Question

Let \( \{X_n\}_{n \geq 1} \) be a sequence of i.i.d. random variables with the common probability density function \[ f(x) = \frac{1}{\pi(1 + x^2)}, \quad -\infty<x<\infty. \] Define \[ Y_n = \frac{1}{2} + \frac{1}{\pi} \tan^{-1}(X_n) { for } n = 1, 2, \dots. \] Then which one of the following options is correct?

• A. \( \frac{1}{n} \sum_{i=1}^{n} Y_i \xrightarrow{P} \frac{1}{2} { as } n \to \infty \)
• B. \( \frac{1}{n} \sum_{i=1}^{n} Y_i \xrightarrow{P} 0 { as } n \to \infty \)
• C. \( \frac{1}{n} \sum_{i=1}^{n} X_i \xrightarrow{P} 0 { as } n \to \infty \)
• D. \( \frac{1}{n} \sum_{i=1}^{n} X_i \xrightarrow{P} \frac{1}{2} { as } n \to \infty \)

Hint:

For i.i.d. random variables, the sample average converges to the expected value as \( n \to \infty \) by the Strong Law of Large Numbers.

Answer 1

The Correct Option is A

Step 1: The given random variable \( X_n \) follows the Cauchy distribution, which has the probability density function \[ f(x) = \frac{1}{\pi(1 + x^2)}. \] The expected value of \( X_n \) does not exist because the Cauchy distribution has an undefined mean. 
Step 2: Consider the transformation \[ Y_n = \frac{1}{2} + \frac{1}{\pi} \tan^{-1}(X_n). \] Since \( X_n \) has the Cauchy distribution, the transformation \( \tan^{-1}(X_n) \) maps \( X_n \) to a random variable that is uniformly distributed on \( \left( 0, \pi \right) \). The expectation of \( Y_n \) is \[ \mathbb{E}[Y_n] = \frac{1}{2} + \frac{1}{\pi} \mathbb{E}[\tan^{-1}(X_n)]. \] It is known that \[ \mathbb{E}[\tan^{-1}(X_n)] = \frac{\pi}{4}. \] Thus, \[ \mathbb{E}[Y_n] = \frac{1}{2} + \frac{1}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}. \] Step 3: By the Strong Law of Large Numbers (SLLN), the sample average of i.i.d. random variables converges to the expected value almost surely. Therefore, \[ \frac{1}{n} \sum_{i=1}^{n} Y_i \xrightarrow{P} \mathbb{E}[Y_n] = \frac{1}{2} { \text{ as } n \to \infty}. \]