Excess Kurtosis of Gamma Distribution/Proof 2
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Theorem
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the excess kurtosis $\gamma_2$ of $X$ is given by:
- $\gamma_2 = \dfrac 6 \alpha$
Proof
From the definition of excess kurtosis, we have:
- $\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where:
- $\mu = \expect X$ is the expectation of $X$
- $\sigma = \sqrt {\var X}$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have:
- $\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have:
- $\sigma^2 = \dfrac \alpha {\beta^2}$
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:
\(\ds \gamma_2\) | \(=\) | \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3\) | Kurtosis in terms of Non-Central Moments | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\dfrac {\alpha^{\overline 4} } {\beta^4} - 4 \paren {\dfrac \alpha \beta} \dfrac {\alpha^{\overline 3} } {\beta^3} + 6 \paren {\dfrac \alpha \beta}^2 \dfrac {\alpha^{\overline 2} } {\beta^2} - 3 \paren {\dfrac \alpha \beta}^4} {\dfrac {\alpha^2} {\beta^4} } - 3\) | Expectation of Gamma Distribution, Variance of Gamma Distribution, Expectation of Power of Gamma Distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\beta^4} - 4 \paren {\dfrac \alpha \beta} \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} + 6 \paren {\dfrac \alpha \beta}^2 \dfrac {\alpha \paren {\alpha + 1} } {\beta^2} - 3 \paren {\dfrac \alpha \beta}^4} \dfrac {\beta^4} {\alpha^2} - 3\) | Definition of Rising Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} - 4 \alpha^2 \paren {\alpha + 1} \paren {\alpha + 2} + 6 \alpha^3 \paren {\alpha + 1} - 3 \alpha^4} {\alpha^2} - 3\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {\alpha^4 + 6 \alpha^3 + 11 \alpha^2 + 6 \alpha} - 4 \paren {\alpha^4 + 3 \alpha^3 + 2 \alpha^2} + 6 \paren {\alpha^4 + \alpha^3 } - 3 \alpha^4} {\alpha^2} - 3\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - 4 + 6 - 3} \alpha^4 + \paren {6 - 12 + 6} \alpha^3 + \paren {11 - 8} \alpha^2 + 6 \alpha} {\alpha^2} - 3\) | gathering terms | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {6 \alpha + 3 \alpha^2} {\alpha^2} - 3\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 6 \alpha\) | simplifying |
$\blacksquare$