Excess Kurtosis of Gamma Distribution/Proof 1
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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:
- $\mu$ is the expectation of $X$.
- $\sigma$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have:
- $\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have:
- $\sigma = \dfrac {\sqrt \alpha} \beta$
So:
\(\ds \gamma_2\) | \(=\) | \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3\) | Kurtosis in terms of Non-Central Moments | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\expect {X^4} - 4 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} } + 6 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } - 3 \paren {\dfrac \alpha \beta}^4} {\paren {\dfrac {\alpha^2} {\beta^4} } } - 3\) | Skewness of Gamma Distribution, Variance of Gamma Distribution |
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$.
By Moment Generating Function of Gamma Distribution, this is given by:
- $\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha} = \beta^\alpha \paren {\beta - t}^{-\alpha}$
From Moment in terms of Moment Generating Function:
- $\expect {X^4} = \map {M_X'} 0$
So:
\(\ds \map {M_X'} t\) | \(=\) | \(\ds \beta^\alpha \paren {-\alpha } \paren {\beta - t}^{-\alpha - 1} \paren {-1}\) | Chain Rule for Derivatives, Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \beta^\alpha \alpha \paren {\beta - t}^{-\alpha - 1}\) | ||||||||||||
\(\ds \map {M_X} t\) | \(=\) | \(\ds \beta^\alpha \alpha \paren {-\alpha - 1} \paren {\beta - t}^{-\alpha - 2} \paren {-1}\) | Chain Rule for Derivatives, Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \beta^\alpha \alpha \paren {\alpha + 1} \paren {\beta - t}^{-\alpha - 2}\) | ||||||||||||
\(\ds \map {M_X} t\) | \(=\) | \(\ds \paren {-\alpha - 2} \beta^\alpha \alpha \paren {\alpha + 1} \paren {\beta - t}^{-\alpha - 3} \paren {-1}\) | Chain Rule for Derivatives, Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\beta - t}^{-\alpha - 3}\) | ||||||||||||
\(\ds \map {M_X'} t\) | \(=\) | \(\ds \paren {-\alpha - 3} \alpha \paren {\alpha + 1} \paren {\alpha + 2} \beta^\alpha \paren {\beta - t}^{-\alpha - 4} \paren {-1}\) | Chain Rule for Derivatives, Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} \beta^\alpha \paren {\beta - t}^{-\alpha - 4}\) |
Setting $t = 0$:
\(\ds \expect {X^4}\) | \(=\) | \(\ds \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} \beta^\alpha \paren {\beta - 0}^{-\alpha - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\beta^4}\) |
Plugging this result back into our equation above:
\(\ds \gamma_2\) | \(=\) | \(\ds \frac {\expect {X^4} - 4 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} } + 6 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } - 3 \paren {\dfrac \alpha \beta}^4} {\paren {\dfrac {\alpha^2} {\beta^4} } } - 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + 3} } {\beta^4} - 4 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} } + 6 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } - 3 \paren {\dfrac \alpha \beta}^4} {\paren {\dfrac {\alpha^2} {\beta^4} } } - 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\alpha^4 + 6 \alpha^3 + 11 \alpha^2 + 6 \alpha} - 4 \paren {\alpha^4 + 3 \alpha^3 + 2 \alpha^2} + 6 \paren {\alpha^4 + \alpha^3 } - 3 \alpha^4} {\paren {\alpha^2 } } - 3\) | $\beta^4$ cancels | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {1 - 4 + 6 - 3} \alpha^4 + \paren {6 - 12 + 6} \alpha^3 + \paren {11 - 8} \alpha^2 + 6 \alpha } {\paren {\alpha^2 } } - 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 6 \alpha\) |
$\blacksquare$