Discussion
Suppose \(\underline{X}_{3 \times 1} \sim \text{MNormal}_{3}\)
\[ \left(\underline{0} \begin{bmatrix} 1 & 0.5 & 0 \\ 0.5 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right). \]
Hence \(\text{var}(2X_{2} \mid X_{1} = x)\) is equal to :
Answer: D
\(\text{var}(2X_{2} \mid X_{1} = x) = 15\).
For a multivariate normal, the conditional variance of \(X_{2}\) given \(X_{1}\) does not depend on \(x\):
\[ \begin{aligned} \text{var}(X_{2} \mid X_{1}) &= \sigma_{22} - \frac{\sigma_{21}^{2}}{\sigma_{11}} \\ &= 4 - \frac{0.5^{2}}{1} = 3.75 \\ \text{var}(2X_{2} \mid X_{1}) &= 2^{2} \times 3.75 = 15 \end{aligned} \]
\(X_{3}\) is uncorrelated with both, so it plays no part. The option 3.75 forgets the factor \(2^{2}\) from scaling, 16 is the unconditional variance \(4 \times 4\), and 8 doubles instead of squaring the coefficient.
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