Proving the balancing score property of propensity score

$\newcommand{\P}{\mathbb{P}}$$\newcommand{\E}{\mathbb{E}}In proving that X \bot A | ps(x) where X denote the baseline covariates, A is binary denoted the Treatment/Exposure and ps(x) denotes the propensity score based on X. I do not understand a part of the proof. The first part was to prove: $$\P(A = 1 \vert ps(x), x) = ps(x), \tag{1}$$ which I understood. The second part was to prove: $$\P(A = 1 \vert ps(x)) = ps(x), \tag{2}$$ then from the second equation, to prove (1) = (2) and that they are thus, conditionally independent. I do not know how:$$\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) ] .$$I understand there was something to do with Law of Iterated Expectations, but do not see how they equate. 3 Answers The proof works by showing that conditioning on the propensity score is "sufficient" when you are conditioning on X=x already and you want an expression for the propensity score. For the first part of the proof you refer to, you can immediately see that since ps(x) is a function of x, ps(x) is redundant. For the second part we just have to show by conditioning on x that we then condition "too much", and that we just get the propensity score. First, note that the propensity score is defined as$$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)],$which is exactly the expression you were looking for. To show--rather elaborately--that this is equal to the propensity score: \begin{align} & \mathbb{E}[\mathbb{E}(A\mid ps(x),x)\mid ps(x)]\\ &=\mathbb{E}[\mathbb{E}(A\mid x)\mid ps(x)]\\ &=\mathbb{E}[\Pr(A=1\mid x)\mid ps(x)]\\ &=\mathbb{E}[ps(x)\mid ps(x)]\\ &=ps(x). \end{align} • I don't get the critical part: "condition the expression byps(x)$" : if you replace$A$by$A|ps(x)$I don't see where the final$|ps(x)$comes from ? Commented Nov 6, 2017 at 16:29 • ok, got it, see e.g. stats.stackexchange.com/questions/307967/… Commented Nov 6, 2017 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 $$ps(x)$$, sounds not self evident. I am very confused. •$P(Q|f(x), x) = P(Q|x)$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)\$.
– Noah
Commented Aug 12, 2019 at 5:50