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Semiparametric Efficiency · Tutorial 797 of 952

The Missing-Data Mean Influence Function

Learn how missingness changes the efficient influence function for a population mean, and how MAR, positivity, and conditional means determine identification and precision.

Advanced 12 min read

What You'll Learn

  • State the MAR, positivity, and moment conditions for identifying a population outcome mean.
  • Derive the missing-data mean’s efficient influence function from observed-data scores.
  • Interpret the influence function’s outcome-residual and covariate components.
  • Calculate influence-function values and the corresponding efficiency bound.
  • Explain why complete-case averaging can target the wrong mean.

Why Missing Outcomes Change the Mean’s Influence Function

When some outcomes are not recorded, the mean among observed cases is not generally the population mean. The difference matters even if the chance of observing an outcome is unrelated to its value after conditioning on observed covariates. In that setting, the covariates help recover both the target mean and the additional uncertainty caused by missingness.

The efficient influence function makes this adjustment explicit. It combines a prediction based on covariates with a weighted residual for people whose outcomes are observed. Its form also reveals why observations with a low response probability can contribute substantial uncertainty.

Model, Target, and Assumptions

Let \(Y\) be an outcome of interest, \(X\) a vector of fully observed covariates, and \(R\) an indicator equal to 1 when \(Y\) is observed and 0 otherwise. The observed data for one unit are \(O=(X,R,RY)\); when \(R=0\), the recorded value \(RY=0\) is only a placeholder, not the outcome. The target is the population mean \(\mu=E[Y]\).

Definition: The missing-at-random (MAR) mean model assumes \(R\) is conditionally independent of \(Y\) given \(X\), written \(R\perp Y\mid X\). Define the response probability \(\pi(X)=P(R=1\mid X)\) and the conditional outcome mean \(m(X)=E[Y\mid X]\). The identifying conditions are MAR, positivity \(\pi(X)>0\) almost surely, and integrability \(E[|Y|]<\infty\). In particular, MAR implies the conditional-mean equality \(E[Y\mid X,R=1]=E[Y\mid X]=m(X)\) wherever \(\pi(X)>0\).

The equality of conditional means is the feature needed to identify this particular target: the observed outcomes reveal \(m(X)\) for covariate values with positive response probability. The full MAR assumption is stronger: it states that the entire conditional distribution of \(Y\), not only its mean, is the same for responders and nonresponders at a given \(X\). A common sufficient positivity condition is that \(\pi(X)\) is bounded below by a positive constant. Pointwise positivity identifies the mean, but may not be enough for a finite efficiency bound.

For the usual nonparametric MAR model, assume independent, identically distributed observations, \(E[Y^2]<\infty\), and a sufficiently regular model for the observed-data distribution. The square-integrability condition needed for a finite bound is more specifically \(E[\operatorname{Var}(Y\mid X)/\pi(X)]<\infty\). These conditions ensure that the influence function below has finite variance.

Definition: An influence function for \(\mu\) is a mean-zero, square-integrable function of the observed data whose inner product with every regular observed-data score equals the derivative of \(\mu\) along the corresponding model path. The efficient influence function is the influence function in the observed-data tangent space; its variance is the semiparametric efficiency bound per observation.

The Efficient Influence Function

Under the stated nonparametric MAR model, the efficient influence function is

$$ \phi(O)=m(X)-\mu+\frac{R}{\pi(X)}\{Y-m(X)\}. $$

The expression is evaluated only when \(R=1\) for its residual term; when \(R=0\), that term is zero and the unobserved \(Y\) need not be supplied. The first component, \(m(X)-\mu\), records how the covariate-specific predicted mean differs from the population mean. The second component corrects that prediction using the observed outcome residual \(Y-m(X)\), reweighted by the inverse response probability.

Here is a score calculation that both verifies the derivative representation and explains the weighting. In a regular path through the nonparametric MAR model, an observed-data score can be decomposed as

$$ S_O=S_X(X)+(R-\pi(X))a(X)+R S_Y(Y,X), $$

where \(E[S_X(X)]=0\), \(a(X)\) is a response-mechanism score component, and \(E[S_Y(Y,X)\mid X]=0\). The target’s pathwise derivative is the full-data mean derivative

$$ \dot{\mu} =E[(Y-\mu)\{S_X(X)+S_Y(Y,X)\}] =E[(m(X)-\mu)S_X(X)]+E[(Y-m(X))S_Y(Y,X)]. $$

The second equality follows by conditioning on \(X\): the part of \(Y-\mu\) that varies around \(m(X)\) has conditional mean zero, and the conditional score \(S_Y\) has mean zero given \(X\). Now take the inner product of \(\phi\) with the observed-data score. Its pairing with \(S_X\) is \(E[(m(X)-\mu)S_X(X)]\), because the weighted residual has conditional expectation zero. Its pairing with \((R-\pi(X))a(X)\) is zero: the covariate component is orthogonal to this response score, while the residual component has conditional expectation zero given \(X\). Finally, pairing the residual component with \(R S_Y\) gives

$$ E\left[\frac{R}{\pi(X)}\{Y-m(X)\}R S_Y(Y,X)\right] =E[(Y-m(X))S_Y(Y,X)], $$

because \(R^2=R\) and, under MAR, \(E[R\mid X,Y]=\pi(X)\). Thus \(E[\phi(O)S_O]=\dot{\mu}\) for every such score. Also, \(E[\phi(O)]=0\), since \(E[m(X)]=\mu\) and \(E[R\{Y-m(X)\}/\pi(X)\mid X]=0\). The covariate component is a mean-zero function of \(X\), and the residual component is a conditional-mean-zero function of the observed outcome among responders; both lie in the nonparametric observed-data tangent space. The derivative representation is therefore the canonical gradient, not merely an influence function with an unnecessary extra component.

Worked Examples

Worked Example: Evaluating the Influence Function

Suppose \(X\) has two groups, A and B, with probabilities 0.6 and 0.4. Assume the hypothetical conditional means are \(m(A)=10\) units and \(m(B)=20\) units, and response probabilities are \(\pi(A)=0.8\) and \(\pi(B)=0.5\). Under MAR, the population mean is

$$ \mu=0.6(10)+0.4(20)=14\text{ units}. $$

For an observed person in group A with \(Y=12\), substitute the values into the influence function:

$$ \phi=10-14+\frac{1}{0.8}(12-10) =-4+2.5=-1.5\text{ units}. $$

For an observed person in group B with \(Y=17\), the calculation is

$$ \phi=20-14+\frac{1}{0.5}(17-20) =6-6=0\text{ units}. $$

For a person in group B whose outcome is missing, the residual term is zero, so \(\phi=20-14=6\) units. This is a value of the influence function, not a claim that the person’s missing outcome equals 20.

Worked Example: Calculating the Efficiency Bound

Continue with the same hypothetical groups, and suppose \(\operatorname{Var}(Y\mid A)=4\) square units and \(\operatorname{Var}(Y\mid B)=9\) square units. The two components of the influence function are uncorrelated: the residual has conditional mean zero given \(X\). Consequently,

$$ \operatorname{Var}\{\phi(O)\} =\operatorname{Var}\{m(X)\} +E\left[\frac{\operatorname{Var}(Y\mid X)}{\pi(X)}\right]. $$

The variance of the conditional mean is

$$ \operatorname{Var}\{m(X)\} =0.6(10-14)^2+0.4(20-14)^2 =0.6(16)+0.4(36)=24\text{ square units}. $$

The missingness contribution is

$$ E\left[\frac{\operatorname{Var}(Y\mid X)}{\pi(X)}\right] =0.6\left(\frac{4}{0.8}\right)+0.4\left(\frac{9}{0.5}\right) =3+7.2=10.2\text{ square units}. $$

Thus the efficiency bound is \(24+10.2=34.2\) square units per observation. For \(n\) independent observations, the corresponding first-order lower bound on the variance of a regular estimator is \(34.2/n\). The response-weighted residual contribution is larger in group B because outcomes there are less often observed.

Worked Example: Why Complete Cases Can Miss the Target

Consider a hypothetical population with equal proportions in groups A and B. Let \(m(A)=10\) and \(m(B)=20\) units, so \(\mu=0.5(10)+0.5(20)=15\) units. Suppose response probabilities are \(\pi(A)=0.75\) and \(\pi(B)=0.25\), with response independent of \(Y\) within each group. The MAR and positivity conditions hold, but responders are more likely to come from A.

Among all population members, the proportions who are both in group A and observed, and both in group B and observed, are \(0.5(0.75)=0.375\) and \(0.5(0.25)=0.125\). The total observed proportion is 0.5, so the group shares among responders are \(0.375/0.5=0.75\) and \(0.125/0.5=0.25\). The expected complete-case mean is therefore

$$ 0.75(10)+0.25(20)=12.5\text{ units}, $$

which is not the target of 15 units. In contrast, the inverse-probability weighted outcome contribution has conditional expectation \(E[RY/\pi(X)\mid X]=m(X)\). Averaging over the two groups gives \(0.5(10)+0.5(20)=15\) units. The augmented expression \(m(X)+R\{Y-m(X)\}/\pi(X)\) has the same expectation, because its residual term has conditional expectation zero. This example demonstrates the population calculation; it does not guarantee that a finite-sample estimate will equal the target exactly.

Why This Form Matters

The formula separates two sources of sampling variation. Variation in \(m(X)\) reflects differences in predicted outcomes across covariate values. The residual term reflects outcome variation not explained by \(X\), amplified when outcomes are rarely observed. In particular, the variance identity

$$ \operatorname{Var}\{\phi(O)\} =\operatorname{Var}\{m(X)\} +E\left[\frac{\operatorname{Var}(Y\mid X)}{\pi(X)}\right] $$

shows how limited response can reduce precision. If \(\pi(X)\) is small where the conditional outcome variance is substantial, the efficiency bound can be large or even infinite. Positivity therefore has a precision implication as well as an identification role.

In applications, the function \(m(X)\) is a conditional mean prediction and \(\pi(X)\) is a response probability. The influence function specifies the first-order contribution of each observation when these functions are known; when they are estimated, additional conditions are needed for valid inference. The expression alone does not establish that MAR is credible. Whether missingness is plausibly independent of the outcome after conditioning on \(X\) is a substantive assumption that cannot generally be verified from observed outcomes alone.

Common Pitfall

  • Wrong reasoning: “The outcomes are missing at random, so averaging the observed outcomes estimates the population mean.”
  • Correction: MAR means independence from \(Y\) conditional on \(X\), not necessarily independence without conditioning. If response probabilities vary with \(X\), the observed cases can have a different distribution of \(X\) from the population. The complete-case mean then generally targets a response-weighted mean. The influence function corrects this through covariate-specific prediction and the inverse-probability-weighted residual.
  • Another error: Treating positivity as sufficient for a stable estimator. Positive response probabilities identify the mean under MAR, but probabilities close to zero can inflate the efficiency bound; finite variance requires the weighted conditional variance to be integrable.
Key takeaway: Under MAR, positivity, and the stated moment conditions, the efficient influence function for the population outcome mean is \(m(X)-\mu+R\{Y-m(X)\}/\pi(X)\). Its variance quantifies the best attainable first-order precision in the nonparametric MAR model.

Check Your Understanding

Answer each question using the definitions and conditions stated above.

  1. State the MAR condition and explain how it implies the conditional-mean equality needed to identify \(\mu\).
  2. Why does positivity matter for identifying the population mean, and what additional issue arises when \(\pi(X)\) is close to zero?
  3. For an observed person with \(m(X)=8\), \(\mu=11\), \(\pi(X)=0.5\), and \(Y=10\), calculate the influence-function value and state its units if \(Y\) is measured in kilograms.
  4. Explain why \(m(X)-\mu\) and the weighted residual have zero covariance.
  5. In what situation can the complete-case mean differ systematically from \(\mu\), even when MAR holds?
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