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