Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Gamma distribution ▷ Mean

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

\[\label{eq:gam-mean-s3} \begin{split} \mathrm{E}(X) &= \frac{a}{b} \int_{0}^{\infty} \mathrm{Gam}(x; a+1, b) \, \mathrm{d}x \\ &= \frac{a}{b} \; . \end{split}\]

Metadata: ID: P108 | shortcut: gam-mean | author: JoramSoch | date: 2020-05-19, 06:54.