Why Missingness Can Hide Distinct Full-Data Distributions
When data are missing, the observed records do not directly reveal the values that were not recorded. If the probability of a value being missing depends on that value itself, the observed cases may give a misleading picture of the full population. More fundamentally, distinct population distributions and missingness processes can produce exactly the same observed records. In that case, collecting more records under the same process does not distinguish the explanations.
The issue is one of identification, not merely of precision. Identification asks whether the observed-data distribution uniquely determines a target feature of the full-data distribution. We will use a binary outcome to show the issue transparently, then extend the reasoning to an observed covariate.
Definitions: Missingness and Observational Equivalence
For a vector \(Y\), the response indicator is a vector \(R\) marking which components are observed, and the record includes the pattern \(R\) and the observed components \(Y_{\mathrm{obs}}\). For each possible pattern, nonignorability means that the probability of that pattern can vary with values of components unobserved under that pattern, after conditioning on the observed components and any fully observed covariates. In the binary examples below there are no covariates, so nonignorability simply means that \(P(R=1\mid Y=1)\) and \(P(R=1\mid Y=0)\) differ.
Write \(a=P(Y=1,R=1)\), \(b=P(Y=0,R=1)\), and \(c=P(R=0)\) for a binary outcome. These probabilities describe everything available in the observed records: the observed positive cases, observed negative cases, and missing cases. They satisfy \(a+b+c=1\). In particular, the records with \(R=0\) do not disclose whether their outcome was 0 or 1.
Worked Example: Two Models With the Same Observed Records
Suppose the outcome \(Y\) is binary. Model A has \(P(Y=1)=0.5\), with response probabilities \(P(R=1\mid Y=1)=0.4\) and \(P(R=1\mid Y=0)=0.8\). Its observed cells are
The missing probability is \(P(R=0)=0.5(0.6)+0.5(0.2)=0.4\). Thus the observed-data probabilities are \((a,b,c)=(0.2,0.4,0.4)\), while the full-data prevalence is \(0.5\).
Model B instead has \(P(Y=1)=0.4\), \(P(R=1\mid Y=1)=0.5\), and \(P(R=1\mid Y=0)=2/3\). Its observed cells are
Its missing probability is \(0.4(0.5)+0.6(1/3)=0.2+0.2=0.4\), so it produces the same \((a,b,c)=(0.2,0.4,0.4)\). Yet its full-data prevalence is \(0.4\), not \(0.5\). Both models are nonignorable: the response probabilities differ between \(Y=1\) and \(Y=0\). No feature of the observed records distinguishes these two models.
A Continuum of Compatible Full-Data Distributions
The two models above are not an isolated coincidence. For any observed cells with \(a>0\), \(b>0\), and \(c>0\), let \(u=P(Y=1,R=0)\), the probability mass of missing cases whose unobserved outcome is 1. Then \(P(Y=0,R=0)=c-u\). The observed data do not tell us \(u\), but every choice \(0\leq u\leq c\) assigns the same observed cells. For \(0<u<c\), the implied full-data prevalence and response probabilities are
These expressions follow by adding the observed and missing mass for each outcome, then dividing each observed cell by its full-data outcome probability. Because \(a,b>0\), the denominators are positive throughout this interval. Whenever the two response probabilities differ, this compatible model is nonignorable.
Proof. For any \(u\in[0,c]\), assign probabilities \(a+u\) to \(Y=1\) and \(b+c-u\) to \(Y=0\). Assign response probabilities \(a/(a+u)\) and \(b/(b+c-u)\), respectively, whenever the denominators are positive. The resulting observed joint probabilities are
The remaining probability is \(u+(c-u)=c\), all assigned to \(R=0\). Thus each allocation reproduces the same observed-data law, and the prevalence \(a+u\) ranges from \(a\) to \(a+c\). If a denominator is zero at an endpoint, the corresponding outcome has probability zero, so its response probability can be assigned arbitrarily without changing the law. This proves the result. \(\square\)
The interval is an identification bound, not a confidence interval: it describes values compatible with the population observed-data distribution itself. Sampling uncertainty would add a separate layer. Even if \(a\), \(b\), and \(c\) were known exactly, this argument would not select one value within the interval without additional information or assumptions.
Worked Example: Calculating the Compatible Prevalence Range
Suppose an observed-data distribution has \(a=0.24\), \(b=0.36\), and \(c=0.40\). The observed cases therefore include 24% positive outcomes and 36% negative outcomes, while 40% of outcomes are missing. The prevalence must lie in
For example, choosing \(u=0.10\) gives prevalence \(0.24+0.10=0.34\). The implied response probabilities are
The missing mass is \(0.10\) for \(Y=1\) and \(0.30\) for \(Y=0\), totaling \(0.40\). Choosing instead \(u=0.30\) gives prevalence \(0.54\), with
Here the missing masses are \(0.30\) and \(0.10\), again totaling \(0.40\). Both models reproduce the observed probabilities exactly, but their prevalences differ. In each, response probabilities depend on the outcome, as required for nonignorability in this setting.
Why More Observations Cannot Resolve Equivalence
The examples demonstrate failure of this definition for the full-data prevalence: the same observed law is compatible with different prevalences. This also gives a precise limit on what an estimator based only on observed records can achieve.
Proof. Since the observed-record distributions agree, the joint distribution of any sample of size \(n\) agrees under the two models. Therefore any estimator \(\hat{\theta}_n\) computed from that sample has the same distribution under both. Let \(\delta=|\theta_1-\theta_2|>0\) and choose \(\varepsilon=\delta/3\). The intervals \((\theta_1-\varepsilon,\theta_1+\varepsilon)\) and \((\theta_2-\varepsilon,\theta_2+\varepsilon)\) are disjoint. If the estimator were consistent under both models, its probability of falling in the first interval would tend to 1 under the first model, and its probability of falling in the second interval would tend to 1 under the second model. But the estimator has the same distribution under both, so these two probabilities are probabilities of disjoint events under one and the same distribution. Their sum cannot exceed 1, contradicting convergence of both to 1. Hence simultaneous consistency is impossible. \(\square\)
This is not a claim that every estimate from incomplete data is useless. It says that, without a restriction or source of information that distinguishes the equivalent models, the observed records alone cannot guarantee the correct full-data target. Increasing the sample size can estimate the observed-data probabilities more precisely, but it cannot separate models that agree on those probabilities exactly.
Worked Example: A Fully Observed Covariate Does Not Always Resolve Ambiguity
Let \(Z\) be a fully observed binary covariate with \(P(Z=0)=P(Z=1)=0.5\), and let \(Y\) be a binary outcome that may be missing. Suppose the conditional observed cells \((a,b,c)\), given \(Z\), are \((0.2,0.4,0.4)\) for \(Z=0\) and \((0.3,0.3,0.4)\) for \(Z=1\). Consider two models with these same observed cells in each stratum.
For \(Z=0\), Model A assigns \(u=0.3\), so its full-data prevalence is \(0.2+0.3=0.5\). Its response probabilities are \(0.2/0.5=0.4\) for \(Y=1\) and \(0.4/(0.4+0.1)=0.8\) for \(Y=0\). Model B assigns \(u=0.2\), giving prevalence \(0.4\), with response probabilities \(0.2/0.4=0.5\) and \(0.4/(0.4+0.2)=2/3\).
For \(Z=1\), Model A assigns \(u=0.1\), giving prevalence \(0.3+0.1=0.4\) and response probabilities \(0.3/0.4=0.75\) and \(0.3/(0.3+0.3)=0.5\). Model B assigns \(u=0.3\), giving prevalence \(0.6\) and response probabilities \(0.3/0.6=0.5\) and \(0.3/(0.3+0.1)=0.75\). Thus both models retain the stated observed cells within both covariate groups, with nonconstant response probabilities in each group.
The population prevalence is \(0.5(0.5)+0.5(0.4)=0.45\) in Model A and \(0.5(0.4)+0.5(0.6)=0.50\) in Model B. Because the models agree on the observed covariate distribution and on the observed cells within each covariate group, they induce the same observed-data law, despite these different population prevalences. Recording a covariate can be useful, but it does not automatically identify outcomes that remain missing.
Interpretation and Common Pitfalls
The practical importance of observational equivalence is that conclusions about full-data quantities can depend on assumptions that are not testable from the observed records alone. In the binary calculation, the data determine \(a\), \(b\), and \(c\), but not \(u\), the allocation of missing cases across outcome values. A proposed single prevalence therefore reflects some additional restriction, external information, or modeling choice beyond the observed-data distribution.
- Mistake: “With a large enough sample, the true prevalence will become clear.” A larger sample can estimate the observed probabilities accurately. If two models give those probabilities exactly, the sample size does not distinguish them.
- Mistake: “The observed fraction of positive outcomes is the population prevalence.” The observed fraction targets \(P(Y=1\mid R=1)=a/(a+b)\), whereas the full-data prevalence is \(a+u\). These quantities need not agree.
- Mistake: “A fully observed covariate always fixes the problem.” The stratified example shows that distinct allocations of missing outcomes can remain compatible within every covariate stratum.
Additional information may narrow or remove the ambiguity—for example, a defensible restriction on the missingness process or outcome information obtained through a separate measurement effort. The essential check is whether that added information rules out one of the otherwise observationally equivalent models. A precise answer under a chosen assumption is not the same as identification from observed data alone.
Check Your Understanding
Use the definitions and binary-outcome calculations above to answer these questions.
- For \(a=0.15\), \(b=0.55\), and \(c=0.30\), what is the compatible range of the full-data prevalence?
- In the first worked example, what are the probabilities of \(Y=1\) among the missing cases under Models A and B?
- Explain why two models with the same observed-data law have the same distribution for every sample statistic computed from observed records.
- What does the term “identification bound” describe, and how does it differ from a confidence interval?
- Why does observing a covariate not, by itself, guarantee identification of the full-data outcome distribution?