Question

Let 𝑋 and 𝑌 be random variables such that 𝑋~𝑁(1, 2) and 𝑃 (𝑌=\(\frac{𝑋}{2}\)+1)=1. Let 𝛼=𝐶𝑜𝑣(𝑋, 𝑌), 𝛽=𝐸(𝑌) and 𝛾=𝑉𝑎𝑟(𝑌). Then, the value of 𝛼+2𝛽+4𝛾 equals

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The Correct Option is B

To solve the given problem, we need to understand the definitions and relationships between the random variables \(X\) and \(Y\), as well as the statistical properties involving covariance, expectation, and variance. Given the setup:

• \(X \sim N(1, 2)\): This means \(X\) is normally distributed with a mean \( \mu_X = 1 \) and a variance \( \sigma_X^2 = 2 \).
• \(Y = \frac{X}{2} + 1\): This represents a linear transformation of \(X\).

Step 1: Calculate \(E(Y)\) 

The expectation of a random variable that is a linear transformation of another is calculated as follows:

\[ E(Y) = E\left(\frac{X}{2} + 1\right) = \frac{1}{2}E(X) + 1 = \frac{1}{2} \times 1 + 1 = \frac{1}{2} + 1 = \frac{3}{2} \]

Thus, \(\beta = E(Y) = \frac{3}{2}\).

Step 2: Calculate \(\text{Var}(Y)\)

The variance of \(Y\) when \(Y\) is a linear transformation of \(X\) can be calculated as:

\[ \text{Var}(Y) = \text{Var}\left(\frac{X}{2} + 1\right) = \left(\frac{1}{2}\right)^2 \text{Var}(X) = \frac{1}{4} \times 2 = \frac{1}{2} \]

Thus, \(\gamma = \text{Var}(Y) = \frac{1}{2}\).

Step 3: Calculate \(\text{Cov}(X, Y)\)

Since \(Y = \frac{X}{2} + 1\), the covariance of \(X\) and \(Y\) is:

\[ \text{Cov}(X, Y) = \text{Cov}\left(X, \frac{X}{2} + 1\right) = \frac{1}{2} \text{Cov}(X, X) = \frac{1}{2} \text{Var}(X) = \frac{1}{2} \times 2 = 1 \]

So, \(\alpha = \text{Cov}(X, Y) = 1\).

Step 4: Calculate \( \alpha + 2\beta + 4\gamma \)

We substitute the values of \(\alpha\), \(\beta\), and \(\gamma\) we found:

\[ \alpha + 2\beta + 4\gamma = 1 + 2 \times \frac{3}{2} + 4 \times \frac{1}{2} = 1 + 3 + 2 = 6 \]

Thus, the value of \(\alpha + 2\beta + 4\gamma\) is 6.

Conclusion

Therefore, the correct answer is 6.