Question

Let \(\{X_n\}_{n \ge 1}\) be i.i.d. random variables with \[ f(x) = \begin{cases} 1, & 0<x<1 \\ 0, & \text{otherwise} \end{cases} \] Then, the value of the limit \[ \lim_{n \to \infty} P\left( -\frac{1}{n}\sum_{i=1}^{n} \ln X_i \le 1 + \frac{1}{\sqrt{n}} \right) \] is equal to:

• A. \(\frac{1}{2}\)
• B. \(\Phi(1)\)
• C. 0
• D. \(\Phi(2)\)

Hint:

When sums of i.i.d. random variables are normalized, apply the Central Limit Theorem to approximate the distribution using the standard normal variable.

Answer 1

The Correct Option is B

Step 1: Transform the variable. 
If \( X_i \sim U(0,1) \), then \( Y_i = -\ln X_i \) follows an exponential distribution with mean \( 1 \) and variance \( 1 \). 
Step 2: Apply Central Limit Theorem (CLT). 
For large \( n \), \[ \frac{\frac{1}{n}\sum_{i=1}^{n} Y_i - 1}{1/\sqrt{n}} \sim N(0,1) \] 
Step 3: Express the probability. 
\[ P\left(-\frac{1}{n}\sum_{i=1}^{n} \ln X_i \le 1 + \frac{1}{\sqrt{n}}\right) = P\left(\frac{\frac{1}{n}\sum Y_i - 1}{1/\sqrt{n}} \le 1\right) \] 
Step 4: Use standard normal distribution. 
By CLT, the probability approaches \(\Phi(1)\), the cumulative distribution function of the standard normal distribution at \(1\). 
Final Answer: \[ \boxed{\Phi(1)} \]