Originally Posted by

**NoFace** First the question:

Let $\displaystyle X$ and $\displaystyle Y$ be two independent variables. If $\displaystyle X$ and $\displaystyle Y$ are both standard normal, then what is the distribution of $\displaystyle \frac{1}{2} (X^2-Y^2)$?

So I've tried to use the fact that X and Y are standard normal to tell me that $\displaystyle \mu_x=\mu_y=0$ and $\displaystyle {\sigma}_{x}=\sigma_y=1$.

So,

$\displaystyle {\sigma}^2=E[X^2]-E[X]^2$

$\displaystyle 1=E[X^2]-0^2 \implies E[X^2]=1$

I found that $\displaystyle E[\frac{1}{2} (X^2-Y^2)]= \frac{1}{2} (E[X^2] -E[Y^2]) = \frac{1}{2}(1^2-1^2) = 0$

I'm not really sure what to do with this.

Maybe I am wrongly assuming something.

Thanks in advance!