From Dropout Hazards to Observed-Data Contributions
In a longitudinal selection model, the observed-data likelihood combines two ingredients: a model for the outcomes over time and a model for the dropout decisions. The key bookkeeping task is to include the dropout probability for the observed response pattern while summing or integrating over every outcome that was not observed. This becomes especially visible in a three-visit study: a participant may complete follow-up, drop out after visit one, or drop out after visit two.
We use \(Y_t\) for the outcome at visit \(t\). Let \(D_t=1\) indicate dropout at the decision associated with visit \(t\), among participants still under follow-up. Thus \(D_2=1\) means dropout after visit one, and \(D_3=1\) means dropout after visit two. We assume no dropout before visit one. Under the timing convention used here, the outcome at the decision visit is generated before the dropout decision, even if dropout means that outcome is not recorded.
For the examples, the outcome model has joint distribution \(f(y_1,y_2,y_3)=f_1(y_1)f_2(y_2\mid y_1)f_3(y_3\mid y_1,y_2)\). The dropout model has hazards at visits two and three. The hazard may depend on the current outcome and past history; in the numerical example, history terms are omitted to keep the arithmetic transparent. These are model assumptions, not consequences of the observed response pattern. In particular, the current outcome may be latent when dropout occurs, but it still enters the selection probability under this timing convention.
Three Response Patterns and Their Likelihood Contributions
A likelihood contribution is the joint probability or density of the observed outcomes and the response pattern. It is not generally just the outcome density for the recorded values: the selection factors also describe why those outcomes were observed. For continuous outcomes, the sums below become integrals.
The first expression contains two probabilities of continued follow-up: the participant must remain through the visit-two decision and then through the visit-three decision. For dropout after visit one, \(Y_1\) is observed but \(Y_2\) is latent at the dropout decision. The contribution therefore includes \(q_2\) evaluated at each possible \(Y_2\), weighted by its outcome probability. \(Y_3\) is also missing, but there is no visit-three dropout factor for this pattern: the participant already dropped out at visit two. Summing its conditional outcome distribution over \(Y_3\) gives one. For dropout after visit two, \(Y_2\) is observed and the participant has survived the visit-two decision; \(Y_3\) is latent when the visit-three dropout decision occurs, so it is summed together with \(q_3\).
Worked Example: Setting the Dropout Hazards
Worked Example: Setting the Dropout Hazards
Suppose, hypothetically, the conditional dropout hazards depend only on the current binary outcome \(Y_t\), with \(Y_t=0\) or \(1\). Let each hazard follow a logistic model, \(\operatorname{logit}(q_t(y))=\alpha_t+\beta_t y\). Choose coefficients that produce \(q_2(0)=0.10,\;q_2(1)=0.40,\;q_3(0)=0.20,\) and \(q_3(1)=0.50\).
For visit two, \(\alpha_2=\operatorname{logit}(0.10)=\log(1/9)\approx-2.1972\). The coefficient for \(Y_2\) is the difference between the two logits:
Substitution gives \(\operatorname{logit}\{q_2(1)\}=-2.1972+1.7918\approx-0.4054\), whose inverse-logit is \(0.400\); at \(Y_2=0\), the inverse-logit of \(-2.1972\) is \(0.100\). Similarly, \(\alpha_3=\operatorname{logit}(0.20)=\log(1/4)\approx-1.3863\), and \(\beta_3=\operatorname{logit}(0.50)-\operatorname{logit}(0.20)=\log(4)\approx1.3863\). Thus the hazards at visit three are \(0.200\) when \(Y_3=0\) and \(0.500\) when \(Y_3=1\). These coefficients specify hazards conditional on current outcome; they do not by themselves specify the distribution of the outcomes.
Worked Example: Calculating Contributions for Observed Records
Worked Example: Calculating Contributions for Observed Records
Continue with those hazards. Suppose, hypothetically, that the outcomes are independent Bernoulli variables with \(P(Y_1=1)=0.40,\;P(Y_2=1)=0.30,\) and \(P(Y_3=1)=0.20\). This independence assumption makes \(f_1f_2f_3\) a product of the corresponding Bernoulli probabilities. Calculate contributions for a complete record \((1,0,1)\), a record with \(Y_1=0\) followed by dropout after visit one, and a record \((Y_1,Y_2)=(1,1)\) followed by dropout after visit two.
For the complete record, the outcome probability is \(0.40(0.70)(0.20)\). Its visit-two non-dropout probability is \(1-q_2(0)=0.90\), and its visit-three non-dropout probability is \(1-q_3(1)=0.50\). Therefore
For dropout after visit one with \(Y_1=0\), the probability of the observed first outcome is \(0.60\). The latent \(Y_2\) can equal zero or one. Averaging its dropout hazard over its Bernoulli distribution gives \(0.70(0.10)+0.30(0.40)=0.070+0.120=0.190\). The distribution of \(Y_3\) sums to one and contributes no further dropout factor to this response pattern. Thus
For dropout after visit two with \(Y_1=1,Y_2=1\), the observed outcomes have probability \(0.40(0.30)\), and the participant remains through the visit-two decision with probability \(1-q_2(1)=0.60\). The unobserved \(Y_3\) is zero with probability \(0.80\) and one with probability \(0.20\), so its averaged visit-three dropout probability is \(0.80(0.20)+0.20(0.50)=0.160+0.100=0.260\). Hence
Each value is a probability for the specified outcomes together with the specified response pattern under this fully specified hypothetical model. These individual contributions are not expected to add to one, because they refer to different observed values as well as different patterns.
Why the Summation Is Essential
The preceding calculations illustrate a general justification for integrating over missing outcomes. The full-data model assigns probability to a complete outcome path and a response path. For an observed-data contribution, the unobserved outcomes are not known, so the joint probability must be summed over all their possible values. This operation preserves the dropout hazard at the decision time while removing outcomes that the data do not reveal.
For the relevant response pattern, multiply the longitudinal outcome probability by the sequential probabilities of dropout or continued follow-up.
A complete record has two continued-follow-up factors. A participant who drops out after visit one has a visit-two dropout factor, while one who drops out after visit two has a visit-two continuation factor and a visit-three dropout factor.
For each latent outcome, sum the product of its outcome probability and any dropout factor that depends on it. If no factor depends on a later outcome, its conditional probabilities sum to one.
For example, with the numerical outcome model above, the overall probability of complete follow-up is the average of the two continuation probabilities. Independence of \(Y_2\) and \(Y_3\) gives
The total probability of dropout after visit one is \(0.190\), the average visit-two dropout hazard. The total probability of dropout after visit two is \(0.810(0.260)=0.2106\): the participant must first remain through visit two and then drop out at visit three. The three pattern probabilities check:
This normalization is a useful check on the hazards and factorization. It is a check on the total response-pattern probabilities, not on the sum of likelihood contributions for selected records.
Diagnostics, Alternatives, and Limitations
A common mistake is to treat dropout after visit one as depending only on the observed \(Y_1\), then omit the visit-two selection factor because \(Y_2\) was not recorded. That reasoning is incorrect under the stated timing: \(Y_2\) is generated before the visit-two dropout decision, and the hazard may depend on its latent value. The correction is to calculate \(f_2(y_2\mid y_1)q_2(y_2\mid y_1)\) for each possible \(y_2\), then sum. Conversely, adding a visit-three hazard to this pattern is also wrong: a participant who already dropped out at visit two does not face the later decision.
Practical checks include confirming that every observed outcome appears in the outcome factors, every dropout or continuation event appears once, and each missing outcome is integrated over its full support. For discrete outcomes, probabilities in each conditional outcome distribution should sum to one. Summing the resulting total probabilities over all possible response patterns should also yield one. A total above or below one often signals a missing continuation factor, a dropout factor included twice, or an outcome summed out without weighting by the hazard that depends on it.
The selection-model description makes the dropout probability conditional on current and possibly past outcomes. A missing-at-random model instead expresses the response mechanism as not depending on the unobserved current outcome after conditioning on observed information. These are distinct assumptions about the relationship between outcomes and response, and the observed records alone do not generally establish which is appropriate. A pattern-mixture description organizes the model by response pattern rather than factoring outcome and response distributions in this order. These are alternative parameterizations of incomplete-data assumptions, not a reason to omit selection factors from a likelihood once a selection model has been specified.
The numerical example uses independent binary outcomes and fixed hazards solely to make every sum explicit. In an application, outcome distributions may depend on earlier outcomes, and dropout hazards may depend on observed history as well as the current outcome. The same summation logic applies, but the calculations can become more demanding. Moreover, dependence on a latent outcome introduces information that cannot generally be learned from the observed-data distribution without additional assumptions or sensitivity choices. A correctly calculated likelihood contribution does not, by itself, resolve that identification issue or validate the assumed outcome and dropout models.
Check Your Understanding
Answer each question using the timing convention and definitions in this tutorial. Brief solutions follow.
- For dropout after visit one, which outcome is latent at the dropout decision, and which hazard belongs in the contribution?
- Suppose \(q_2(0)=0.20,\;q_2(1)=0.50\), and \(P(Y_2=1\mid Y_1=y_1)=0.40\). What is the dropout probability averaged over \(Y_2\), conditional on \(Y_1=y_1\)?
- For a record observed through visit two and then lost, why must the contribution include \(1-q_2\) as well as a visit-three dropout probability?
- In the numerical example, verify the total probability of dropout after visit two.
- What does a pattern-probability sum different from one suggest, and what does it not establish?
Solutions. (1) \(Y_2\) is latent at the visit-two decision, so the contribution averages \(q_2(Y_2\mid Y_1)\) over the conditional distribution of \(Y_2\); it does not include a visit-three hazard. (2) The average is \(0.60(0.20)+0.40(0.50)=0.12+0.20=0.320\). (3) The participant must remain under follow-up at visit two before reaching the visit-three dropout decision, giving the factor \(1-q_2\); the visit-three factor is then averaged over latent \(Y_3\). (4) The probability is \(0.810(0.260)=0.2106\). (5) It suggests a factorization or summation error, such as omitting or duplicating a decision factor. A sum of one does not establish that the outcome model, dropout model, or identifying assumptions are correct.