Proof: Mean of the gamma distribution
Index:
The Book of Statistical Proofs ▷
Probability Distributions ▷
Univariate continuous distributions ▷
Gamma distribution ▷
Mean
Metadata: ID: P108 | shortcut: gam-mean | author: JoramSoch | date: 2020-05-19, 06:54.
Theorem: Let $X$ be a random variable following a gamma distribution:
\[\label{eq:gam} X \sim \mathrm{Gam}(a, b) \; .\]Then, the mean or expected value of $X$ is
\[\label{eq:gam-mean} \mathrm{E}(X) = \frac{a}{b} \; .\]Proof: The expected value is the probability-weighted average over all possible values:
\[\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .\]With the probability density function of the gamma distribution, this reads:
\[\label{eq:gam-mean-s1} \begin{split} \mathrm{E}(X) &= \int_{0}^{\infty} x \cdot \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{b^a}{\Gamma(a)} x^{(a+1)-1} \exp[-b x] \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{1}{b} \cdot \frac{b^{a+1}}{\Gamma(a)} x^{(a+1)-1} \exp[-b x] \, \mathrm{d}x \; . \end{split}\]Employing the relation $\Gamma(x+1) = \Gamma(x) \cdot x$, we have
\[\label{eq:gam-mean-s2} \mathrm{E}(X) = \int_{0}^{\infty} \frac{a}{b} \cdot \frac{b^{a+1}}{\Gamma(a+1)} x^{(a+1)-1} \exp[-b x] \, \mathrm{d}x\]and again using the density of the gamma distribution, we get
\[\label{eq:gam-mean-s3} \begin{split} \mathrm{E}(X) &= \frac{a}{b} \int_{0}^{\infty} \mathrm{Gam}(x; a+1, b) \, \mathrm{d}x \\ &= \frac{a}{b} \; . \end{split}\]∎
Sources: - Turlapaty, Anish (2013): "Gamma random variable: mean & variance"; in: YouTube, retrieved on 2020-05-19; URL: https://www.youtube.com/watch?v=Sy4wP-Y2dmA.
Metadata: ID: P108 | shortcut: gam-mean | author: JoramSoch | date: 2020-05-19, 06:54.