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 $X$ and $Y$ have the same distribution, they have the same density functions $f_X$ and $f_Y$. So $$\int_{-\infty}^{\infty} t^2 f_X(t)\ dt=\int_{-\infty}^{\infty} t^2 f_Y(t)\ dt$$
(If your probability measures are't absolutely continuous with respect to Lebesgue measure, just replace $f_X(t)\ dt$ with the appropriate measure.)
