Coding Monotone Dropout Patterns

Why a Single Pattern Code Can Help

With repeated measurements, the missingness pattern can be written as a vector of response indicators: one indicator for each visit. When dropout is monotone, these vectors have a special structure. Once a visit is missed because the participant has dropped out, all later visits are also missed. Instead of carrying a whole vector for every participant, we can encode that pattern with one number: the last visit observed.

This coding is useful for describing how often each dropout pattern occurs and for organizing outcome distributions by pattern. It does not, by itself, explain why people dropped out, establish that outcomes are comparable across patterns, or reveal outcomes that were never observed.

Definition and Coding Rule

Consider \(T\) scheduled visits and a complete outcome vector \(Y=(Y_1,\ldots,Y_T)\). Let \(R_j=1\) if the outcome at visit \(j\) is observed and \(R_j=0\) otherwise. We allow participants to have no observed outcomes at these visits; if a study guarantees that visit 1 is observed, the zero-observation pattern is simply excluded.

Definition: Dropout is monotone if the response indicators satisfy \(R_1\ge R_2\ge \cdots \ge R_T\). Define \(D=\sum_{j=1}^T R_j\), the number of observed visits. Under monotone dropout, \(D\) is also the index of the last observed visit: \(D=d\) means visits \(1,\ldots,d\) are observed and visits \(d+1,\ldots,T\) are missing. The case \(D=0\) means no outcomes are observed; \(D=T\) means all are observed.

The equality between the sum and the last observed visit depends on monotonicity. For example, the vector \((1,1,0,0)\) has sum 2 and last observed visit 2. The vector \((1,0,1,0)\) has sum 2 too, but it is not monotone: visit 3 is observed after visit 2 was missed. Its sum does not encode its full pattern.

R_j=\mathbb{1}(D\ge j), \qquad j=1,\ldots,T.

Here \(\mathbb{1}(\cdot)\) equals 1 when its condition is true and 0 otherwise. Thus each allowed value of \(D\) maps to exactly one response vector: \(D=0\) gives all zeros, while \(D=d\) for \(1\le d\le T\) gives \(d\) ones followed by \(T-d\) zeros. This one-to-one correspondence is why a single code is sufficient for monotone patterns.

Worked Example: Coding Individual Response Histories

In a hypothetical study with four follow-up visits, three participants have response vectors \((1,1,1,0)\), \((1,0,0,0)\), and \((1,1,1,1)\). Find each participant’s last-observed-visit code.

For \((1,1,1,0)\), the sum is \(1+1+1+0=3\), so \(D=3\). The participant’s last observed visit is visit 3, and only the outcome at visit 4 is missing.

For \((1,0,0,0)\), the sum is \(1\), so \(D=1\). Only the first outcome is observed.

For \((1,1,1,1)\), the sum is \(4\), so \(D=4\). All outcomes are observed. Each vector has the required prefix-of-ones form, so each is compatible with monotone dropout.

Pattern Probabilities and Pattern-Specific Distributions

The random variable \(D\) takes values \(0,\ldots,T\). Write \(\pi_d=P(D=d)\) for the probability of pattern \(d\). The probabilities are nonnegative and sum to one. For outcomes, let \(q_d(y_1,\ldots,y_T)\) denote the conditional distribution of the full outcome vector \(Y\) given \(D=d\). It is a probability mass function for discrete outcomes or a density for continuous outcomes.

\pi_d=P(D=d), \qquad \sum_{d=0}^{T}\pi_d=1, \qquad q_d(y_1,\ldots,y_T)=f(Y=y\mid D=d).

The notation \(f(Y=y\mid D=d)\) is shorthand for the appropriate conditional probability or density. When \(\pi_d>0\), the joint distribution of outcomes and pattern can be written as \(P(D=d)q_d(y)=\pi_dq_d(y)\). The pattern-specific distribution concerns the full outcome vector, including outcomes after dropout. Those later outcomes are part of the statistical model even though they are not observed for people with \(D=d<T\).

For a person in pattern \(d\), the observed outcomes are the prefix \(Y_1,\ldots,Y_d\). Their conditional distribution is the marginal of \(q_d\) over the unobserved suffix:

g_d(y_1,\ldots,y_d) = \int q_d(y_1,\ldots,y_T)\,dy_{d+1}\cdots dy_T.

For discrete outcomes, the integral is replaced by a sum over the possible values of \(Y_{d+1},\ldots,Y_T\). For \(d=T\), there is no missing suffix and \(g_T=q_T\). For \(d=0\), the observed prefix is empty: the pattern probability is \(\pi_0\), but no outcome values are observed to inform the outcome distribution \(q_0\) directly.

Within any pattern, the chain rule can also express the full outcome distribution sequentially. For example, when the conditional distributions exist,

q_d(y_1,\ldots,y_T) = \prod_{j=1}^{T} f(y_j\mid y_1,\ldots,y_{j-1},D=d).

This factorization is a way to describe \(q_d\); it does not turn unobserved outcomes into observed data. For a participant with \(D=d\), only the first \(d\) factors concern observed outcomes, while the later factors describe outcomes that are missing for that participant.

Worked Example: Calculating Pattern Probabilities

A hypothetical sample of 100 participants has four scheduled visits. The observed monotone patterns occur with counts \(10, 20, 30, 25,\) and \(15\) for \(D=0,1,2,3,4\), respectively. Calculate the empirical pattern probabilities and check their total.

For each pattern, divide its count by 100:

\hat{\pi}_0=\frac{10}{100}=0.10,\quad \hat{\pi}_1=\frac{20}{100}=0.20,\quad \hat{\pi}_2=\frac{30}{100}=0.30,\quad \hat{\pi}_3=\frac{25}{100}=0.25,\quad \hat{\pi}_4=\frac{15}{100}=0.15.

The total is \(0.10+0.20+0.30+0.25+0.15=1.00\), as required. For example, the empirical probability of observing exactly the first two visits is 0.30. These are sample proportions describing this hypothetical sample, not proof that its population pattern probabilities have those exact values.

Worked Example: A Pattern-Specific Mean and a Marginal Mean

Suppose a hypothetical three-visit population has pattern probabilities \(\pi_1=0.20,\ \pi_2=0.30,\ \pi_3=0.50\), with \(\pi_0=0\). Suppose the conditional means of the full outcome at visit 1 are \(E(Y_1\mid D=1)=8\), \(E(Y_1\mid D=2)=7\), and \(E(Y_1\mid D=3)=6\), measured in symptom-score points. Find \(E(Y_1)\).

The patterns \(D=1,2,3\) partition the population, so the law of total expectation gives a weighted average of the pattern-specific means:

E(Y_1) = \sum_{d=1}^{3}P(D=d)E(Y_1\mid D=d) =(0.20)(8)+(0.30)(7)+(0.50)(6) =1.60+2.10+3.00 =6.70.

The implied population mean is 6.70 symptom-score points under the stated probabilities and conditional means. In practice, the visit-1 outcome is observed in all three patterns in this example, so those conditional means are in principle informed by observed visit-1 data. A corresponding mean at visit 3 for people with \(D=1\) would concern an unobserved outcome and would not be directly identified by their records.

Justification, Checks, and Interpretation

The central coding claim follows directly from the monotonicity condition. If \(R_1\ge\cdots\ge R_T\), any indicator equal to zero forces all later indicators to be zero. Therefore the vector must consist of \(d\) ones followed by \(T-d\) zeros for some \(d\). Its sum is \(d\), and its last observed index is also \(d\) when \(d>0\). Conversely, a vector with that prefix form satisfies the monotonicity inequalities. This proves that the possible monotone vectors are exactly the vectors represented by \(D=0,\ldots,T\).

A practical diagnostic is to tabulate response histories and check whether any participant has a zero followed by a one. Such a history violates the monotone-dropout premise for the visit indicators as defined here. Before applying the coding, check that visits have been ordered correctly and distinguish an outcome that was not scheduled from one that was scheduled but missing. A visit that was not part of a participant’s follow-up schedule should not automatically be treated as a missed visit in a dropout sequence.

The last-visit code is more compact than storing the full response vector, but it is appropriate only when the histories have the prefix form. Keeping the full vector is safer when intermittent missingness is possible, because it preserves which particular visits were observed. Coding with \(D\) also does not impose a common outcome distribution across patterns: each \(q_d\) may differ. Nor does observing pattern frequencies determine the missing outcomes or the reasons for dropout.

A particularly important limitation concerns pattern \(D=0\). Its probability can be estimated from the frequency of participants with no observed visits, but none of their outcomes are available to estimate \(q_0\) from those records. More generally, for \(d<T\), the observed data in pattern \(d\) provide information about the prefix distribution \(g_d\), not the full conditional distribution \(q_d\) of the unobserved suffix as well. Additional assumptions or information are needed to determine those unobserved parts.

Common Mistakes

Using the sum as a pattern code without checking monotonicity. The vector \((1,0,1,0)\) sums to 2, but it is not the same response pattern as \((1,1,0,0)\). The sum uniquely identifies a pattern only under the monotone condition.
Calling \(q_d\) the distribution of observed outcomes only. By definition here, \(q_d\) is the distribution of the full outcome vector conditional on pattern \(d\). The observed-prefix distribution is its marginal \(g_d\).
Treating an estimated \(\hat{\pi}_d\) as an outcome model. Pattern frequencies describe how common each response pattern is; they do not specify the outcomes within a pattern.
Assuming a compact code solves missing-outcome identification. The code organizes patterns but cannot reveal the unobserved suffix for dropouts.

Key takeaway: Under monotone dropout, \(D=\sum_jR_j\) uniquely records the last observed visit, and \(\pi_d\) and \(q_d\) describe the pattern probability and full outcome distribution within pattern \(d\). The observed data within that pattern contain only the prefix distribution \(g_d\).

Check Your Understanding

Answer each question, then compare with the solutions below.

For five visits, what response vector corresponds to \(D=3\)?
Does \((1,0,1,1)\) satisfy monotone dropout? Explain why its sum is not a sufficient pattern code here.
In a study with four visits, the pattern counts for \(D=0,1,2,3,4\) are \(5,15,20,30,30\) among 100 participants. Find the empirical probability of each pattern.
For \(D=2\) with four visits, which outcomes are observed, and how is their conditional distribution related to \(q_2\)?

Solutions. (1) \(D=3\) corresponds to \((1,1,1,0,0)\). (2) No. The zero at visit 2 is followed by ones at visits 3 and 4, violating monotonicity. Its sum is 3, but a monotone pattern with \(D=3\) would be \((1,1,1,0)\) for four visits, not this vector. (3) Divide each count by 100: the probabilities are \(0.05, 0.15, 0.20, 0.30,\) and \(0.30\); they sum to 1.00. (4) Outcomes \(Y_1\) and \(Y_2\) are observed, while \(Y_3\) and \(Y_4\) are missing. Their conditional distribution is the marginal of \(q_2\) obtained by summing or integrating over \(Y_3,Y_4\).