What the Joint Law Describes
With repeated follow-up, missingness is not always adequately summarized by one number such as the proportion of people with any missing outcome. A participant observed at visits 1 and 3 but not visit 2 has a different response history from one observed only at visit 2. The response pattern records which scheduled outcomes were observed, and its probability can differ across baseline covariate groups.
Let \(T\) be the number of scheduled follow-up visits. Write \(Y=(Y_1,\ldots,Y_T)\) for the outcome vector and \(R=(R_1,\ldots,R_T)\) for its response-indicator vector, where \(R_j=1\) means \(Y_j\) was observed and \(R_j=0\) means it was not. Let \(X\) denote baseline covariates. The response pattern is the realized value \(r\) of \(R\). This tutorial concerns the probability distribution of those patterns, especially conditional on \(X\).
The vector representation preserves the visit-by-visit detail. The same information can be stored as one categorical variable, say \(C\), whose categories are in one-to-one correspondence with the elements of \(\mathcal R\). For example, when \(T=2\), the category labelled “observed at visit 1 only” corresponds to \(r=(1,0)\). The category label is just a convenient name: it does not mean the pattern is ordered or measured on a numerical scale.
Patterns and Their Probabilities
There are \(2^T\) possible patterns because each of the \(T\) indicators has two possible values. With two visits, the four patterns are \((1,1)\), \((1,0)\), \((0,1)\), and \((0,0)\). They mean, respectively, observed at both visits, observed only at visit 1, observed only at visit 2, and observed at neither visit. With three visits there are eight patterns; for instance, \((1,0,1)\) means that visits 1 and 3 were observed but visit 2 was not.
The conditional distribution \(P(R=r\mid X=x)\) assigns one probability to each complete pattern. These probabilities need not be equal, and some may be zero. Their required properties are simply \(\pi_r(x)\geq 0\) and \(\sum_{r\in\mathcal R}\pi_r(x)=1\) for each \(x\) where the conditional distribution is defined. A zero probability can reflect a structural restriction or a pattern not seen in a finite sample; these possibilities should not be treated as equivalent without substantive information.
The joint law of the baseline covariates and the response pattern follows directly from this conditional distribution. For any set \(A\) of baseline-covariate values,
where \(F_X\) is the distribution of \(X\). If \(X\) is discrete, this is a sum: \(P(X=x,R=r)=P(X=x)\pi_r(x)\). Thus the joint probabilities combine how common a covariate value is with how likely each pattern is at that value. The marginal probability of a pattern is obtained by averaging its conditional probability over the baseline-covariate distribution:
Worked Example: Four Patterns in Two Covariate Groups
Hypothetical data. Suppose a follow-up study has two visits and records a binary baseline covariate \(X\), labelled Group A or Group B. Among 40 participants in Group A, the observed counts for patterns \((1,1),(1,0),(0,1),(0,0)\) are 16, 8, 10, and 6. Among 60 participants in Group B, the counts in that same order are 36, 12, 6, and 6. Assume each participant contributes one recorded pattern and a known baseline-group label.
Within each group, estimate a conditional pattern probability by dividing that pattern’s count by the group total. For Group A, the estimates are \(16/40=0.4000\), \(8/40=0.2000\), \(10/40=0.2500\), and \(6/40=0.1500\). They sum to \(0.4000+0.2000+0.2500+0.1500=1.000\).
For Group B, the estimates are \(36/60=0.6000\), \(12/60=0.2000\), \(6/60=0.1000\), and \(6/60=0.1000\). These also sum to \(1.000\). In this sample, the pattern \((1,1)\) is more frequent in Group B, while \((0,1)\) is more frequent in Group A. The estimates describe the observed groups; by themselves they do not establish why the distributions differ.
The assumed conditions for these calculations are that the pattern categories are mutually exclusive and exhaustive, each participant is counted once, and baseline group is recorded for everyone included. The counts satisfy these conditions by construction in this hypothetical example. In an actual analysis, inconsistent coding or missing baseline-group labels would need to be addressed before interpreting these within-group proportions.
From Conditional to Joint Probabilities
The conditional probabilities answer “which pattern is likely among participants with this baseline covariate value?” The joint probabilities answer “how likely is this covariate value together with this pattern?” Keeping those questions distinct prevents a common error: comparing raw pattern counts across groups of different sizes as if the counts were conditional probabilities.
Worked Example: Recovering Joint and Marginal Pattern Probabilities
Use the hypothetical counts above. The sample proportions in Groups A and B are \(40/100=0.4000\) and \(60/100=0.6000\). For the pattern \((1,0)\), the estimated conditional probabilities are \(0.2000\) in both groups. The estimated joint probabilities are therefore
Adding across groups gives the estimated marginal probability \(0.0800+0.1200=0.2000\) for \((1,0)\). This agrees with the overall count: there are \(8+12=20\) such participants among 100, and \(20/100=0.2000\).
For a second check, the estimated marginal probability of \((1,1)\) is
The direct overall count gives \((16+36)/100=0.5200\), as it should. This agreement is the finite-sample version of averaging conditional pattern probabilities over the distribution of \(X\).
Worked Example: Coding a Three-Visit Pattern
Worked Example: Coding a Three-Visit Pattern
Hypothetical data. A participant has a recorded baseline covariate value \(X=x\) and follow-up outcomes scheduled at three visits. The outcomes at visits 1 and 3 are recorded, while the visit-2 outcome is not. The response vector is \(r=(1,0,1)\). Suppose the conditional pattern probabilities for this covariate value are given below, in the stated pattern order.
\((1,1,1):0.35\), \((1,1,0):0.10\), \((1,0,1):0.20\), \((1,0,0):0.05\), \((0,1,1):0.10\), \((0,1,0):0.05\), \((0,0,1):0.10\), and \((0,0,0):0.05\).
The participant belongs to the category corresponding to \((1,0,1)\), whose conditional probability is \(0.20\). The distribution is valid because all entries are nonnegative and \(0.35+0.10+0.20+0.05+0.10+0.05+0.10+0.05=1.00\). The probability of observing visit 1 is the sum of the probabilities of all patterns with first entry 1: \(0.35+0.10+0.20+0.05=0.70\). This marginal indicator probability does not identify the full pattern distribution: it does not say whether visits 2 or 3 were observed.
This calculation assumes the list covers all \(2^3=8\) possible patterns and uses a consistent order for the indicators. Both checks are met here. If one pattern were omitted, the probabilities would not constitute a complete conditional distribution even if the listed values looked plausible.
Diagnostics, Comparisons, and Limitations
A useful first diagnostic is a pattern table with one row per pattern and one column per baseline group. Check that each participant is assigned to exactly one row and group, that every possible pattern is accounted for, and that the within-group proportions sum to 1. In a dataset, sums that differ from 1 can indicate an omitted category, duplicate records, an incorrect denominator, or missing covariate values handled inconsistently.
Next, compare conditional distributions rather than relying only on pooled frequencies. In the two-group example, the overall proportion for \((1,1)\) is \(0.5200\), but the within-group estimates are \(0.4000\) and \(0.6000\). The pooled value summarizes this sample’s mix of baseline groups; it does not describe either group on its own. For a continuous baseline covariate, grouping or modelling choices affect how the conditional distribution is summarized, and sparse data can make estimates unstable.
| Representation or summary | What it preserves | What to watch for |
|---|---|---|
| Indicator vector \(R\) | Which visits were observed for each participant | Use the same visit order for every vector |
| Categorical pattern label | The same complete pattern information in one variable | Use labels that map unambiguously to vectors |
| Any-missing indicator | Whether at least one visit is missing | Collapses distinct patterns into one category |
| Pattern distribution pooled over \(X\) | Marginal pattern frequencies in the sample or population | Can conceal differences across baseline covariates |
The vector and categorical forms are equivalent when the mapping between them is one-to-one; the choice is mainly about clarity and convenience. A summary such as “any visit missing” is not equivalent because it merges patterns. Likewise, a marginal distribution \(P(R=r)\) is not a substitute for \(\pi_r(x)\) when the question concerns pattern probabilities within baseline groups.
These probabilities describe how response patterns are distributed, not the reasons for missing outcomes. A pattern table alone cannot establish that missingness is independent of unobserved outcomes, nor can it explain the substantive process that generated the patterns. It is also descriptive of the specified population and covariates: different covariate distributions can produce different marginal pattern probabilities even when the conditional probabilities are unchanged.
Common Mistakes
- Merging patterns that differ by visit. Treating \((1,0)\) and \((0,1)\) as the same because each has one observed visit discards which outcome was observed. Keep the visit order explicit.
- Confusing counts with probabilities. A group with more participants can have larger pattern counts without having a larger conditional probability. Divide by the relevant group total.
- Pooling before checking covariates. An overall pattern proportion is an average over the baseline-covariate distribution, not necessarily the probability for any particular covariate value.
- Assuming every pattern must occur. A valid distribution may assign probability zero to a pattern. Conversely, a zero sample count does not by itself prove that the population probability is zero.
- Reading a pattern probability as an explanation. \(P(R=r\mid X=x)\) quantifies a conditional distribution; it does not, on its own, identify a causal effect or a missingness mechanism.
Check Your Understanding
Answer each question, then compare with the solutions below.
- With four follow-up visits, how many response patterns are possible? What does \((1,0,0,1)\) mean?
- For two patterns with conditional probabilities \(0.25\) and \(0.50\) at a given \(x\), what must the probabilities of the remaining patterns sum to?
- Suppose \(P(X=A)=0.30\) and \(P(R=(1,1)\mid X=A)=0.40\). Find \(P(X=A,R=(1,1))\).
- In a sample, 12 of 30 participants in Group A have pattern \((0,1)\), and 20 of 50 in Group B have that pattern. Find each conditional sample proportion and the pooled proportion.
- Why does the probability of “at least one missing visit” fail to specify the full response-pattern distribution?
Solutions. (1) There are \(2^4=16\) patterns. The vector \((1,0,0,1)\) means visits 1 and 4 were observed, while visits 2 and 3 were not. (2) All pattern probabilities sum to 1, so the remaining patterns sum to \(1-0.25-0.50=0.25\). (3) By the multiplication rule, \(0.30(0.40)=0.1200\). (4) Group A: \(12/30=0.4000\); Group B: \(20/50=0.4000\). The pooled proportion is \((12+20)/(30+50)=32/80=0.4000\). (5) It combines distinct patterns, such as missing only visit 1 and missing only visit 2, so it does not reveal which visits were observed or missing.