Excess Kurtosis of Geometric Distribution/Formulation 2

Theorem
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
 * $\map X \Omega = \set {0, 1, 2, \ldots} = \N$
 * $\map \Pr {X = k} = p \paren {1 - p}^k$

Then the excess kurtosis $\gamma_2$ of $X$ is given by:


 * $\gamma_2 = 6 + \dfrac {p^2} {1 - p}$

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 Geometric Distribution: Formulation 2, we have:


 * $\mu = \dfrac {1 - p} p$

By Variance of Geometric Distribution: Formulation 2, we have:


 * $\sigma = \dfrac {\sqrt {1 - p} } p$

So:

To calculate $\gamma_2$, we must calculate $\expect {X^4}$.

We find this using the moment generating function of $X$, $M_X$.

From Moment in terms of Moment Generating Function:


 * $\expect {X^4} = \map { {M_X}^{\paren 4} } 0$

From Moment Generating Function of Geometric Distribution: Fourth Moment:


 * $\map { {M_X}^{\paren 4} } t = p \paren {1 - p } e^t \paren {\dfrac {1 + 11 \paren {1 - p} e^t + 11 \paren {1 - p}^2 e^{2t} + \paren {1 - p}^3 e^{3t} } {\paren {1 - \paren {1 - p} e^t}^5 } }$

Setting $t = 0$ and from Exponential of Zero, we have:


 * $\expect {X^4} = \paren {1 - p } \paren {\dfrac {1 + 11 \paren {1 - p} + 11 \paren {1 - p}^2 + \paren {1 - p}^3 } {p^4} }$

So: