Uncorrelatedness and conditional expectation

Question

Two random variables $X$ and $Y$. How are the following two statements related:

• $E(XY) = E(X)E(Y)$, ($X$ and $Y$ are called uncorrelated)
• $E(X\mid Y)= E(X)$ a.s., (what is this case called?)

Does one imply the other, and/or are there counterexample to such implications, or are there some condition that can make one imply the other? Thanks!

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From a deleted reply, there is an interesting statement

• $E(X\mid Y) E(Y) = E(XY)$ a.s.

I don't quite remember it correctly. Can anyone who can see it (with 10k reputation) verify that? I wonder when it is true? Any implication with the previous two statements?

Answer 1

Since $cov(X,Y)=E(XY)-E(X)E(Y),$ then for $E(XY)=E(X)E(Y)$to be true, $X,Y$ are uncorrelated.

Answer 2

With the help of Did's comment "from the characterization of the conditional expectation"

$$ E[Y E(X|Y)] = E[E(XY|Y)] = E(XY) $$ So the second implies the first.

A deleted comment said that the tower property of conditional expectation can help, but I don't know how.

How about the reverse direction?

Answer 3

1. Following the answer of @Tim, let $E(X|Y)=E(X)$. By the tower rule we have $$E(XY)=E(E(XY|Y)).$$

   But $E(E(XY|Y))=E(YE(X|Y))=E(YE(X))=E(X)E(Y)$ and so

   $$E(XY)=E(Y)E(X).$$

   As a result, we have that $X,Y$ are uncorrelated when $E(X|Y)=E(X)$.

2. Following these notes by Prof. Amir Dembo, we can show that the reverse is not true in general: let $\Omega=\{-1,0,1\}$ with $P(\{\omega\})=\frac{1}{3}$ for each $\omega \in \Omega$. Let also $X(\omega)=I_{\{0\}}(\omega)$ and $Y(\omega)=\omega$.

   Then $XY=0$ and so $E(XY)=0$. In addition, $E(Y)=0$ and so $E(XY)=0=E(X)E(Y)$, meaning that $X$ and $Y$ are uncorrelated. However, $E(X|Y)=X$ which is never equal to $E(X)=\frac{1}{3}$.

Answer 4

$\newcommand{\e}{\operatorname{E}}$In some cases $\e(X\mid Y) \ne \e(X)$ but $\e(XY) = \e(X)\e(Y).$ The simplest example I know is this: $$ Y = \begin{cases} +1 \\ \phantom{+}0 & \text{each with probability } 1/3, \\ -1 \end{cases} \quad \text{and } X= Y^2. $$ In that case we have $$ \e(X\mid Y) = \begin{cases} 1 & \text{if } Y=1 \text{ or } Y=-1, \\ 0 & \text{if } Y=0. \end{cases} $$ But $$ \e(XY) = 0 = \tfrac 2 3 \cdot 0 = \e(X)\e(Y). $$