Why “Associated” Does Not Mean “Observed”
In a multiple-decrement model, several causes can remove a person from the cohort during the same interval. An observed decrement probability records the chance that a particular cause is the one that removes the person, with the competing causes present. An associated single-decrement probability asks a different question: what would the probability of that cause be over the interval if every competing cause were removed?
The distinction matters because an observed exit count describes the cohort that actually experienced competition, whereas an associated probability describes a hypothetical cause-only model. The two quantities answer different questions and are not interchangeable. In particular, an observed cause-specific probability does not, by itself, tell us the associated probability.
The prime in \(q_j'\) is a label for the associated quantity; it does not mean a derivative. The phrase “single-decrement” refers to the hypothetical model having only one decrement cause. It does not mean that the person is guaranteed to exit, or that \(q_j'\) is the observed fraction of the original cohort exiting from cause \(j\).
Two Different Probability Questions
Consider a person in the cohort at the beginning of the year. Under competition, the person might exit through cause \(j\), exit through another cause, or remain in the cohort through the end of the year. The observed probability \(q_j\) concerns the first of these outcomes. To define \(q_j'\), instead imagine suppressing every cause other than \(j\), while retaining the specified cause-\(j\) process. The probability of exit through \(j\) in that hypothetical setting is \(q_j'\).
A useful way to see the distinction is to describe potential event times. Suppose \(T_j\) is the time at which cause \(j\) would occur if considered on its own, and \(T_k\) is the corresponding time for each competing cause \(k\). In a model where suppressing competitors leaves the cause-\(j\) event process unchanged, the associated probability is the chance that \(T_j\) falls within the year. The observed probability also requires that no competing cause preempt it.
This event description gives a short justification for an important comparison. Under the stated unchanged-process interpretation, every outcome counted in \(q_j\) is also an outcome in which \(T_j\leq 1\). The observed event is therefore a subset of the event defining \(q_j'\), so \(q_j\leq q_j'\). This comparison depends on the interpretation that removing competitors does not itself change the cause-\(j\) mechanism. If a model allows the mechanism for cause \(j\) to change when another cause is removed, the comparison need not hold.
This event argument explains the meaning of the associated quantity; it does not provide a universal conversion formula from observed probabilities to associated probabilities. Such a conversion requires additional assumptions about how decrements operate during the interval. The specific within-interval assumptions and conversion methods are separate topics.
Worked Example: Separating an Observed Exit Rate From a Cause-Only Risk
A hypothetical cohort has 2,000 people at the start of a year. During the year, 120 exit through withdrawal and 60 exit through disability; the remaining people stay in the cohort to year-end. A separate hypothetical model that retains withdrawal as the only decrement assigns an associated withdrawal probability of \(0.075\). Find the observed probabilities, the associated withdrawal probability, and the corresponding counts implied by the cause-only model.
Method. Divide observed exits by the actual starting cohort to obtain the observed multiple-decrement probabilities. Treat the separately supplied \(0.075\) as the cause-only probability; do not substitute an observed probability for it.
Observed withdrawal probability. \(q_{\text{withdrawal}}=120/2{,}000=0.060\).
Observed disability probability. \(q_{\text{disability}}=60/2{,}000=0.030\). The total observed decrement probability is \(0.060+0.030=0.090\). Thus the number remaining is \(2{,}000(1-0.090)=1{,}820\), consistent with \(2{,}000-120-60=1{,}820\).
Associated withdrawal probability. The problem supplies \(q_{\text{withdrawal}}'=0.075\) for the hypothetical model with disability and all other competing causes removed. It is not the observed value \(0.060\).
Cause-only count equivalent. If the cause-only probability were applied to 2,000 comparable starting lives, the expected withdrawal count would be \(2{,}000(0.075)=150\). This 150 is a hypothetical cause-only count, not the 120 withdrawals observed in the competing cohort.
Check and trap avoided. The actual exits total \(120+60=180\), or \(180/2{,}000=0.090\), matching the sum of observed probabilities. The common mistake is to call \(150\) the observed withdrawal count or to report \(0.075\) as the cohort’s observed withdrawal probability; the actual observed count is 120 and the actual observed probability is \(0.060\).
Observed Probabilities Do Not Determine Associated Probabilities
Knowing the observed probabilities tells us how many people exited through each cause in the competing model. It does not reveal exactly how many would have exited through each cause had the other causes been absent. Different descriptions of how causes operate within the year can produce the same observed probabilities but different associated probabilities.
The next example uses deliberately specified sequential mechanisms to demonstrate that point. These are hypothetical constructions, not a general conversion rule. The stage order and the conditional probabilities are part of each model’s assumptions.
Worked Example: Same Observed Exits, Different Associated Risks
A hypothetical two-cause model has observed probabilities \(q_A=0.12\) and \(q_B=0.08\). Compare two possible within-year mechanisms. In Model 1, cause A is assessed first with probability \(0.12\); if A does not occur, cause B is assessed with a conditional probability chosen to reproduce \(q_B=0.08\). In Model 2, cause B is assessed first with probability \(0.08\); if B does not occur, cause A is assessed with a conditional probability chosen to reproduce \(q_A=0.12\). For each model, calculate both associated probabilities by suppressing the other cause while retaining the stated stage probabilities.
Method. First solve for the conditional probability at the second stage from the observed exit probability. Then suppress the competing cause in turn and use the remaining cause’s specified probability. This calculation depends on the sequential assumptions in the question.
Model 1: A is assessed first. Cause A’s observed probability is \(0.12\). Cause B can occur only when A has not occurred, which has probability \(1-0.12=0.88\). Let \(r_B\) be B’s conditional probability at its stage. Then \(0.88r_B=0.08\), so \(r_B=0.08/0.88=0.090909\ldots\). To three decimals, \(r_B=0.091\).
If B is suppressed, A retains its first-stage probability, so \(q_A'=0.12\). If A is suppressed, B’s specified second-stage probability applies, so \(q_B'=0.090909\ldots\), or \(0.091\) to three decimals. The observed probabilities are recovered: \(0.12+0.88(0.090909\ldots)=0.12+0.08=0.20\).
Model 2: B is assessed first. Cause B’s observed probability is \(0.08\). Cause A can occur only when B has not occurred, which has probability \(1-0.08=0.92\). Let \(r_A\) be A’s conditional probability at its stage. Then \(0.92r_A=0.12\), so \(r_A=0.12/0.92=0.130434\ldots\). To three decimals, \(r_A=0.130\).
If A is suppressed, B retains its first-stage probability, so \(q_B'=0.08\). If B is suppressed, A’s specified second-stage probability applies, so \(q_A'=0.130434\ldots\), or \(0.130\) to three decimals. Again, the observed probabilities are recovered: \(0.08+0.92(0.130434\ldots)=0.08+0.12=0.20\).
Check and trap avoided. Both models produce the same observed pair \((0.12,0.08)\), but Model 1 gives associated probabilities \((0.12,0.091)\), while Model 2 gives \((0.130,0.08)\), with the pairs ordered as A then B. Thus the observed pair alone does not determine the associated pair. The common mistake is to assert that an observed probability uniquely specifies its associated probability without stating a within-year model.
Interpreting the Associated Table
An associated single-decrement table is best read as a family of cause-specific hypothetical tables. For each cause, the other decrement causes are removed for the purpose of defining that cause’s associated probabilities. Consequently, the different associated tables do not describe several causes operating simultaneously in one cohort.
For this reason, associated probabilities cannot generally be added to obtain the observed probability of any decrement. Nor should the associated probability for cause \(j\) be multiplied by the actual cohort size and presented as the observed number of exits from \(j\). Such a multiplication can be useful only as a clearly labelled hypothetical expectation under the corresponding cause-only model.
A reliable interpretation proceeds in three steps:
Confirm that the probabilities refer to the same one-year interval and the same type of starting life.
Ask whether it is the observed probability under competition, \(q_j\), or the associated cause-only probability, \(q_j'\).
Observed probabilities alone are not a conversion method. Any calculation connecting observed and associated quantities must use an explicitly stated model for the causes within the year.
Worked Example: Keeping Actual Counts and Cause-Only Expectations Apart
A hypothetical program begins the year with 500 members. Its observed multiple-decrement probabilities are \(q_A=0.11\) and \(q_B=0.07\). Separately, a cause-only assessment reports \(q_A'=0.16\) when only A remains and \(q_B'=0.10\) when only B remains. Find the actual observed exit counts by cause and the cause-only expected counts for groups of 500 under each associated probability. State whether the two associated probabilities give the actual total exit probability.
Method. Apply observed probabilities to the actual competing cohort. Apply each associated probability only to its own hypothetical cause-only comparison group.
Actual exits through A. \(500(0.11)=55\) members.
Actual exits through B. \(500(0.07)=35\) members. The actual total number of exits is \(55+35=90\), and the observed total probability is \(90/500=0.18\). The number remaining is \(500-90=410\).
Cause-only expectation for A. \(500(0.16)=80\) expected exits in a hypothetical group of 500 exposed only to A.
Cause-only expectation for B. \(500(0.10)=50\) expected exits in a separate hypothetical group of 500 exposed only to B.
The sum of associated probabilities is \(0.16+0.10=0.26\), but \(0.26\) is not the actual total exit probability in the competing cohort. The actual total is \(0.11+0.07=0.18\). The associated counts 80 and 50 are likewise not the actual counts 55 and 35.
Check and trap avoided. The actual count check is \(55+35+410=500\). The common mistake is to add the associated probabilities and report \(0.26\) as the observed total risk; the observed figures supplied in the problem give \(0.18\).
Common Exam Traps
- Dropping the hypothetical condition. “With all other causes removed” is part of the definition of \(q_j'\), not an optional detail.
- Using the wrong denominator or scenario. Observed exits are measured against the actual starting cohort. An associated probability belongs to a cause-only hypothetical model.
- Assuming a universal conversion. The relationship between \(q_j\) and \(q_j'\) depends on how decrements operate during the interval. Do not infer a conversion formula from the definitions alone.
- Adding associated probabilities. Each cause-only probability comes from a different hypothetical setting. Their sum does not generally equal the observed probability of any exit.
- Overstating the inequality. Under an unchanged cause-specific process, competition can preempt a cause, giving \(q_j\leq q_j'\). If removing competitors also changes cause \(j\)’s mechanism, that conclusion requires further justification.
Check Your Understanding
For each problem, distinguish the observed competing-decrement quantity from the associated cause-only quantity. Treat supplied probabilities as exact unless stated otherwise.
- A cohort of 800 has 48 observed exits from cause R and 32 from cause S during a year. A separate cause-only model gives \(q_R'=0.09\). Find \(q_R\), \(q_S\), the number remaining, and the expected cause-R count in a cause-only group of 800. (Answer: \(q_R=0.060\), \(q_S=0.040\), 720 remain, and 72 expected cause-R exits in the cause-only group.)
- Under an unchanged cause-specific event process, explain the event-set reason that the observed probability \(q_M\) cannot exceed the associated probability \(q_M'\). (Answer: the observed event requires \(T_M\leq1\) and no competitor to preempt M, so it is a subset of the event \(T_M\leq1\) defining \(q_M'\); hence \(q_M\leq q_M'\).)
- A sequential model assesses cause C first with probability \(0.10\), then assesses D if C does not occur. The observed probability of D is \(0.09\). Find the conditional D probability at its stage and both associated probabilities under the stated sequential mechanism. (Answer: \(0.09/0.90=0.10\); \(q_C'=0.10\) and \(q_D'=0.10\).)
- Two causes have observed probabilities \(q_U=0.15\) and \(q_V=0.05\). In a model with U assessed first, calculate the conditional V probability if V is assessed only after U does not occur. Then state the associated probabilities under that model. (Answer: conditional V probability \(0.05/0.85=0.0588\) to four decimals; \(q_U'=0.15\), \(q_V'=0.0588\) to four decimals.)
- A cohort of 1,200 has observed probabilities \(q_E=0.08\) and \(q_F=0.06\), while separate cause-only assessments give \(q_E'=0.12\) and \(q_F'=0.09\). Find the actual exits by cause, the actual number remaining, and the two cause-only expected exit counts for comparison groups of 1,200. (Answer: 96 actual exits from E, 72 from F, 1,032 remain; cause-only expected counts are 144 for E and 108 for F.)