If $X_1$ and $X_2$ are independent random variables with distribution $N(0,1)$, then proof the sum of $X_{1}^2 + X_{2}^2$ has gamma distribution.

Question

I know $Y=X^2$ has density function $\frac{1}{\sqrt{2y\pi}}e^{\frac{-y}{2}} $ and this is gamma distribution $ G(\frac{1}{2},\frac{1}{2})$. My question is how can I do the sum of $X_{1}^2 + X_{2}^2$ and that is equal to $G(1, \frac{1}{2})$.

Answer 1

One way is polar coordinates: \begin{align} & F_Y(y) = \Pr(Y\le y) \\[8pt] = {} & \iint\limits_{x_1,x_2\,:\,x_1^2\,+\,x_2^2\,\le\,y} \frac 1 {2\pi} e^{-x_1^2/2} e^{-x_2^2/2} \, d(x_1,x_2) \\[8pt] = {} & \iint\limits_{x_1,x_2\,:\,x_1^2\,+\,x_2^2\,\le\,y} \frac 1 {2\pi} e^{-(x_1^2+x_2^2)/2} \, d(x_1,x_2) \\[8pt] = {} & \int_0^{2\pi} \left( \int_0^{\sqrt y} \frac 1 {2\pi} e^{-r^2/2} (r\,dr) \right) \, d\theta \end{align} The inner integral does not depend on $\theta,$ i.e. it is constant as $\theta$ goes from $0$ to $2\pi,$ so this expression is the length, $2\pi-0$ of the interval times that constant. Thus it is $$ \int_0^{\sqrt y} e^{-r^2/2} (r\,dr) = \int_0^{y/2} e^{-u} \, du $$ (When $r = \sqrt y$ then $u = r^2/2 = y/2.$)

Also it is sometimes useful to know that if independent random variables have distributions $$ \frac 1 {\Gamma(\alpha_i)} (\lambda y)^{\alpha_i-1} e^{-\lambda y} (\lambda\,dy) \text{ for } y>0\quad \text{for } i=1,2, $$ then the sum of those two random variables have distribution $$ \frac 1 {\Gamma(\alpha_1+\alpha_2)} (\lambda y)^{\alpha_1+\alpha_2-1} e^{-\lambda y} (\lambda\,dy) \text{ for } y>0. $$ In your example, you have $\alpha_1=\alpha_2 = \frac 1 2,$ so $\alpha_1+\alpha_2-1=0.$ And $\lambda=1/2.$