Question

Suppose that thenrandom variablesX1, . . . , Xnform a random sample from a continuous distribution for which the p.d.f. isf. Determine the probability that at leastk of thesenrandom variables will lie in a specified intervala≤x≤b.

Short answer

The probability that at least \(k\) is \(\frac{2}{{\left( {b - a} \right)}}\) .

Step-by-step solution

Given Information

Here given distribution is continuous distribution for \(n\) random variables.

State the random variables 

For \(n\) random variables we take let \({x_1}, \ldots ,{x_n} \sim u\left( {a,b} \right)\) . Here we assume that the random variables follow uniform distribution.

Compute the probability at least \(k\) 

For \({x_1}, \ldots ,{x_n}\) the pdf is given by

\(f\left( {{x_1}, \ldots ,{x_n}} \right) = \frac{k}{{{{\left( {b - a} \right)}^n}}}\) where the range is \(a \le {x_1}, \ldots ,{x_n} \le b\) and \(k\) is any constant .

We know that sum of all pdf is\(1\). So here we first calculate the constant value is

\(\begin{align}\int_{ - \infty }^\infty {f\left( {{x_1}, \ldots ,{x_n}} \right) = 1} \\\int_a^b {\frac{k}{{{{\left( {b - a} \right)}^n}}}dx = 1} \\k \times \frac{{\left( {b - a} \right)}}{{{{\left( {b - a} \right)}^n}}} &= 1\\k &= {\left( {b - a} \right)^{n - 1}}\end{align}\)

Then find the probability at least\(k\)is

\(\begin{align}p\left( {x \ge k} \right) &= p\left( {x \ge {{\left( {b - a} \right)}^{n - 1}}} \right)\\ &= p\left( {x = {{\left( {b - a} \right)}^n}} \right) + p\left( {x = \frac{1}{{\left( {b - a} \right)}}} \right)\\ &= \int_a^b {{{\left( {b - a} \right)}^{n - 1}}dx + \int_a^b {dx} } \\ &= \frac{2}{{\left( {b - a} \right)}}\end{align}\)

Hence the value is \(\frac{2}{{\left( {b - a} \right)}}\) .