Skewness of Poisson Distribution

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

Then the skewness of $X$, $\gamma_1$ 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:


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

From Moment in terms of Moment Generating Function:


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

In Variance of Poisson Distribution, it is shown that:


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

So:

Setting $t = 0$:

So: