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Suppose $Y \in \mathbb{R}^n$ and $X \in \mathbb{R}^n$ are random vectors. By the law of iterated expectation, does the following hold?

$$E[g(X,Y)] = E[E[g(X,Y)| \{Y: AY \geq b\}]],$$

where $g(\cdot, \cdot)$ is a function, and $A \in \mathbb{R}^{n \times n}$ and $b \in \mathbb{R}^n$ are a matrix and vector of constants, respectively.

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No, it is not correct in general because the event $AY\geq b$ is not a partition of the whole space. Let's say the $Y$ vectors satisfying this relationship constitute the set $\mathcal Y$. Then, the correct relation is $$E[g(X,Y)]=E[g(X,Y)|Y\in \mathcal Y]P(Y\in \mathcal Y)+E[g(X,Y)|Y\notin \mathcal Y]P(Y\notin \mathcal Y)$$

A simple counter-example: let $g(X,Y)=Y$, and let $Y$ be a Bernoulli RV (with parameter $p$), so $n=1$. Furthermore, let $A=b=1$. Then, we have $$E[E[g(X,Y)|\{Y:Y\geq 1\}]]=E[E[g(X,Y)]|Y=1]=E[1]=1$$

However, the correct answer is $p$, i.e. $E[g(X,Y)]=E[Y]=p$

P.S. I'm assuming $\geq$ operation is defined properly between $n\times1$ vectors.

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  • $\begingroup$ Just to clarify, you're defining $\mathcal{Y} = \{Y: AY \geq b\}$, correct? $\endgroup$
    – Adrian
    Jan 24, 2021 at 20:25
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    $\begingroup$ Correct, the set of $Y$ satisfying $AY\geq b$. $\endgroup$
    – gunes
    Jan 24, 2021 at 20:26
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    $\begingroup$ Well, it's a condition on $Y$, and will most probably decrease the amount of possibilities for it. $\endgroup$
    – gunes
    Jan 24, 2021 at 20:30
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    $\begingroup$ Yes, my example was discrete just because of convenience. $\endgroup$
    – gunes
    Jan 24, 2021 at 20:57
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    $\begingroup$ No, it's scalar because condition is like $Y\geq 1$ $\endgroup$
    – gunes
    Jan 25, 2021 at 15:49

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