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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.

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    $\begingroup$ $\frac {X^{2}} {Y^{2}}$ has infinite mean and (the variance is not defined). $\endgroup$ Mar 12, 2022 at 0:05
  • $\begingroup$ Thanks a lot! Would you mind pointing me to some references? $\endgroup$
    – Gavin
    Mar 12, 2022 at 0:06
  • $\begingroup$ $E\frac {X^{2}} {Y^{2}}=(EX^{2})( E\frac 1 {Y^{2}})$ and it is easy to see that $E\frac 1 {Y^{2}}=\infty$. $\endgroup$ Mar 12, 2022 at 0:09
  • $\begingroup$ It may be obvious, but why $E\frac{1}{Y^2} = \infty$? Thank you. $\endgroup$
    – Gavin
    Mar 12, 2022 at 1:11
  • $\begingroup$ $\int \phi (y)/y^{2}dy=\infty$. Just look at the integral in a neighborhood of $0$. ($\phi$ is the standard normal density). $\endgroup$ Mar 12, 2022 at 4:47

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