When is E(A*B) = E(A * E(B))? I'm looking at the documentation for the econml python package. On this page it is stated that:
\begin{split}E[Y_{i, t}^{IPS} | X, W] =~& E\left[\frac{Y_i 1\{T_i=t\}}{p_t(X_i, W_i)} | X_i, W_i\right] = E\left[\frac{Y_i^{(t)} 1\{T_i=t\}}{p_t(X_i, W_i)} | X_i, W_i\right]\\
=~&  E\left[\frac{Y_i^{(t)} E[1\{T_i=t\} | X_i, W_i]}{p_t(X_i, W_i)} | X_i, W_i\right] = E\left[Y_i^{(t)} | X_i, W_i\right]\end{split}
where $Y_i$ is some dependent variable, $T_i$ is a treatment variable, $X_i, W_i$ are covariates, and $p_t(X_i, W_i)$ is the probability that $T = t$.
I don't understand the third equality (from top line at the right to bottom line at the left). It looks like something of the form:
$$E[A*B | X] = E[A * E[B | X] | X]$$
where $A = \frac{Y_i}{p_t(X_i, W_i)}$ and $B=1\{T_i=t\}$. But unless I'm mistaken this equality does not hold in general.
Can someone explain to me why this equality holds in this case? Thanks!
 A: The issue here is not simply one of expectations, but of conditional expectation given a random variable.
I learnt this stuff a few years ago and don't remember it too well, but I'll give it a shot.
The basic intuition is when we say $E[A | X]$ we are saying what is the distribution of $A$ if we know $X$. It then is a function of $X$.
For example if $A = X + U$, $U \sim N(0,1)$ then $$E[A | X] = X + E[U|X] = X + E[U] = X$$ as $U$ is independent of $X$ and has mean 0. So the expectation of $A$ given $X$ is just $X$.
We have a few basic properties we can use:
\begin{align}
(A) && E[A|X] = E\left[E\left[A|X\right]|X\right]\\
(B) &&  E\left[B * E\left[A|X\right]|X\right] = E[ B | X] E[A|X]\\
\text{for $A$ and $B$ independent, } \; (C) && E[A * B |X] = E[A|X] E[B|X]
\end{align}
To understand these things more deeply you need to learn about sigma algebras, conditional expectation etc. See Capinski and Kopp, Measure, integral and probability, 2004 for a textbook on that. Unfortunately I can't remember it well enough to give a good intuitive explanation for why (A), (B) and (C) hold if they're not clear to you.
We can then apply these to get the result
\begin{split}
E\left[\frac{Y_i^{\left(t\right)} 1\{T_i = t\}}{p_t\left(X_i , W_i \right)} | X_i , W_i \right] & = E\left[ E\left[\frac{ Y_i^{\left(t\right)} 1\{T_i = t\}}{p_t\left(X_i , W_i \right)} | X_i , W_i \right] | X_i , W_i \right] (1)\\
& = E\left[\frac{ E\left[Y_i^{\left(t\right)} 1\{T_i = t\} | X_i , W_i \right]}{p_t\left(X_i , W_i \right)} | X_i , W_i \right] (2) \\
& = E\left[\frac{ E\left[Y_i^{\left(t\right)}| X_i , W_i \right] E\left[ 1\{T_i = t\}| X_i , W_i \right] }{p_t\left(X_i , W_i \right)} | X_i , W_i \right] (3) \\
& = E\left[\frac{ E\left[Y_i^{\left(t\right)}| X_i , W_i \right] p_t\left(X_i , W_i \right) }{p_t\left(X_i , W_i \right)} | X_i , W_i \right] (4) \\
& = E\left[E\left[Y_i^{\left(t\right)}| X_i , W_i \right] | X_i , W_i \right] (5) \\
& = E\left[Y_i^{\left(t\right)}| X_i , W_i \right] (6)
\end{split}
(1) holds by (A)
(2) holds because $p_t\left(X_i , W_i \right) = E\left[ 1\{T_i = t\}| X_i , W_i \right]$, so we can apply (B)
(3) holds by (C)
(4) holds by definition in the link you said
(5) is just cancelling numerator and denominator
(6) holds by (A)
