Augmented Estimating Equations for MAR

Why Augment Inverse-Response Weighting?

When outcomes are missing, an estimating equation defined for fully observed data cannot usually be evaluated for every individual. Inverse-response weighting addresses this by giving more weight to observed outcomes from groups that are less likely to respond. Outcome regression offers another piece of information: it predicts the estimating-equation contribution from variables observed for everyone. An augmented estimating equation combines these approaches. Under suitable conditions, it has the right population expectation if either the response-probability model or the outcome-regression model is correct.

That “either” statement is a population consistency result, not a guarantee that every finite-sample estimate will be accurate. It depends on missingness assumptions, positive response probabilities, identification of the target, and adequate estimation of the nuisance models.

Setup and Definition

Let $Z=(X,Y)$ denote full data, where $X$ is observed for every individual and $Y$ is an outcome that may be missing. Let $R=1$ if $Y$ is observed and $R=0$ otherwise. The target parameter $\beta_0$ is defined by a full-data estimating equation $E[U(Z;\beta_0)]=0$, where $U$ may be scalar- or vector-valued. Assume observations of $(X,R,RY)$ are independent and identically distributed.

For this tutorial, missing at random means $R$ and $Y$ are conditionally independent given $X$. Write the true response probability as $\pi_0(X)=P(R=1\mid X)$. The positivity condition is $\pi_0(X)>0$ almost surely: within every covariate group represented in the target population, there must be some chance of observing the outcome. Define the conditional mean of the full-data estimating function by $m_0(X;\beta)=E[U(X,Y;\beta)\mid X]$.

Definition: Given working models $\pi^*(X)>0$ for the response probability and $m^*(X;\beta)$ for the conditional mean of the estimating function, the augmented estimating function is $$ \psi(O;\beta,\pi^*,m^*)=m^*(X;\beta)+\frac{R}{\pi^*(X)}\{U(X,Y;\beta)-m^*(X;\beta)\}, $$ where $O=(X,R,RY)$ denotes the observed data. When $R=0$, the term multiplied by $R$ is set to zero and does not require observing $Y$. The augmented estimating equation sets the sample mean of $\psi$ equal to zero.

The response-weighted residual $U-m^*$ corrects the regression prediction using observed outcomes. With the true response probability, the observed residuals are reweighted to represent the full population. With a correct outcome regression, their conditional mean is zero, so the correction has expectation zero even if the working response model is wrong.

Two Expectation Identities

The key calculation uses MAR. For any integrable function $h(X,Y)$, conditional independence gives $E[R h(X,Y)\mid X]=\pi_0(X)E[h(X,Y)\mid X]$. Indeed, conditional on $X,Y$, the probability of response is $\pi_0(X)$; taking the conditional expectation over $Y$ yields the identity. Apply it to $h=U-m^*$ to obtain the augmented equation’s conditional expectation.

Result: Suppose MAR holds, $U(X,Y;\beta)$ and $m^*(X;\beta)$ are integrable, and $\pi^*(X)>0$. Then $$ E[\psi(O;\beta,\pi^*,m^*)\mid X] =m^*(X;\beta)+\frac{\pi_0(X)}{\pi^*(X)} \{m_0(X;\beta)-m^*(X;\beta)\}. $$

To verify the result, condition on $X$. The first term is fixed given $X$, and the MAR identity implies $E[R\{U-m^*\}\mid X]=\pi_0(X)E[U-m^*\mid X]$. By the definition of $m_0$, $E[U-m^*\mid X]=m_0-m^*$. Dividing by $\pi^*(X)$ gives the displayed expression.

Theorem: Suppose $E[U(X,Y;\beta_0)]=0$, MAR and positivity hold, and the expectations below exist. If either (i) $\pi^*(X)=\pi_0(X)$ almost surely, or (ii) $m^*(X;\beta_0)=m_0(X;\beta_0)$ almost surely, then $$ E[\psi(O;\beta_0,\pi^*,m^*)]=0. $$

Proof when the response model is correct. Substitute $\pi^*=\pi_0$ into the conditional expectation identity. It becomes $m^*+(m_0-m^*)=m_0$. Taking expectations gives $E[m_0(X;\beta_0)]=E[U(X,Y;\beta_0)]=0$, by iterated expectation and the defining full-data equation.

Proof when the outcome regression is correct. Substitute $m^*=m_0$ into the identity. The residual term vanishes because $m_0-m^*=0$, leaving $m_0$, regardless of whether $\pi^*$ equals $\pi_0$. Again, $E[m_0(X;\beta_0)]=E[U(X,Y;\beta_0)]=0$. Thus either correctly specified nuisance model makes the population augmented equation unbiased at $\beta_0$. $\square$

From the Population Equation to an Estimate

For observations $i=1,\ldots,n$, a practical estimator solves $$ \frac{1}{n}\sum_{i=1}^n \left[ \hat m(X_i;\beta)+\frac{R_i}{\hat\pi(X_i)} \{U(X_i,Y_i;\beta)-\hat m(X_i;\beta)\} \right]=0. $$ Here $\hat\pi$ and $\hat m$ are fitted nuisance models. This equation is directly computable: an individual with $R_i=0$ contributes only $\hat m(X_i;\beta)$, while a respondent contributes both the prediction and a weighted residual.

The theorem concerns the population equation when the nuisance functions have the stated values. To transfer it to an estimator, additional regularity is needed. One sufficient set of conditions is that the fitted nuisance functions converge to fixed limits, the sample equation converges uniformly in $\beta$ to its population limit, and that limiting equation has a unique root at $\beta_0$. Then a root of the sample equation converges to $\beta_0$. For example, bounded contributions, response probabilities bounded away from zero, regular finite-dimensional nuisance models, and a uniform law of large numbers can support these requirements. If a nuisance model is misspecified, its fitted limit need not equal the corresponding true function; the theorem requires the *other* nuisance model to be correct.

Worked Calculations

Worked Example: A Sample Augmented Mean

Suppose $Y$ is a scalar outcome and the target is its population mean $\theta=E[Y]$. Use $U(Y;\theta)=Y-\theta$. For this target, let $\hat\mu(X)$ estimate $E[Y\mid X]$. The augmented equation is $$ \frac{1}{n}\sum_{i=1}^n \left[\hat\mu(X_i)+\frac{R_i}{\hat\pi(X_i)} \{Y_i-\hat\mu(X_i)\}-\theta\right]=0, $$ so its solution is the sample average of the bracketed predicted-and-corrected outcome, before subtracting $\theta$.

Consider two covariate groups, each containing half the sample. In group $X=0$, half respond, their mean is 10, the fitted response probability is $0.5$, and the fitted outcome mean is 12. The group contribution to the estimated population mean is $12+(0.5/0.5)(10-12)=12-2=10$. In group $X=1$, three quarters respond, their mean is 20, the fitted response probability is $0.75$, and the fitted outcome mean is 18. Its contribution is $18+(0.75/0.75)(20-18)=18+2=20$. Weighting the two equally sized groups gives $\hat\theta=(10+20)/2=15$.

The response fraction enters because the sample averages the respondent residual over all individuals in the group. Equivalently, the group's average correction is its response fraction times the inverse-probability-weighted respondent residual.

Worked Example: Correct Response Probabilities, Imperfect Predictions

Let $P(X=0)=P(X=1)=0.5$, with true conditional outcome means $\mu_0=10$ and $\mu_1=20$. Suppose the true response probabilities are $\pi_0(0)=0.5$ and $\pi_0(1)=0.75$, and the fitted response model gets both exactly right. Use imperfect outcome predictions $m^*(0)=12$ and $m^*(1)=18$.

In group $X=0$, the conditional expectation of the augmented outcome is $12+(0.5/0.5)(10-12)=12-2=10$. In group $X=1$, it is $18+(0.75/0.75)(20-18)=18+2=20$. Therefore the population expectation is $0.5(10)+0.5(20)=15$, the target mean.

A tempting but incorrect calculation for $X=0$ is $12+2(10-12)=8$, using the inverse response probability but omitting the response probability in the conditional expectation. The residual term includes $R$; its conditional expectation contributes the factor $P(R=1\mid X=0)=0.5$. Thus the ratio is $0.5/0.5=1$, not 2, and the correct group value is 10.

Worked Example: Correct Outcome Regression, Imperfect Response Probabilities

Keep the same two equally likely groups, true means 10 and 20, and true response probabilities 0.5 and 0.75. Now suppose the outcome regression is correct, so $m^*(0)=10$ and $m^*(1)=20$, but the working response probabilities are wrong: $\pi^*(0)=0.25$ and $\pi^*(1)=1$. Both working probabilities are positive.

For $X=0$, the conditional expectation is $10+(0.5/0.25)(10-10)=10+2(0)=10$. For $X=1$, it is $20+(0.75/1)(20-20)=20+0.75(0)=20$. Consequently, the overall expectation is $0.5(10)+0.5(20)=15$. The response-model errors do not affect this population calculation because the conditional residual has mean zero in each group.

When the Result Matters—and Common Pitfalls

Augmentation is useful when response probabilities and outcome predictions provide complementary ways to represent the full-data estimating equation. It can protect consistency against misspecification of one nuisance model, provided the other is correct and the assumptions connecting observed and full data hold. The same construction works with vector-valued estimating functions, such as those defining several parameters jointly; the conditional means and residuals are then vectors.

Confusing MAR with unconditional response. MAR here is conditional independence given the fully observed $X$. If response still depends on $Y$ after conditioning on $X$, the conditional-expectation identity used in the proof does not follow.
Ignoring positivity. If a group has zero response probability, its outcome distribution cannot be recovered by inverse weighting from that group’s respondents. Very small probabilities can also produce unstable finite-sample corrections.
Calling the method automatically robust to everything. “Doubly robust” refers to consistency if either one of the two nuisance models is correct, under the stated assumptions. It does not mean both can be wrong, nor does it guarantee low variance or reliable finite-sample behavior.
Dropping the response factor in an expectation. The term $R/\pi^*(X)$ has conditional expectation involving $\pi_0(X)/\pi^*(X)$. Omitting the numerator response probability gives incorrect calculations whenever one works out a conditional expectation.
Using a full-data estimating function when it is not observed. The residual involving $U(X,Y;\beta)$ is evaluated only for respondents. With partially observed multivariate outcomes, the relevant observed-data pattern and its response mechanism must be represented appropriately; the single-indicator formula cannot be applied blindly to a contribution that is unavailable for some respondents.

Key takeaway: Under MAR and positivity, the augmented estimating function has the full-data estimating equation’s expectation if the response-probability model is correct or if the conditional mean of the estimating function is correct. Estimation still requires suitable identification and regularity conditions.

Check Your Understanding

Use the conditional expectation identity to answer each question.

State the MAR condition used here and explain what $X$ must provide for it to be useful.
For a group with true response probability $0.4$, working response probability $0.4$, outcome mean 8, and prediction 11, calculate the conditional expectation of the augmented outcome.
If the outcome-regression model is correct, why does the conditional expectation of the weighted residual vanish even when the working response probability is incorrect?
What does positivity rule out, and why is it needed for this inverse-probability construction?
Does consistency under either correct nuisance model imply that the estimator is unbiased in every finite sample? Explain briefly.