Skewness of Poisson Distribution

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Theorem

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

Then the skewness $\gamma_1$ of $X$ is given by:

$\gamma_1 = \dfrac 1 {\sqrt \lambda}$


Proof

From Skewness in terms of Non-Central Moments:

$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.

We have, by Expectation of Poisson Distribution:

$\expect X = \lambda$

By Variance of Poisson Distribution:

$\var X = \sigma^2 = \lambda$

so:

$\sigma = \sqrt \lambda$


To now calculate $\gamma_1$, we must calculate $\expect {X^3}$.

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:

$\expect {X^3} = \map {M_X'''} 0$

In Variance of Poisson Distribution: Proof 3, it is shown that:

$\ds \map {M_X''} t = \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}$

So:

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

Setting $t = 0$:

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

So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\lambda^3 + 3 \lambda^2 + \lambda - 3 \lambda^2 - \lambda^3} {\lambda^{3/2} }\)
\(\ds \) \(=\) \(\ds \frac \lambda {\lambda^{3/2} }\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt \lambda}\)

$\blacksquare$