Question:

For \( n \in \mathbb{N} \), let \( X_1, X_2, \ldots, X_n \) be a random sample from the Cauchy distribution having probability density function
\[f(x) = \frac{1}{\pi(1 + x^2)}, \quad -\infty < x < \infty.\]
Let \( g: \mathbb{R} \to \mathbb{R} \) be defined by
\[g(x) = \begin{cases} x, & \text{if } -1000 \leq x \leq 1000 \\0, & \text{otherwise}\end{cases}.\]
Let
\[\alpha = \lim_{n \to \infty} P\left( \frac{1}{n^{3/4}} \sum_{i=1}^n g(X_i) > \frac{1}{2} \right).\]
Then \( 100\alpha \) is equal to __________ (answer in integer).

Updated On: Jan 25, 2025
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Solution and Explanation

1. Properties of the Cauchy Distribution: - The Cauchy distribution is heavy-tailed, meaning that the expected value and variance of the random variable \( X \) do not exist. - The truncation function \( g(X) \) limits \( X \) to the interval \( [-1000, 1000] \). 2. Effect of Truncation and Scaling: - The truncation ensures \( g(X_i) \) is bounded, but the Cauchy distribution does not exhibit the typical behavior of convergence in sums. - For a scaling factor \( \frac{1}{n^{3/4}} \), the sum \( \sum_{i=1}^n g(X_i) \) divided by \( n^{3/4} \) diminishes for large \( n \) because the normalization is faster than the growth of \( g(X_i) \). 3. Limiting Probability \( \alpha \): - For large \( n \), the probability: \[ P\left( \frac{1}{n^{3/4}} \sum_{i=1}^n g(X_i) > \frac{1}{2} \right) \] approaches 0 as the numerator grows slower than the denominator. 
4. Value of \( 100\alpha \): - Since \( \alpha = 0 \), we have: \[ 100\alpha = 0. \]

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