Variance of Gamma Distribution
Theorem
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The variance of $X$ is given by:
- $\var X = \dfrac \alpha {\beta^2}$
Proof 1
From the definition of the Gamma distribution, $X$ has probability density function:
- $\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$
From Variance as Expectation of Square minus Square of Expectation:
- $\ds \var X = \int_0^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$
So:
\(\ds \var X\) | \(=\) | \(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha + 1} e^{-\beta x} \rd x - \paren {\frac \alpha \beta}^2\) | Expectation of Gamma Distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \paren {\frac t \beta}^{\alpha + 1} e^{-t} \frac {\d t} \beta - \frac {\alpha^2} {\beta^2}\) | substituting $t = \beta x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha} {\beta^{\alpha + 2} \map \Gamma \alpha} \int_0^\infty t^{\alpha + 1} e^{-t} \rd t - \frac {\alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma {\alpha + 2} } {\beta^2 \map \Gamma \alpha} - \frac {\alpha^2} {\beta^2}\) | Definition of Gamma Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma {\alpha + 2} - \alpha^2 \map \Gamma \alpha} {\beta^2 \map \Gamma \alpha}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha \paren {\alpha + 1} \map \Gamma \alpha - \alpha^2 \map \Gamma \alpha} {\beta^2 \map \Gamma \alpha}\) | Gamma Difference Equation | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha \map \Gamma \alpha \paren {\alpha + 1 - \alpha} } {\beta^2 \map \Gamma \alpha}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \alpha {\beta^2}\) |
$\blacksquare$
Proof 2
By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by:
- $\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$
for $t < \beta$.
From Variance as Expectation of Square minus Square of Expectation:
- $\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Gamma Distribution:
- $\expect X = \dfrac \alpha \beta$
From Moment Generating Function of Gamma Distribution: Second Moment:
- $\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$
From Moment in terms of Moment Generating Function, we also have:
- $\expect {X^2} = \map { {M_X}} 0$
Setting $t = 0$, we obtain the second moment:
\(\ds \map {M_X} 0\) | \(=\) | \(\ds \expect {X^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - 0}^{\alpha + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\beta^{\alpha + 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha \paren {\alpha + 1} } {\beta^2}\) |
So:
\(\ds \var X\) | \(=\) | \(\ds \frac {\alpha \paren {\alpha + 1} } {\beta^2} - \frac {\alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \alpha {\beta^2}\) |
$\blacksquare$
Proof 3
From Expectation of Power of Gamma Distribution‎, we have:
- $\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$
where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.
Hence:
\(\ds \var X\) | \(=\) | \(\ds \expect {X^2} - \paren {\expect X}^2\) | Variance as Expectation of Square minus Square of Expectation | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\alpha^{\overline 2} } {\beta^2} - \paren {\dfrac {\alpha^{\overline 1} } \beta}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\alpha \paren {\alpha + 1} } {\beta^2} - \paren {\dfrac \alpha \beta}^2\) | Definition of Rising Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \alpha {\beta^2}\) |
$\blacksquare$
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $13$: Probability distributions
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $15$: Probability distributions