Deriving the Complete-Case Restriction Sequentially

Why the Restriction Must Be Applied Visit by Visit

In a monotone-dropout study, a participant who misses a visit is assumed to miss every later visit. A pattern can therefore be indexed by the number of visits observed. The complete cases supply observed outcomes at every visit, but they may be a selective subset of the study. To use them to describe missing outcomes, a restriction is needed: for participants in an incomplete pattern, their next unobserved outcome is assigned the same conditional distribution as that outcome among complete cases with the same preceding outcome history.

For more than two visits, this is not just one comparison between an incomplete pattern and completers. The restriction is applied successively. At the first missing visit, condition on the outcomes observed so far. At a later missing visit, condition additionally on the intervening outcomes, even though those outcomes are themselves missing for that pattern. This recursive conditioning is what turns a collection of conditional restrictions into a joint distribution for all future outcomes.

Definition: Let $Y_1,\ldots,Y_K$ be the outcomes at $K$ scheduled visits. Let $G=s$ mean that visits $1,\ldots,s$ are observed and visits $s+1,\ldots,K$ are missing; $G=K$ denotes complete cases. The sequential complete-case restriction is, for every incomplete pattern $s<K$ and every visit $j>s$, $$ \mathcal{L}(Y_j\mid Y_1,\ldots,Y_{j-1},G=s) = \mathcal{L}(Y_j\mid Y_1,\ldots,Y_{j-1},G=K), $$ on histories where the complete-case conditional distribution is defined. Here $\mathcal{L}$ denotes a conditional distribution. For $j>s$, the conditioning history includes outcomes that are missing in pattern $G=s$ whenever their visit precedes $j$.

This is a restriction on conditional outcome distributions, not on the pattern frequencies. It does not require $P(G=s)$ to equal $P(G=K)$, and it does not require the observed outcome histories in the two patterns to have the same distribution. A pattern can be more common among participants with particular observed outcomes. The restriction instead says that, after conditioning on the full preceding outcome history, the next outcome follows the complete-case conditional law.

Deriving the Joint Distribution Sequentially

For a participant in pattern $G=s$, the missing outcomes are $Y_{s+1},\ldots,Y_K$. The chain rule factors their joint conditional distribution into one-visit conditional distributions. Applying the restriction to each factor gives the identifying formula:

\begin{aligned} &f(y_{s+1},\ldots,y_K\mid y_1,\ldots,y_s,G=s)\\ &\quad=\prod_{j=s+1}^{K} f(y_j\mid y_1,\ldots,y_{j-1},G=s)\\ &\quad=\prod_{j=s+1}^{K} f(y_j\mid y_1,\ldots,y_{j-1},G=K). \end{aligned}

The first equality is the chain rule. In the second equality, the restriction is applied separately to each visit $j$. The result is a sequential recipe: draw or calculate $Y_{s+1}$ from its complete-case conditional distribution given $Y_1,\ldots,Y_s$; then use that value when forming the complete-case conditional distribution for $Y_{s+2}$; continue through visit $K$. Matching only the first missing visit would not determine the joint law of all later outcomes.

This derivation also exposes the assumption’s reach. For the first missing visit, the conditioning history is observed in pattern $s$. For later visits, it contains one or more outcomes unavailable in that pattern. The complete-case law must therefore be available over histories that the incomplete pattern cannot itself reveal. Where the complete cases have little or no information for a history, the restriction requires extrapolation rather than a data-supported conditional comparison.

Worked Example: Two Missing Binary Visits

Suppose there are three visits and binary outcomes. Consider a hypothetical pattern $G=1$, in which only $Y_1$ is observed. Among complete cases with $Y_1=1$, suppose the estimated conditional probabilities are $P(Y_2=1\mid Y_1=1,G=3)=0.60$, $P(Y_3=1\mid Y_1=1,Y_2=1,G=3)=0.80$, and $P(Y_3=1\mid Y_1=1,Y_2=0,G=3)=0.30$. Assume these histories have complete-case support and apply the sequential restriction.

First, the restriction sets $P(Y_2=1\mid Y_1=1,G=1)=0.60$. Thus the probabilities of $Y_2=1$ and $Y_2=0$ are $0.60$ and $0.40$. The second missing visit is then conditioned on both $Y_1$ and the value of $Y_2$, giving:

\begin{aligned} P(Y_2=1,Y_3=1\mid Y_1=1,G=1)&=0.60(0.80)=0.48,\\ P(Y_2=1,Y_3=0\mid Y_1=1,G=1)&=0.60(0.20)=0.12,\\ P(Y_2=0,Y_3=1\mid Y_1=1,G=1)&=0.40(0.30)=0.12,\\ P(Y_2=0,Y_3=0\mid Y_1=1,G=1)&=0.40(0.70)=0.28. \end{aligned}

The four probabilities sum to $0.48+0.12+0.12+0.28=1.00$, as required. The implied probability of $Y_3=1$ is $0.48+0.12=0.60$. This calculation uses the conditional law at visit 3 separately for each possible value of the intervening outcome $Y_2$; simply carrying forward a single marginal probability for $Y_3$ would not implement the sequential restriction.

Sequential Propagation of Continuous Outcomes

With continuous outcomes, the same factorization applies to conditional densities. A model for each complete-case conditional distribution supplies the next step. If those conditional distributions are normal and linear in the preceding outcomes, sequential prediction can also propagate uncertainty from earlier missing outcomes into later ones. Replacing an earlier outcome by its conditional mean alone generally gives a mean prediction for the next visit, but does not by itself give the correct uncertainty at that visit.

Worked Example: Carrying Uncertainty to a Later Visit

Consider a hypothetical four-visit study and pattern $G=2$, with $Y_1=2$ and $Y_2=3$ observed. Suppose complete-case conditional models, in the outcome’s measurement units, are:

\begin{aligned} Y_3\mid Y_1,Y_2,G=4&\sim N(2+0.2Y_1+0.4Y_2,\ 9),\\ Y_4\mid Y_1,Y_2,Y_3,G=4&\sim N(0.5Y_1+0.3Y_2+0.6Y_3,\ 16). \end{aligned}

Assume these conditional laws are supported over the histories used here. By the restriction, the first model is used for $Y_3$ in pattern $G=2$. Substitution gives a conditional mean of $2+0.2(2)+0.4(3)=3.6$, so $Y_3\mid Y_1=2,Y_2=3,G=2\sim N(3.6,9)$.

For $Y_4$, the restriction uses its complete-case conditional law given the realized $Y_3$. Its conditional mean, as a function of $Y_3$, is $0.5(2)+0.3(3)+0.6Y_3=1.9+0.6Y_3$. Therefore, $E(Y_4\mid Y_1=2,Y_2=3,G=2)=1.9+0.6(3.6)=4.06$. Under the stated normal sequential model, the variance includes both the conditional variance at visit 4 and the variation in its conditional mean induced by $Y_3$:

\operatorname{Var}(Y_4\mid Y_1=2,Y_2=3,G=2) =16+(0.6)^2(9)=19.24.

Thus the implied distribution is $Y_4\mid Y_1=2,Y_2=3,G=2\sim N(4.06,19.24)$. The variance is in squared outcome units. Reporting only the conditional mean $4.06$ would hide the uncertainty passed forward from $Y_3$. This variance calculation relies on the specified normal conditional distributions; the restriction alone does not impose normality or these particular variances.

Why Conditional Matching Is Not Marginal Matching

A complete-case conditional distribution can depend strongly on prior outcomes. Consequently, participants in an incomplete pattern can have different implied marginal distributions for a missing visit even though the restriction matches their conditional laws to those of completers. The following example shows why an unconditional complete-case average is not a substitute for sequential conditioning.

Worked Example: A Different History Gives a Different Marginal

Suppose hypothetical complete-case data give $P(Y_1=1\mid G=3)=0.50$, with $P(Y_2=1\mid Y_1=0,G=3)=0.20$ and $P(Y_2=1\mid Y_1=1,G=3)=0.80$. For visit 3, suppose the conditional probabilities of $Y_3=1$, listed for histories $(Y_1,Y_2)=(0,0),(0,1),(1,0),(1,1)$, are $0.10,0.50,0.60,0.90$, respectively. Assume all these histories have support among complete cases.

For pattern $G=1$, suppose participants have $Y_1=1$. The restriction gives $P(Y_2=1\mid Y_1=1,G=1)=0.80$. Applying it again at visit 3 yields:

P(Y_3=1\mid Y_1=1,G=1) =0.80(0.90)+0.20(0.60)=0.84.

By contrast, the complete-case marginal probability at visit 3 is $0.51$. Among complete cases with $Y_1=0$, the conditional probability is $0.80(0.10)+0.20(0.50)=0.18$; with $Y_1=1$, it is $0.20(0.60)+0.80(0.90)=0.84$. Averaging over the complete-case distribution of $Y_1$ gives $0.50(0.18)+0.50(0.84)=0.51$. The value $0.84$, not $0.51$, is the restriction-based probability for this pattern and history. The two values differ because the histories are distributed differently.

Assumptions, Diagnostics, and Alternatives

The restriction is an assumption about distributions that are not fully observed for dropouts. Its use requires care about both the conditional models and their support. In particular, for each pattern and visit, the complete cases must provide information about the preceding histories to which the fitted conditional law is applied. Matching histories only approximately, or fitting a model that smooths across histories, can make the result depend on modeling choices.

Check support and sparsity. Tabulate or inspect how many complete cases contribute to each relevant history. For continuous histories, assess whether dropout histories lie far outside the complete-case range. Sparse or unsupported histories make conditional estimates unstable or extrapolative.
Check conditional model fit. Use appropriate distributional and residual diagnostics for the complete-case conditional models, and assess whether functional forms or interactions are adequate. A well-fitting model among completers does not establish that its conditional law is correct for dropouts.
Separate observable checks from the identifying assumption. The distribution of observed outcomes within each pattern can be described and modeled. But the equality of a missing outcome’s conditional law in an incomplete pattern and in completers cannot generally be verified from the observed data alone: one side of that comparison is unobserved.
Track uncertainty across steps. Uncertainty in a conditional model at one visit affects predictions at later visits. Plugging in estimated probabilities or means without accounting for this uncertainty can understate the uncertainty of reconstructed quantities.

The complete-case restriction is not the same as discarding everyone except completers; complete-case analysis uses only observed complete cases and does not reconstruct missing outcomes in incomplete patterns. It is also not equality of unconditional distributions, which would ignore the histories that determine the visit-specific laws. A pattern-mixture analysis without this restriction can leave the missing conditional distributions unspecified and examine alternative assumptions instead. Selection models express assumptions through a model for observation or dropout conditional on outcomes; they are a different factorization, not a justification for the complete-case restriction.

A common mistake is to say, “The incomplete-pattern participants resemble completers because the outcomes they did report look similar, so their future outcomes have the complete-case marginal distribution.” This does not follow. The restriction is conditional on the entire preceding history, and even under that restriction the marginal future distribution depends on the incomplete pattern’s history distribution. A second pitfall is to apply the restriction only to the first missing visit. The chain-rule derivation shows that every later visit needs its own conditional equality, including histories that contain intervening missing outcomes.

Key takeaway: For an incomplete pattern, reconstruct the joint law of future outcomes by applying the complete-case conditional law at each visit, given all preceding outcomes. The construction is explicit, but its validity depends on an untestable conditional restriction and adequate complete-case support for the histories used.

Check Your Understanding

Work through each question; concise solutions follow.

In a four-visit study, a pattern has $G=2$. Which outcomes condition the restriction for $Y_3$, and which condition it for $Y_4$?
For a three-visit binary outcome, suppose $P(Y_2=1\mid Y_1=0,G=3)=0.25$, $P(Y_3=1\mid Y_1=0,Y_2=1,G=3)=0.60$, and $P(Y_3=1\mid Y_1=0,Y_2=0,G=3)=0.20$. Under the restriction, calculate $P(Y_2=1,Y_3=1\mid Y_1=0,G=1)$.
Does the sequential complete-case restriction require the incomplete pattern and complete cases to have the same distribution of $Y_1$? Explain.
What does it mean if an incomplete-pattern history has no comparable complete cases?

Solutions. (1) For $Y_3$, condition on $Y_1,Y_2$; for $Y_4$, condition on $Y_1,Y_2,Y_3$. (2) The chain rule and restriction give $0.25(0.60)=0.15$. (3) No. The restriction matches conditional distributions of future outcomes given preceding histories, not the distributions of those histories across patterns. (4) The relevant conditional law is unsupported by complete-case data there; applying it requires extrapolation, so the resulting reconstruction is especially dependent on modeling assumptions.