Double Robustness of the AIPW Estimator

Why Two Models Can Provide One Robust Estimator

The augmented inverse-probability-weighted (AIPW) estimator combines a propensity-score model with an outcome-regression model. Its important property is not that both models must be correct: under suitable identification and regularity conditions, the estimator is consistent for the average treatment effect if either model is correctly specified. This is called double robustness.

The qualification matters. Double robustness is a statement about a large-sample limit, not a promise that a finite-sample estimate is unbiased or that arbitrary model errors cancel. It also does not remove the need for conditional independence and positivity. This tutorial derives the population result and shows how to interpret it.

Setup and Estimator

Let \(T\in\{0,1\}\) denote treatment, \(X\) the observed covariates, and \(Y(1)\) and \(Y(0)\) the potential outcomes under treatment and control. The observed outcome is \(Y=T Y(1)+(1-T)Y(0)\). The target is the average treatment effect (ATE), \(E[Y(1)-Y(0)]\), in the population represented by the sample.

Identification assumptions: (1) Consistency: the observed outcome equals the potential outcome under the treatment actually received, and one unit’s treatment does not change another unit’s outcome. (2) Conditional independence: \((Y(1),Y(0))\) is independent of \(T\) conditional on \(X\). (3) Positivity: \(0<e_0(X)<1\) with probability one, where \(e_0(X)=P(T=1\mid X)\). These assumptions identify the ATE from the observed-data distribution.

Define the true propensity score \(e_0(x)=P(T=1\mid X=x)\) and the two conditional outcome means \(\mu_t(x)=E[Y\mid T=t,X=x]\), for \(t=0,1\). Under the assumptions above, \(\mu_t(x)=E[Y(t)\mid X=x]\), so the ATE equals \(E[\mu_1(X)-\mu_0(X)]\).

The fitted propensity-score function is denoted by \(\hat e(X)\), and the fitted outcome-regression functions by \(\hat m_1(X)\) and \(\hat m_0(X)\). For an independent and identically distributed sample of size \(n\), the AIPW estimator is:

\hat{\tau}_{\mathrm{AIPW}} = \frac{1}{n}\sum_{i=1}^{n} \left[ \hat m_1(X_i)-\hat m_0(X_i) +\frac{T_i}{\hat e(X_i)}\{Y_i-\hat m_1(X_i)\} -\frac{1-T_i}{1-\hat e(X_i)}\{Y_i-\hat m_0(X_i)\} \right].

The first term averages predicted treatment-control outcome differences. The next term adjusts the treated predictions by weighted treated residuals; the last adjusts the control predictions by weighted control residuals. The weights use the fitted probability of the treatment actually observed. The augmentation terms correct the outcome-regression contrast when the propensity model is right; when the outcome regressions are right, their conditional residual means are zero.

The Population Argument for Double Robustness

To see what happens if either model is misspecified, consider the probability limits of the fitted functions: \(e^*(X)\), \(m_1^*(X)\), and \(m_0^*(X)\). A model is called correctly specified here when its fitted function converges to its corresponding true function. For the result below, assume the nuisance estimates converge suitably, the estimated propensities stay bounded away from zero and one, and sample averages converge to their expectations. These are sufficient regularity conditions for replacing the estimator’s fitted functions by their limits in the large-sample calculation.

Theorem: Under consistency, conditional independence, positivity, and the stated convergence conditions, the AIPW estimator is consistent for the ATE if either \(e^*(X)=e_0(X)\) or both \(m_1^*(X)=\mu_1(X)\) and \(m_0^*(X)=\mu_0(X)\). The other nuisance model may converge to a misspecified limit.

The key calculation is the conditional expectation of the limiting AIPW summand given \(X\). By conditional independence, \(E[Y\mid T=1,X]=\mu_1(X)\) and \(E[Y\mid T=0,X]=\mu_0(X)\). Also \(P(T=1\mid X)=e_0(X)\). Therefore:

\begin{aligned} E[\psi^*\mid X] ={}&m_1^*(X)-m_0^*(X)\\ &+\frac{e_0(X)}{e^*(X)}\{\mu_1(X)-m_1^*(X)\}\\ &-\frac{1-e_0(X)}{1-e^*(X)}\{\mu_0(X)-m_0^*(X)\}, \end{aligned}

where \(\psi^*\) is the summand with fitted functions replaced by their probability limits. The factors \(e_0/e^*\) and \((1-e_0)/(1-e^*)\) arise because the corresponding residual is observed only in the treated or control group.

If the propensity model is correct.
Set \(e^*(X)=e_0(X)\). Both ratios in the conditional expectation become one, giving \(m_1^*-m_0^*+\mu_1-m_1^*-(\mu_0-m_0^*)=\mu_1-\mu_0\). The outcome models cancel, even if misspecified.

If both outcome regressions are correct.
Set \(m_t^*(X)=\mu_t(X)\). Both residual differences are zero, so the conditional expectation is \(\mu_1(X)-\mu_0(X)\), for any valid limiting propensity \(e^*(X)\).

Average over covariates.
In either case, \(E[\psi^*]=E[\mu_1(X)-\mu_0(X)]\), which equals the ATE under the identification assumptions. Convergence of the sample average to this expectation gives consistency.

This argument is a population calculation. It does not say that the two correction terms are zero in each sample, or that every fitted model yields the target. It says that one correctly specified nuisance model is enough to make the limiting expectation of the AIPW summand equal the identified ATE.

Worked Examples

Worked Example: Correct Propensity, Misspecified Outcome Regressions

Suppose \(X\) takes values A and B, with \(P(X=\mathrm{A})=0.6\) and \(P(X=\mathrm{B})=0.4\). The true treatment probabilities are \(e_0(\mathrm{A})=0.25\) and \(e_0(\mathrm{B})=0.75\). Let the true outcome means be \((\mu_1,\mu_0)=(8,5)\) in A and \((12,8)\) in B. A fitted propensity model converges to the truth, but the outcome models converge to \(m_1^*(\mathrm{A})=7,\ m_0^*(\mathrm{A})=6,\ m_1^*(\mathrm{B})=10,\ m_0^*(\mathrm{B})=9\).

In stratum A, the limiting conditional expectation is:

(7-6)+\frac{0.25}{0.25}(8-7)-\frac{0.75}{0.75}(5-6) =1+1-(-1)=3.

In stratum B it is:

(10-9)+\frac{0.75}{0.75}(12-10)-\frac{0.25}{0.25}(8-9) =1+2-(-1)=4.

Thus the population expectation is \(0.6(3)+0.4(4)=3.4\) outcome units. The identified ATE is also \(0.6(8-5)+0.4(12-8)=3.4\). The outcome regressions are wrong, but the correct propensity model makes their errors cancel in expectation.

Worked Example: Correct Outcome Regressions, Misspecified Propensity

Use the same two strata and true means, so the ATE is 3.4 outcome units. Suppose now that both outcome regressions are correct, but the fitted propensity converges to \(e^*(\mathrm{A})=0.4\) and \(e^*(\mathrm{B})=0.6\), rather than the true probabilities 0.25 and 0.75.

In stratum A, the outcome residual differences in the population calculation are \(8-\mu_1(\mathrm{A})=0\) and \(5-\mu_0(\mathrm{A})=0\). Consequently:

(\mu_1-\mu_0)+\frac{0.25}{0.4}(8-\mu_1) -\frac{0.75}{0.6}(5-\mu_0)=8-5=3.

In stratum B, the residual differences are also zero, giving \(12-8=4\), regardless of the propensity ratios. Averaging over strata gives \(0.6(3)+0.4(4)=3.4\) outcome units. The propensity model is wrong, but correct conditional outcome means make the weighted residual corrections have expectation zero.

Worked Example: Calculating a Finite-Sample AIPW Estimate

For illustration, suppose four hypothetical observations have the following outcomes, treatment indicators, fitted propensities, and fitted outcome means. The values are in outcome units; this small calculation alone does not establish that either nuisance model is correct.

Unit	\(T_i\)	\(Y_i\)	\(\hat e(X_i)\)	\(\hat m_1(X_i)\)	\(\hat m_0(X_i)\)
1	1	10	0.50	8	5
2	0	4	0.50	8	5
3	1	7	0.25	6	4
4	0	3	0.75	6	4

The four summands, substituting each row into the estimator, are:

\begin{aligned} \psi_1&=(8-5)+\frac{10-8}{0.50}=7,\\ \psi_2&=(8-5)-\frac{4-5}{1-0.50}=5,\\ \psi_3&=(6-4)+\frac{7-6}{0.25}=6,\\ \psi_4&=(6-4)-\frac{3-4}{1-0.75}=6. \end{aligned}

Therefore \(\hat{\tau}_{\mathrm{AIPW}}=(7+5+6+6)/4=6\) outcome units. This is the estimate for these four observations and these fitted values. It need not equal the true ATE, even if a nuisance model is correct: double robustness guarantees consistency under its conditions as the sample grows, not exact equality in a particular sample.

What Double Robustness Does Not Guarantee

The result depends on the identification assumptions as well as on model specification. A correct propensity model does not control unmeasured confounding if conditional independence fails. If some covariate profiles have no chance of receiving one treatment, positivity fails and the necessary weighting information is unavailable. Very small fitted propensities can also produce unstable estimates in finite samples, even when the formal population conditions hold.

Nor is double robustness a guarantee when both nuisance models are misspecified. In the conditional expectation formula, errors in both models generally leave terms that do not cancel. The fact that an estimate looks plausible, or that two fitted models were used, does not establish consistency. Model specification and the assumptions supporting identification must be considered separately.

A common mistake is to reason, “AIPW uses two models, so at least one of them must be correct in practice.” That does not follow. The theorem is conditional: if one model is correctly specified and the other assumptions and convergence conditions hold, consistency follows. It does not determine which model is correct from the data, and it cannot guarantee correctness when both are wrong.

Key takeaway: The AIPW summand has conditional expectation equal to \(\mu_1(X)-\mu_0(X)\) when either the propensity limit is correct or both outcome-regression limits are correct. Averaging this result identifies the ATE, but only under conditional independence, positivity, and suitable estimation conditions.

Check Your Understanding

Use the population conditional-expectation formula to answer these questions.

When the propensity model is correct, which terms in the conditional expectation cancel, and why does the result not depend on the outcome-model limits?
When both outcome regressions are correct, what happens to the two residual terms, even if the propensity model is misspecified?
State the three identification assumptions used to connect \(E[\mu_1(X)-\mu_0(X)]\) to the ATE.
Why does double robustness not imply that a finite-sample AIPW estimate equals the true ATE?
What does the double-robustness result imply if both nuisance models converge to incorrect limits?