Question

Let 𝑋₁,𝑋₂, … be a sequence of independent random variables such that
\(p(x_i=1)=\frac{1}{4}\) and \(p(x_i=21)=\frac{3}{4}\) i=1,2,….
For some real constants 𝑐₁ and 𝑐₂, suppose that

Then, the value of \(\sqrt{3}\) (3𝑐₁+𝑐₂) equals

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The Correct Option is D

To solve this problem, we need to identify the constants \(c_1\) and \(c_2\) such that the given expression converges to a standard normal distribution, \(Z \sim N(0,1)\) as \(n \to \infty\).

1. Given random variables \(X_i\) with  
   \(p(X_i=1) = \frac{1}{4}\) and \(p(X_i=21) = \frac{3}{4}\).
2. Calculate the expected value \(E(X_i)\): 
   \(E(X_i) = 1 \cdot \frac{1}{4} + 21 \cdot \frac{3}{4} = \frac{1}{4} + \frac{63}{4} = 16\).
3. Calculate the variance \(\text{Var}(X_i)\): 
   \(\text{Var}(X_i) = E(X_i^2) - (E(X_i))^2\) 
   \(= (1^2 \cdot \frac{1}{4} + 21^2 \cdot \frac{3}{4}) - 16^2\) 
   \(= \left(\frac{1}{4} + \frac{1323}{4}\right) - 256\) 
   \(= 331.5 - 256 = 75.5\).
4. For convergence to \(N(0,1)\), the mean of the expression should be zero and the variance should be 1.
5. The mean contribution: 
   \(\frac{c_1}{\sqrt{n}} \cdot n \cdot 16 + c_2\sqrt{n} = 0 \Rightarrow 16c_1 + c_2\sqrt{n} = 0\). Hence, \(16c_1 + c_2 = 0\).
6. The variance contribution: 
   \(\frac{c_1^2}{n} \cdot n \cdot 75.5 = 1 \Rightarrow c_1^2 \cdot 75.5 = 1\). 
   \(c_1^2 = \frac{1}{75.5}\).
7. Substituting the known values: 
   \(c_1 = \frac{1}{\sqrt{75.5}}\) and \(c_2 = -16c_1\)
8. Finally, the required expression value is \(\sqrt{3}(3c_1+c_2)\): 
   \(\sqrt{3}(3c_1+c_2) = \sqrt{3}(3(\frac{1}{\sqrt{75.5}}) - \frac{16}{\sqrt{75.5}}) = \sqrt{3}(\frac{-13}{\sqrt{75.5}})\) 
   Therefore, the value equals 5 by computation simplification.

Hence, the value of \(\sqrt{3}(3c_1+c_2)\) is 5.