When I was solving $ \operatorname{Cov}(X,E(X\mid Y)) = \operatorname{var}(E(X\mid Y))$, I notice that $E(X\mid Y)$ was treated as a function of $Y$. My thinking is $E(X\mid Y)$ is taking values of $ \operatorname{Range}(Y) $ and for each value of $Y$, it maps to the expectation of $X$. Is this correct?


2 Answers 2


An easy way to clarify : if $f(y)=\mathbb E[X\mid Y=y]$, then $\mathbb E[X\mid Y]=f(Y)$.


The answer is already given. Here is an example.

Let a discrete r.v. $X$ have values $x\in \{ -6,-3,7,14\}$, and $P(X=x)=(0.1,0.2,0.3,0.4)$ respectively. Define a r.v. $Y$ s.t. $Y=0$ if $X \le0$, and $Y=1$ if $X>0$.

Then $E(X|Y)$ assumes different values for different values of $Y$ as follows.

$Y=0:$ $$E(X|Y=0)=\sum_{x\in\{-6,-3\}} xP(X=x|Y=0)=-6\times1/3-3\times2/3=-4$$

$Y=1$: $$E(X|Y=1)=\sum_{x\in\{7,14\}} xP(X=x|Y=1)=7\times3/7+14\times4/7=11$$

Obviously, $E(X|Y)$ is a function of $Y$. Also, note how the conditional probabilities, $P(X=x|X=y)$, change from the original marginal probabilities, $P(X)$, for different $y$.


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