Constructing Compatible Missing-Outcome Laws

Why Construct Compatible Laws?

When an outcome is missing, the observed data contain no direct record of its value for that unit. A useful way to study what the observed data can support is to construct multiple complete-data distributions that agree on everything observed but assign different distributions to those missing outcomes. This makes the source of the uncertainty explicit: the observed-data law fixes some conditional distributions, while the distribution among nonrespondents must be supplied by an additional assumption or model.

Let \(Y\) be an outcome, \(R\) its response indicator, and \(X\) fully observed covariates. Set \(R=1\) when \(Y\) is observed and \(R=0\) when it is missing. The observed record is \(X\), \(R\), and \(Y\) when \(R=1\); when \(R=0\), the record contains no value of \(Y\). The construction below does not require a particular missingness mechanism. It starts from the observed-data distribution and fills in a conditional law for the unobserved values.

Definition: A complete-data law is compatible with a given observed-data law if observing only the variables recorded under the response pattern produces exactly that observed-data law. Two complete-data laws are observationally equivalent when they are both compatible with the same observed-data law.

The earlier material on observed-data distributions and the binary-outcome proof of MNAR nonidentification motivate this construction. Here the emphasis is on the general recipe: specify the observed part, choose a valid conditional distribution for the missing part, and verify that marginalizing over the missing values reproduces the same observed distribution.

The Construction for One Partly Observed Outcome

Write \(g(x,r,y)\) for a complete-data density or mass function. When \(R=1\), the observed data include \(Y\), so the observed law supplies \(g(x,1,y)\). When \(R=0\), only \(X\) and \(R\) are observed, so the observed law supplies the marginal \(g(x,0)=\int g(x,0,y)\,dy\), with a sum replacing the integral for a discrete outcome.

For covariate values with \(P(R=0\mid X=x)>0\), choose any conditional distribution \(q(y\mid x,R=0)\) on the allowed outcome support. Then define the complete-data law by retaining the observed part and completing the missing part as follows:

g_q(x,r,y)= \begin{cases} g_{\mathrm{obs}}(x,1,y), & r=1,\\[4pt] g_{\mathrm{obs}}(x,0)\,q(y\mid x,R=0), & r=0. \end{cases}

Here \(g_{\mathrm{obs}}(x,1,y)\) is the observed joint density or mass for \(X=x,R=1,Y=y\), and \(g_{\mathrm{obs}}(x,0)\) is the observed marginal for \(X=x,R=0\). The condition that \(q\) integrates or sums to one ensures that the second line has the observed \(X,R\) marginal.

Key result: Every valid choice of \(q(y\mid x,R=0)\) gives the same observed-data law, provided it leaves the observed part unchanged and is a normalized distribution on the allowed outcome support. If \(q_0\) and \(q_1\) are two such choices, then \(q_t=(1-t)q_0+tq_1\), for \(0\leq t\leq1\), also gives a compatible law for every \(t\).

This result is a direct consequence of marginalization. For \(R=0\), integrating the completed law over \(Y\) gives \(g_{\mathrm{obs}}(x,0)\int q(y\mid x,R=0)\,dy=g_{\mathrm{obs}}(x,0)\). For \(R=1\), the construction keeps the observed joint distribution \(g_{\mathrm{obs}}(x,1,y)\) unchanged. Thus every observable probability is preserved.

\int g_q(x,0,y)\,dy = g_{\mathrm{obs}}(x,0)\int q(y\mid x,R=0)\,dy = g_{\mathrm{obs}}(x,0).

For the mixture claim, \(q_t\) is normalized because \(\int q_t\,dy=(1-t)\int q_0\,dy+t\int q_1\,dy=1\). It is nonnegative because it is a convex combination of nonnegative distributions. Substituting \(q_t\) into the completion therefore preserves the same observed marginal. If \(q_0\) and \(q_1\) differ, varying \(t\) gives a continuum of distinct completions.

Worked Constructions

Worked Example: A Binary Outcome With Different Missing-Outcome Rates

Suppose a hypothetical study has \(P(R=1)=0.60\), and among respondents \(P(Y=1\mid R=1)=0.25\). The outcome \(Y\) is binary. Construct compatible complete-data laws indexed by \(q=P(Y=1\mid R=0)\), and compare their implied overall outcome probabilities.

The observed probabilities are \(P(R=1,Y=1)=0.60(0.25)=0.15\), \(P(R=1,Y=0)=0.60(0.75)=0.45\), and \(P(R=0)=0.40\). These three quantities are fixed by the observed-data law.

For any \(q\) between zero and one, assign \(P(Y=1\mid R=0)=q\) and \(P(Y=0\mid R=0)=1-q\). For example, with \(q=0.10\), the completed probabilities for \(R=0\) are \(P(R=0,Y=1)=0.40(0.10)=0.04\) and \(P(R=0,Y=0)=0.40(0.90)=0.36\). With \(q=0.90\), they are \(0.36\) and \(0.04\), respectively. In both cases, summing over \(Y\) when \(R=0\) gives \(0.40\); the respondent probabilities remain \(0.15\) and \(0.45\).

The implied full-population proportion with \(Y=1\) is \(P(Y=1)=P(R=1,Y=1)+P(R=0,Y=1)=0.15+0.40q\). It is \(0.19\) for \(q=0.10\) and \(0.51\) for \(q=0.90\). Every value between these follows from an intermediate \(q\), while the observed-data probabilities remain exactly the same. The choices of \(q\) are hypothetical completions, not estimates supplied by the observed data.

Worked Example: Allowing the Missing Law to Depend on Covariates

Suppose a hypothetical population has a binary covariate \(X\), with \(P(X=A)=P(X=B)=0.50\). For \(X=A\), the response rate is \(0.80\) and \(P(Y=1\mid X=A,R=1)=0.20\). For \(X=B\), the response rate is \(0.40\) and \(P(Y=1\mid X=B,R=1)=0.60\). Compare two possible conditional laws for \(Y\) among nonrespondents: completion I uses \(q_A=0.10,q_B=0.30\), while completion II uses \(q_A=0.90,q_B=0.80\).

Both completions leave the observed probabilities within each covariate group unchanged. For group A, \(P(R=1\mid A)=0.80\), with respondent outcome probability \(0.20\); for group B, \(P(R=1\mid B)=0.40\), with respondent outcome probability \(0.60\). In each group, the missing-outcome probabilities sum to one after assigning \(P(Y=0\mid X,R=0)=1-q_X\).

Under completion I, the conditional full-population outcome probabilities are \(P(Y=1\mid A)=0.80(0.20)+0.20(0.10)=0.18\) and \(P(Y=1\mid B)=0.40(0.60)+0.60(0.30)=0.42\). Therefore \(P(Y=1)=0.50(0.18)+0.50(0.42)=0.30\).

Under completion II, the corresponding values are \(P(Y=1\mid A)=0.80(0.20)+0.20(0.90)=0.34\) and \(P(Y=1\mid B)=0.40(0.60)+0.60(0.80)=0.72\). Thus \(P(Y=1)=0.50(0.34)+0.50(0.72)=0.53\). The observed response rates and respondent outcome distributions are identical across the two completions. Allowing \(q\) to vary separately by \(X\) makes clear that the missing-outcome law can depend on observed covariates.

Worked Example: A Continuum for a Continuous Outcome

In a hypothetical study, suppose \(P(R=1)=0.70\), and the observed outcome distribution is \(Y\mid R=1\sim N(10,4)\), where the second normal parameter is the variance. Consider the missing-outcome distributions \(q_0=N(8,9)\) and \(q_1=N(14,9)\). Construct a continuum between them and calculate the implied overall mean.

For \(0\leq t\leq1\), let \(q_t=(1-t)q_0+tq_1\), a mixture distribution. It is normalized because both component distributions integrate to one. The completed joint law assigns probability \(0.70\) to the respondent pattern with the stated \(N(10,4)\) outcome distribution, and probability \(0.30\) to the nonrespondent pattern with conditional outcome law \(q_t\). Integrating over \(Y\) in the nonrespondent pattern gives \(0.30\), regardless of \(t\), so the observed-data law is unchanged.

The mean of the mixture \(q_t\) is \((1-t)8+t14=8+6t\). The full-population mean under this completion is \(0.70(10)+0.30(8+6t)=9.4+1.8t\). It ranges from \(9.4\) at \(t=0\) to \(11.2\) at \(t=1\), in the outcome’s measurement units. The observed respondent distribution remains \(N(10,4)\) throughout. A mixture of normal distributions need not itself be normal; no normality assumption for the completed population has been made.

More Than One Missing Outcome

The same construction applies when there are multiple response patterns and multiple outcomes. Let \(M\) denote a pattern, and write \(Y_{\mathrm{obs}(M)}\) and \(Y_{\mathrm{mis}(M)}\) for the components observed and missing under that pattern. The observed-data law supplies the joint distribution of \(M\), covariates, and the components observed under \(M\). For each pattern, choose a conditional distribution for the missing components given the pattern, covariates, and observed components.

p_q(x,m,y_{\mathrm{obs}},y_{\mathrm{mis}}) = p_{\mathrm{obs}}(x,m,y_{\mathrm{obs}}) q(y_{\mathrm{mis}}\mid x,m,y_{\mathrm{obs}}).

The compatibility check is to integrate or sum over all components missing under that pattern. Since \(q\) is normalized for every \(x,m,y_{\mathrm{obs}}\), this yields \(p_{\mathrm{obs}}(x,m,y_{\mathrm{obs}})\). The conditional law can differ across patterns and can depend on the observed components, as long as it remains a valid distribution and respects any genuine support restrictions. If a pattern has probability zero, its conditional law is not learned from that pattern; it also contributes no probability mass to the observed law.

Assumptions, Diagnostics, and Limits

The construction relies on a few explicit conditions. The response pattern and observed values must be represented consistently in the complete and observed laws. Each chosen conditional distribution must be normalized, nonnegative, and supported on allowed values of the outcome. For a conditional law given \(X=x,R=0\) to describe units in that stratum, the stratum must have positive probability. The recipe imposes no equality between the outcome distributions of respondents and nonrespondents; imposing such equality would be an additional restriction, not a consequence of compatibility.

In practice, check the construction by verifying three things: the observed response-pattern probabilities have not changed; the distribution of observed outcomes within each pattern is unchanged; and each proposed missing-outcome law has total probability one. A purported completion fails if, for example, it changes a recorded covariate distribution, assigns positive probability to impossible outcome values, or uses conditional probabilities that do not sum to one. In a multi-pattern setting, check each pattern separately rather than assuming that a valid completion for one pattern automatically completes all the others.

A pattern-mixture representation makes the choice of \(q\) direct: it specifies outcome distributions within response patterns. A selection representation instead specifies response probabilities conditional on outcomes and covariates. Either representation must respect the same observed-data constraints, but neither representation by itself supplies the missing conditional law from observations that were never recorded. Parametric restrictions or substantive assumptions may select or limit proposed completions; they should be stated as assumptions, not presented as observed facts.

Common pitfall: “The observed respondent distribution is known, so the distribution among nonrespondents must match it.” This conclusion does not follow from the observed data. The correction is to treat the nonrespondent distribution as a separate conditional law unless an explicit assumption equates it with the respondent law. Compatibility requires preserving the observed margins, not copying the observed outcome distribution into every missing pattern.

This construction is a way to display what remains unspecified by the observed-data law, not a rule for choosing a preferred completion. Its scope is also limited by the restrictions one is willing to impose: a known outcome range, a structural constraint, or a scientific model can rule out some candidate laws. Such restrictions must be incorporated into the allowed support or conditional distributions before claiming that a completion is valid.

Check Your Understanding

Answer each question, then compare with the solutions below.

If \(P(R=0)=0.25\), why does replacing one normalized law \(q_0\) for \(Y\mid R=0\) with another normalized law \(q_1\) preserve the observed probability \(P(R=0)\)?
For the binary example, calculate \(P(Y=1)\) when \(q=0.40\).
In the covariate example, why must the missing-outcome law be specified separately for \(X=A\) and \(X=B\) if it is allowed to depend on \(X\)?
For the continuous example, what is the overall mean at \(t=0.50\), and does that mean imply that the completed outcome distribution is normal?

Solutions. (1) The completed missing-pattern mass is \(P(R=0)q(y\mid R=0)\); summing or integrating over \(y\) gives \(P(R=0)\int q(y\mid R=0)\,dy=0.25\). (2) \(P(Y=1)=0.15+0.40(0.40)=0.31\). (3) The conditional distributions \(P(Y\mid X=A,R=0)\) and \(P(Y\mid X=B,R=0)\) are distinct parts of the completion; one common value would impose an extra restriction not required by the observed law. (4) The mean is \(9.4+1.8(0.50)=10.3\) outcome units. No: the missing distribution is a mixture of two normals, and combining it with the respondent distribution does not generally produce a normal distribution.