Two Different Strengths of Restriction
In a pattern-mixture analysis, the distribution of an outcome can be described separately for each missingness pattern. A restriction then relates the outcome distribution in one pattern to the distribution in another. The key distinction here is whether the restriction equates the entire conditional distribution or only its conditional mean. Both can supply information about an unobserved outcome, but they make different assumptions and support different conclusions.
Let \(G\) denote a dropout pattern, let \(Y\) be the outcome at the visit of interest, and let \(H\) denote the history and covariates on which the restriction conditions. For a pattern \(g\), write \(F_g(y\mid h)=P(Y\leq y\mid H=h,G=g)\) for the conditional distribution function and \(\mu_g(h)=E(Y\mid H=h,G=g)\) for the conditional mean. The comparison between patterns must use the same outcome, conditioning variables, and value \(h\).
These are assumptions about pattern-specific outcome laws, not statements that the patterns are identical in every respect. In particular, they do not assert that dropout is random or that pattern membership causes an outcome difference. A restriction specifies which features of a conditional law are treated as equal for the purpose of analysis.
What Full Equality Implies
When the conditional laws are equal and the conditional means are finite, their means must also be equal. This follows because the mean is determined by the distribution. For a continuous outcome with density \(f_g(y\mid h)\), for example, the conditional mean is the integral of \(y\) against that density. For a discrete outcome, it is the corresponding sum.
The reverse implication does not hold. Many distinct distributions can have the same expectation. Knowing one feature—the location expressed by the mean—does not determine how probability is distributed around that mean. A mean restriction therefore leaves additional distributional features unspecified unless the analysis supplies further assumptions.
The implication can also be seen by integrating the outcome against its conditional law. Equality of the laws gives the same integral for every integrable function of the outcome, including the function \(y\mapsto y\). Equality of means gives equality only for that one function; it does not force equal integrals for other functions, such as \(y\mapsto y^2\) or an indicator of a tail event.
This implication requires finite first moments for the means to exist. A comparison also requires that the conditioning value \(h\) be meaningful in both patterns. If a history occurs in the target pattern but has no support in the reference pattern, the reference conditional law at that history cannot be obtained directly from observed cases without an additional modeling or extrapolation assumption.
Worked Example: Equal Means With Different Normal Spreads
Suppose, purely hypothetically, that at a fixed history \(h\), the reference pattern has \(Y\mid H=h,G=r\sim N(10,4)\). Consider two candidate laws for a second pattern. A full-distribution restriction assigns that pattern the same law, \(N(10,4)\). A mean-only restriction could instead allow \(N(10,16)\). The second argument of \(N(\mu,\sigma^2)\) is the variance, so these laws have standard deviations 2 and 4, respectively.
Both candidate laws have conditional mean 10 units. To compare their upper-tail probabilities above 12 units, standardize under each law. Under full-distribution equality,
Under the mean-only candidate \(N(10,16)\), the standard deviation is 4, giving
The means agree at 10 units, yet the probability of exceeding 12 units is about 0.159 under the borrowed reference distribution and 0.309 under the wider mean-only candidate. These calculations illustrate what a mean restriction leaves open; they do not establish which candidate law is correct for any real data.
Equal Means Can Conceal Different Shapes
The distinction is not limited to normal distributions or to differences in spread. An asymmetric distribution and a symmetric distribution can share a mean while assigning different probabilities to particular outcomes. This matters whenever the analysis concerns more than an average—for example, a threshold, a quantile, or a probability of an unusually high or low outcome.
Worked Example: Two Discrete Laws With the Same Mean
At the same fixed history \(h\), suppose the reference conditional law assigns probability \(0.5\) to \(Y=0\) and probability \(0.5\) to \(Y=2\). Its mean is
Consider a different candidate law in another pattern: probability \(0.5\) at 0, probability \(0.25\) at 1, and probability \(0.25\) at 3. Its probabilities sum to 1, and its mean is
Thus, the two laws satisfy the same conditional mean restriction. But the probability of an outcome above 2 is 0 under the reference law and 0.25 under the alternative law. Their variances differ as well. The reference second moment is \(0^2(0.5)+2^2(0.5)=2\), so its variance is \(2-1^2=1\). The alternative second moment is \(0^2(0.5)+1^2(0.25)+3^2(0.25)=2.5\), so its variance is \(2.5-1^2=1.5\). Equal means do not determine either the tail probability or the variance.
Why Pattern Weights Do Not Remove the Distinction
A pattern-mixture distribution combines pattern-specific distributions using the proportions of individuals in each pattern. If all pattern-specific means are equal, changing the shape in one pattern need not change the overall mean. Nevertheless, the resulting mixture distribution can change. The next calculation separates these two conclusions.
Worked Example: A Stable Mixture Mean With a Changing Tail
Suppose two patterns have weights 0.7 and 0.3. In pattern 1, let the outcome have the reference law from the previous example: probability 0.5 at 0 and 0.5 at 2. In pattern 2, compare two possibilities: under full-distribution borrowing it has the same law; under a mean-only restriction it has probabilities 0.5 at 0, 0.25 at 1, and 0.25 at 3. Each pattern-specific law in either possibility has mean 1. The mixture mean is therefore, in both cases,
Under full-distribution borrowing, neither pattern has any probability above 2, so the mixture probability of \(Y>2\) is \(0.7(0)+0.3(0)=0\). Under the mean-only alternative, pattern 2 has probability 0.25 above 2, so the mixture probability is
The overall mean remains 1 unit, but the mixture probability above 2 units changes from 0 to 0.075. Pattern weights average the pattern-specific distributions; they do not make different distributions equivalent just because their means match.
Choosing and Interpreting the Restriction
A full-distribution restriction is stronger than a mean restriction. It can provide a complete conditional law for an outcome in a pattern where that outcome is unobserved, but only by assuming equality of all distributional features being borrowed. A mean restriction targets the conditional average and is less restrictive about other features. It does not, without further assumptions, specify a complete predictive distribution or determine tail probabilities.
| Feature | Full conditional distribution restriction | Conditional mean restriction |
|---|---|---|
| What is equated? | The entire conditional law at each relevant history | The conditional expectation at each relevant history |
| What follows for means? | Means are equal when finite | Means are equal by assumption |
| What follows for spread and shape? | These conditional distributional features are equal | They may differ |
| What does it supply? | A conditional distribution for the target pattern, if the reference law is available | A conditional mean, not a full law by itself |
Neither choice is automatically justified by the observed data. In a pattern where the outcome is missing, the relevant conditional distribution or mean cannot generally be checked directly from that pattern's observed outcomes. Observed data can help assess whether a proposed borrowing rule is compatible with observed portions of the data, but compatibility is not proof that the unobserved portion obeys the restriction.
A practical check is to state exactly what is equated, at what visit, and conditional on which history and covariates. Then ask whether the scientific target is only a mean or also depends on other distributional features. If only a mean is needed, a full-law equality assumption may be stronger than necessary. If a full predictive distribution is needed, mean equality alone is insufficient: the remaining distributional assumptions must be stated rather than silently inferred.
Common Pitfall and Limitations
- Wrong reasoning: “The two patterns have the same conditional mean, so their outcome distributions are the same.” Correction: Equal means constrain one expectation only. The normal and discrete examples show that spread, shape, and tail probabilities can still differ.
- Wrong comparison: Equating means calculated at different histories or with different covariate conditioning sets. Correction: A conditional restriction must compare the same outcome given the same \(H=h\). Equality after averaging over different histories is a different claim.
- Overstating evidence: Treating a restriction as empirically verified because observed data do not contradict it. Correction: A condition involving unobserved outcomes is generally not directly testable from those outcomes; the restriction remains an identifying assumption.
- Overreading a mean-only restriction: Using it as though it supplied tail probabilities or a complete imputation distribution. Correction: It supplies a conditional mean. A full distribution requires additional modeling choices or restrictions.
The central limitation is that restrictions trade assumptions for information about missing outcomes. A stronger restriction can identify more distributional features, but those features are identified only conditional on the stronger assumption. Results should therefore be interpreted as consequences of the specified restriction, not as facts established merely by equality of observed means.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If two conditional outcome distributions are equal and both have finite means, what follows about their conditional means? Does the converse follow?
- Two patterns have laws \(P(Y=0)=0.5,\ P(Y=2)=0.5\) and \(P(Y=0)=0.5,\ P(Y=1)=0.25,\ P(Y=3)=0.25\). Verify that their means are equal and give one distributional feature that differs.
- Suppose two pattern-specific conditional means are both 6 units and the pattern weights are 0.4 and 0.6. What is their weighted mixture mean? Does this determine the mixture probability of exceeding 8 units?
- Why must a comparison of conditional means use the same history and covariate conditioning set in both patterns?
Solutions. (1) Equal full distributions imply equal means when the means are finite, because integration of \(y\) against equal laws gives equal results. The converse does not follow: distinct laws can have the same mean. (2) The first mean is \(0(0.5)+2(0.5)=1\); the second is \(0(0.5)+1(0.25)+3(0.25)=1\). Their probabilities of \(Y>2\) are 0 and 0.25, respectively. (3) The mixture mean is \(0.4(6)+0.6(6)=6\) units. It does not determine the probability above 8 units, which depends on the pattern-specific distributions beyond their means. (4) Conditional means refer to the distribution at a particular history. If the histories or conditioning sets differ, an apparent equality may compare different subpopulations rather than impose equality for the same \(h\).