Question

For the conditions of Exercise 9, determine the probabilitythat the interval from \({Y_1}\;to\;{Y_n}\) will not contain thepoint 1/3.

Short answer

\({\left( {\frac{1}{3}} \right)^n} + {\left( {\frac{2}{3}} \right)^n}\)

Step-by-step solution

Given information

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

Therefore, the PDF of X is expressed as,

\({f_x} = \left\{ \begin{array}{l}\frac{1}{{1 - 0}} = 1\;\;\;\;\;\;\;\;\;\;0 \le x \le 1\\0;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{array} \right.\)

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

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

Defining the probability

We have to find,

\(\begin{aligned}\\Pr \left( {{Y_1} > \frac{1}{3}\;and\;Y_n^{} < \frac{1}{3}} \right)\ &= P\left( {{Y_1} > \frac{1}{3}} \right) + P\left( {{Y_n} < \frac{1}{3}} \right)\\ &= P\left( {{X_1} > \frac{1}{3}, \ldots ,{X_n} > \frac{1}{3}} \right) + P\left( {{X_1} < \frac{1}{3}, \ldots ,{X_n} < \frac{1}{3}} \right)\\ &= {\left( {1 - F\left( {\frac{1}{3}} \right)} \right)^n} - F{\left( {\frac{1}{3}} \right)^n}\\ &= {\left( {\frac{2}{3}} \right)^n} + {\left( {\frac{1}{3}} \right)^n}\end{aligned}\)

Therefore, the answer is \({\left( {\frac{2}{3}} \right)^n} + {\left( {\frac{1}{3}} \right)^n}\)