Question

Suppose that the n variables\({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the uniform distribution on the interval [0, 1]and that the random variables \({{\bf{Y}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{Y}}_{\bf{n}}}\) are defined as in Eq. (3.9.8). Determine the value of \({\bf{Pr}}\left( {{{\bf{Y}}_{\bf{1}}} \le {\bf{0}}{\bf{.1}}\;{\bf{and}}\;{\bf{Y}}_{\bf{n}}^{} \le {\bf{0}}{\bf{.8}}} \right)\)

Short answer

\[{0.8^n} - {0.7^n}\]

Step-by-step solution

Given information

Here,\({X_1} \ldots {X_n}\)is a random sample from 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,

\(\begin{aligned}\Pr \left( {{Y_1} \le 0.1\;{\rm{and}}\;Y_n^{} \le 0.8} \right) &= P\left( {Y_n^{} \le 0.8} \right) - P\left( {{Y_1} > 0.1,Y_n^{} \le 0.8} \right)\\ &= P\left( {{X_1} \le 0.8, \ldots ,{X_n} \le 0.8} \right) - P\left( {0.1 < {X_1} \le 0.8, \ldots ,0.1 < {X_n} \le 0.8} \right)\\ &= {\left[ {P\left( {{X_1} \le 0.8} \right)} \right]^n} - {\left[ {P\left( {0.1 < {X_1} \le 0.8} \right)} \right]^n}\\ &= {\left[ {F\left( {0.8} \right)} \right]^n} - {\left[ {F\left( {0.8} \right) - F\left( {0.1} \right)} \right]^n}\\ &= {0.8^n} - {\left( {0.8 - 0.1} \right)^n}\\ &= {0.8^n} - {0.7^n}\end{aligned}\)

Hence, the answer is\({\bf{0}}{\bf{.}}{{\bf{8}}^{\bf{n}}}{\bf{ - 0}}{\bf{.}}{{\bf{7}}^{\bf{n}}}\)