Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Gamma distribution ▷ Scaling of a gamma random variable

Theorem: Let $X$ be a random variable following a gamma distribution with shape $a$ and rate $b$:

$\label{eq:gam} X \sim \mathrm{Gam}(a,b) \; .$

Then, the quantity $Y = c X$ will also be gamma-distributed with shape $a$ and rate $b/c$:

$\label{eq:gam-scal} Y = b X \sim \mathrm{Gam}\left( a, \frac{b}{c} \right) \; .$

Proof: Note that $Y$ is a function of $X$

$\label{eq:Y-X} Y = g(X) = c X$

with the inverse function

$\label{eq:X-Y} X = g^{-1}(Y) = \frac{1}{c} Y \; .$

Because the parameters of a gamma distribution are positive, $c$ must also be positive. Thus, $g(X)$ is strictly increasing and we can calculate the probability density function of a strictly increasing function as

$\label{eq:pdf-sifct} f_Y(y) = \left\{ \begin{array}{rl} f_X(g^{-1}(y)) \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \; , & \text{if} \; y \in \mathcal{Y} \\ 0 \; , & \text{if} \; y \notin \mathcal{Y} \end{array} \right.$ $\label{eq:gam-pdf} f_X(x) = \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \; .$

Applying \eqref{eq:pdf-sifct} to \eqref{eq:gam-pdf}, we have:

$\label{eq:Y-pdf} \begin{split} f_Y(y) &= \frac{b^a}{\Gamma(a)} [g^{-1}(y)]^{a-1} \exp[-b g^{-1}(y)] \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \\ &= \frac{b^a}{\Gamma(a)} \left( \frac{1}{c} y \right)^{a-1} \exp\left[-b \left( \frac{1}{c} y \right) \right] \, \frac{\mathrm{d}\left( \frac{1}{c} y \right)}{\mathrm{d}y} \\ &= \frac{b^a}{\Gamma(a)} \left( \frac{1}{c} \right)^{a} \left( \frac{1}{c} \right)^{-1} y^{a-1} \exp\left[- \frac{b}{c} y \right] \, \frac{1}{c} \\ &= \frac{(b/a)^a}{\Gamma(a)} y^{a-1} \exp\left[- \frac{b}{c} y \right] \end{split}$

which is the probability density function of a gamma distribution with shape $a$ and rate $b/c$.

Sources:

Metadata: ID: P426 | shortcut: gam-scal | author: JoramSoch | date: 2023-11-24, 13:08.