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Likelihood Inference and Instability · Tutorial 118 of 499

Likelihood Contributions Under Monotone Dropout in Missing-Data Models

Learn to construct and evaluate an individual observed-data likelihood contribution for each monotone dropout time, including a dropout event that may depend on a missing outcome.

Advanced 12 min read

What You'll Learn

  • Define the dropout time as the last visit at which a subject’s outcome is observed.
  • Factor an individual likelihood into longitudinal outcome densities and sequential response probabilities.
  • Include the dropout-event probability at the first missing visit.
  • Integrate over missing current and future outcomes when the dropout probability depends on them.
  • Simplify a contribution when dropout depends only on observed history.
  • Check that retention and dropout factors match the subject’s observed pattern.

Why the Dropout Event Belongs in the Likelihood

With monotone dropout, a subject’s outcomes are observed through some visit and then missing at every later scheduled visit. The observed outcomes alone do not describe the entire observed data: the fact that the subject stopped contributing measurements is also observed. A likelihood contribution must therefore account both for the observed outcome values and for the response pattern that ended in dropout.

The key calculation is at the first missing visit. The subject was retained through the last observed visit, then dropped out at the next one. If dropout at that next visit can depend on its outcome, that outcome is missing precisely when its dropout probability is needed. The event probability must then be averaged over the model for the missing outcome, rather than evaluated at a substituted value.

Notation and Model Factorization

Definition: For one subject with \(T\) scheduled visits, let \(Y_t\) be the outcome at visit \(t\), and let \(R_t=1\) if that outcome is observed and \(R_t=0\) otherwise. Monotone dropout means that \(R_t=0\) implies \(R_s=0\) for every \(s>t\). Define \(K\) as the last observed visit: \(K=k<T\) means outcomes are observed at visits \(1,\ldots,k\), and dropout occurs at visit \(k+1\). A completer has \(K=T\). We use \(K=0\) for dropout before any scheduled outcome is observed.

Condition throughout on the subject’s covariates \(X\). Specify the longitudinal outcome model by conditional densities \(f_t(y_t\mid y_1,\ldots,y_{t-1},X)\), with \(f_1(y_1\mid X)\) for the first visit. These densities factor the full outcome distribution; the model need not be Markov, because each density may condition on the full previous outcome history.

For a subject still under observation at visit \(t\), let \(\pi_t(y_1,\ldots,y_t,X)=P(R_t=1\mid R_{t-1}=1,Y_1=y_1,\ldots,Y_t=y_t,X)\) be the probability of being observed at that visit. Set \(R_0=1\). Thus \(1-\pi_t\) is the probability of dropping out at visit \(t\), conditional on having remained through visit \(t-1\) and on the outcome history through \(t\). This is a selection-model specification: dropout may depend on the current outcome, including an outcome that is missing because dropout occurs at that visit.

Assumption: Conditional on \(X\), the full outcome distribution has the stated sequential factorization, and the response pattern has the sequential probabilities \(\pi_t\). Dropout is absorbing, and subjects contribute independently to the likelihood. The likelihood for an observed pattern is obtained by multiplying its outcome and response factors, then integrating over every unobserved outcome. A response probability after dropout is not included: once a subject has dropped out, later visits are not new retention opportunities in this pattern.

Contribution for Each Dropout Time

Suppose outcomes \(y_1,\ldots,y_k\) are observed and \(k<T\). At visits \(1,\ldots,k\), the subject both has the observed outcome and remains under observation. At visit \(k+1\), the subject drops out. Write \(u_{k+1},\ldots,u_T\) for the unobserved outcomes. The individual observed-data likelihood contribution is

$$ L_k(y_1,\ldots,y_k\mid X) = \int \left\{ \prod_{t=1}^{k} f_t(y_t\mid y_1,\ldots,y_{t-1},X)\, \pi_t(y_1,\ldots,y_t,X) \right\} f_{k+1}(u_{k+1}\mid y_1,\ldots,y_k,X) \left\{1-\pi_{k+1}(y_1,\ldots,y_k,u_{k+1},X)\right\} \prod_{t=k+2}^{T} f_t(u_t\mid y_1,\ldots,y_{t-1},X) \,du_{k+1}\cdots du_T . $$

The product through \(k\) records both the observed outcomes and the retention events. The factor \(1-\pi_{k+1}\) records the dropout event itself. All outcomes after the last observed visit are integrated out. If a dropout probability depends on an unobserved outcome at a later visit, that outcome cannot simply be omitted: it remains inside the integral wherever it enters the response model.

For a completer, every outcome is observed and the contribution has no missing-outcome integral:

$$ L_T(y_1,\ldots,y_T\mid X) = \prod_{t=1}^{T} f_t(y_t\mid y_1,\ldots,y_{t-1},X)\, \pi_t(y_1,\ldots,y_t,X). $$

For dropout before the first outcome, use the same pattern with \(k=0\): integrate the full outcome distribution multiplied by the probability of dropout at visit 1. For independent subjects, the full likelihood is the product of their individual contributions, or equivalently the sum of their log contributions on the log-likelihood scale.

Worked Example: Dropout Depending on the Missing Current Outcome

Consider a hypothetical two-visit study with binary outcomes. For one subject, \(Y_1=1\) is observed and the subject drops out at visit 2. Suppose \(P(Y_1=1)=0.40\), \(\pi_1(Y_1=1)=0.80\), and \(P(Y_2=1\mid Y_1=1)=0.70\). At visit 2, the dropout probabilities are \(0.10\) if \(Y_2=0\) and \(0.30\) if \(Y_2=1\). Calculate the individual contribution.

The observed first outcome contributes \(0.40\), and being observed at visit 1 contributes \(0.80\). Since \(Y_2\) is missing, sum over its two possible values, multiplying each outcome probability by the corresponding dropout probability:

$$ \begin{aligned} L_1 &=P(Y_1=1)\,\pi_1(1) \left[ P(Y_2=0\mid Y_1=1)\{1-\pi_2(1,0)\} + P(Y_2=1\mid Y_1=1)\{1-\pi_2(1,1)\} \right] \\ &=(0.40)(0.80)\left[(0.30)(0.10)+(0.70)(0.30)\right]\\ &=(0.40)(0.80)(0.24)=0.0768. \end{aligned} $$

Thus the contribution is \(0.0768\). The bracketed term is the probability of dropout at visit 2, averaged over the missing \(Y_2\) under its conditional outcome distribution. The event probability is not just \(0.10\) or \(0.30\): neither outcome value was observed.

How the Integration Can Simplify

Sometimes dropout at the first missing visit depends only on the observed history, not on its missing current outcome. Then its event probability can be taken outside the integral. The missing outcome densities integrate to one, provided the response model has no further factors depending on those missing outcomes. This yields a simpler pattern probability while retaining the observed outcome densities and the retention probabilities up to the last observed visit.

Worked Example: Dropout Depending Only on Observed History

In a hypothetical three-visit study, a subject has \(Y_1=1\) observed and drops out before visit 2. Suppose \(P(Y_1=1)=0.40\), \(\pi_1(1)=0.90\), and the probability of dropping out at visit 2, given \(Y_1=1\), is \(0.25\). Assume this probability does not depend on \(Y_2\) or later outcomes. Find the contribution.

The general expression includes the outcome densities for visits 2 and 3 inside an integral. Because the dropout probability is constant over those outcomes, their conditional densities integrate to one:

$$ \begin{aligned} L_1 &=(0.40)(0.90)(0.25) \int f_2(u_2\mid 1)f_3(u_3\mid 1,u_2)\,du_2\,du_3\\ &=(0.40)(0.90)(0.25)(1)=0.0900. \end{aligned} $$

The contribution is \(0.0900\). No value is imputed for either missing outcome; their outcome densities disappear only because they are integrated over their full ranges and the dropout event does not depend on them.

A Later Dropout Event and a Complete Pattern

For dropout at a later visit, each earlier observed visit contributes both an outcome density and a retention probability. The next example makes that sequence explicit.

Worked Example: Observed Through Visit 2, Dropout at Visit 3

Consider a hypothetical three-visit study with binary outcomes. A subject has \(Y_1=0\) and \(Y_2=1\) observed, then drops out at visit 3. Suppose \(P(Y_1=0)=0.60\), \(\pi_1(0)=0.90\), \(P(Y_2=1\mid Y_1=0)=0.50\), and \(\pi_2(0,1)=0.80\). Let \(P(Y_3=1\mid Y_1=0,Y_2=1)=0.80\). The visit-3 dropout probability is \(0.20\) if \(Y_3=0\) and \(0.50\) if \(Y_3=1\). Calculate the contribution.

The first two visits contribute the observed outcome probabilities and the probabilities of remaining observed. At visit 3, sum the dropout event probability over the missing binary outcome:

$$ \begin{aligned} P(\text{dropout at 3}\mid Y_1=0,Y_2=1) &=(0.20)(0.20)+(0.80)(0.50)\\ &=0.04+0.40=0.44,\\ L_2 &=(0.60)(0.90)(0.50)(0.80)(0.44)\\ &=0.09504. \end{aligned} $$

The individual contribution is \(0.09504\). In this example the event probability depends on the missing \(Y_3\), so it is averaged over its conditional distribution. There are no later outcomes to integrate because visit 3 is the final scheduled visit.

For comparison, a hypothetical subject observed at all three visits contributes the product of three outcome densities and three retention probabilities. The completer’s contribution contains \(\pi_3\), not \(1-\pi_3\); the subject remained observed at visit 3. This distinction is a useful check on the pattern being encoded.

Derivation and Practical Checks

The formula follows directly from the joint model for outcomes and response. For fully specified outcomes, the outcome factor is the product of the \(f_t\)'s. Under the sequential response model, the probability of a pattern observed through \(k\) and then dropping out is the product of retention probabilities through \(k\), followed by the dropout probability at \(k+1\). Multiplying these gives the joint density for the full outcomes and that response pattern. The observed-data likelihood is the marginal density of what was recorded, so integrate that joint density over every missing outcome. This establishes both why the dropout-event factor appears and why it must remain inside the integral when it depends on missing outcomes.

A practical construction check is to read the subject’s pattern from left to right:

1
Record observed outcomes.
Include one outcome density for each observed visit, evaluated at its recorded value.
2
Record retention.
Multiply by \(\pi_t\) at every visit through the last observed visit.
3
Record dropout.
For a dropout pattern, include \(1-\pi_{k+1}\) at the first missing visit, using the model’s full conditioning history.
4
Marginalize missing outcomes.
Integrate all missing outcomes, keeping any outcome-dependent response factor inside the integral.

Several failures can be detected by checking whether a factor matches the observed pattern. A dropout pattern must not contain a retention factor at its first missing visit; a completer must not contain a dropout factor there. The contribution should also be a probability or density on the appropriate scale, and the integrated missing-outcome model should be normalized. Numerically, an outcome-dependent dropout term can be especially sensitive to the tail behavior of the missing-outcome distribution, so integration limits, quadrature accuracy, and probability calculations should be checked.

Relation to Other Likelihood Representations

The sequential expression is one way to calculate the observed-data likelihood under a specified selection model. A joint-pattern formulation can instead write the probability of each complete response pattern conditional on the outcomes, then integrate missing outcomes. It gives the same contribution when it represents the same joint model. The sequential form is useful for monotone dropout because it makes each retention event and the first dropout event visible.

A pattern-mixture formulation instead models outcomes conditional on the missingness pattern. It organizes the same observed data by pattern, but it is not obtained by simply deleting the dropout factor from the selection-model contribution: the conditional outcome distributions and pattern probabilities must together represent a coherent joint distribution. Complete-case analysis is different again; it discards dropout patterns and does not calculate their likelihood contributions. These alternatives do not remove the need to specify assumptions about unobserved outcomes.

Common Mistake and Limitations

Common mistake: “The subject’s last outcome is missing, so I will use the dropout probability evaluated at the last observed outcome,” or “I can omit the dropout factor because the outcome is missing.” Neither is generally correct. If the response model depends on the missing current outcome, the correct event probability is integrated over that outcome under the specified outcome model. Omitting the event factor changes the response pattern being modeled.

These contributions are valid only relative to the specified outcome and response models, the visit schedule, and the monotone-pattern assumption. They do not establish that the selection model is correct or that its unobserved-outcome assumptions are empirically identified from observed data. If dropout is nonmonotone, the response-pattern factorization must represent the actual return and missingness process rather than treating the first absence as absorbing. In all cases, a likelihood can be calculated precisely while still depending strongly on assumptions about outcomes that were never observed.

Check Your Understanding

Answer each question, then compare with the brief solutions.

  1. A subject is observed at visits 1 and 2, then drops out before visit 3. Which response factors appear in the contribution?
  2. Why must a visit-3 dropout probability depending on \(Y_3\) remain inside an integral when \(Y_3\) is missing?
  3. If dropout at visit \(k+1\) depends only on the observed history through visit \(k\), what happens to the integrated outcome densities for visits \(k+1,\ldots,T\), assuming no later response factors enter?
  4. For a subject observed through the final visit, should the final response factor be \(\pi_T\) or \(1-\pi_T\)? Explain.

Solutions. (1) Include \(\pi_1\) and \(\pi_2\), followed by \(1-\pi_3\); the observed outcome densities at visits 1 and 2 also appear. (2) The dropout event probability varies with the missing value, so the observed-data contribution averages its product with the outcome density over possible \(Y_3\). (3) They integrate to one when the event probability is constant over those outcomes and the conditional densities are normalized. (4) Use \(\pi_T\), because the subject was observed at the final visit; \(1-\pi_T\) would encode dropout there.

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