We need to prove that under unconfoundedness and positivity,
\(EY(t) = E\frac{\mathbb{1}(T=t)Y}{P(t|W)}.\)
\[ \begin{split} EY(t) &= E(Y|do(t))\\ &= EE(Y|do(t), W)\\ &= EE(Y|t, W)\\ &= \Sigma_w E(Y|t, w) P(w)\\ &= \Sigma_w \Sigma_y y P(y|t,w) P(w)\\ &= \Sigma_w \Sigma_y y \frac{P(y,t,w)}{P(t,w)} P(w)\\ &= \Sigma_w \Sigma_y y P(y,t,w)\frac{1}{P(t|w)}\\ &= \Sigma_w \Sigma_y y E(\mathbb{1}(Y=y,T=t,W=w))\frac{1}{P(t|w)}\\ &= \Sigma_w E(\Sigma_y y \mathbb{1}(Y=y) \mathbb{1}(T=t,W=w))\frac{1}{P(t|w)}\\ &= \Sigma_w E(Y \mathbb{1}(T=t,W=w))\frac{1}{P(t|w)}\\ &= E(Y \mathbb{1}(T=t) \Sigma_w \mathbb{1}(W=w)\frac{1}{P(t|w)})\\ &= E(Y \mathbb{1}(T=t) \frac{1}{P(t|W)}) \end{split} \]
Note that we used the following properties of the indicator function:
\[ E\mathbb{1}(X=x) = P(X=x) \]
\[ \mathbb{1}(X=x, Y=y) = \mathbb{1}(X=x) \mathbb{1}(Y=y) \]
\[ \Sigma_x \mathbb{1}(X=x)f(x) = f(X) \]
Citation
@online{zhang2023,
author = {Jia Zhang},
title = {Detailed {Proof} of {IPW} {Identification}},
date = {2023-01-05},
url = {https://jiazhang42.github.io/mysite/blog/ipw/},
langid = {en}
}