I currently have a problem in deriving the distribution (or moments) of the random variable $T = \frac{X_1X_2}{\sum_{i=1}^n X_i^2}, \text{where }X_i \sim N(0, 1).$
I attempted to utilize the fact that $\sum_{i=1}^n X_i^2 \sim \chi_{(n)}.$ However, because the numerator and the denominator is not independent, I have encountered difficulty in proceeding.
As far as I am aware, I have not found any known distribution (or moments) for the above random variable $T$.
You can obtain a close expression for the distribution through
\begin{align} f_T(t)&=\int_{(x_1,x_2)\in \mathbb{R}^2 } f_T(t|X_1=x_1,X_2=x_2) f_{X_1}(x_1) f_{X_2}(x_2) dx_1 dx_2 \\ &= \int_{(x_1,x_2)\in \mathbb{R}^2 } f_T(t|X_1=x_1,X_2=x_2) e^{-\frac{1}{2}(x_1^2+x_2^2)} dx_1 dx_2 \end{align}
Let $Q \sim \chi_{n-2}^2$, then
\begin{equation} f_T(t|X_1=x_1,X_2=x_2)=f_Q\left(\frac{x_1x_2}{t}-x_1^2-x_2^2\right)=\frac{1}{2^{\frac{n}{2}-1}\Gamma(\frac{n}{2}-1)}\left(\frac{x_1x_2}{t}-x_1^2-x_2^2\right)^{\frac{n}{2}-2} e^{\frac{-x_1 x_2}{t}} e^{\frac{1}{2}(x_1^2+x_2^2)} \end{equation}
yielding
\begin{equation} f_T(t)=\frac{1}{2^{\frac{n}{2}-1}\Gamma(\frac{n}{2}-1)} \int_{(x_1,x_2)\in \mathbb{R}^2 } \left(\frac{x_1x_2}{t}-x_1^2-x_2^2\right)^{\frac{n}{2}-2} e^{\frac{-x_1 x_2}{t}} dx_1 dx_2 \end{equation}
I was thinking about a nicer closed-form expression, but I didn't find a way to simplify it.
The rv $U=\frac{1}{X_1^2+\cdots+X_n^2}(X_1^2,\ldots,X_n^2)=(U_1,\ldots,U_n)$ is Dirichlet distributed $D(1/2,\ldots,1/2)$ and therefore the density of $(U_1,U_2)$ is $$\frac{\Gamma(n/2)}{\Gamma(1/2)^2\Gamma((n-2)/2)}u_1^{-1/2}u_2^{-1/2}(1-u_1-u_2)^{(n-4)/2}$$ You are interested in the distribution of $\epsilon \sqrt{U_1U_2}$ where $\Pr(\epsilon =\pm 1)=1/2$ and $\epsilon$ is independent of $(U_1,U_2).$ The calculation of the moments is easy since
$$E(U_1^aU_2^b)=\frac{\Gamma(n/2)}{\Gamma(1/2)^2}\times \frac{\Gamma(a+\frac{1}{2})\Gamma(b+\frac{1}{2})}{\Gamma(a+b+\frac{n}{2})}$$
