Question

Let \(X\) and \(Y\) be two random variables with mean \(0\), variance \(1\), and correlation coefficient \(\tfrac{1}{3}\). Then the value of \(\mathrm{Var}(X+3Y)\) is equal to:

• A. 9
• B. 10
• C. 11
• D. 12

Hint:

For any linear combination \(\mathbf{a}^\top\mathbf{Z}\), use the compact rule \(\mathrm{Var}(\mathbf{a}^\top\mathbf{Z})=\mathbf{a}^\top\Sigma\,\mathbf{a}\) with the covariance matrix \(\Sigma\). It avoids manual expansion.

Answer 1

The Correct Option is D

To find the variance of \(X+3Y\), we use the formula for the variance of a linear combination of random variables:
\[ \mathrm{Var}(aX + bY) = a^2\mathrm{Var}(X) + b^2\mathrm{Var}(Y) + 2ab\mathrm{Cov}(X,Y) \] Here, \(a = 1\) and \(b = 3\). Given that \(\mathrm{Var}(X) = 1\), \(\mathrm{Var}(Y) = 1\), and the correlation coefficient \( \rho_{XY} = \frac{1}{3} \), we know: \(\mathrm{Cov}(X,Y) = \rho_{XY} \sqrt{\mathrm{Var}(X) \mathrm{Var}(Y)} = \frac{1}{3} \times 1 \times 1 = \frac{1}{3}\). Substitute these into the formula:
\[ \mathrm{Var}(X+3Y) = 1^2 \cdot 1 + 3^2 \cdot 1 + 2 \cdot 1 \cdot 3 \cdot \frac{1}{3} \] \[ = 1 + 9 + 2 \] \[ = 12 \] Therefore, \(\mathrm{Var}(X+3Y) = 12\).