Prove that $T=\frac{2X}{\sqrt{Y}} \sim t_4$ where $X \sim N(0,1)$ and $Y \sim \chi^2_4$ Question. Let $X,Y$ be independent random variables. Suppose $X \sim N(0,1)$ and $Y \sim \chi^2_4$, use a bivariate transformation to show that $T=\frac{2X}{\sqrt{Y}} \sim t_4$.
Attempt. Use the bivariate transformation $T=\frac{2X}{\sqrt{Y}}, S=\sqrt{Y}$. So the inverse transform is given by $X=\frac{1}{2}TS, Y=S^2$. The Jacobian is thus $s^2$. Then I used the transform formula to try to derive the joint density function of $f_{T,S}$ and thus differentiate w.r.t. $s$ to get the marginal density function of $T$. But the expression just got too messy so I wonder if my initial transform was the best way.
Would appreciate any help/hint.

If $U,\,V$ are independent continuous random variables of respective pdfs $f_U,\,f_V$ with $V$ of support $[0,\,\infty)$, $W:=U/V$ has pdf $f_W(w):=\int_0^\infty vf_U(wv)f_V(v)dv$. With the choice$$f_U(u)=\frac{1}{\sqrt{2\pi}}\exp-\frac{u^2}{2},\,f_V(v)=\frac12 v^3\exp -\frac{v^2}{2}$$so $U\sim N(0,\,1),\,V\sim\chi_4$, $$f_{U/V}(w)=\int_0^\infty\frac{1}{\sqrt{8\pi}}v^4\exp-\frac{(1+w^2)v^2}{2}dv\propto(1+w^2)^{-5/2}.$$Similarly, $2U/V$ has a pdf $\propto\left(1+\frac{w^2}{4}\right)^{-5/2}$, so is $t_4$-distributed. You're welcome as an exercise to double-check this method gets the right coefficients for unitarity. 
