$\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.


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

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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