Excess Kurtosis of Gamma Distribution/Proof 1

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

\(\ds \)

\(=\)

\(\ds \frac {\paren {1 - 4 + 6 - 3} \alpha^4 + \paren {6 - 12 + 6} \alpha^3 + \paren {11 - 8} \alpha^2 + 6 \alpha } {\paren {\alpha^2 } } - 3\)

\(\ds \)

\(=\)

\(\ds \dfrac 6 \alpha\)

$\blacksquare$