The Role of Estimated Propensity Scores in AIPW

Why Estimating the Propensity Can Matter

AIPW combines outcome predictions with inverse-propensity-weighted residuals. Its propensity score is often estimated from the same observations used to estimate the treatment effect. That is not merely a practical way to fill in an unknown quantity: when the propensity model is correct, fitting it can change the estimator’s first-order variability. In particular, the fitted score equation can remove a component of the AIPW estimator that is correlated with treatment assignment.

The variance adjustment is not a universal bonus from estimating more parameters. It depends on a correctly specified propensity model and on the relationship between the AIPW estimating function and the propensity-model score. If the outcome predictions are correct, that relationship is zero and estimating the propensity has no first-order effect on AIPW variance. The result below explains both cases.

Setup and Assumptions

For each independent observation, let $O_i=(X_i,A_i,Y_i)$, where $A_i$ is binary, $Y_i$ is the observed outcome, and $X_i$ is a vector of pretreatment covariates. Let $Y_i(1)$ and $Y_i(0)$ denote potential outcomes. The target is the average treatment effect $\psi=E\{Y(1)-Y(0)\}$. Assume consistency, conditional exchangeability $\{Y(1),Y(0)\}\perp A\mid X$, and positivity: $0<P(A=1\mid X)<1$ on the covariate support relevant to the target.

Write $\mu_a(X)=E(Y\mid A=a,X)$ for the true conditional outcome mean and $m_a(X)$ for a fitted or fixed prediction. The propensity model is $e_\gamma(X)=P_\gamma(A=1\mid X)$, with true parameter $\gamma_0$ and true score $e_0(X)=e_{\gamma_0}(X)$. Assume this parametric model is correctly specified, its parameter is identifiable and interior, and the usual smoothness and finite-moment conditions for likelihood estimation and Taylor expansion hold.

Definition: For a specified propensity $e$ and outcome predictions $m_1,m_0$, the AIPW estimating function for the ATE is $$ \psi_i(e,m_1,m_0)=m_1(X_i)-m_0(X_i) +\frac{A_i}{e(X_i)}\{Y_i-m_1(X_i)\} -\frac{1-A_i}{1-e(X_i)}\{Y_i-m_0(X_i)\}. $$ The sample AIPW estimator is the sample average of this function, using the fitted nuisance quantities when they are estimated.

The key expectation identity is easiest to see by conditioning on $X$. Under a law with propensity $e_\gamma$, while the distribution of $X$ and the conditional outcome distributions given $A,X$ remain fixed,

E_{P_\gamma}\left[\frac{A}{e_\gamma(X)}\{Y-m_1(X)\}\mid X\right] =\mu_1(X)-m_1(X),

and

E_{P_\gamma}\left[-\frac{1-A}{1-e_\gamma(X)}\{Y-m_0(X)\}\mid X\right] =m_0(X)-\mu_0(X).

Adding $m_1(X)-m_0(X)$ leaves $\mu_1(X)-\mu_0(X)$. Averaging over $X$ gives $E_{P_\gamma}\{\psi_i(e_\gamma,m_1,m_0)\}=\psi$, for every $\gamma$ in this constructed family. This identity holds for fixed predictions $m_a$; the two residual terms cancel their prediction errors in expectation.

Worked Example: Keep the Data Law Fixed When Varying the Score

Suppose there are no covariates, $e_0=0.5$, $Y=1$ for every treated unit, and $Y=0$ for every control. Thus $\mu_1=1$, $\mu_0=0$, and the ATE is $\psi=1$. Set $m_1=m_0=0$, and evaluate the AIPW function using a candidate propensity $e_\gamma$ while keeping the actual data law fixed at $e_0=0.5$.

The control residual contribution is zero because control outcomes are zero. The treated contribution is $AY/e_\gamma$, so its expectation under the fixed true law is $$ E_{P_0}\{\psi_i(e_\gamma,0,0)\} =\frac{P_0(A=1)E(Y\mid A=1)}{e_\gamma} =\frac{0.5\times 1}{e_\gamma}. $$

At $e_\gamma=0.5$, this equals $1$, the target. At $e_\gamma=0.4$, it equals $0.5/0.4=1.25$; at $e_\gamma=0.6$, it equals $0.5/0.6=0.8333$. Therefore, the expectation under the fixed law does not remain equal to the target as the candidate propensity changes. The constant-expectation identity instead compares each propensity $e_\gamma$ with its corresponding law $P_\gamma$, in which the treatment probability also changes.

From the Score Equation to the Variance Adjustment

Let $s_i$ be the column score for one observation in the propensity model, evaluated at the truth:

s_i=\left.\frac{\partial}{\partial\gamma}\log P_\gamma(A_i\mid X_i)\right|_{\gamma=\gamma_0}, \qquad I=E_{P_0}(s_i s_i^{T}).

Assume $I$ is nonsingular. Because the propensity model describes treatment assignment conditional on $X$, $E_{P_0}(s_i\mid X_i)=0$, and hence $E_{P_0}(s_i)=0$. Differentiate the constant-expectation identity under $P_\gamma$ at $\gamma_0$. The derivative has two parts: the change in the estimating function and the change in the probability law. The latter contributes its likelihood score.

0=E_{P_0}\left[\left.\frac{\partial}{\partial\gamma} \psi_i(e_\gamma,m_1,m_0)\right|_{\gamma_0}\right] +E_{P_0}\left[\{\psi_i(e_0,m_1,m_0)-\psi\}s_i^{T}\right].

Define $D_\gamma$ as the row-vector derivative in the first expectation, taken with the data law held fixed at $P_0$. Then

D_\gamma=-E_{P_0}\left[\{\psi_i(e_0,m_1,m_0)-\psi\}s_i^{T}\right].

This is the score-equation condition behind the result: the fixed-law derivative is the negative covariance between the AIPW estimating function and the propensity score. The distinction between the fixed law in $D_\gamma$ and the changing law in the differentiated identity is essential.

For a correctly specified propensity model fitted by maximum likelihood, the score equation is $\sum_{i=1}^n s_i(\hat\gamma)=0$. Its first-order expansion gives

\hat\gamma-\gamma_0 =I^{-1}\frac{1}{n}\sum_{i=1}^n s_i+o_p(n^{-1/2}).

A Taylor expansion of the AIPW average around $\gamma_0$ therefore gives the influence function

\phi_{\mathrm{est},i} =\phi_i+D_\gamma I^{-1}s_i =\phi_i-c^{T}I^{-1}s_i, \qquad \phi_i=\psi_i(e_0,m_1,m_0)-\psi, \qquad c=E_{P_0}(s_i\phi_i).

This expansion treats the outcome predictions as fixed or as regular estimates whose first-order contribution vanishes. At the true propensity, the expectation derivative of the AIPW function with respect to either prediction is zero: for example, the coefficient on a perturbation to $m_1(X)$ has conditional expectation $1-E(A/e_0(X)\mid X)=0$. Thus regular parametric outcome fits do not add a first-order term under these conditions.

Since $E(s_i)=0$ and $E(s_i s_i^T)=I$, the asymptotic variance of the estimated-propensity influence function is

\operatorname{Var}(\phi_{\mathrm{est},i}) =\operatorname{Var}(\phi_i)-c^T I^{-1}c.

The subtracted quantity is nonnegative because $I$ is positive definite. Thus, relative to the same AIPW estimating function with the true propensity held fixed, estimating a correctly specified propensity cannot increase its first-order variance under these conditions. The reduction is strict when $c\ne 0$, and it is zero when the AIPW influence function is uncorrelated with the propensity score.

Worked Example: A Score Fit Removes a Correlated Component

Continue with no covariates, but now use the data law $P_0(A=1)=0.5$, $Y=1$ for treated units, $Y=0$ for controls, and $m_1=m_0=0$. Fit the propensity by an intercept-only logistic model. At $\gamma_0=0$, $e_0=0.5$, and the logistic score is $s_i=A_i-0.5$.

The AIPW function is $\psi_i=2A_i$, so its mean is $1$, and its fixed-propensity influence function is $\phi_i=2A_i-1$. The information is $$ I=E\{(A_i-0.5)^2\}=0.25. $$

The score covariance is $$ c=E\{(A_i-0.5)(2A_i-1)\}=0.5. $$

Consequently the variance with the true propensity fixed is $E(\phi_i^2)=1$, while the estimated-propensity variance is $$ 1-\frac{(0.5)^2}{0.25}=0. $$ Indeed, the fitted propensity is the sample treated fraction. With the stated deterministic outcomes and both treatment groups represented, the AIPW estimator reduces to the treated mean, which is exactly $1$. The zero asymptotic variance here is a special feature of these hypothetical outcomes, not a general property of AIPW.

Worked Example: Correct Outcome Predictions Give No Score Adjustment

Suppose again that $e_0=0.5$ and there are no covariates, but now $m_a=\mu_a$. Let the treated residual $Y-\mu_1$ have conditional mean zero and variance $4$, and the control residual $Y-\mu_0$ have conditional mean zero and variance $1$. Then $$ \phi_i=\frac{A_i}{0.5}(Y_i-\mu_1) -\frac{1-A_i}{0.5}(Y_i-\mu_0). $$

The score $s_i=A_i-0.5$ depends only on treatment assignment. Conditional on either treatment arm, the corresponding residual has mean zero. Hence $c=E(s_i\phi_i)=0$: each arm’s contribution to this expectation is its treatment-score value times a zero residual mean. The variance is $$ \operatorname{Var}(\phi_i) =0.5\frac{4}{0.5^2}+0.5\frac{1}{0.5^2} =8+2=10 $$ in squared outcome units. Estimating the propensity does not change this first-order variance; the covariance term being subtracted is zero.

Interpretation, Diagnostics, and Alternatives

The result is a comparison between two versions of the same AIPW procedure: one inserts the true propensity, and the other estimates it by a correctly specified score model. It is not a claim that estimating unknown quantities always improves an analysis, nor does it establish that an estimated-propensity AIPW estimator is more precise than every estimator that could use a known assignment probability. The projection adjustment uses the fitted score equation; it does not replace the assumptions needed to identify the ATE.

Useful checks include confirming that the propensity likelihood fit converged, that the fitted model is not separated or nearly unidentified, and that fitted probabilities do not create extreme residual weights. Inspect overlap and covariate balance as well: a score equation can be solved even when the model is substantively inadequate or the data provide little support in parts of the covariate space. A finite-sample variance estimate need not be smaller just because the asymptotic formula subtracts a nonnegative term.

With correct outcome predictions, the usual AIPW influence function has zero covariance with any propensity score that depends only on $A,X$, so fitting the propensity gives no first-order gain. When outcome predictions are imperfect, a nonzero covariance can make the adjustment helpful. If the propensity model is misspecified, however, the score identity derived above need not hold at the fitted parameter; the variance comparison is no longer guaranteed. A correct outcome regression may still support consistency, but that does not restore this particular variance result.

A fixed-propensity analysis is appropriate when the assignment probability is genuinely known, as in some randomized designs, but its variance should be justified for that design and estimator. Outcome-regression and weighting estimators make different uses of the data and may have different finite-sample behavior. For flexible nuisance fits, additional conditions and sample-splitting methods can matter; the parametric score expansion here should not be assumed to cover every data-adaptive fitting procedure.

Common Mistake

A frequent error is to change $e_\gamma$ inside the AIPW function while silently keeping the true data law fixed, then claim its expectation remains $\psi$. That is false in general: in the first worked example, the expectation is $0.5/e_\gamma$, not $1$ except at the true propensity. The constant-expectation argument changes the assignment law along with the propensity, and differentiating it produces both the fixed-law derivative and the score term. A second error is to reverse the residual signs: at the true propensity, their conditional expectations are $\mu_1-m_1$ for treatment and $m_0-\mu_0$ for control.

Key takeaway: Under a correctly specified, regular propensity model, the score equation changes the AIPW influence function by subtracting its projection onto the propensity score. The variance reduction is $c^T I^{-1}c$; it is zero when that score covariance is zero and is not a blanket guarantee for misspecified or unstable fits.

Check Your Understanding

Answer each question, then compare with the solutions below.

With a correct propensity model and fixed outcome predictions, what two terms appear when differentiating the constant-expectation identity?
If $c=0$, how does estimating the propensity affect the first-order AIPW variance?
In the fixed-law example with $e_0=0.5$, $\mu_1=1$, $m_1=m_0=0$, and zero control outcomes, what is the expectation of the AIPW function at candidate propensity $e=0.25$?
Why does a correct outcome regression imply zero covariance between the AIPW influence function and a propensity score depending on $A,X$?

Solutions. (1) The derivative of the estimating function under the fixed true law, plus the expectation of the centered estimating function multiplied by the propensity score. (2) The subtracted variance term $c^T I^{-1}c$ is zero, so there is no first-order change. (3) It is $0.5/0.25=2$, not the target $1$; the expectation is being taken under the fixed true law. (4) With $m_a=\mu_a$, the influence function consists of weighted outcome residuals plus no outcome-prediction error term. Each residual has conditional mean zero given $A,X$, while the propensity score is a function of $A,X$, so their expected product is zero.