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 $\sigma \in \R_{> 0}$:
- $X \sim \Gaussian \mu {\sigma^2}$
Then the kurtosis $\alpha_4$ of $X$ is equal to $3$.
That is, $\Gaussian \mu {\sigma^2}$ is mesokurtic.
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:
\(\ds \alpha_4\) | \(=\) | \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4}\) | Kurtosis in terms of Non-Central Moments | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\expect {X^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}\) | Skewness of Gaussian Distribution: Proof 2, Variance of Gaussian Distribution: Proof 2 |
To calculate $\alpha_4$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$.
From Moment in terms of Moment Generating Function:
- $\expect {X^4} = \map { {M_X}^{\paren 4} } 0$
From Moment Generating Function of Gaussian Distribution: Fourth Moment:
- $\map { {M_X}^{\paren 4} } t = \paren {3 \sigma^4 + 6 \sigma^2 \paren {\mu + \sigma^2 t}^2 + \paren {\mu + \sigma^2 t}^4} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
Setting $t = 0$ and from Exponential of Zero, we have:
\(\ds \expect {X^4}\) | \(=\) | \(\ds \paren {3 \sigma^4 + 6 \sigma^2 \paren {\mu + \sigma^2 0}^2 + \paren {\mu + \sigma^2 0}^4} \map \exp {\mu 0 + \dfrac 1 2 \sigma^2 0^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4\) | Exponential of Zero |
Plugging this result back into our equation above:
\(\ds \alpha_4\) | \(=\) | \(\ds \frac {\expect {X^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {1 - 4 + 6 - 3 } \mu^4 + \paren {6 - 12 + 6 } \mu^2 \sigma^2 + 3 \sigma^4 } {\sigma^4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 \sigma^4 } {\sigma^4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3\) |
$\blacksquare$
Also see
- Definition:Mesokurtic: a random variable with a kurtosis equal to $3$
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): kurtosis
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): kurtosis
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): kurtosis
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): kurtosis
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): kurtosis