Excess Kurtosis of Exponential Distribution

Theorem

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.

Then the excess kurtosis $\gamma_2$ of $X$ is equal to $6$.

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 Exponential Distribution we have:

$\mu = \beta$

By Variance of Exponential Distribution we have:

$\sigma = \beta$

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

\(\ds \)

\(=\)

\(\ds \frac {4! \beta^4 - 4 \beta \paren {3! \beta^3} + 6 \beta^2 \paren {2! \beta^2} - 3 \beta^4} {\beta^4} - 3\)

Raw Moment of Exponential Distribution

\(\ds \)

\(=\)

\(\ds \frac {24 \beta^4 - 24 \beta^4 + 12 \beta^4 - 3 \beta^4} {\beta^4} - 3\)

\(\ds \)

\(=\)

\(\ds \frac {9 \beta^4} {\beta^4} - 3\)

\(\ds \)

\(=\)

\(\ds 6\)

$\blacksquare$