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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 (general) law of iterated expectations?

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Without further assumption it is not correct. From the definition of a conditional expectation and the properties of a density we have (for $Y$ continuous):

$$ \mathbb{E}(Y_i|T_i=1, D_i=1) - \mathbb{E}(Y_i|T_i=0, D_i=1) \\ = \int y f_Y(y|T_i=1, D_i=1) dy - \int y f_Y(y|T_i=0, D_i=1)dy \\ = \int y \int f_{Y|M}(y|T_i=1, M_i=m, D_i=1)f_M(m|T_i=1, D_i=1)dm dy \\ - \int y \int f_{Y|M}(y|T_i=0, M_i=m, D_i=1)f_M(m|T_i=0, D_i=1)dm dy $$

The density of $M$ is conditional to $T_i=1$ in the first case but conditional to $T_i=0$ in the last line, and cannot be factorized. However, if $M_i|D_i$ and $T_i$ are statistically independent, then it works, because $f_M(m|T_i=1, D_i=1)=f_M(m|T_i=0, D_i=1)$ in that case.

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