# Expectations of Independent/Dependent Random Variables

A die is rolled two times

Let $X =$ sum of the two numbers that turn up

and $Y =$ difference of the numbers

Now, I need to show that $E(XY) = E(X)\cdot E(Y)$ and are the events independent?

This is my attempt:

let $$X = Z_1 + Z_2$$ $$Y = Z_1 - Z_2$$

$E[Z_1]$ = $1(\frac{1}{6})+2(\frac{1}{6})+3(\frac{1}{6})+4(\frac{1}{6})+5(\frac{1}{6})+6(\frac{1}{6}) = 3.5$

$E[Z_1] = E[Z_2]$ $$E[X] = 3.5 + 3.5 = 7$$ $$E[Y] = 3.5 - 3.5 = 0$$ $$E[X]E[Y] = 0$$ $$XY = (Z_1+Z_2)(Z_1-Z_2) = Z_1^2 - Z_2^2$$ $$E[XY] = (3.5)^2 - (3.5)^2 = 0$$ $$E[XY] = E[X]E[Y] = 0$$

Now, from this result, how do I determine if the events are independent/dependent? Any idea how to approach this? Also in general, does it mean anything if the expected value is 0? It's true to say that if $E[XY] \neq E[X]E[Y]$ then the events are not independent, thus dependent?

I suspect that X and Y cannot be independent. Because, if I know if X = 2 or 12, I know that Y must be 0. Hence, I think I must use conditional probability here?

• Where did you show that E[XY] = E[X]E[Y] or that E[XY]=0 already?
– Did
Feb 18, 2016 at 21:50
• @Did just edited it Feb 18, 2016 at 21:54

You calculated $E[X]$ and $E[Y]$, but you didn't show the calculation of $E[XY] = 0$.
It's true that if $E[XY] \ne E[X] E[Y]$ the events are dependent, but in this case you do have $E[XY] = E[X] E[Y]$, so that's not going to help you.
Your suspicion is correct: if $X$ and $Y$ are independent and values $X=x$ and $Y=y$ are individually possible (i.e. $P(X=x) > 0$ and $P(Y=y) > 0$), then $(X=x,Y=y)$ must be possible ($P(X=x,Y=y) = P(X=x) P(Y=y) >0$).
• It's not true that $E[Z_1^2] = 3.5^2$, though of course it's true that $E[Z_1^2] = E[Z_2^2]$ (so you don't really need to compute the correct value of $E[Z_1^2]$). Feb 18, 2016 at 22:19