Let $X, Y$ be two random variables, defined on some probability space $(\Omega , A , P)$., each only has two district values: $X \to \{x_1,x_2\}$ , $Y \to \{y_1,y_2\}$.
Recall that, in this case , $X$ and $Y$ are independent if, of any $i , j (i , j = 1,2)$:
P(X= $x_i$ , Y = $y_i$) = P(X = $x_i$)P(Y =$y_i$)
Show that, in this vase, X and Y are independent if and only if
E(XY) = E(X)E(Y)
My proving Part( if ) Show that
Suppose X and Y are independent. Then for any (x,y) $\in$ $\mathbb{R}$
p(x,y) = P(X=x , Y=y) = P(X=x) P(Y=y) = p$_x$(x) p$_y$(y)
When p(x,y) = p$_x$(x) p$_y$(y) for all ($x,y)$ $in$ Then for any A $\subset$ $\mathbb{R}$ and B$\subset$ $\mathbb{R}$ So that
P(X $\in$A , Y $\in$ B) $=\sum_{X \in A} \sum_{y \in B} $ P(X=x , Y=y)
$=\sum_{X \in A} \sum_{y \in B} $ p$_x$(x) p$_y$(y)
$=\sum_{X \in A} $.p$_x$(x) $\sum_{y \in B}$ p$_y$(y)
=P(X $\in$A , Y $\in$ B)
Hence X and Y are independent
**** My proving True or False in part "if"**
** If can't proving in part "only if" . Please help me to proving that.
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Here is a sketch of a proof. Normalize the random variables by letting $\tilde{X} = \frac{X - x_1}{x_2-x_1}$ and $\tilde{Y} = \frac{Y - y_1}{y_2 - y_1}$. Show that $X$ and $Y$ are independent if and only if $\tilde{X}$ and $\tilde{Y}$ are, and use linearity of expectation to show $$ \E[XY] = \E[X]\E[Y] \iff \E[\tilde{X}\tilde{Y}] = \E[\tilde{X}]\E[\tilde{Y}] $$
So it is sufficient to prove the claim for two 0-1 random variables. If $X$ and $Y$ are 0-1 random variables, $$ \E[XY] = \E[X]\E[Y] \iff \Pr(X=1, Y=1)=\Pr(X=1)\Pr(X=1) $$
It is easy to confirm that the latter equality is equivalent to independence of $X$ and $Y$.
Note: Since all the implications in this proof are bidirectional, this proves both the "if" and "only if" parts of the question.
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