6  Gamma distribution

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Definition 6.1 Let \(\mathcal Y=\set{0,1,2,\ldots}\). The uncertain quantity \(Y\in\mathcal Y\) has a (denoted by \(Y\sim\distgamma(a,b)\) or \(Y\sim\Gamma(a,b)\)) if \[ \Pr\qty(Y=\theta\mid a,b)=\pdfgamma(\theta,a, b)=\dfrac{b^a}{\Gamma(a)}\theta^{a-1}\me^{-b\theta},\quad \text{for }\theta,a,b>0. \]

Here

• \(a\) is the shape.
• \(b\) is the rate.
• \(\lambda=\frac1b\) is the scale.

Python code

from scipy.stats import gamma
import matplotlib.pyplot as plt
import numpy as np

rv = gamma(a=20, scale=1/2)

plt.plot(rv.pdf(np.arange(100)))

rv.interval(.95)

(6.108259792701973, 14.835426785792794)

Theorem 6.1 If \(Y\sim \distgamma(a, b)\), then

• \(\Exp\qty[Y\mid a, b]=a/b\),
• \(\Var\qty[Y\mid a, b]=a/b^2\),
• \(\Mode\qty[Y\mid a, b]=\)

Proof

\[ \begin{split} \Exp\qty[Y\mid a,b]&= \int_{0}^{\infty}\theta\frac{b^a}{\Gamma(a)}\theta^{a-1}\me^{-b\theta}\dl3\theta\\ &=\int_{0}^{\infty}\frac{b^a}{\Gamma(a)}\theta^{(a+1)-1}\me^{-b\theta}\dl3\theta\\ &=\int_{0}^{\infty}\frac{\Gamma(a+1)}{b\Gamma(a)}\frac{b^{a+1}}{\Gamma(a+1)}\theta^{(a+1)-1}\me^{-b\theta}\dl3\theta\\ &=\frac{\Gamma(a+1)}{b\Gamma(a)}\int_{0}^{\infty}\pdfgamma(\theta,a+1,b)\dl3\theta\\ &=\frac{a}{b},\\ \Exp\qty[Y^2\mid a,b]&= \int_{0}^{\infty}\theta^2\frac{b^a}{\Gamma(a)}\theta^{a-1}\me^{-b\theta}\dl3\theta\\ &=\int_{0}^{\infty}\frac{b^a}{\Gamma(a)}\theta^{(a+2)-1}\me^{-b\theta}\dl3\theta\\ &=\int_{0}^{\infty}\frac{\Gamma(a+2)}{b^2\Gamma(a)}\frac{b^{a+2}}{\Gamma(a+2)}\theta^{(a+2)-1}\me^{-b\theta}\dl3\theta\\ &=\frac{\Gamma(a+2)}{b^2\Gamma(a)}\int_{0}^{\infty}\pdfgamma(\theta,a+2,b)\dl3\theta\\ &=\frac{a(a+1)}{b^2},\\ \Var\qty[Y\mid a, b]&=\Exp\qty[Y^2\mid a, b]-\qty(\Exp\qty[Y\mid a, b])^2\\ &=\frac{a(a+1)}{b^2}-\frac{a^2}{b^2}=\frac{a}{b^2}. \end{split} \]