91Ó°ÊÓ

For all numbers a and b, \(a < b\).

Let X be a random variable such that \(X \sim Uniform\left( {a,b} \right)\).

The PDF of the random variable X is:

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{\frac{1}{{b - a}};{\rm{ if a}} < x < b}\\{0;{\rm{ otherwise}}}\end{aligned}} \right.\)

\(\begin{aligned}{c}E\left( X \right) = \int\limits_a^b {xf\left( x \right)dx} \\ = \int\limits_a^b {x\frac{1}{{b - a}}dx} \\ = \frac{1}{{b - a}}\int\limits_a^b {xdx} \\ = \frac{1}{{b - a}}\left( {\frac{{{x^2}}}{2}} \right)_a^b\\ = \frac{{b + a}}{2}\end{aligned}\)

Therefore, \(E\left( X \right) = \frac{{b + a}}{2}\).

\(\begin{aligned}{c}E\left( {{X^2}} \right) &= \int\limits_a^b {{x^2}f\left( x \right)dx} \\ &= \int\limits_a^b {{x^2}\frac{1}{{b - a}}dx} \\ &= \frac{1}{{b - a}}\int\limits_a^b {{x^2}dx} \\ &= \frac{1}{{b - a}}\left( {\frac{{{x^3}}}{3}} \right)_a^b\\ &= \frac{{{b^2} + ab + {a^2}}}{3}\end{aligned}\)

Therefore, \(E\left( {{X^2}} \right) = \frac{{{b^2} + ab + {a^2}}}{3}\).

Now, we know that

\(\begin{aligned}{}Var\left( X \right) &= E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}\\ &= \frac{{{b^2} + ab + {a^2}}}{3} - {\left( {\frac{{b + a}}{2}} \right)^2}\\ &= \frac{{{b^2} + ab + {a^2}}}{3} - \frac{{{b^2} + 2ab + {a^2}}}{4}\\ &= \frac{{{b^2} - 2ab + {a^2}}}{{12}}\\ = \frac{{{{\left( {b - a} \right)}^2}}}{{12}}\end{aligned}\)

Therefore, \(Var\left( X \right) = \frac{{{{\left( {b - a} \right)}^2}}}{{12}}\).