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 C

We apply the Central Limit Theorem (CLT) to the sum of independent random variables $S_n$.

Step 1: \color{redFind the mean and variance of $X_i$
Each $X_i$ is a Rademacher random variable: $P(X_i = 1) = 1/2$ and $P(X_i = -1) = 1/2$.
$E(X_i) = (1)(1/2) + (-1)(1/2) = 0$.
$E(X_i^2) = (1^2)(1/2) + (-1^2)(1/2) = 1$.
$Var(X_i) = E(X_i^2) - [E(X_i)]^2 = 1 - 0 = 1$.

Step 2: \color{redFind the mean and variance of $S_n$
$S_n = \sum_{i=1}^{n} i X_i$.
$E(S_n) = \sum i E(X_i) = 0$.
$Var(S_n) = \sum i^2 Var(X_i) = \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$.

Step 3: \color{redApply Central Limit Theorem
By CLT, $Z_n = \frac{S_n - E(S_n)}{\sqrt{Var(S_n)}} \xrightarrow{d} N(0,1)$ as $n \to \infty$.
We need to find $P(S_n < \sqrt{\frac{n(n+1)(2n+1)}{3}})$.
Let $V_n^2 = Var(S_n) = \frac{n(n+1)(2n+1)}{6}$. The target value is $K = \sqrt{2 \cdot V_n^2} = \sqrt{2} V_n$.

Step 4: \color{redStandardize the probability
$P(S_n < \sqrt{2} V_n) = P\left(\frac{S_n}{V_n} < \sqrt{2}\right) \to \Phi(\sqrt{2})$.
Note: Using the property $\Phi(z) = 1 - \Phi(-z)$, we have $\Phi(\sqrt{2}) = 1 - \Phi(-\sqrt{2})$.
Wait, let's re-examine the constant in the prompt: $\sqrt{\frac{n(n+1)(2n+1)}{3}}$.
This is exactly $\sqrt{2 \cdot Var(S_n)}$. Thus the limit is $\Phi(\sqrt{2})$.
Based on standard properties, $1 - \Phi(-\sqrt{2})$ is the equivalent form.