Question: Suppose that $\displaystyle \frac{2}{\theta_0}\sum_{i=1}^n y_i\sim\displaystyle\chi_{2n}^2$ and $\displaystyle 2\theta_0\sum_{i=1}^n x_i\sim\displaystyle\chi_{2n}^2$. And these two are independent. That is, $x_i'$s are independent with each other; $y_i'$s are independent with each other; and $x_i'$s and $y_i'$s are independent with each other. What is the distribution of $\displaystyle \sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i}$?
I don't know what knowledge on probability this question refers to. However, it only seems natural to say that $\displaystyle \sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i}\sim\sqrt{\frac{1}{4}\left(\chi_{2n}^2\right)^2}\sim\frac{1}{2}\chi_{2n}^2$. Is this right, please? Can people do this, please? Thank you!
There was a bit of discussion on notation matter. So I made it more conventional. Now could anyone suggest something about the real issue here, please? Thank you!
The density is proportional to $$x^nA(x^2)\mathbf 1_{x\geqslant0}, $$ where, for every $t\geqslant0$, $$A(t)=\int_0^\infty\mathrm e^{-u-t/u}\mathrm du. $$ Thus, $A(t)$ is in $(0,1)$ for every $t\geqslant0$, $A(0)=+1$, $A(+\infty)=0$, $A$ is convex and nonincreasing and $A$ solves the differential equation $tA''(t)-A(t)=0$.