Question

For \( n \in \mathbb{N} \), let \( X_1, X_2, \ldots, X_n \) be a random sample from the \( F_{20,40} \) distribution. Then, as \( n \to \infty \), \( \frac{1}{n} \sum_{i=1}^n \frac{1}{X_i} \) converges in probability to __________ (round off to 2 decimal places).

Answer 1

Correct Answer: 1 - 1.2

1. Expectation of \( \frac{1}{X_i} \): - For \( X \sim F_{20,40} \), the expectation \( E\left(\frac{1}{X}\right) \) exists when the numerator degrees of freedom (\( d_1 \)) satisfies \( d_1 > 2 \). - Using properties of the \( F \)-distribution: \[ E\left(\frac{1}{X}\right) = \frac{d_2}{d_2 - 2}, \quad \text{where } d_1 = 20, \, d_2 = 40. \] - Substituting \( d_2 = 40 \): \[ E\left(\frac{1}{X}\right) = \frac{40}{40 - 2} = \frac{40}{38} \approx 1.05. \] 

2. Convergence in Probability: - By the Law of Large Numbers, the sample mean converges in probability to the expected value: \[ \frac{1}{n} \sum_{i=1}^n \frac{1}{X_i} \overset{P}{\to} E\left(\frac{1}{X}\right). \]