[proofplan] We first pass from strong ignorability given the full covariate $X$ to strong ignorability given the scalar propensity score $e(X)$ using the [Rosenbaum-Rubin propensity score theorem](/theorems/9660). Positivity ensures that the treated and control conditional means given the propensity score are defined on the relevant strata. For each treatment value $t$, conditional ignorability removes the conditioning on $T=t$ from the potential-outcome mean, and consistency replaces the potential outcome by the observed outcome on the event $\{T=t\}$. Averaging the resulting conditional contrast over the distribution of $e(X)$ gives the average treatment effect by the tower property. [/proofplan]

[step:Introduce the propensity-score sigma-algebra and record positivity on its strata] For any [random variable](/page/Random%20Variable) or finite tuple of random variables, write $\sigma(\cdot)$ for the sub-$\sigma$-algebra of $\mathcal F$ that it generates. Define the propensity-score random variable \begin{align*} S:(\Omega,\mathcal F)\to([0,1],\mathcal B([0,1])) \end{align*} by $S=e(X)$. Let $\mathcal G:=\sigma(S)$ be the sub-$\sigma$-algebra generated by $S$. Since $e(X)=\mathbb P(T=1\mid X)$ $\mathbb P$-a.s. and $0<\mathbb P(T=1\mid X)<1$ $\mathbb P$-a.s., we have \begin{align*} 0<S<1 \quad \mathbb P\text{-a.s.} \end{align*} Moreover, by the tower property for [conditional expectation](/page/Conditional%20Expectation) applied first with respect to $\sigma(X)$ and then with respect to $\mathcal G\subseteq\sigma(X)$, \begin{align*} \mathbb P(T=1\mid \mathcal G)=\mathbb E[\mathbb 1_{\{T=1\}}\mid \mathcal G]=\mathbb E[\mathbb E[\mathbb 1_{\{T=1\}}\mid X]\mid \mathcal G]=\mathbb E[S\mid \mathcal G]=S \end{align*} $\mathbb P$-a.s. Hence \begin{align*} \mathbb P(T=0\mid \mathcal G)=1-S \end{align*} $\mathbb P$-a.s. Therefore $\mathbb P(T=t\mid \mathcal G)>0$ $\mathbb P$-a.s. for both $t\in\{0,1\}$. [/step]

[step:Use Rosenbaum-Rubin to obtain conditional ignorability given $e(X)$]By the [Rosenbaum-Rubin Propensity Score Theorem][citetheorem:9660], strong ignorability of $T$ given $X$ and the definition of the propensity score imply \begin{align*} (Y(1),Y(0))\perp\!\!\!\perp T\mid \mathcal G. \end{align*} In particular, for each $t\in\{0,1\}$, the integrable random variable $Y(t)$ is conditionally independent of $T$ given $\mathcal G$. Since $\mathbb P(T=t\mid\mathcal G)>0$ $\mathbb P$-a.s. by the preceding step, the conditional mean on the treatment stratum is well defined and satisfies \begin{align*} \mathbb E[Y(t)\mid T=t,S]=\mathbb E[Y(t)\mid S] \end{align*} $\mathbb P$-a.s.[/step]

[guided]The purpose of the propensity score is to replace adjustment for the full covariate $X$ by adjustment for the one-dimensional score $S=e(X)$. The result that justifies this replacement is the [Rosenbaum-Rubin Propensity Score Theorem][citetheorem:9660]. Its hypotheses match the present setting: $T$ is binary, $e(X)$ is a propensity score because $e(X)=\mathbb P(T=1\mid X)$ $\mathbb P$-a.s., and strong ignorability given $X$ is assumed. Therefore the theorem gives \begin{align*} (Y(1),Y(0))\perp\!\!\!\perp T\mid \sigma(S). \end{align*} We also record positivity inside this guided argument. Since $e(X)=\mathbb P(T=1\mid X)$ $\mathbb P$-a.s. and $S=e(X)$, the assumed positivity gives \begin{align*} 0<S<1 \quad \mathbb P\text{-a.s.} \end{align*} Moreover, because $S$ is $\sigma(S)$-measurable and $\sigma(S)\subseteq\sigma(X)$, the tower property gives \begin{align*} \mathbb P(T=1\mid \sigma(S))=\mathbb E[\mathbb 1_{\{T=1\}}\mid \sigma(S)]=\mathbb E[\mathbb E[\mathbb 1_{\{T=1\}}\mid X]\mid \sigma(S)]=\mathbb E[S\mid \sigma(S)]=S \end{align*} $\mathbb P$-a.s. Therefore $\mathbb P(T=0\mid\sigma(S))=1-S$ $\mathbb P$-a.s., and $\mathbb P(T=t\mid\sigma(S))>0$ $\mathbb P$-a.s. for both $t\in\{0,1\}$. Fix $t\in\{0,1\}$. Since $Y(t)$ is one coordinate of the tuple $(Y(1),Y(0))$, conditional independence of the tuple implies \begin{align*} Y(t)\perp\!\!\!\perp T\mid \sigma(S). \end{align*} This conditional independence says that, after conditioning on $S$, learning whether $T=t$ does not change the conditional law of $Y(t)$. Because $Y(t)\in L^1(\Omega,\mathcal F,\mathbb P)$, conditional expectations of $Y(t)$ exist. Because we have just proved $\mathbb P(T=t\mid\sigma(S))>0$ $\mathbb P$-a.s., conditioning further on the event $\{T=t\}$ is legitimate on almost every propensity-score stratum. Hence \begin{align*} \mathbb E[Y(t)\mid T=t,S]=\mathbb E[Y(t)\mid S] \end{align*} $\mathbb P$-a.s. This is the exact point where ignorability is used: it converts an unobservable conditional mean of the potential outcome among subjects assigned treatment $t$ into the unconditional propensity-score-specific potential-outcome mean.[/guided]

[step:Use consistency to replace potential outcomes by observed outcomes within treatment strata] Fix $t\in\{0,1\}$. Consistency gives \begin{align*} Y=Y(t) \quad \mathbb P\text{-a.s. on } \{T=t\}. \end{align*} Let $H$ be a bounded $\sigma(T,S)$-measurable real-valued random variable. Since $\mathbb 1_{\{T=t\}}H$ is $\sigma(T,S)$-measurable and vanishes outside $\{T=t\}$, consistency implies \begin{align*} \mathbb E[\mathbb 1_{\{T=t\}}HY]=\mathbb E[\mathbb 1_{\{T=t\}}HY(t)]. \end{align*} This is precisely the testing identity for conditional expectations on the sub-$\sigma$-algebra $\sigma(T,S)$ restricted to the stratum $\{T=t\}$. Since both random variables are integrable, it gives \begin{align*} \mathbb E[Y\mid T=t,S]=\mathbb E[Y(t)\mid T=t,S] \end{align*} $\mathbb P$-a.s. on the treatment stratum, hence for the chosen versions of the stratum conditional means. Combining this identity with conditional ignorability from the preceding step yields \begin{align*} \mathbb E[Y\mid T=t,S]=\mathbb E[Y(t)\mid S] \end{align*} $\mathbb P$-a.s. for each $t\in\{0,1\}$. [/step]

[step:Average the propensity-score-specific contrast to obtain the ATE] Applying the previous step with $t=1$ and $t=0$ gives \begin{align*} \mathbb E[Y\mid T=1,S]-\mathbb E[Y\mid T=0,S]=\mathbb E[Y(1)\mid S]-\mathbb E[Y(0)\mid S] \end{align*} $\mathbb P$-a.s. By linearity of conditional expectation for integrable random variables, \begin{align*} \mathbb E[Y(1)\mid S]-\mathbb E[Y(0)\mid S]=\mathbb E[Y(1)-Y(0)\mid S] \end{align*} $\mathbb P$-a.s. Taking expectations and using the tower property gives \begin{align*} \mathbb E\big[\mathbb E[Y\mid T=1,S]-\mathbb E[Y\mid T=0,S]\big]=\mathbb E[\mathbb E[Y(1)-Y(0)\mid S]]=\mathbb E[Y(1)-Y(0)]. \end{align*} By the definition of $\operatorname{ATE}$, the right-hand side is $\operatorname{ATE}$. Since $S=e(X)$, this is precisely \begin{align*} \operatorname{ATE} = \mathbb E\big[\mathbb E[Y\mid T=1,e(X)]-\mathbb E[Y\mid T=0,e(X)]\big]. \end{align*} [/step]