Question

Let $X_{1}, X_{2}, \dots$ be independent variables each taking values $+1$ or $-1$ with equal probability respectively. If $S_{n}=\sum_{i=1}^{n} i X_{i}$ then $\lim_{n \rightarrow \infty} P\left(S_{n} < \sqrt{\frac{n(n+1)(2 n+1)}{3}}\right)$ where $\Phi$ is distribution function of standard normal variate, is

• A. $\Phi(-\sqrt{3})$
• B. $\Phi(-\sqrt{2})$
• C. $1-\Phi(-\sqrt{3})$
• D. $1-\Phi(-\sqrt{2})$

Hint:

For any sum $S_n$, the limit of $P(S_n < k \sigma_n)$ is always $\Phi(k)$. Identify the variance $\sigma_n^2$ first to see how many standard deviations the bound represents.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:

\( S_n = \sum i X_i \) where each \( X_i = \pm 1 \) with probability 1/2, independent. We need the limiting probability that \( S_n \) is below \( \sqrt{\frac{n(n+1)(2n+1)}{3}} \).

Step 2: Detailed Explanation:

• \( E(S_n) = 0 \) and \( Var(S_n) = \sum i^2 \cdot Var(X_i) = \sum i^2 = \frac{n(n+1)(2n+1)}{6} \).

• So the threshold is \( \sqrt{\frac{n(n+1)(2n+1)}{3}} = \sqrt{2\, Var(S_n)} = \sqrt{2}\, \sigma_n \).

• By the Lindeberg central limit theorem, \( \frac{S_n}{\sigma_n} \to N(0,1) \), so the limit is \( \Phi(\sqrt{2}) \).

• Using \( \Phi(z) = 1 - \Phi(-z) \), \( \Phi(\sqrt{2}) = 1 - \Phi(-\sqrt{2}) \).

Step 3: Final Answer:
The correct option is (D) \( 1 - \Phi(-\sqrt{2}) \).