Theorem: Excess Kurtosis of Log Normal Distribution

Let $X$ be a continuous random variable with the Log Normal distribution with $\mu \in \R, \sigma \in \R_{> 0}$.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \map \exp {4 \sigma^2} + 2 \map \exp {3 \sigma^2} + 3 \map \exp {2 \sigma^2} - 6$

Proof:
From Kurtosis in terms of Non-Central Moments, we have: 
:$\gamma_2 = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Log Normal Distribution we have: 
:$\mu = \map \exp {\mu + \dfrac {\sigma^2} 2}$ 
By Variance of Log Normal Distribution we have: 
:$\sigma = \map \exp {\mu + \dfrac {\sigma^2} 2} \sqrt {\paren {\map \exp {\sigma^2} - 1} }$ 
From Raw Moment of Log Normal Distribution, we have:
:$\expect {X^n} = \map \exp {n \mu + \dfrac {n^2 \sigma^2} 2}$
Hence:
{{begin-eqn}}
{{eqn | l = \gamma_2
      | r = \frac {\map \exp {4 \mu + \dfrac {16 \sigma^2} 2} - 4 \map \exp {\mu + \dfrac {\sigma^2} 2} \paren {\map \exp {3 \mu + \dfrac {9 \sigma^2} 2} } + 6 \paren {\map \exp {\mu + \dfrac {\sigma^2} 2} }^2 \paren {\map \exp {2 \mu + \dfrac { 4 \sigma^2} 2} } - 3 \paren {\map \exp {\mu + \dfrac {\sigma^2} 2} }^4} {\paren {\map \exp {\mu + \dfrac {\sigma^2} 2} \sqrt {\paren {\map \exp {\sigma^2} - 1} } }^4} - 3
      | c = 
}}
{{eqn | r = \frac {\map \exp {4 \mu + 8 \sigma^2} - 4 \map \exp {4 \mu + 5 \sigma^2} + 6 \map \exp {2 \mu + \sigma^2} \map \exp {2 \mu + 2 \sigma^2} - 3 \map \exp {4 \mu + 2 \sigma^2} } {\map \exp {4 \mu + 2 \sigma^2} \paren {\map \exp {\sigma^2} - 1}^2} - 3
      | c = Power of Power and Product of Powers
}}
{{eqn | r = \frac {\map \exp {4 \mu + 8 \sigma^2} - 4 \map \exp {4 \mu + 5 \sigma^2} + 6 \map \exp {4 \mu + 3 \sigma^2} - 3 \map \exp {4 \mu + 2 \sigma^2} } {\map \exp {4 \mu + 2 \sigma^2} \paren {\map \exp {\sigma^2} - 1}^2} - 3
      | c = Product of Powers
}}
{{eqn | r = \frac {\map \exp {4 \mu + 2 \sigma^2} \paren {\map \exp {6 \sigma^2} - 4 \map \exp {3 \sigma^2} + 6 \map \exp {\sigma^2} - 3} } {\map \exp {4 \mu + 2 \sigma^2} \paren {\map \exp {\sigma^2} - 1}^2} - 3
      | c = factoring out $\map \exp {4 \mu + 2 \sigma^2}$
}}
{{eqn | r = \frac {\map \exp {6 \sigma^2} - 4 \map \exp {3 \sigma^2} + 6 \map \exp {\sigma^2} - 3} {\paren {\map \exp {\sigma^2} - 1}^2} - 3
      | c = cancelling the $\map \exp {4 \mu + 2 \sigma^2}$
}}
{{eqn | r = \frac {\map \exp {6 \sigma^2} - 4 \map \exp {3 \sigma^2} + 3 \map \exp {2 \sigma^2} - 3 \map \exp {2 \sigma^2} + 6 \map \exp {\sigma^2} - 3} {\paren {\map \exp {\sigma^2} - 1}^2} - 3
      | c = adding and subtracting $3 \map \exp {2 \sigma^2}$
}}
{{eqn | r = \frac {\map \exp {6 \sigma^2} - 4 \map \exp {3 \sigma^2} + 3 \map \exp {2 \sigma^2} - 3 \paren {\map \exp {\sigma^2} - 1}^2} {\paren {\map \exp {\sigma^2} - 1}^2} - 3
      | c = Square of Sum
}}
{{eqn | r = \frac {\map \exp {6 \sigma^2} - 4 \map \exp {3 \sigma^2} + 3 \map \exp {2 \sigma^2} } {\paren {\map \exp {\sigma^2} - 1}^2} - 6
      | c = 
}}
{{eqn | r = \frac {\paren {\map \exp {2 \sigma^2} - 2 \map \exp {\sigma^2} + 1} \paren {\map \exp {4 \sigma^2}  + 2 \map \exp {3 \sigma^2} + 3 \map \exp {2 \sigma^2} } } {\paren {\map \exp {\sigma^2} - 1}^2} - 6
      | c = rewriting
}}
{{eqn | r = \map \exp {4 \sigma^2} + 2 \map \exp {3 \sigma^2} + 3 \map \exp {2 \sigma^2} - 6
      | c = cancelling $\paren {\map \exp {\sigma^2} - 1}^2$
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Log Normal Distribution
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