Excess Kurtosis of Poisson Distribution

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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:

\(\ds \gamma_2\) \(=\) \(\ds \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3\) Kurtosis in terms of Non-Central Moments
\(\ds \) \(=\) \(\ds \frac {\expect {X^4} - 4 \lambda \paren {\lambda^3 + 3 \lambda^2 + \lambda} + 6 \lambda^2 \paren {\lambda^2 + \lambda} - 3 \lambda^4} {\lambda^2} - 3\) Skewness of Poisson Distribution, Variance of Poisson Distribution: Proof 1


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:

\(\ds \map {M_X} t\) \(=\) \(\ds \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1}^2 e^{\lambda \paren {e^t - 1} + t}\)
\(\ds \) \(=\) \(\ds \lambda^3 e^{\lambda \paren {e^t - 1} + 3t} + 3 \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda e^{\lambda \paren {e^t - 1} + t}\) expanding

So:

\(\ds \map {M_X'} t\) \(=\) \(\ds \lambda^3 \paren {\lambda e^t + 3} e^{\lambda \paren {e^t - 1} + 3t} + 3 \lambda^2 \paren {\lambda e^t + 2} e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}\) Chain Rule for Derivatives, Derivative of Exponential Function, Derivative of Power
\(\ds \) \(=\) \(\ds \lambda^4 e^{\lambda \paren {e^t - 1} + 4t} + 6 \lambda^3 e^{\lambda \paren {e^t - 1} + 3t} + 7 \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda e^{\lambda \paren {e^t - 1} + t}\) Exponential of Sum

Setting $t = 0$:

\(\ds \expect {X^4}\) \(=\) \(\ds \lambda^4 e^{\lambda \paren {e^0 - 1} + 0} + 6 \lambda^3 e^{\lambda \paren {e^0 - 1} + 0} + 7 \lambda^2 e^{\lambda \paren {e^0 - 1} + 0} + \lambda e^{\lambda \paren {e^0 - 1} + 0}\)
\(\ds \) \(=\) \(\ds \lambda^4 + 6 \lambda^3 + 7 \lambda^2 + \lambda\) Exponential of Zero

Plugging this result back into our equation above:


\(\ds \gamma_2\) \(=\) \(\ds \frac {\expect {X^4} - 4 \lambda \paren {\lambda^3 + 3 \lambda^2 + \lambda} + 6 \lambda^2 \paren {\lambda^2 + \lambda} - 3 \lambda^4} {\lambda^2} - 3\)
\(\ds \) \(=\) \(\ds \frac {\paren {\lambda^4 + 6 \lambda^3 + 7 \lambda^2 + \lambda} - 4 \lambda \paren {\lambda^3 + 3 \lambda^2 + \lambda} + 6 \lambda^2 \paren {\lambda^2 + \lambda} - 3 \lambda^4} {\lambda^2} - 3\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 - 4 + 6 - 3 } \lambda^4 + \paren {6 - 12 + 6 } \lambda^3 + \paren {7 - 4 } \lambda^2 + \lambda } {\lambda^2} - 3\)
\(\ds \) \(=\) \(\ds \frac {3 \lambda^2 + \lambda } {\lambda^2} - 3\)
\(\ds \) \(=\) \(\ds \frac 1 \lambda\)

$\blacksquare$

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