# 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:

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

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.

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

$\blacksquare$