Excess Kurtosis of Poisson Distribution

Theorem
Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$.

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


 * $\gamma_2 = \dfrac 1 \lambda$

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 Poisson Distribution, we have:


 * $\mu = \lambda$

By Variance of Poisson Distribution, we have:


 * $\sigma = \sqrt {\lambda}$

So:

To calculate $\gamma_2$, we must calculate $\expect {X^4}$.

We find this using the moment generating function of $X$, $M_X$.

By Moment Generating Function of Poisson Distribution, this is given by:


 * $\ds \map {M_X} t = e^{\lambda \paren {e^t - 1} }$

From Moment in terms of Moment Generating Function:


 * $\expect {X^4} = \map {M_X} 0$

In Skewness of Poisson Distribution, it is shown that:

So:

Setting $t = 0$:

Plugging this result back into our equation above: