The Kurtosis of a random variable $X$ is the fourth moment of $X$. For a normal distribution, we often see it in the textbook as:

$$ \begin{align}\label{kurt} Kurt(X) = E\left[\left(\frac{X-\mu}{\sigma}\right)^4\right] - 3 \end{align} $$

My textbook explains:

“We subtract $3$ to make any Normal distribution have kurtosis $0$ "

I’ve always wondered how $E\left[\left(\frac{X-\mu}{\sigma}\right)^4\right]$ is calculated to be 3. I will attempt to derive this solution.

Assumptions

The Normal Distribution has Probability Density Function (PDF):

$$ \begin{align}\label{eq1} f(x | \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \end{align} $$

We are dealing with standard normal, so mean $\mu = 0$, and variance $\sigma^2 = 1$. We can now write Eq $\ref{eq1}$ as

$$ \begin{align}\label{eq2} f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(x)^2}{2}} \end{align} $$

We need to find $E\left[\left(\frac{X-\mu}{\sigma}\right)^4\right]$, which can now be written as $E[X^4]$.

Derivation

$$ \begin{align*} E[X^4] &= \int^{\infty}{-\infty}x^4 f(x)dx \\ &= \frac{1}{\sqrt{2\pi}}\int^{\infty}{-\infty}x^4 e^{-\frac{x^2}{2}}dx \\ \end{align*} $$

We will evaluate this integral via integration by parts. Recall that

$$ \begin{align*} \int udv = uv - \int vdu \end{align*} $$

Let $u = x^3$, $du = 3x^2dx$, $dv = xe^{-\frac{x^2}{2}}dx$, $v = -e^{-\frac{x^2}{2}}$:

$$ \begin{align*} E[X^4] &= \frac{1}{\sqrt{2\pi}}\left(x^3 \left(-e^{-\frac{x^2}{2}}\right) - \int^{\infty}{-\infty}\left(-e^{-\frac{x^2}{2}}\right)3x^2dx\right) \\ &= \frac{1}{\sqrt{2\pi}}\left(-x^3e^{-\frac{x^2}{2}} + 3\int{-\infty}^{\infty}x^2e^{-\frac{x^2}{2}}dx\right) \end{align*} $$

We can begin some evaluations of the terms here. We’ll use L’Hôpital’s rule to evaluate the first term in the parentheses, $-x^3e^{-\frac{x^2}{2}}$. We know that $e^{-\frac{x^2}{2}}$ goes to zero faster than $x^3$ goes to infinity.

$$ \begin{align*} \lim_{x\to\infty} \frac{-x^3}{e^{\frac{x^2}{2}}} &= \lim_{x\to\infty} \frac{-3x^2}{xe^{\frac{x^2}{2}}} \\ &= \lim_{x\to\infty} \frac{-6x}{x^2e^{\frac{x^2}{2}}} \\ &= \lim_{x\to\infty} \frac{-6}{x^3e^{\frac{x^2}{2}}} \\ &= 0 \end{align*} $$

Therefore, this term is $0$. We are now left with:

$$ \begin{align*} E[X^4] &= \frac{1}{\sqrt{2\pi}}\left(3\int_{-\infty}^{\infty}x^2e^{-\frac{x^2}{2}}dx\right) \\ &= 3\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{x^2}{2}}dx\right) \end{align*} $$

Notice that $\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{x^2}{2}}dx\right)$ is actually the second moment $E[X^2]$, or as we know it, the variance of the standard normal distribution. You may recognize $Var(X) = E[X^2] - [EX]^2$, with the second term being the mean ($\mu$) squared. For a standard normal distribution, with $\mu = 0$, $Var(x) = E[X^2]$. Additionally, we know that for a standard normal, the variance $\sigma^2 = 1$. Therefore:

$$ \begin{align*} E[X^4] &= 3\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{x^2}{2}}dx\right) \\ &= 3 \cdot E[X^2] \\ &= 3 \cdot Var(X) \\ &= 3 \cdot (1) \\ &= 3 \end{align*} $$

$E[X^4]$ is shown to be 3, and we are now done. As a sanity check, we can use R to do a Monte Carlo simulation of a standard normal distribution and calculate the kurtosis:

# Standard Normal Distribution
set.seed(123)
n.sample <- rnorm(n = 100000, mean = 0, sd = 1)

# Kurtosis
library(moments)
kurtosis(n.sample) # we get 2.990068

Close enough! $\blacksquare$