Question

Let $\{X_n\}_{n \ge 1}$ be a sequence of i.i.d. random variables such that \[ P(X_1 = 0) = \frac{1}{4}, P(X_1 = 1) = \frac{3}{4}. \] Define \[ U_n = \frac{1}{n} \sum_{i=1}^n X_i \text{and} V_n = \frac{1}{n} \sum_{i=1}^n (1 - X_i)^2, n = 1, 2, \ldots \] Then which of the following statements is/are TRUE? 
 

• A. $\lim_{n \to \infty} P\left(|U_n - \frac{3}{4}| < \frac{1}{100}\right) = 1$
• B. $\lim_{n \to \infty} P\left(|U_n - \frac{3}{4}| > \frac{1}{100}\right) = 0$
• C. $\lim_{n \to \infty} P\left(\sqrt{n}\left(U_n - \frac{3}{4}\right) \le 1\right) = \Phi(2)$
• D. $\lim_{n \to \infty} P\left(\sqrt{n}\left(V_n - \frac{1}{4}\right) \le 1\right) = \Phi\left(\frac{4}{\sqrt{3}}\right)$

Hint:

LLN deals with convergence in probability, while CLT describes the distribution of scaled deviations from the mean.

Answer 1

The Correct Option is A, B, D

Step 1: Mean and variance of $X_i$.
\[ E(X_i) = 0 \times \frac{1}{4} + 1 \times \frac{3}{4} = \frac{3}{4}, \mathrm{Var}(X_i) = \frac{3}{4}\left(1 - \frac{3}{4}\right) = \frac{3}{16}. \]

Step 2: Apply Law of Large Numbers (LLN).
By LLN, $U_n \xrightarrow{p} E(X_i) = \frac{3}{4}$. Hence, \[ \lim_{n \to \infty} P\left(|U_n - \frac{3}{4}| < \epsilon\right) = 1, \] so (A) and (B) are both true.

Step 3: Apply Central Limit Theorem (CLT).
By CLT, \[ \sqrt{n}\left(U_n - \frac{3}{4}\right) \xrightarrow{d} N\left(0, \frac{3}{16}\right). \] Hence, \[ P\left(\sqrt{n}\left(U_n - \frac{3}{4}\right) \le 1\right) \approx \Phi\left(\frac{1}{\sqrt{3/16}}\right) = \Phi(2.309) \approx \Phi(2). \] Thus, (C) is true.

Step 4: For $V_n$.
Since $(1 - X_i)^2 = 1 - 2X_i + X_i^2$, expectation depends on $E(X_i)$, not on variance. Hence $V_n$ does not follow the same limiting distribution; (D) is incorrect.

Step 5: Conclusion.
\[ \boxed{(A), (B), \text{ and } (C) \text{ are correct.}} \]