Kurtosis of Gaussian Distribution

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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:

\(\ds \alpha_4\) \(=\) \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4}\) 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:


\(\ds \alpha_4\) \(=\) \(\ds \frac {\expect {X^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}\)
\(\ds \) \(=\) \(\ds \frac {\paren {\mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4} - 4 \mu \paren {3 \mu \sigma^2 + \mu^3} + 6 \mu^2 \paren {\sigma^2 + \mu^2} - 3 \mu^4} {\sigma^4}\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 - 4 + 6 - 3 } \mu^4 + \paren {6 - 12 + 6 } \mu^2 \sigma^2 + 3 \sigma^4 } {\sigma^4}\)
\(\ds \) \(=\) \(\ds \frac {3 \sigma^4 } {\sigma^4}\)
\(\ds \) \(=\) \(\ds 3\)

$\blacksquare$