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Pattern-Mixture Models for Dropout · Tutorial 198 of 499

Observed-Data Contributions by Dropout Pattern

Learn to derive and calculate the observed-data likelihood contribution for each dropout pattern, including the integration over outcomes that were not observed.

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What You'll Learn

  • Define a monotone dropout pattern by the last observed visit.
  • Derive the observed-data contribution from the joint pattern-outcome distribution.
  • Distinguish a pattern-specific marginal density from the full outcome density.
  • Calculate subject-level and cohort likelihoods for three-visit examples.
  • Identify common errors in likelihood construction and checks for pattern coding.

Why the Pattern-Specific Contribution Matters

A longitudinal record with dropout contains two kinds of observed information: which visits were observed, and the outcome values recorded at those visits. A likelihood contribution must account for both. In a pattern-mixture formulation, the contribution for a participant who drops out after visit 2 is not just a density for that participant’s observed outcomes. It also includes the probability of belonging to the dropout pattern, and the outcome density must be the appropriate marginal within that pattern.

This distinction is easy to miss when writing a likelihood from a complete-data model. Future outcomes are not observed for a participant who drops out, so their values cannot be inserted into a complete-data density. They must be integrated out. For monotone dropout over three visits, this gives a simple set of pattern-specific contributions.

Patterns and the Observed-Data Likelihood

Let \(Y=(Y_1,Y_2,Y_3)\) denote the outcomes at three scheduled visits, and let \(D\) be the last visit with an observed outcome. Thus \(D=0\) means no outcome visit was observed, \(D=1\) means only \(Y_1\) was observed, \(D=2\) means \(Y_1,Y_2\) were observed, and \(D=3\) means all three outcomes were observed. This encoding assumes monotone dropout: once a visit is missed because of dropout, later scheduled outcomes are also missing.

Write \(\pi_d=P(D=d)\) for the probability of pattern \(d\). For each pattern, write \(f_d(y_1,y_2,y_3)\) for the joint density of the full outcome vector conditional on \(D=d\). This pattern-specific distribution may describe outcomes that are unobserved for members of that pattern. Define \(f_d(y_{1:d})\) as its marginal density for the outcomes observed through visit \(d\), obtained by integrating out the later outcomes. For \(d=0\), the observed outcome vector is empty and its marginal density is 1.

Definition: For a participant with pattern \(D=d\) and observed outcomes \(y_{1:d}\), the observed-data likelihood contribution is \(\pi_d f_d(y_{1:d})\). The factor \(\pi_d\) describes the probability of the observed pattern; \(f_d(y_{1:d})\) describes the observed outcomes conditional on that pattern.

For \(d=1\) or \(2\), the marginal density is defined by integration over the missing outcomes:

$$ f_d(y_{1:d})= \int f_d(y_1,y_2,y_3)\,dy_{d+1}\cdots dy_3. $$

The integral is interpreted as a sum for discrete outcomes and as the corresponding combination of sums and integrals for mixed outcomes. At \(d=3\), there is nothing to integrate; at \(d=0\), integrating the full conditional density over all outcomes gives 1. If covariates \(X=x\) are included, the same expression applies conditionally on \(x\), with \(\pi_d(x)=P(D=d\mid X=x)\) and \(f_d(y_{1:d}\mid x)\).

For independent participants \(i=1,\ldots,n\), with observed patterns \(d_i\) and observed outcomes \(y_{i,1:d_i}\), the observed-data likelihood is

$$ L_{\mathrm{obs}} = \prod_{i=1}^{n}\pi_{d_i}f_{d_i}(y_{i,1:d_i}). $$

This product is a likelihood for the observed patterns and observed outcomes. It does not evaluate a density at a fabricated value for any missing outcome. The pattern-specific full distributions are part of the model, but the likelihood contribution for an incomplete record uses only their observed-outcome marginals.

Why Marginalization Gives This Contribution

By definition of the pattern-mixture factorization, the joint density of \(D=d\) and the full outcome vector is \(\pi_d f_d(y_1,y_2,y_3)\). The observed record contains \(D=d\) and \(Y_1,\ldots,Y_d\); the remaining outcomes are unobserved. The observed-data density is therefore the joint density integrated over every unobserved outcome:

$$ \begin{aligned} p(D=d,Y_{1:d}=y_{1:d}) &=\int \pi_d f_d(y_1,y_2,y_3)\,dy_{d+1}\cdots dy_3\\ &=\pi_d\int f_d(y_1,y_2,y_3)\,dy_{d+1}\cdots dy_3\\ &=\pi_d f_d(y_{1:d}). \end{aligned} $$

The pattern probability can be taken outside the integral because it does not depend on the outcome values being integrated. For a participant with \(D=0\), the same reasoning integrates over the entire outcome vector, leaving \(\pi_0\int f_0(y)\,dy=\pi_0\). For a complete participant, the contribution is \(\pi_3 f_3(y_1,y_2,y_3)\).

A useful special case arises when a pattern-specific full density is written as a product of proper sequential conditional densities. Integrating out the post-dropout outcomes removes those future factors one at a time, because each conditional density integrates to 1. That simplification is valid only when the factors are normalized conditional distributions and the relevant marginalization is done correctly; it does not justify evaluating missing outcomes at arbitrary values.

Worked Example: One Participant With Dropout After Visit 2

Suppose the hypothetical pattern probabilities are \(\pi_0=0.10\), \(\pi_1=0.20\), \(\pi_2=0.30\), and \(\pi_3=0.40\). Consider a participant with \(D=2\), whose observed binary outcomes are \(Y_1=1\) and \(Y_2=0\). Within pattern 2, suppose \(Y_1\) and \(Y_2\) are independent Bernoulli outcomes with \(P(Y_1=1\mid D=2)=0.80\) and \(P(Y_2=1\mid D=2)=0.75\).

The observed outcomes have conditional probability \(f_2(1,0)=0.80(1-0.75)=0.80(0.25)=0.20\). The participant’s observed-data contribution is therefore

$$ \pi_2 f_2(1,0)=0.30(0.20)=0.060. $$

This is the joint probability of observing pattern 2 and the outcome pair \((1,0)\) under the specified model. No probability for \(Y_3\) is multiplied into this contribution: \(Y_3\) was not observed and has already been marginalized out.

Calculating a Cohort Likelihood

Each record contributes according to its own pattern. To construct a cohort likelihood, identify the pattern and observed outcome vector for every participant, calculate the corresponding \(\pi_d f_d(y_{1:d})\), and multiply the contributions. The following example uses binary outcomes so that the likelihood can be checked directly.

Worked Example: Four Records Across Three Visits

Suppose four hypothetical participants have patterns and outcomes as follows: one has \(D=0\), one has \(D=1\) with \(Y_1=1\), one has \(D=2\) with \((Y_1,Y_2)=(1,0)\), and one has \(D=3\) with \((Y_1,Y_2,Y_3)=(0,1,1)\). Use pattern probabilities \((\pi_0,\pi_1,\pi_2,\pi_3)=(0.10,0.20,0.30,0.40)\).

For the \(D=1\) participant, suppose \(Y_1\mid D=1\) is Bernoulli with success probability \(0.70\). The contribution is \(0.20(0.70)=0.14\).

For the \(D=2\) participant, use the pattern-2 model from the preceding example. Its contribution is \(0.30[0.80(0.25)]=0.060\).

For the \(D=3\) participant, suppose the three outcomes are independent within pattern 3, with success probabilities \(0.40,0.60,0.75\). The probability of observing \((0,1,1)\) conditional on \(D=3\) is \((1-0.40)(0.60)(0.75)=0.60(0.60)(0.75)=0.27\). The contribution is \(0.40(0.27)=0.108\).

For the \(D=0\) participant, there is no observed outcome vector, so the contribution is \(\pi_0=0.10\). Multiplying all four contributions gives

$$ L_{\mathrm{obs}} =(0.10)(0.14)(0.060)(0.108) =0.00009072. $$

This number is the likelihood of these four observed records under the specified model. It is not a probability that the model is correct; likelihood values compare model parameter values for the same observed data.

When the Full Density Must Be Integrated

A pattern-specific full density can be written with conditional distributions for all three visits, including visits after dropout. The observed contribution still depends on the marginal distribution of the visits that were actually observed. The next calculation shows why substituting a value for a missing outcome gives the wrong contribution.

Worked Example: Integrating Out Visit 3

For a hypothetical participant with \(D=2\), suppose the pattern-2 full distribution factors as \(f_2(y_1,y_2,y_3)=f_2(y_1)f_2(y_2\mid y_1)f_2(y_3\mid y_1,y_2)\). Let \(P(Y_1=1\mid D=2)=0.60\), \(P(Y_2=0\mid Y_1=1,D=2)=0.30\), and \(P(Y_3=1\mid Y_1=1,Y_2=0,D=2)=0.80\). The observed outcomes are \(Y_1=1,Y_2=0\).

The observed-outcome marginal is obtained by summing over both possible values of the missing binary outcome:

$$ \begin{aligned} f_2(1,0) &=\sum_{y_3\in\{0,1\}} f_2(1,0,y_3)\\ &=(0.60)(0.30)\big[(1-0.80)+0.80\big]\\ &=(0.18)(1)=0.18. \end{aligned} $$

If \(\pi_2=0.25\), the observed-data contribution is \(0.25(0.18)=0.045\). Evaluating the full density only at \(Y_3=0\) would instead give \(0.25(0.60)(0.30)(0.20)=0.009\), which is not the observed-data contribution. That calculation treats one unobserved value as though it had been observed.

Assumptions, Checks, and Common Errors

The derivation relies on a defined pattern variable, a probability model for the pattern and a proper outcome distribution within each pattern. In the monotone three-visit setup, the observed outcomes for pattern \(d\) must be exactly the prefix \(Y_1,\ldots,Y_d\). If records have intermittent missingness rather than monotone dropout, this last-observed-visit encoding does not represent their patterns; the pattern definition and corresponding observed subvectors must be revised.

Before using a likelihood, check that the pattern probabilities are valid: \(\pi_d\geq 0\) for every \(d\), and \(\sum_{d=0}^3\pi_d=1\). Check that each \(f_d\) is a normalized distribution and that its observed-data marginal is taken over exactly the missing components. If a pattern has no or very few participants, its pattern-specific outcome distribution may be weakly informed by the data. A likelihood formula can be correct even when its parameters are difficult to estimate reliably.

  • Omitting the pattern probability. Using only \(f_d(y_{1:d})\) ignores the fact that the observed record also includes \(D=d\). Include \(\pi_d\) when the likelihood is for the joint observed record.
  • Using the full density without marginalizing. A density for \((Y_1,Y_2,Y_3)\) is not the density for \((Y_1,Y_2)\). Integrate or sum over the unobserved outcome.
  • Assigning a value to a missing outcome. Neither zero nor a guessed value represents integration over uncertainty. Such substitution changes the likelihood.
  • Misencoding a pattern. Confirm that \(D=d\) means the first \(d\) visits are observed. A different coding convention requires matching changes to the observed subvector and formula.

The observed-data contribution is the pattern-mixture way to express the likelihood. A selection-model likelihood instead factors the joint distribution using an outcome model and a model for missingness conditional on outcomes; it is an alternative factorization, not a reason to omit the pattern term here. A complete-data likelihood would use the full outcome vector, but for incomplete records its missing components must be integrated out to obtain an observed-data likelihood. The pattern-mixture representation also does not, by itself, make unobserved outcomes empirically known: distributions for outcomes after dropout require modeling assumptions beyond the observed record.

Key takeaway: For pattern \(d\), the observed-data contribution is the probability of pattern \(d\) multiplied by the pattern-specific marginal density of the outcomes observed through visit \(d\). Derive that marginal by integrating over every unobserved outcome.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. For three visits, what observed outcome vector corresponds to \(D=1\), and what is the likelihood contribution for a record with value \(Y_1=y_1\)?
  2. Suppose \(\pi_2=0.40\) and \(f_2(y_1,y_2)=0.15\) for a participant’s observed outcomes. What is the contribution?
  3. A participant has \(D=0\). What outcome-density factor appears in the contribution, and why?
  4. Why is evaluating \(f_2(y_1,y_2,y_3)\) at a guessed \(y_3\) not a valid contribution for a participant with \(D=2\)?

Solutions. (1) The observed vector is \(Y_1\), and the contribution is \(\pi_1 f_1(y_1)\). (2) The contribution is \(0.40(0.15)=0.060\). (3) The outcome factor is 1, because there is no observed outcome and the full conditional density integrates to 1; the contribution is \(\pi_0\). (4) \(Y_3\) is unobserved, so the observed-data density requires integrating over its possible values. Substitution selects one unobserved value rather than marginalizing it.

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