If $X \sim N(0,1)$ and $Y \sim N(0,1)$ are two random variables that may or may not be independent, what is $E(XY)$? If $X \sim N(0,1)$ and $Y \sim N(0,1)$ are two random variables that may or may not be independent, I am wondering what $\operatorname{E}(XY)$ might be. It appears they have the same distribution, but does that imply anything about independence?
 A: Since $\mathbb{E}[X]=\mathbb{E}[Y]=0$, $\mathbb{E}[XY]$ is the covariance of $X$ and $Y$, which in this case could be any real number in $[-1,1]$. You can't say anything more based on the information you've given.
A: One possibility is that $X$ and $Y$ are both the same random variable, so always equal to each other. In that case $\operatorname{E}(XY) = \operatorname{E}(X^2) = 1.$
If $X\sim N(0,1),$ then $-X\sim N(0,1),$ so one could have $Y$ being $-X$, in which case $\operatorname{E}(XY) = \operatorname{E}(-X^2) = -1.$
Another possibility is that $X$ and $Y$ are independent, so $\operatorname{E}(XY) = \operatorname{E}(X)\operatorname{E}(Y) = 0.$
Another is that after observing $X,$ you toss a coin and then decide $Y = X$ or $Y=-X$ according to the outcome. In that case it can be shown that $\operatorname{E}(XY) = 0$ even though $X$ and $Y$ are far from independent.
Another is that $Y=(X+Z)/\sqrt 2$ where $X,Z$ are independent random variables distributed as $N(0,1).$  In that case $Y\sim N(0,1)$ and $\operatorname{E}(XY) = 1/\sqrt 2.$
