91Ó°ÊÓ

X has a discrete uniform distribution on the integers \(1, \ldots \ldots ,n\).

The probability mass function of X is \(f\left( x \right) = \left\{ {\begin{aligned}{*{20}{l}}{\frac{1}{n};x \in \left\{ {1,2,....,n} \right\}}\\{0;{\rm{ otherwise}}}\end{aligned}} \right.\).

Now,

\(\begin{aligned}{}E\left( X \right) &= \sum\limits_{x = 1}^n {xf\left( x \right)} \\ &= \sum\limits_{x = 1}^n {x\frac{1}{n}} \\ &= \frac{1}{n}\sum\limits_{x = 1}^n x \\ &= \frac{1}{n}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)\\ &= \frac{{n + 1}}{2}\end{aligned}\)

\(\begin{aligned}{}E\left( {{X^2}} \right) &= \sum\limits_{x = 1}^n {{x^2}f\left( x \right)} \\ &= \sum\limits_{x = 1}^n {{x^2}\frac{1}{n}} \\ &= \frac{1}{n}\sum\limits_{x = 1}^n {{x^2}} \\ &= \frac{1}{n}\left\{ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right\}\\ &= \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\end{aligned}\)

\(\begin{aligned}{}Var\left( X \right) &= E\left( {{X^2}} \right) - {\left\{ {E\left( X \right)} \right\}^2}\\ &= \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - {\left( {\frac{{n + 1}}{2}} \right)^2}\\ &= \frac{{n + 1}}{2}\left( {\frac{{2n + 1}}{3} - \frac{{n + 1}}{2}} \right)\\ &= \frac{{n + 1}}{2}\left( {\frac{{n - 1}}{6}} \right)\\ &= \frac{{{n^2} - 1}}{{12}}\end{aligned}\)

Thus, \(Var\left( X \right) = \frac{{{n^2} - 1}}{{12}}\).