> [!definition] > > Let $\alpha, \lambda > 0$, define the $Gamma(\alpha, \lambda)$ [[Density|density]] as > $ > \gamma_{\alpha, \lambda}(x) = \frac{\lambda^\alpha x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)} \cdot \one_{\bracs{x \ge 0}} > $ > where $\Gamma(x) = \int_{x \ge 0}x^{\alpha - 1}e^{-x}dx$ is the [[Gamma Function|gamma function]]. A real-valued [[Random Variable|random variable]] $X$ is $Gamma(\alpha, \lambda)$-distributed if it has density $\gamma_{\alpha, \lambda}$ with respect to the [[Lebesgue Measure|Lebesgue measure]]. > [!theorem] > > If $X$ is $Gamma(\alpha, \lambda)$-distributed, then $\lambda X$ is $Gamma(\alpha, 1)$-distributed. > [!theorem] > > Let $X$ and $Y$ be [[Probabilistic Independence|independent]] random variables with $X \sim Gamma(\alpha, \lambda)$ and $Y \sim Gamma(\beta, \lambda)$, then $Z = X + Y$ is $Gamma(\alpha + \beta, \lambda)$-distributed. > > *Proof*. Via a [[Change of Variables|change of variables]], we may assume $\lambda = 1$. In which case, > $ > \begin{align*} > f_Z(x) &= \int_{[0, x]}\gamma_{\alpha, 1}(y)\gamma_{\beta, 1}(x - y)dy \\ > &= \int_0^x \frac{y^{\alpha - 1}e^{-y}}{\Gamma(\alpha)} \frac{(x - y)^{\beta - 1}e^{-(x - y)}}{\Gamma(\beta)} dy \\ > &= \frac{e^{-x}}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x y^{\alpha - 1}(x - y)^{\beta - 1}dy > \end{align*} > $ > Using the change of variables $u = y/x$, > $ > \begin{align*} > f_Z(x) &= \frac{x^{\alpha + \beta - 1}e^{-x}}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1 u^{\alpha - 1}(1 - u)^{\beta - 1}du > \end{align*} > $ > where integrating $f_Z$ over $[0, \infty)$ gives $1$, and by definition of the gamma function, $\int_{0}^\infty x^{\alpha + \beta - 1}e^{-x} = \Gamma(\alpha)\Gamma(\beta)$. So > $ > \begin{align*} > 1 &= \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1u^{\alpha - 1}(1 - u)^{\beta - 1}du \\ > f_Z(x) &= \frac{x^{\alpha + \beta - 1}e^{-x}}{\Gamma(\alpha + \beta)} > \end{align*} > $