Question

Let 𝑋₁,𝑋₂, … be a sequence of 𝑖. 𝑖. 𝑑. random variables such that
\(P(X_1=0)=p(X_1=1)=P(X_1=2)=\frac{1}{3}\)

Then, the value of 𝛼₁+2𝛼₂+3𝛼₃+4𝛼₄ equals

• A. 6
• B. 5
• C. 4
• D. 3

Answer 1

The Correct Option is A

To find the value of \( \alpha_1 + 2\alpha_2 + 3\alpha_3 + 4\alpha_4 \), we start by analyzing the provided sequences of 𝑖.𝑖.𝑑. random variables and the corresponding limits.

\(P(X_1 = 0) = \frac{1}{3}\)

\(P(X_1 = 1) = \frac{1}{3}\)

\(P(X_1 = 2) = \frac{1}{3}\)

Given:

• \( S_n = \frac{1}{n} \sum_{i=1}^{n} X_i \) 
• \( T_n = \frac{1}{n} \sum_{i=1}^{n} X_i^2 \)

Step 1: Analyze the Limit for \( \alpha_1 \)

The central limit theorem indicates \( S_n \to E[X] \) as \( n \to \infty \). Here,

\(E[X] = \frac{0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3}}{1} = 1\).

We have:

\(\alpha_1 = \lim_{n \to \infty} P\left( \frac{1}{2} < S_n < \frac{3}{4} \right) = 0\)since these bounds do not encompass 1.

Step 2: Analyze the Limit for \( \alpha_2 \)

\(\alpha_2 = \lim_{n \to \infty} P\left( \left| S_n - \frac{1}{3} \right| < 1 \right) = 1\)because this encompasses the mean value 1 as \( n \to \infty \).

Step 3: Analyze the Limit for \( \alpha_3 \)

Calculate \(E[X^2]\):

\(E[X^2] = \frac{0^2 \cdot \frac{1}{3} + 1^2 \cdot \frac{1}{3} + 2^2 \cdot \frac{1}{3}}{1} = \frac{5}{3}\).

\(\alpha_3 = \lim_{n \to \infty} P\left( \left| T_n - \frac{1}{3} \right| < \frac{3}{2} \right) = 1\)because it encompasses the expected value \(\frac{5}{3}\).

Step 4: Analyze the Limit for \( \alpha_4 \)

\(\alpha_4 = \lim_{n \to \infty} P\left( \left| T_n - \frac{2}{3} \right| < \frac{1}{2} \right) = 1\)since it includes \(\frac{5}{3}\).

Final Calculation:

Thus, the required value is:

\(\alpha_1 + 2\alpha_2 + 3\alpha_3 + 4\alpha_4 = 0 + 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 1 = 6\)

The correct answer is 6.

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