Excess Kurtosis of Hat-Check Distribution

Theorem
Let $X$ be a discrete random variable with a Hat-Check distribution with parameter $n$. ($n \gt 3$)

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


 * $\gamma_2 = 1$

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 $X$.

We have, by Expectation of Hat-Check Distribution:


 * $\expect X = n - 1$

By Variance of Hat-Check Distribution:


 * $\var X = \sigma^2 = 1$
 * $\expect {X^2} = \paren {n - 1 }^2 + 1$

By Skewness of Hat-Check Distribution:


 * $\expect {X^3} = n^3 - 3n^2 + 6n - 5$

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

To help complete the sum above, recall that:

Therefore:

So: