TITLE: Sum of independent Gamma distributions is a Gamma distribution
QUESTION [31 upvotes]: If $X\sim \Gamma(a_1,b)$ and $Y \sim \Gamma(a_2,b)$, I need to prove $X+Y\sim\Gamma(a_1+a_2,b)$ if $X$ and $Y$ are independent.
I am trying to apply formula for independence integral and just trying to multiply the gamma function but stuck ?

REPLY [15 votes]: It's easier to use Moment Generating Functions to prove that.
$$
M(t;\alpha,\beta ) = Ee^{tX} = \int_{0}^{+\infty} e^{tx} f(x;\alpha,\beta)dx 
 = \int_{0}^{+\infty} e^{tx} \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1}e^{-\beta x}dx \\
 = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_{0}^{+\infty} x^{\alpha-1}e^{-(\beta - t) x}dx = \frac{\beta^\alpha}{\Gamma(\alpha)} \frac{\Gamma(\alpha)}{(\beta - t)^\alpha} = \frac{1}{(1- \frac{t}{\beta})^\alpha}
$$
By using the property of independent random variables, we know
$$M_{X + Y}(t) = M_{X}(t)M_{Y}(t) $$
So
if $X \sim Gamma(\alpha_1,\beta), Y \sim Gamma(\alpha_2,\beta), $
$$M_{X + Y}(t) = \frac{1}{(1- \frac{t}{\beta})^{\alpha_1}} 
\frac{1}{(1- \frac{t}{\beta})^{\alpha_2}} = \frac{1}{(1- \frac{t}{\beta})^{\alpha_1 + \alpha_2}}$$
You can see the MGF of the product is still in the format of Gamma distribution. Finally we can get $X + Y \sim Gamma(\alpha_1 + \alpha_2, \beta)$