If $X,Y$ are independent standard normal variables, then $X^2, Y^2$ are independent chi-squared r.v with 1 degree of freedom, i.e. $X^2 \in \chi^2_1, Y^2 \in \chi^2_1$.
I am trying to find the mean and variance of $\frac{X^2}{Y^2}$, i.e. $E[\frac{X^2}{Y^2}], Var[[\frac{X^2}{Y^2}]$
After a quick research on the Internet, I found that $\frac{X^2}{Y^2}$ is F-distribution with 1 degree of freedom, i.e. $\frac{X^2}{Y^2} \sim F(1,1)$. But I couldn't find the mean and variance for it. What I found is that for $F(d_1,d_2)$ $$E[\frac{X^2}{Y^2}]= \frac{d_2}{d_2-2}, \text{for } d_2>2$$ $$Var[\frac{X^2}{Y^2}] = \frac{2d_2^2(d_1+d_2-2)}{d_1(d_2-2)(d_2-4)}, \text{for }d_2>4 $$ But here $$d_1=d_2=1$$
Thanks all in advance.