Excess Kurtosis of Geometric Distribution/Formulation 2

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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:

\(\ds \gamma_2\) \(=\) \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3\) Kurtosis in terms of Non-Central Moments
\(\ds \) \(=\) \(\ds \frac {\expect {X^4} - 4 \paren {\dfrac {1 - p} p} \dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3 } + 6 \paren {\dfrac {1 - p} p}^2 \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 } - 3 \paren {\dfrac {1 - p} p}^4 } {\paren {\dfrac {\sqrt {1 - p} } p}^4 } - 3\) Skewness of Geometric Distribution: Formulation 2


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:

\(\ds \gamma_2\) \(=\) \(\ds \frac {\expect {X^4} - 4 \paren {\dfrac {1 - p} p} \dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3 } + 6 \paren {\dfrac {1 - p} p}^2 \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 } - 3 \paren {\dfrac {1 - p} p}^4 } {\paren {\dfrac {\sqrt {1 - p} } p}^4 } - 3\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 - p } \paren {\dfrac {1 + 11 \paren {1 - p} + 11 \paren {1 - p}^2 + \paren {1 - p}^3 } {p^4} } - 4 \paren {\dfrac {1 - p} p} \dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3 } + 6 \paren {\dfrac {1 - p} p}^2 \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 } - 3 \paren {\dfrac {1 - p} p}^4 } {\paren {\dfrac {\sqrt {1 - p} } p}^4 } - 3\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 + 11 \paren {1 - p} + 11 \paren {1 - p}^2 + \paren {1 - p}^3} - 4 \paren {1 - p} \paren {6 - 6 p + p^2} + 6 \paren {1 - p}^2 \paren {2 - p} - 3 \paren {1 - p}^3 } {\paren {1 - p } } - 3\) $\dfrac {\paren {1 - p} } {p^4}$ cancels
\(\ds \) \(=\) \(\ds \frac {\paren {-1 + 4 - 6 + 3} p^3 + \paren {11 + 3 -4 -24 + 12 + 12 -9} p^2 + \paren {-11 -22 -3 + 24 + 6 -24 - 12 + 9} p + \paren{1 + 11 + 11 + 1 - 24 + 12 - 3} } {\paren {1 - p } } - 3\)
\(\ds \) \(=\) \(\ds \frac {p^2 - 9 p + 9 } {\paren {1 - p } } - \frac {3 - 3 p} {1 - p}\)
\(\ds \) \(=\) \(\ds \frac {6 - 6 p + p^2 } {\paren {1 - p } }\)
\(\ds \) \(=\) \(\ds 6 + \dfrac {p^2} {1 - p}\)

$\blacksquare$

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