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

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

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    $\begingroup$ 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. $\endgroup$ – Xi'an Aug 27 '20 at 5:22
  • $\begingroup$ Though the solution doesn't use change of variables, as the OP wanted... $\endgroup$ – Thomas Lumley Aug 27 '20 at 6:44
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    $\begingroup$ Yes, the fact that I didn't use the intended method sort of satisfies that self-study issue. At least they have the correct answer. $\endgroup$ – Adrian Keister Aug 27 '20 at 12:07

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