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:

We have:


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

so we can see the integrand is odd.

So, by Definite Integral of Odd Function:


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