Theorem: Excess Kurtosis of Geometric Distribution/Formulation 2

:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
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: 
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
	| c = Kurtosis in terms of Non-Central Moments
}}
{{eqn	| r = \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
	| c = Skewness of Geometric Distribution: Formulation 2
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn	| l = \gamma_2
	| r = \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
	| c = 
}}
{{eqn	| r = \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
	| c = 
}}
{{eqn	| r = \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
        | c = $\dfrac {\paren {1 - p} } {p^4}$ cancels
}}
{{eqn	| r = \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
        | c = 
}}
{{eqn	| r = \frac {p^2 - 9 p + 9 } {\paren {1 - p } } - \frac {3 - 3 p} {1 - p}
        | c = 
}}
{{eqn	| r = \frac {6 - 6 p + p^2 } {\paren {1 - p } }
        | c = 
}}
{{eqn	| r = 6 + \dfrac {p^2} {1 - p}
}}
{{end-eqn}}
{{qed}}
Category:Excess Kurtosis of Geometric Distribution
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