Expectations of Independent/Dependent Random Variables

Question

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?

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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?

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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?

Answer 1

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 ($P(X=x,Y=y) = P(X=x) P(Y=y) >0$).