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.


1 Answer 1


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.


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