Question

Let \( \{X_n\}_{n \ge 1} \) be i.i.d. random variables distributed as \( N(0,1) \). Then find \[ \lim_{n \to \infty} P\left( \frac{\sum_{i=1}^{n} X_i^2 - 3n}{\sqrt{32n}} \le \sqrt{6} \right) \] is equal to:

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

Hint:

For sums of chi-square distributed variables, use the Central Limit Theorem to approximate probabilities for large \( n \).

Answer 1

The Correct Option is B

Step 1: Distribution of \( X_i^2 \).
Since \( X_i \sim N(0,1) \), each \( X_i^2 \) follows a chi-square distribution with mean \( 1 \) and variance \( 2 \).
Step 2: Mean and variance of sum.
\[ E\left(\sum_{i=1}^{n} X_i^2\right) = n, \quad Var\left(\sum_{i=1}^{n} X_i^2\right) = 2n \]
Step 3: Apply Central Limit Theorem.
\[ \frac{\sum_{i=1}^{n} X_i^2 - n}{\sqrt{2n}} \xrightarrow{d} N(0,1) \] We can rewrite the given expression as: \[ \frac{\sum_{i=1}^{n} X_i^2 - 3n}{\sqrt{32n}} = \frac{1}{4\sqrt{2}} \cdot \frac{\sum_{i=1}^{n} X_i^2 - n}{\sqrt{2n}} - \frac{1}{\sqrt{2}} \]
Step 4: Simplify and find probability.
The transformed variable is normally distributed with mean \(-\frac{1}{\sqrt{2}}\) and variance \( \frac{1}{16} \). Thus, the probability becomes: \[ P(Z \le \sqrt{6}) = \Phi(\sqrt{2}) \] Final Answer: \[ \boxed{\Phi(\sqrt{2})} \]