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$\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.

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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}

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  • $\begingroup$ I don't get the critical part: "condition the expression by $ps(x)$" : if you replace $A$ by $A|ps(x)$ I don't see where the final $|ps(x)$ comes from ? $\endgroup$ – oDDsKooL Nov 6 '17 at 16:29
  • $\begingroup$ ok, got it, see e.g. stats.stackexchange.com/questions/307967/… $\endgroup$ – oDDsKooL Nov 6 '17 at 16:35
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Could anybody give a clearer proof of P(A=1|ps(x),x)=ps(x)?

Verbally, P(A=1|ps(x),x)=ps(x) is the probability of A=1 given x such that it's propensity score is ps(x)

Then the argument that P(A=1) given x alone is equal to P(A=1) given a particular x that satisfies a particular value of ps, sounds not self evident.

I am very confused.

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  • $\begingroup$ $P(Q|f(x), x) = P(Q|x)$ for any event $Q$ and any function $f(.)$. $f(x)$ doesn't provide more informstion than $x$ does because it depends solely on $x$. $P(A=1|x)$ is the propensity score by definition. $P(A=1|ps(x), x) = P(A=1|x)= ps(x)$. $\endgroup$ – Noah Aug 12 '19 at 5:50
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    $\begingroup$ Also, answers are not to be used for asking new questions. You should have made your own question. $\endgroup$ – Noah Aug 12 '19 at 5:51
  • $\begingroup$ @Noah, thank you for kind answering. You pointed out that f(x) cannot have more than one value. It makes sense now. BTW, I would have added my post as a comment if I had known that I could use math expression in the comment. :( $\endgroup$ – Royalblue Aug 12 '19 at 5:58

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