Kurtosis of Gaussian Distribution

Theorem
Let $X$ be a continuous random variable with a Gaussian distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.

Then the kurtosis $\alpha_4$ of $X$ is equal to $3$.

Proof
From the definition of kurtosis, we have:


 * $\alpha_4 = \expect {\paren {\dfrac {X - \mu} \sigma}^4}$

where:
 * $\mu$ is the expectation of $X$.
 * $\sigma$ is the standard deviation of $X$.

By Expectation of Gaussian Distribution, we have:


 * $\mu = \mu$

By Variance of Gaussian Distribution, we have:


 * $\sigma = \sigma$

So:

To calculate $\alpha_4$, we must calculate $\expect {X^4}$.

We find this using the moment generating function of $X$, $M_X$.

By Moment Generating Function of Gaussian Distribution, this is given by:


 * $\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

From Moment in terms of Moment Generating Function:


 * $\expect {X^4} = \map { {M_X}^{\paren 4} } 0$

In Skewness of Gaussian Distribution: Proof 2, it is shown that:

So:


 * $\map { {M_X}^{\paren 4} } t = \paren {\mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4 + \paren {4 \mu^3 \sigma^2 + 12 \mu \sigma^4 } t + \paren {6 \mu^2 \sigma^4 + 6 \sigma^6 } t^2 + 4 \mu \sigma^6 t^3 + \sigma^8 t^4} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

Setting $t = 0$:

Plugging this result back into our equation above: