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: Find distributions

Since $X_i \in {0, 1}$:

• $E(X_i) = 0 \cdot \frac{1}{4} + 1 \cdot \frac{3}{4} = \frac{3}{4}$
• $\text{Var}(X_i) = E(X_i^2) - [E(X_i)]^2 = \frac{3}{4} - \frac{9}{16} = \frac{3}{16}$

For $(1-X_i)^2$:

• When $X_i = 0$: $(1-0)^2 = 1$
• When $X_i = 1$: $(1-1)^2 = 0$

So $E((1-X_i)^2) = 1 \cdot \frac{1}{4} + 0 \cdot \frac{3}{4} = \frac{1}{4}$

And $\text{Var}((1-X_i)^2) = E((1-X_i)^4) - [E((1-X_i)^2)]^2 = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$

Step 2: Apply Law of Large Numbers

By SLLN: $$U_n \xrightarrow{P} E(X_1) = \frac{3}{4}$$ $$V_n \xrightarrow{P} E((1-X_1)^2) = \frac{1}{4}$$

Step 3: Analyze each option

(A) $\lim_{n \to \infty} P\left(\left|U_n - \frac{3}{4}\right| < \frac{1}{100}\right) = 1$

Since $U_n \xrightarrow{P} \frac{3}{4}$, for any $\epsilon > 0$: $$\lim_{n \to \infty} P\left(\left|U_n - \frac{3}{4}\right| < \epsilon\right) = 1$$

With $\epsilon = \frac{1}{100}$, this is TRUE 

(B) $\lim_{n \to \infty} P\left(\left|U_n - \frac{3}{4}\right| > \frac{1}{100}\right) = 0$

This is the complement of (A), so it's also TRUE 

(C) $\lim_{n \to \infty} P\left(\sqrt{n}\left(U_n - \frac{3}{4}\right) \leq 1\right) = \Phi(2)$

By CLT: $\sqrt{n}\left(U_n - \frac{3}{4}\right) \xrightarrow{d} N\left(0, \frac{3}{16}\right)$

So: $\frac{\sqrt{n}(U_n - 3/4)}{\sqrt{3/16}} \xrightarrow{d} N(0,1)$

$$P\left(\sqrt{n}\left(U_n - \frac{3}{4}\right) \leq 1\right) = P\left(\frac{\sqrt{n}(U_n - 3/4)}{\sqrt{3/16}} \leq \frac{1}{\sqrt{3/16}}\right) \to \Phi\left(\frac{4}{\sqrt{3}}\right) = \Phi\left(\frac{4\sqrt{3}}{3}\right)$$

This does NOT equal $\Phi(2)$, so FALSE 

(D) $\lim_{n \to \infty} P\left(\sqrt{n}\left(V_n - \frac{1}{4}\right) \leq 1\right) = \Phi\left(\frac{4}{\sqrt{3}}\right)$

By CLT: $\sqrt{n}\left(V_n - \frac{1}{4}\right) \xrightarrow{d} N\left(0, \frac{3}{16}\right)$

$$P\left(\sqrt{n}\left(V_n - \frac{1}{4}\right) \leq 1\right) \to \Phi\left(\frac{1}{\sqrt{3/16}}\right) = \Phi\left(\frac{4}{\sqrt{3}}\right)$$

This is TRUE 

Answer: (A), (B), and (D) are true