Calculating transformation of normal random variables.

Question

Let's say you have 4 i.i.d $N(0, 1)$ random variables $X_1 ,X_2, X_3, X_4$, how would you compute the pdf of $\frac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2 + X_4^2}}$. I am also interested in the general approach for doing these kind of calculations and if there is a way to do it without running into really messy integrals.

Answer 1

Note, $X_i^2\sim\chi^2_1$ for $i=1,2,3,4$. Hence, $T=X_2^2+X_3^2+X_4^2\sim\chi^2_3=\Gamma(\dfrac{3}{2},\dfrac{1}{2})$

Also you know that $X_1^2\sim\chi^2_1=\Gamma(\dfrac{1}{2},\dfrac{1}{2})$.

Therefore, invoking the result that if $U\sim \Gamma(\alpha,\theta)$ and $V\sim\Gamma(\beta,\theta)$ are independent then $\dfrac{U}{U+V}\sim Beta(\alpha,\beta)$, we conclude that $\dfrac{X_1^2}{X_1^2+T}\sim Beta(\dfrac{1}{2},\dfrac{3}{2})$.

So the square of your statistic follows Beta with the above parameters. Also note that your statistic is symmetric about $0$.

So, calling your statistic $S$, you know $P(S^2\leq x)=F(x)$ where $F$ is the cdf of $Beta(\dfrac{1}{2},\dfrac{3}{2})$. Equivalently, $P(-\sqrt{x}\leq S\leq\sqrt{x})=F(x)$ implying $P(S\leq\sqrt{x})-P(S<-\sqrt{x})=F(x)$ implying $2P(S\leq\sqrt{x})-1=F(x)$, where in the last step we have used the fact that $S$ is symmetric about $0$.

Can you conclude from here?