Excess Kurtosis of Chi-Squared Distribution

Theorem
Let $n$ be a strictly positive integer.

Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

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


 * $\gamma_2 = \dfrac {12} n$

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 Chi-Squared Distribution:


 * $\mu = n$

By Variance of Chi-Squared Distribution:


 * $\sigma = \sqrt {2 n}$

As is shown in Skewness of Chi-Squared Distribution, we have:


 * $\expect {X^3} = n^3 + 6 n^2 + 8 n$

We also have:

So: