Excess Kurtosis of Pareto Distribution

Theorem
Let $X$ be a continuous random variable with the Pareto distribution with $a, b \in \R_{> 0}$.

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


 * $\gamma_2 = \begin {cases} \dfrac {6 \paren { a^3 + a^2 - 6 a - 2} } {a \paren {a - 3} \paren {a - 4} } & 4 < a \\ \text {does not exist} & 4 \ge a \end {cases}$

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 Pareto Distribution we have:


 * $\mu = \dfrac {a b } {\paren {a - 1} }$

By Variance of Pareto Distribution we have:


 * $\sigma = \dfrac {\sqrt a b } {\sqrt {\paren {a - 2} } \paren {a - 1} }$

From Raw Moment of Pareto Distribution, we have:


 * $\ds \expect {X^n} = \begin {cases} \dfrac {a b^n} {a - n} & n < a \\ \text {does not exist} & n \ge a \end {cases}$

Hence: