91Ó°ÊÓ

From the given information the random variables\({X_1}, \ldots ,{X_n}\)form a random sample of size\(n\)from the uniform distribution on the interval\(\left[ {0,1} \right]\). So the pdf of\({X_1}, \ldots ,{X_n}\)is given by

\(\begin{array}{c}f\left( {{X_1}, \ldots ,{X_n}} \right) = \frac{1}{{1 - 0}}\\ = 1\end{array}\)

Where the range is given by

\(\begin{array}{l}0 \le {X_1} \ldots \le {X_n} \le 1\\0 \le {X_{\left( 1 \right)}} \ldots \le {X_{\left( n \right)}} \le 1\end{array}\)

Then\({X_{\left( 1 \right)}} = 0\) and\({X_{\left( n \right)}} = 1\)

And also given \({Y_1} = \min \left\{ {{X_1}, \ldots ,{X_n}} \right\}\) and \({Y_n} = \max \left\{ {{X_1}, \ldots ,{X_n}} \right\}\)

From the given distribution the mean will be calculated by order statistics. So by calculating it to get

\(\begin{array}{c}E\left( {{Y_1}} \right) = E\left[ {\min \left\{ {{X_1}, \ldots ,{X_n}} \right\}} \right]\\ = \min \left[ {E\left\{ {{X_1}, \ldots ,{X_n}} \right\}} \right]\\ = E\left[ {{X_{\left( 1 \right)}}} \right]\\ = \frac{1}{{n + 1}}\end{array}\)

Previous method apply to calculating the distribution to get

\(\begin{array}{c}E\left( {{Y_n}} \right) = E\left[ {\max \left\{ {{X_1}, \ldots ,{X_n}} \right\}} \right]\\ = \max \left[ {E\left\{ {{X_1}, \ldots ,{X_n}} \right\}} \right]\\ = E\left[ {{X_{\left( n \right)}}} \right]\\ = \frac{n}{{n + 1}}\end{array}\)