# Skewness of Continuous Uniform Distribution

## Theorem

Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$.

Then the skewness $\gamma_1$ of $X$ is equal to $0$.

## Proof

From the definition of skewness:

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

where:

- $\mu$ is the mean of $X$
- $\sigma$ is the standard deviation of $X$.

From the definition of the continuous uniform distribution, $X$ has probability density function:

- $\map {f_X} x = \dfrac 1 {b - a}$

So, from Expectation of Function of Continuous Random Variable:

- $\displaystyle \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x$

Then:

\(\displaystyle \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x\) | \(=\) | \(\displaystyle \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \frac {a + b} 2}^3 \rd x\) | Expectation of Continuous Uniform Distribution | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\paren {b - a} \sigma^3} \int_{a - \frac {a + b} 2}^{b - \frac {a + b} 2} u^3 \rd u\) | substituting $u = x - \dfrac {a + b} 2$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u\) |

We have:

- $\paren {-u}^3 = -u^3$

so we can see the integrand is odd.

So, by Definite Integral of Odd Function:

- $\displaystyle \gamma_1 = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u = 0$

$\blacksquare$