I have the following question at hand:
If $X_1,\cdots,X_n$ is a random sample from $N(0,\sigma^2)$. What is the distribution of the statistic $${\bar X^2\over {1\over n}X_i^2}$$
As far as I found out, ${\bar X^2\over \sigma^2/n}$ and ${X_i^2\over \sigma^2}$ have $\chi^2(1)$ distribution.. HOWEVER had these 2 been independent, I would easily have easily found the joint pdf (and hence by the convolution formula-Done!). How to tackle these dependent cases?
For the time being, suppose $\sigma = 1$. Then $$W_n = \frac{n \bar X^2}{X_1^2} = \frac{1}{n} \left( 1 + \frac{X_2 + \cdots + X_n}{X_1} \right)^2.$$ Then $$Y = \frac{X_2 + \cdots + X_n}{X_1} \sim \operatorname{Cauchy}(0, \sqrt{n-1})$$ with $$F_Y(y) = \frac{1}{2} + \frac{1}{\pi} \tan^{-1} \frac{y}{\sqrt{n-1}}.$$ Then $$\begin{align*} \Pr[W_n \le w] &= \Pr[-\sqrt{nw} - 1 \le Y \le \sqrt{nw} - 1] \\ &= \frac{1}{\pi} \left( \tan^{-1} \frac{\sqrt{nw} - 1}{\sqrt{n-1}} - \tan^{-1} \frac{-\sqrt{nw} - 1}{\sqrt{n-1}} \right) \end{align*}$$ and $$f_{W_n}(w) = \frac{(1+w)\sqrt{n(n-1)}}{\pi \sqrt{w}(n(1+w)^2 - 4w)}, \quad w > 0, \; n > 1.$$ I leave it as an exercise to see what happens in the general $\sigma > 0$ case.