Question

Let W denote the range of a random sample of nobservations from the uniform distribution on the interval[0, 1]. Determine the value of

\({\bf{Pr}}\left( {{\bf{W > 0}}{\bf{.9}}} \right)\).

Short answer

\[1 - \left( {{{0.9}^{n - 1}}n - {{0.9}^n}n + {{0.9}^n}} \right)\]

Step-by-step solution

Given information

The random variable X follows uniform distribution on the interval [0,1], i.e.,\(X \sim U\left[ {0,1} \right]\)Here \({X_1} \ldots {X_n}\)is a random sample from a uniform distribution \(U\left( {0,1} \right)\).

\(\begin{aligned}{Y_1} &= \min \left\{ {{X_1} \ldots {X_n}} \right\}\\{Y_n} &= \max \left\{ {{X_1} \ldots {X_n}} \right\}\end{aligned}\)

Obtain the PDF and CDF of X

The pdf of a uniform distribution is obtained by using the formula: \(\frac{1}{{b - a}};a \le x \le b\)

Here, \(a = 0,b = 1\)

\({f_x} = \frac{1}{{1 - 0}};0 \le x \le 1\)

\(\begin{aligned}{f_x} &= 1,0 \le x \le 1\\ &= 0,o.w.\end{aligned}\)

The CDF of a uniform distribution is obtained by using the formula:

\(\begin{aligned}{F_X}\left( x \right) &= P\left( {X \le x} \right)\\ &= \frac{{x - a}}{{b - a}}\\ &= \frac{{x - 0}}{{1 - 0}}\\ &= x\,{\rm{for}}\,\,0 < x < 1\end{aligned}\)

Obtain the joint p.d.f of a random variable

The joint distribution of \({Y_1},{Y_n}\)is,

\(\begin{aligned}G\left( {{y_1},{y_n}} \right) &= \Pr \left( {{Y_1} \le {y_1}\,and\,{Y_n} \le {y_n}} \right)\\ &= \Pr \left( {{Y_n} \le {y_n}} \right) - \Pr \left( {\,{Y_n} \le {y_n}\,and\,{Y_1} > {y_1}} \right)\\ &= \Pr \left( {{Y_n} \le {y_n}} \right) - \Pr \left( {{y_1} < {X_1} \le {y_n},{y_1} < {X_2} \le {y_n}, \ldots ,{y_1} < {X_n} \le {y_n}} \right)\\ &= {G_n}\left( {{y_n}} \right) - \prod\limits_{i = 1}^n {\Pr \left( {{y_1} < {X_i} \le {y_n}} \right)} \\ &= {\left( {F\left( {{y_n}} \right)} \right)^n} - {\left( {F\left( {{y_n}} \right) - F\left( {{y_1}} \right)} \right)^n}\end{aligned}\)

The bivariate joint p.d.f is found by differentiating the joint CDF

\(\begin{aligned}g\left( {{y_1},{y_n}} \right) &= \frac{{{\partial ^2}G\left( {{y_1},{y_n}} \right)}}{{\partial {y_1}\partial {y_n}}}, - \infty < {y_1} < {y_n} < \infty \\ &= n\left( {n - 1} \right){\left( {F\left( {{y_n}} \right) - F\left( {{y_1}} \right)} \right)^{n - 2}}f\left( {{y_1}} \right)f\left( {{y_n}} \right)\\ &= \left\{ \begin{aligned}n\left( {n - 1} \right){\left( {\left( {{y_n}} \right) - \left( {{y_1}} \right)} \right)^{n - 2}},0 < {y_1} < {y_n} < 1\\0,\,otherwise\end{aligned} \right.\end{aligned}\)

Obtain the Distribution of the Range

The random variable \(W = {Y_n} - {Y_1}\) is called the range of the sample.

Let \(Z = {Y_n} - W\)

\(\begin{aligned}h\left( w \right) &= \int\limits_0^{1 - w} {n\left( {n - 1} \right)} {w^{n - 2}}dz\\ &= \left\{ \begin{aligned}n\left( {n - 1} \right){w^{n - 2}}\left( {1 - w} \right),0 < w < 1\\0,\,otherwise\end{aligned} \right.\end{aligned}\)

Calculate the required probability

\(\begin{aligned}\Pr \left( {W > 0.9} \right) &= \int\limits_{0.9}^1 {n\left( {n - 1} \right){w^{n - 2}}\left( {1 - w} \right)} dw\\ &= n\left( {n - 1} \right)\int\limits_{0.9}^1 {\left( {{w^{n - 2}} \times {w^{n - 1}}} \right)} dw\\ &= \left( {n{w^{n - 1}} \times \left( {n - 1} \right){w^n}} \right)_{0.9}^1\\ &= 1 - \left( {{{0.9}^{n - 1}}n - {{0.9}^n}n + {{0.9}^n}} \right)\end{aligned}\)

Therefore, the answer is\({\bf{1 - }}\left( {{\bf{0}}{\bf{.}}{{\bf{9}}^{{\bf{n - 1}}}}{\bf{n - 0}}{\bf{.}}{{\bf{9}}^{\bf{n}}}{\bf{n + 0}}{\bf{.}}{{\bf{9}}^{\bf{n}}}} \right)\)