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Definition

Given an i.i.d. sequence of random variables \(X_1, X_2, \dots, X_n,\) define \begin{align} & X_{(1)} = \text{min}(\{X_1, X_2, \dots, X_n\}) \\ & X_{(2)} = \text{min}(\{X_1, X_2, \dots, X_n\} - \{X_{(1)}\}) \\ & X_{(3)} = \text{min}(\{X_1, X_2, \dots, X_n\} - \{X_{(1)}, X_{(2)}\}) \\ & \vdots \\ & X_{(n)} = \text{min}(\{X_1, X_2, \dots, X_n\} - \{X_{(1)}, X_{(2)}, \dots, X_{(n-1)}\}) = \text{max}(\{X_1, X_2, \dots, X_n\}) \end{align} The order statistic of \(X_1, X_2, \dots, X_n\) is the probability that \((X_{(1)} = x_1, X_{(2)} = x_2, \dots, X_{(n)} = x_n)\) for some numbers \(x_1 < x_2 < \dots < x_n.\)

Given that \((X_{(1)} = x_1, X_{(2)} = x_2, \dots, X_{(n)} = x_n)\) occurs, there are \(n!\) possible values for \((X_1, X_2, \dots, X_n)\) since there are \(n!\) ways to order the numbers \(x_1, x_2, \dots, x_n.\) So, if we let \(f(x)\) be the pdf of the random variables in the i.i.d. sequence \(X_1, X_2, \dots, X_n,\) the order statistic has pdf \(f(x_1, \dots, x_n)\) defined by \[f(x_1, x_2, \dots, x_n) = n!f(x_1)f(x_2) \dots f(x_n)\]

Example

Let \(X_1,\) \(X_2\) and \(X_3\) be independent, exponentially distributed random variables with parameter \(1.\) What is the probability that all of the random variables are at least \(1\) apart?

▼ Solution:
If \(X_1,\) \(X_2\) and \(X_3\) are at least one apart, then \(X_{(1)} + 1 < X_{(2)}\) and \(X_{(2)} + 1< X_{(3)}.\) The joint pdf of \(X_{(1)}, X_{(2)}, X_{(3)}\) is \(f(x_1, x_2, x_3) = 3!e^{-x_1}e^{-x_2}e^{-x_3}\) over \(0 < x_1 < x_2 < x_3.\)

Given the conditions, we must integrate over \(x_1 + 1 < x_2\) and \(x_2 + 1< x_3.\) \begin{align} \int_0^\infty \int_{x_1+1}^\infty \int_{x_2+1}^\infty 3!e^{-x_1}e^{-x_2}e^{-x_3} dx_3 dx_2 dx_1 & = 6\int_0^\infty e^{-x_1} \int_{x_1+1}^\infty e^{-x_2} \int_{x_2+1}^\infty e^{-x_3} dx_3 dx_2 dx_1 \\ & = 6\int_0^\infty e^{-x_1} \int_{x_1+1}^\infty e^{-x_2} e^{-x_2-1} dx_2 dx_1 \\ & = \frac{6}{e} \int_0^\infty e^{-x_1} \int_{x_1+1}^\infty e^{-2x_2} dx_2 dx_1 \\ & = \frac{6}{e} \int_0^\infty e^{-x_1} \left(\frac{1}{2} e^{-2x_1-2} \right) dx_1 \\ & = \frac{3}{e^3} \int_0^\infty e^{-3x_1} dx_1 \\ & = \frac{3}{e^3} \left(\frac{1}{3}\right) \\ & = \frac{1}{e^3} \\ & \approx 0.05 \end{align}

1. Let \(U_1, U_2, U_3, U_4\) be independent, continuous uniform random variables over \([0, 5].\) What is the probability that some pair of the random variables are within \(1\) of each other?




Unanswered