Question

Let \( \{X_n\}_{n \geq 1} \) be a sequence of independent and identically distributed random variables, each having mean 4 and variance 9. If \( Y_n = \frac{1}{n} \sum_{i=1}^{n} X_i \) for \( n \geq 1 \), then \[ \lim_{n \to \infty} E \left[ \left( \frac{Y_n - 4}{\sqrt{n}} \right)^2 \right] \text{ (in integer) equals:} \]

• A. 0
• B. 4
• C. 9
• D. 1

Hint:

- The Law of Large Numbers (LLN) implies that sample averages converge to the expected value.
- When normalizing by \( \sqrt{n} \), the variance decreases and tends to zero in the limit.

Answer 1

The Correct Option is A

1) Understanding the problem:
We are given that the sequence \( \{X_n\}_{n \geq 1} \) is a sequence of independent random variables with mean 4 and variance 9. We are asked to find the limit of the expected value of the squared deviation of \( Y_n \) from 4, normalized by \( \sqrt{n} \).
2) Applying the Law of Large Numbers:
By the Weak Law of Large Numbers (WLLN), as \( n \to \infty \), \( Y_n \) converges in probability to 4. Therefore, the quantity \( \frac{Y_n - 4}{\sqrt{n}} \) tends to 0 as \( n \) increases.
3) Evaluating the limit:
As \( n \to \infty \), the expected value of the squared deviation becomes: \[ E \left[ \left( \frac{Y_n - 4}{\sqrt{n}} \right)^2 \right] = \frac{\text{Var}(Y_n)}{n} \] Since \( \text{Var}(Y_n) = \frac{9}{n} \), we get: \[ E \left[ \left( \frac{Y_n - 4}{\sqrt{n}} \right)^2 \right] = \frac{9}{n^2} \to 0 \text{ as } n \to \infty \] Thus, the final result is 0.