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Consider a randomized experiment (AB test), where $n$ units are randomized into the treatment group $T_i=1$ and control group $T_i=0$. Let $M_i\in P$ denote the observed value of a continuous variable that is realized after the exposure to the treatment where $P$ is the support of $M_i$. $D_i$ is a binary variable. $F$ represents the distribution function. Can we re-write the expression:

$x=\int \{\mathbb{E}(Y_i|T_i=1, M_i=m, D_i=1) - \mathbb{E}(Y_i|T_i=0, M_i=m, D_i=1)\}\mathrm{d} F_{M_i|D_i=1}(m),$

into

$ x = \mathbb{E}(Y_i|T_i=1, D_i=1) - \mathbb{E}(Y_i|T_i=0, D_i=1)$

by using the law of iterated expectations?

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I believe, you can; we can think over $E[Y_i|T_i=1,D_i=1]$ first. By the law of iterated expectations, we have $$E[Y_i|T_i=1,D_i=1] = E[ E[Y_i|T_i=1,D_i=1,M_i] ] \\ =\int{E[Y_i|T_i=1,D_i=1,M_i=m]\ \ f_{M_i|D_i=1}(m)dm}$$ assuming independence of $M_i$ and $T_i$ given $D_i$. Finally, we have $f_{M_i|D_i}(m)dm=dF_{M_i|D_i=1}(m)$. So, for the first summand, you have exactly the same expression and for the second summand this follows exactly the same way.

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  • $\begingroup$ how can "independence of Mi and Ti given Di" be reasonably assumed if OPs says "Let Mi∈P denote the observed value of a continuous variable that is realized after the exposure to the treatment"? $\endgroup$ – Krantz Dec 20 '18 at 15:20

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