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An augmented inverse-probability-weighted estimator averages a regression prediction plus an inverse-probability-weighted residual. In the treatment setting of doubly robust estimation, let ${\hat{m}}_a(X_i)$ estimate the conditional outcome mean and ${\hat{e}}_a(X_i)$ the probability of receiving treatment $a$. ThenThe first term predicts outcomes for everyone; the weighted residual corrects prediction errors using observed outcomes from the corresponding treatment group. The conditional-expectation calculation in doubly robust estimation makes the remaining bias a product of outcome-regression and propensity-score errors. An exactly correct outcome regression makes the residual correction mean zero even with an incorrect propensity score; a correct propensity score makes that correction repair the regression bias.

The nuisance estimates are commonly obtained by cross-fitting, with each observation evaluated using functions trained on other observations. Both treatment probabilities must remain positive on the population being studied, and small fitted probabilities can make the variance very large. Double robustness concerns consistency under one correct nuisance model; it does not automatically deliver valid finite-sample or asymptotic confidence intervals.

Doubly robust estimation combines two nuisance functions so that either one can be correctly estimated while the other is misspecified. For example, let $A\in \{0,1\}$ indicate treatment, $X$ covariates, $m_a(X)=\mathbb{E}[Y\mid A=a,X]$, and $e_a(X)=\mathbb{P}(A=a\mid X)$. Under conditional exchangeability, positivity and the usual consistency of observed and potential outcomes, ${\mu }_a=\mathbb{E}[m_a(X)]$ is the mean potential outcome under treatment $a$.

For candidate functions ${\tilde{m}}_a,{\tilde{e}}_a$ with denominators bounded away from zero, the augmented signal isConditioning on $X$ shows thatThe bias vanishes if either nuisance function is correct. This algebra explains Double robustness and the second-order product bias. It does not protect against both models being wrong, failed Causal identification, or nearly zero denominators. cross-fitting helps when flexible functions are estimated from the same sample; valid confidence intervals require further rate and moment assumptions. The average treatment effect is ${\mu }_1-{\mu }_0$.^[ref]

The homogeneous treatment effect and consistency implyInstrument validity gives $Z\perp R\mid X$.

If Assumption 1 holds, put $V=R-{\alpha }_{0}X$. Then $\mathbb{E}[V\mid X]=0$, so conditional instrument independence givesConsequently both population estimating equations vanish at $({\beta }_0,{\alpha }_0)$ for any probability limit of $\hat{\gamma }$.

If Assumption 2 holds, choose the linear-projection coefficientThen $\mathbb{E}[(R-{\alpha }_{\ast }X)X]=0$, whileThe first term is zero by conditional instrument independence and the second by the definition of ${\alpha }_{\ast }$. Thus $({\beta }_0,{\alpha }_{\ast })$ solves the population equations. Under the nonsingularity condition from part b, the root is unique, so standard estimating equation consistency proveswhenever either Assumption 1 or Assumption 2 holds. This is double robustness.

If both outcome models are correct, each residual term in the preceding conditional expectations has mean zero regardless of the working propensity model. Thus ${\tilde{\mu }}_z^{\mathrm{dr}}(X)={\mu }_z(X)$ for $z=0,1$, and their difference is again $\tau (X)$. This proves double robustness.