Is it true that $E[X^2]-E[Y^2] = 0?$

Question

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

Answer 1

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.)

Answer 2

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?