Excess Kurtosis of Gamma Distribution

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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 1

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

$\blacksquare$


Proof 2

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

$\blacksquare$