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: $$\frac{1}{n} \sum_{i=1}^n \frac{1}{X_i} \overset{P}{\to} E\left(\frac{1}{X}\right).$$