Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Gamma distribution ▷ Relationship to standard gamma distribution

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

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

Then, the quantity $Y = b X$ will have a standard gamma distribution with shape $a$ and rate $1$:

$\label{eq:Y-snorm} Y = b X \sim \mathrm{Gam}(a,1) \; .$

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

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

with the inverse function

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

Because $b$ is positive, $g(X)$ is strictly increasing and we can calculate the probability density function of a strictly increasing function as

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

where $\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace$. With the probability density function of the gamma distribution, we have

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

which is the probability density function of the standard gamma distribution.

Sources:

Metadata: ID: P177 | shortcut: gam-sgam2 | author: JoramSoch | date: 2020-10-15, 12:04.