Question

Given that \(X=(X_1\ X_2\ X_3)' \sim N_3(\mu, \Sigma)\), where \(\mu=(0\ 0\ 0)'\) and \(\Sigma = \begin{pmatrix}1 & 0 & 0\\0 & 4 & -1\\0 & -1 & 1\end{pmatrix}\).
Which of the following is true?

• A. The components of \(X\) are independent.
• B. \(X_1\) is independent of \(X_2\) and \(X_3\)
• C. \(X_2\) is independent of \(X_1\) and \(X_3\)
• D. \(X_3\) is independent of \(X_1\) and \(X_2\)

Hint:

For a multivariate normal vector, zero covariance between components (or blocks) implies independence between them; check the off-diagonal entries of \(\Sigma\).

Answer 1

The Correct Option is B

Step 1: Partition $\Sigma$ into blocks corresponding to $X_1$ alone and $(X_2,X_3)'$ jointly: \[\Sigma = \begin{pmatrix}\Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22}\end{pmatrix}, \quad \Sigma_{11}=1,\ \Sigma_{12}=(0\ 0),\ \Sigma_{22}=\begin{pmatrix}4 & -1\\-1 & 1\end{pmatrix}\]
Step 2: Because the off-diagonal block $\Sigma_{12}$ is the zero vector, $\Sigma^{-1}$ is also block diagonal in the same partition, so the joint normal density factors as $f(x_1,x_2,x_3) = f_1(x_1)\cdot f_{23}(x_2,x_3)$. A density that factors this way means $X_1$ is independent of the vector $(X_2,X_3)$.
Step 3: Inside $\Sigma_{22}$, the off-diagonal entry (covariance of $X_2,X_3$) is $-1 \neq 0$, so $X_2$ and $X_3$ do not factor further and remain dependent on each other.
Step 4: So $X_1$ splits off as independent of $(X_2,X_3)$, while $X_2,X_3$ stay dependent, matching option (B).
\[\boxed{X_1 \text{ is independent of } X_2 \text{ and } X_3}\]