$\newcommand{\P}{\mathbb{P}}$$\newcommand{\E}{\mathbb{E}}$In proving that $X \bot A | ps(x)$ where $X$ denote the baseline covariates, $A$ is binary denoted the Treatment/Exposure and $ps(x)$ denotes the propensity score based on $X$.
I do not understand a part of the proof.
The first part was to prove:
\begin{equation}
\P(A = 1 \vert ps(x), x) = ps(x), \tag{1}
\end{equation}
which I understood.
The second part was to prove:
\begin{equation}
\P(A = 1 \vert ps(x)) = ps(x), \tag{2}
\end{equation}
then from the second equation, to prove (1) = (2) and that they are thus, conditionally independent.
I do not know how: $$\P(A = 1 \vert ps(x) ) = \E(A \vert ps(x)) = \E[\E(A \vert ps(x), x) \vert ps(x) ] = ps(x). $$ In particular, I do not understand how: $$\E(A \vert ps(x)) = \E[\E(A \vert ps(x), x) \vert ps(x) ] .$$ I understand there was something to do with Law of Iterated Expectations, but do not see how they equate.
Best Answer
The proof works by showing that conditioning on the propensity score is "sufficient" when you are conditioning on $X=x$ already and you want an expression for the propensity score. For the first part of the proof you refer to, you can immediately see that since $ps(x)$ is a function of $x$, $ps(x)$ is redundant. For the second part we just have to show by conditioning on $x$ that we then condition "too much", and that we just get the propensity score.
First, note that the propensity score is defined as
$$ps(x) = \Pr(A=1\mid x).$$
For ease of notation, you can simply see that
$$\mathbb{E}(A)=\mathbb{E}[\mathbb{E}(A\mid x)],$$
by--as you mention--the law of iterated expectations.
We can make this expression conditional on the propensity score $ps(x)$:
$$\mathbb{E}(A\mid p(x))=\mathbb{E}[\mathbb{E}(A\mid ps(x),x)\mid ps(x)],$$
which is exactly the expression you were looking for. To show--rather elaborately--that this is equal to the propensity score:
\begin{align} & \mathbb{E}[\mathbb{E}(A\mid ps(x),x)\mid ps(x)]\\ &=\mathbb{E}[\mathbb{E}(A\mid x)\mid ps(x)]\\ &=\mathbb{E}[\Pr(A=1\mid x)\mid ps(x)]\\ &=\mathbb{E}[ps(x)\mid ps(x)]\\ &=ps(x). \end{align}