what matters is whether that competition removes exposure to an opportunity for A.
Using the Difference as an Exposure Check
The difference \(q'_j-q_j\) can be interpreted as the probability contribution lost because competing exits reduce exposure to cause \(j\). It is not generally equal to the probability of the competing cause: the amount of exposure removed matters, as does the size of \(p_{j,r}\) in the subintervals no longer reached.
Worked Example: Locating the Exposure Reduction
For a hypothetical interval with three subintervals, suppose the conditional probabilities for cause A are \(0.05, 0.15, 0.25\), and those for cause B are \(0.10, 0.05, 0.15\). Calculate \(q_A\), \(q'_A\), and their difference. Identify how each subinterval contributes to that difference.
Observed exposure. The probabilities of surviving the three subintervals with both causes operating are \(0.85\), \(0.80\), and \(0.60\). Thus
The observed probability for A is
Associated exposure. With B removed, the probabilities of reaching the subintervals are
Therefore
The difference is \(0.394375-0.3475=0.046875\). Its subinterval contributions are zero in the first subinterval, \( (0.95-0.85)(0.15)=0.015\) in the second, and \((0.8075-0.68)(0.25)=0.031875\) in the third. Their sum is \(0.015+0.031875=0.046875\), matching the direct difference. Rounding only at the final step, to four decimal places, gives \(q_A=0.3475\), \(q'_A=0.3944\), and \(q'_A-q_A=0.0469\).
The method identifies exactly where the gap arises: B reduces exposure before the second and third subintervals, and A has positive conditional probabilities in both. It avoids treating the gap as simply the total probability of B.
Common Exam Traps and Checks
A frequent error is to use the observed reaching probabilities \(R_r\) in the associated calculation. That calculates \(q_j\) again, not \(q'_j\). For example, in the first worked example, using \(1,0.70,0.49\) gives \(0.289\), the observed A probability. The associated calculation instead uses the cause-A-only reaching probabilities \(1,0.90,0.72\), giving \(0.352\).
Another error is to add the conditional probabilities for different causes across subintervals without weighting by exposure. The probability \(p_{j,r}\) applies only to those who reach subinterval \(r\), so its contribution to the interval probability is \(R_rp_{j,r}\). A useful final check is that the observed probabilities of all causes, plus the probability of surviving the full interval, sum to 1. The associated probability is from a different model and should not be included in that observed-model total.
For each subinterval, list the probabilities for the target cause and all competing causes, conditional on exposure at its start.
Multiply the survival probabilities under all operating causes to obtain each \(R_r\), then sum \(R_rp_{j,r}\) for \(q_j\).
Remove the other causes, recompute \(R'_r\), and sum \(R'_rp_{j,r}\), or use the associated survival product.
Confirm \(q_j\leq q'_j\), assess strictness using the positive-opportunity condition, and check the observed probabilities sum to one with full-interval survival.
The distinction is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus because it requires careful interpretation of competing decrement probabilities. The central discipline is to keep the exposure model clear: observed probabilities use exposure under all causes, whereas associated probabilities use exposure with only the specified decrement.
Check Your Understanding
For each problem, treat the listed probabilities as conditional on exposure at the start of each subinterval. Causes are mutually exclusive within a subinterval.
- A hypothetical two-subinterval model has cause-A probabilities \(0.30,0.20\) and cause-B probabilities \(0.10,0.40\). Find \(q_A\), \(q'_A\), \(q_B\), and full-interval survival. (Answer: \(q_A=0.420,\ q'_A=0.440,\ q_B=0.340,\ \text{survival}=0.240\))
- Across three subintervals, cause A has probabilities \(0.10,0.20,0.30\), and cause B has probabilities \(0.20,0.10,0.10\). Find the observed and associated probabilities for A, and check observed survival. (Answer: \(q_A=0.387,\ q'_A=0.496,\ \text{survival}=0.294\))
- Cause A has conditional probabilities \(0,1,0.40\), while cause B has probabilities \(0.50,0,0\). Find \(q_A\) and \(q'_A\). Is the inequality strict? (Answer: \(q_A=0.500,\ q'_A=1.000\); yes, competition reduces exposure to the second subinterval, where cause A has positive probability.)
- In a hypothetical three-subinterval model, cause A has probabilities \(0.20,0.10,0.30\), and cause B has probabilities \(0.10,0.20,0.10\). Find the observed probability of A, its associated probability, the probability of B, and full-interval survival. (Answer: \(q_A=0.417,\ q'_A=0.496,\ q_B=0.289,\ \text{survival}=0.294\))
- Suppose a competing cause reduces the probability of reaching subinterval 3, but \(p_{j,3}=0\). It does not affect exposure to any earlier subinterval with positive \(p_{j,r}\). Can this reduction alone make \(q_j<q'_j\)? (Answer: No. It contributes zero to \(q'_j-q_j\) because the exposure difference is multiplied by \(p_{j,3}=0\).)