Is it true that $E[X^2]-E[Y^2] = 0?$ Suppose that $X$ and $Y$ are identically distributed and not necessarily independent. Then clearly $E[X]=E[Y]$.
But is it also the case that $E[X^2]=E[Y^2]$?
Can you think of a counter-example? If not, can you give a quick proof?
Thanks

If $E[X^2]\neq E[Y^2]$, then $X^2$ and $Y^2$ are not identically distributed, which means that $X$ and $Y$ are not identically distributed. Or do I miss something?
