# sum of two gamma distributions with different scales using change of variable [closed]

Let $$X_1\sim\Gamma(r,1)$$ and $$X_2\sim\Gamma(s,1)$$ be independent. Find the distribution $$Y=X_1+X_2.$$

• Those appear to be different shapes not scales – Glen_b Aug 27 '20 at 1:41
• And if they were different scales there isn't a closed-form solution – Thomas Lumley Aug 27 '20 at 6:45

The distribution functions for $$X_1$$ and $$X_2$$ we write out as \begin{align*} f_1(x_1)&=\frac{x_1^{r-1}e^{-x_1}}{\Gamma(r)}\\ f_2(x_2)&=\frac{x_2^{s-1}e^{-x_2}}{\Gamma(s)}, \end{align*} with moment-generating functions \begin{align*} m_{X_1}(t)&=(1-t)^{-r}\\ m_{X_2}(t)&=(1-t)^{-s}. \end{align*} Now it is a theorem that if $$X_1$$ and $$X_2$$ are independent random variables, then $$X_1+X_2$$ has a moment-generating function equal to the product of the moment-generating functions for $$X_1$$ and $$X_2.$$ That is, for our case, $$m_{X_1+X_2}(t)=m_{X_1}(t)\,m_{X_2}(t)=(1-t)^{-r}(1-t)^{-s}=(1-t)^{-(r+s)},$$ which is the moment-generating function for a Gamma distribution with parameters $$r+s$$ and $$1.$$ That is, $$X_1+X_2\sim\Gamma(r+s,1).$$ This last step follows because moment-generating functions are unique.
• This should be treated as a self-study question. Providing the (correct and well-detailed) solution does not necessarily help the OP in the long run. – Xi'an Aug 27 '20 at 5:22