# 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.

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 $$G(1, \frac{1}{2})$$.

• Moment generating functions is a nice method too (if allowed). Plus relationship between standard normal, Chi-squared and gamma distributions. – jdods Dec 2 '20 at 21:11

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 $$\lambda=1/2.$$