I have to show
$$\newcommand\dif{\mathop{}\!\mathrm{d}}
\text{Gam}(\alpha,\beta_1) \ast \text{Gam}(\alpha,\beta_2) = \text{Gam}(\alpha,\beta_1+\beta_2)
$$
where the probability density function for a $\text{Gam}(\alpha,\beta)$-distributed random variable is defined as follows
$$
f_{\alpha,\beta}(x) = \frac{\alpha^\beta}{\Gamma(\beta)} x^{\beta-1} e^{-\alpha x} \;\textbf{1}_{(0,\infty)}(x)
$$
Also let's write down the definition of the $\text{Gamma}$ function
$$
\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \dif{t}
$$
I know this can be proven easily using the probability-generating function but I want to prove it using the defintion of a convolution. This is what I have done so far:
$$
\begin{align*}
& \quad \int_0^\infty \left(\frac{\alpha^{\beta_1}}{\Gamma(\beta_1)} y^{\beta_1-1} e^{-\alpha y} \right)
\left( \frac{\alpha^{\beta_2}}{\Gamma(\beta_2)} (x-y)^{\beta_1-1} e^{-\alpha(x-y)} \right) \dif{y} \\
&= \frac{\alpha^{\beta_1 + \beta_2}}{\Gamma(\beta_1)\Gamma(\beta_2)} \int_0^\infty \left(y^{\beta_1-1} e^{-\alpha y} \right) \left( (x-y)^{\beta_2-1} e^{-\alpha(x-y)} \right) \dif{y} \\
&= \frac{\alpha^{\beta_1 + \beta_2}}{\Gamma(\beta_1)\Gamma(\beta_2)} \int_0^\infty y^{\beta_1-1} (x-y)^{\beta_2-1} e^{-\alpha x} \dif{y} \\
&= \frac{\alpha^{\beta_1+\beta_2} e^{-\alpha x}}{\Gamma(\beta_1)\Gamma(\beta_2)} \int_0^\infty y^{\beta_1-1} (x-y)^{\beta_2-1} \dif{y}
\end{align*}
$$
Now, I know I need to have $\Gamma(\beta_1 + \beta_2)$ in the denominator. The last integral almost looks like the $\text{Gamma}$ function but I don't see how I would arrive at the result.
Best Answer
Strangely, you include the indicator function for the gamma density in your first equation, but then you ignore it when you perform the convolution. You will observe that if $y > x$, the convolution is zero: $$\int_{y=-\infty}^\infty \frac{\alpha^{\beta_1}}{\Gamma(\beta_1)} y^{\beta_1 - 1} e^{-\alpha y} \mathbb 1(y > 0) \frac{\alpha^{\beta_2}}{\Gamma(\beta_2)} (x-y)^{\beta_2 - 1} e^{-\alpha(x-y)} \color{red}{\mathbb 1(x-y > 0)} \, dy \\= \int_{y=0}^x \frac{\alpha^{\beta_1 + \beta_2}}{\Gamma(\beta_1)\Gamma(\beta_2)} e^{-\alpha x} y^{\beta_1 - 1} (x-y)^{\beta_2 - 1} \, dy.$$ Then a simple scaling transformation $$y = xu, \quad dy = x \, du,$$ yields $$\frac{\alpha^{\beta_1 + \beta_2} x^{\beta_1 + \beta_2 - 1} e^{-\alpha x}}{\Gamma(\beta_1) \Gamma(\beta_2)}\int_{u=0}^1 u^{\beta_1 - 1} (1 - u)^{\beta_2 - 1} \, du.$$ The remaining integral is a Beta function integral and equals $$\frac{\Gamma(\beta_1) \Gamma(\beta_2)}{\Gamma(\beta_1 + \beta_2)},$$ from which the desired result follows.