Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
It seems like you're showing that $X$ and $Y$ being independent implies that $X$ and $Y$ is independent. – Stefan Hansen Jul 13 '13 at 7:51


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.

share|cite|improve this answer
An exercise for the original poster: Show that (by an example!) this conclusion is <b>incorrect</b> if the two random variables each can take on three or more values. – kjetil b halvorsen Jul 13 '13 at 13:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.