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

Theorem: Let $X$ be a random variable following a gamma distribution:

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

Then, the variance of $X$ is

$\label{eq:gam-var} \mathrm{Var}(X) = \frac{a}{b^2} \; .$

Proof: The variance can be expressed in terms of expected values as

$\label{eq:var-mean} \mathrm{Var}(X) = \mathrm{E}(X^2) - \mathrm{E}(X)^2 \; .$ $\label{eq:gam-mean} \mathrm{E}(X) = \frac{a}{b} \; .$

With the probability density function of the gamma distribution, the expected value of a squared gamma random variable is

$\label{eq:gam-sqr-mean-s1} \begin{split} \mathrm{E}(X^2) &= \int_{0}^{\infty} x^2 \cdot \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{b^a}{\Gamma(a)} x^{(a+2)-1} \exp[-b x] \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{1}{b^2} \cdot \frac{b^{a+2}}{\Gamma(a)} x^{(a+2)-1} \exp[-b x] \, \mathrm{d}x \; . \end{split}$

Twice-applying the relation $\Gamma(x+1) = \Gamma(x) \cdot x$, we have

$\label{eq:gam-sqr-mean-s2} \mathrm{E}(X^2) = \int_{0}^{\infty} \frac{a \, (a+1)}{b^2} \cdot \frac{b^{a+2}}{\Gamma(a+2)} x^{(a+2)-1} \exp[-b x] \, \mathrm{d}x$

and again using the density of the gamma distribution, we get

$\label{eq:gam-sqr-mean-s3} \begin{split} \mathrm{E}(X^2) &= \frac{a \, (a+1)}{b^2} \int_{0}^{\infty} \mathrm{Gam}(x; a+2, b) \, \mathrm{d}x \\ &= \frac{a^2+a}{b^2} \; . \end{split}$

Plugging \eqref{eq:gam-sqr-mean-s3} and \eqref{eq:gam-mean} into \eqref{eq:var-mean}, the variance of a gamma random variable finally becomes

$\label{eq:gam-var-qed} \begin{split} \mathrm{Var}(X) &= \frac{a^2+a}{b^2} - \left( \frac{a}{b} \right)^2 \\ &= \frac{a}{b^2} \; . \end{split}$
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Metadata: ID: P109 | shortcut: gam-var | author: JoramSoch | date: 2020-05-19, 07:20.