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July 22, 2025 · 6:43 am

Continuous Uniform Random Variable

My Year 12 Mathematics Methods students are doing continuous random variables at the moment and I thought it would be worthwhile deriving the mean and variance formulas for a uniform continuous random variable.

The probability density function for a uniform random variable is

$\begin{equation*}f(x)= \left \{ {\begin{matrix}\frac{1}{b-a} & a\le x\le b \\0 & \text {elsewhere}\end{matrix}}\end{equation}$

and it looks like

Remember, the mean $\mu$ or expected value $E(X)$ of a continuous random variable is

(1) $\begin{equation*}E(X)=\int xp(x) dx\end{equation*}$

and the variance $\sigma^2$ is

(2) $\begin{equation*}\sigma^2=\int (x-\mu)^2p(x) dx\end{equation*}$

We are going to use equations $1$ and $2$ to find formulae for a uniform continuous random variable.

$\begin{equation*}\mu=\int_a^b x (\frac{1}{b-a}) dx\end{equation}$

$\begin{equation*}\mu=\frac{x^2}{2(b-a)}|\begin{matrix}b\\a\end{matrix}\end{equation}$

$\begin{equation*}\mu=\frac{b^2}{2(b-a)}-\frac{a^2}{2(b-a)}=\frac{b^2-a^2}{2(b-a)}\end{equation}$

Factorise the numerator (using difference of squares)

$\begin{equation*}\mu=\frac{(b-a)(b+a)}{2(b-a)}\end{equation}$

Hence,

$\begin{equation*}\mu=\frac{b+a}{2}\end{equation}$

Now for the variance

$\begin{equation*}\sigma^2=\int_a^b (x-(\frac{a+b}{2}))^2(\frac{1}{b-a}) dx\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(x-\frac{a+b}{2})^3}{3})|\begin{matrix}b\\a\end{matrix}\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}((\frac{(b-\frac{a+b}{2})^3}{3})-(\frac{(a-\frac{a+b}{2})^3}{3}))\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3}{12}+\frac{b^3}{12}+\frac{a^2b}{4}-\frac{ab^2}{4})\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3+b^3+3a^2b-3ab^2}{12})\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{b^3-3b^2a+3ba^2-a^3}{12})\end{equation}$

From the binomial expansion theorem, we know

$\begin{equation*}b^3-3b^2a+3ba^2-a^3=(b-a)^3\end{equation}$

Hence

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(b-a)^3}{12}\end{equation}$

and

$\begin{equation*}\sigma^2=\frac{(b-a)^2}{12}\end{equation}$

Category Archives: Uniform

Continuous Uniform Random Variable

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