Question

Let \( \{X_n\}_{n \geq 1} \) be a sequence of i.i.d. random variables with the probability mass function
\[ f(x) = \begin{cases} \frac{1}{4}, & \text{if } x = 4 \\ \frac{3}{4}, & \text{if } x = 8 \\ 0, & \text{otherwise} \end{cases} \] Let \( \bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i, n = 1, 2, \dots \). If \( \lim_{n \to \infty} P(m \leq \bar{X}_n \leq M) = 1 \), then possible values of \( m \) and \( M \) are

• A. \( m = 2.1, M = 3.1 \)
• B. \( m = 3.2, M = 4.1 \)
• C. \( m = 6.1, M = 7.1 \)
• D. \( m = 4.2, M = 5.7 \)

Hint:

The Law of Large Numbers ensures that the sample mean converges to the expected value as the sample size increases.

Answer 1

The Correct Option is C

Step 1: Define the sequence and its distribution.
We are given a sequence of i.i.d. random variables \( \{X_n\}_{n \geq 1} \), with the probability mass function: \[ f(x) = \begin{cases} \frac{1}{4}, & \text{if } x = 4 \\ \frac{3}{4}, & \text{if } x = 8 \\ 0, & \text{otherwise} \end{cases} \] This means each \( X_n \) takes the value 4 with probability \( \frac{1}{4} \) and 8 with probability \( \frac{3}{4} \).

Step 2: Compute the mean and variance of \( X_n \).
The expected value of \( X_n \), denoted as \( E[X_n] \), is calculated as: \[ E[X_n] = 4 \times \frac{1}{4} + 8 \times \frac{3}{4} = 1 + 6 = 7 \] The variance of \( X_n \), denoted as \( \text{Var}(X_n) \), is calculated as: \[ \text{Var}(X_n) = E[X_n^2] - (E[X_n])^2 \] First, calculate \( E[X_n^2] \): \[ E[X_n^2] = 4^2 \times \frac{1}{4} + 8^2 \times \frac{3}{4} = 16 \times \frac{1}{4} + 64 \times \frac{3}{4} = 4 + 48 = 52 \] Now, calculate the variance: \[ \text{Var}(X_n) = 52 - 7^2 = 52 - 49 = 3 \]

Step 3: Apply the Law of Large Numbers.
The sample mean \( \bar{X}_n \) is defined as: \[ \bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i \] By the Law of Large Numbers (LLN), as \( n \to \infty \), the sample mean \( \bar{X}_n \) converges to the expected value \( E[X_n] \), which is 7. Hence: \[ \lim_{n \to \infty} \bar{X}_n = 7 \] This means the probability that \( \bar{X}_n \) falls within any interval that contains 7 approaches 1 as \( n \) increases.

Step 4: Determine the interval \( [m, M] \).
Given that \( \lim_{n \to \infty} P(m \leq \bar{X}_n \leq M) = 1 \), we need to choose \( m \) and \( M \) such that the interval \( [m, M] \) contains 7. Since the distribution is discrete and \( \bar{X}_n \) converges to 7, a reasonable choice for the interval is around the mean value.
Using the standard deviation approximation for large \( n \) and given the discrete nature of the distribution, we can choose: \[ m = 6.1, \quad M = 7.1 \]

Final Answer: \( m = 6.1, M = 7.1 \)