Excess Kurtosis of Bernoulli Distribution

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Theorem

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.

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

$\gamma_2 = \dfrac {1 - 6 p q} {p q}$

where $q = 1 - p$.


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 Bernoulli Distribution, we have:

$\mu = p$

By Variance of Bernoulli Distribution, we have:

$\sigma = \sqrt {p q}$

So:

\(\ds \gamma_2\) \(=\) \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3\) Kurtosis in terms of Non-Central Moments
\(\ds \) \(=\) \(\ds \frac {p - 4p^2 + 6p^3 - 3 p^4} {\paren {p q}^2} - 3\) Raw Moment of Bernoulli Distribution
\(\ds \) \(=\) \(\ds \frac {p \paren {1 - 4p + 6p^2 - 3p^3} } {\paren {p q}^2} - 3\)
\(\ds \) \(=\) \(\ds \frac {p \paren {1 - p} \paren {3p^2 - 3p + 1} } {\paren {p q}^2} - 3\)
\(\ds \) \(=\) \(\ds \frac {p q \paren {1 - 3 p \paren {1 - p} } - 3 \paren {p q}^2} {\paren {p q}^2}\)
\(\ds \) \(=\) \(\ds \frac {1 - 3 p q - 3 p q} {p q}\)
\(\ds \) \(=\) \(\ds \frac {1 - 6 p q} {p q}\)

$\blacksquare$