Question

Let \( X_1, X_2, \ldots, X_{10} \) be a random sample from the \( \text{Exp}(1) \) distribution. Define
\[W = \max\{ e^{-X_1}, e^{-X_2}, \ldots, e^{-X_{10}} \}.\]
Then the value of \( 22E(W) \) is equal to __________ (answer in integer).

Answer 1

Correct Answer: 20

1. Transformation of \( e^{-X_i} \): - For \( X_i \sim \text{Exp}(1) \), \( e^{-X_i} \) is a decreasing function of \( X_i \). Hence, the random variable \( e^{-X_i} \) is uniformly distributed on \( [0, 1] \). 2. Distribution of \( W \): - The maximum \( W = \max\{e^{-X_1}, e^{-X_2}, \dots, e^{-X_{10}}\} \) follows a distribution with cumulative distribution function: \[ P(W \leq w) = P(e^{-X_1} \leq w, \dots, e^{-X_{10}} \leq w) = P(e^{-X_1} \leq w)^{10} = w^{10}, \quad 0 \leq w \leq 1. \] 3. Expectation of \( W \): - The expectation of \( W \) is: \[ E(W) = \int_0^1 w \cdot 10w^9 \, dw = 10 \int_0^1 w^{10} \, dw. \] - Solve the integral: \[ E(W) = 10 \cdot \frac{w^{11}}{11} \Big|_0^1 = \frac{10}{11}. \] 4. Value of \( 22E(W) \): - Multiply \( E(W) \) by 22: \[ 22E(W) = 22 \cdot \frac{10}{11} = 20. \]