= Solution
<Independence> makes the <moment-generating function> of the sum the product of the two <moment-generating functions>. For $t<\lambda$,
$$
M_{X_1+X_2}(t)=\left(\frac{\lambda}{\lambda-t}\right)^{r_1}\left(\frac{\lambda}{\lambda-t}\right)^{r_2}
=\left(\frac{\lambda}{\lambda-t}\right)^{r_1+r_2}.
$$
The <expectations> are finite on, for example, $(-\lambda/2,\lambda/2)$, as is the <moment-generating function> of the candidate <Gamma distribution>. The <uniqueness theorem for moment-generating functions> therefore applies, proving
$$
\boxed{X_1+X_2\sim\operatorname{Gamma}(r_1+r_2,\lambda).}
$$
This is <additivity of independent gamma distributions with a common rate>; a common rate is essential to this specific formula.
Back to article page
