Skewness of Binomial Distribution
Theorem
Let $X$ be a discrete random variable with a Binomial distribution with parameter $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the skewness $\gamma_1$ of $X$ is given by:
- $\gamma_1 = \dfrac {1 - 2 p} {\sqrt {n p q} }$
where $q = 1 - p$.
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 Binomial Distribution:
- $\mu = n p$
By Variance of Binomial Distribution, we also have:
- $\var X = \sigma^2 = n p \paren {1 - p}$
so:
- $\sigma = \sqrt {n p \paren {1 - p} }$
To 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 Binomial Distribution, this is given by:
- $\map {M_X} t = \paren {1 - p + p e^t}^n$
From Moment in terms of Moment Generating Function:
- $\expect {X^3} = \map {M_X} 0$
So:
| \(\ds \map {M_X'} t\) | \(=\) | \(\ds n p e^t \paren {1 - p + p e^t}^{n - 1}\) | Chain Rule for Derivatives, Derivative of Power | |||||||||||
| \(\ds \map {M_X} t\) | \(=\) | \(\ds n p e^t \paren {1 - p + p e^t}^{n - 1} + n \paren {n - 1} p^2 e^{2 t} \paren {1 - p + p e^t}^{n - 2}\) | Product Rule for Derivatives, Chain Rule for Derivatives, Derivative of Power | |||||||||||
| \(\ds \map {M_X} t\) | \(=\) | \(\ds \map {M_X} t + 2 n \paren {n - 1} p^2 e^{2 t} \paren {1 - p + p e^t}^{n - 2} + n \paren {n - 1} \paren {n - 2} p^3 e^{3 t} \paren {1 - p + p e^t}^{n - 3}\) | Product Rule for Derivatives, Chain Rule for Derivatives, Derivative of Power |
Setting $t = 0$:
| \(\ds \expect {X^3}\) | \(=\) | \(\ds \map {M_X} 0 + 2 n \paren {n - 1} p^2 \paren {1 - p + p}^{n - 2} + n \paren {n - 1} \paren {n - 2} p^3 \paren {1 - p + p}^{n - 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n p \paren {1 - p + p}^{n - 1} + n \paren {n - 1} p^2 \paren {1 - p + p}^{n - 2} + 2 n \paren {n - 1} p^2 + n \paren {n - 1} \paren {n - 2} p^3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n p + 3 n \paren {n - 1} p^2 + n \paren {n - 1} \paren {n - 2} p^3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3\) |
Plugging this result back into our equation above:
| \(\ds \gamma_1\) | \(=\) | \(\ds \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n p + 3 n^2 p^2 - 3 n p^2 + n^3 p^3 - 3 n^2 p^3 + 2 n p^3 - 3 n^2 p^2 \paren {1 - p} - n^3 p^3} {\paren {n p \paren {1 - p} }^{3/2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n p - 3 n p^2 + 2 n p^3} {\paren {n p \paren {1 - p} }^{3/2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {n p \paren {1 - p} \paren {1 - 2 p} } {\paren {n p \paren {1 - p} }^{3/2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 - 2 p} {\sqrt {n p \paren {1 - p} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 - 2 p} {\sqrt {n p q} }\) | $q = 1 - p$ |
$\blacksquare$