Question

Suppose a random sample of size 3 is taken from a distribution with the probability density function \[ f(x) = \begin{cases} 2x, & 0<x<1, \\ 0, & \text{elsewhere}. \end{cases} \] If \( p \) is the probability that the largest sample observation is at least twice the smallest sample observation, then the value of \( p \) (rounded off to three decimal places) is ________

Hint:

For problems involving order statistics, use the joint pdf of the smallest and largest observations. Ensure limits match the condition given (e.g., \( X_{(3)} \geq 2X_{(1)} \)).

Answer 1

Correct Answer: 0.436

We are given \( f(x) = 2x \), \( 0<x<1 \), and \( n = 3 \). Let the order statistics be \( X_{(1)} \leq X_{(2)} \leq X_{(3)} \). We need to find: \[ p = P(X_{(3)} \geq 2X_{(1)}). \] The joint pdf of \( X_{(1)} \) and \( X_{(3)} \) for a sample of size \( n = 3 \) is: \[ f_{X_{(1)}, X_{(3)}}(x_1, x_3) = 6 [F(x_3) - F(x_1)]^{1} f(x_1) f(x_3), \quad 0<x_1<x_3<1. \] Since \( F(x) = x^2 \) and \( f(x) = 2x \), we get: \[ f_{X_{(1)}, X_{(3)}}(x_1, x_3) = 6 [x_3^2 - x_1^2] (2x_1)(2x_3) = 24x_1x_3(x_3^2 - x_1^2). \] Now, the required probability is: \[ p = \int_0^{0.5} \int_{2x_1}^{1} 24x_1x_3(x_3^2 - x_1^2) \, dx_3 \, dx_1. \] Evaluating this integral gives approximately: \[ p = 0.4377. \] Thus, the value of \( p \) is \( \boxed{0.438} \).