Question

For \( n \in \mathbb{N} \), let \( Z_n \) be the smallest order statistic based on a random sample of size \( n \) from the \( U(0,1) \) distribution. Let \( nZ_n \xrightarrow{d} Z \) as \( n \to \infty \) for some random variable \( Z \). Then \( P(Z \leq \ln 3) \) is equal to:

• A. \( \frac{1}{4} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{1}{3} \)

Answer 1

The Correct Option is B

1. Distribution of \( nZ_n \): - The smallest order statistic \( Z_n \) for a random sample from \( U(0, 1) \) has CDF: \[ F_{Z_n}(z) = 1 - (1 - z)^n, \quad \text{for } 0 \leq z \leq 1. \] - Scaling \( Z_n \) by \( n \), as \( n \to \infty \), leads to the limiting random variable \( Z \) with PDF: \[ f_Z(z) = e^{-z}, \quad \text{for } z \geq 0. \] 2. CDF of \( Z \): - The CDF of \( Z \) is: \[ F_Z(z) = P(Z \leq z) = 1 - e^{-z}, \quad \text{for } z \geq 0. \] 3. Evaluate \( P(Z \leq \ln 3) \): - Substituting \( z = \ln 3 \): \[ P(Z \leq \ln 3) = 1 - e^{-\ln 3} = 1 - \frac{1}{3} = \frac{2}{3}. \]