Question

Let \( X_1, X_2, X_3 \) be i.i.d. random variables from a continuous distribution having probability density function
\[f(x) = \begin{cases} \frac{1}{2}x^{-3}, & \text{if } x > \frac{1}{2} \\0, & \text{if } x \leq \frac{1}{2} \end{cases}.\]
Let \( X_{(1)} = \min\{X_1, X_2, X_3\} \). Then the value of \( 10E(X_{(1)}) \) is equal to __________ (answer in integer).

Answer 1

Correct Answer: 6

1. Distribution of \( X_{(1)} \): - The cumulative distribution function (CDF) of \( X \) is: \[ F(x) = \begin{cases} 0, & \text{if } x \leq \frac{1}{2}, \\ 1 - \frac{1}{x^2}, & \text{if } x > \frac{1}{2}. \end{cases} \] - For the minimum \( X_{(1)} \), the CDF is: \[ F_{X_{(1)}}(x) = 1 - \left(1 - F(x)\right)^3 = \begin{cases} 0, & \text{if } x \leq \frac{1}{2}, \\ 1 - \left(\frac{1}{x^2}\right)^3, & \text{if } x > \frac{1}{2}. \end{cases} \] 2. PDF of \( X_{(1)} \): - Differentiate the CDF to get the probability density function: \[ f_{X_{(1)}}(x) = \frac{d}{dx}F_{X_{(1)}}(x) = \begin{cases} 0, & \text{if } x \leq \frac{1}{2}, \\ \frac{6}{x^7}, & \text{if } x > \frac{1}{2}. \end{cases} \] 3. Expectation of \( X_{(1)} \): - The expectation is given by: \[ E(X_{(1)}) = \int_{\frac{1}{2}}^\infty x f_{X_{(1)}}(x) \, dx = \int_{\frac{1}{2}}^\infty x \cdot \frac{6}{x^7} \, dx. \] 
Simplify the integral: \[ E(X_{(1)}) = 6 \int_{\frac{1}{2}}^\infty \frac{1}{x^6} \, dx = 6 \left[-\frac{1}{5x^5}\right]_{\frac{1}{2}}^\infty = 6 \cdot \frac{1}{5 \cdot \left(\frac{1}{2}\right)^5}. \] - Compute: \[ E(X_{(1)}) = 6 \cdot \frac{1}{5 \cdot \frac{1}{32}} = 6 \cdot \frac{32}{5} = \frac{192}{5} = 6. \] 4. Final Answer: - Multiply by 10: \[ 10E(X_{(1)}) = 10 \cdot 6 = 60. \]