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?
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?
Which of the following is true?
Show Hint
For a multivariate normal vector, zero covariance between components (or blocks) implies independence between them; check the off-diagonal entries of \(\Sigma\).
Updated On: Jul 4, 2026
- The components of \(X\) are independent.
- \(X_1\) is independent of \(X_2\) and \(X_3\)
- \(X_2\) is independent of \(X_1\) and \(X_3\)
- \(X_3\) is independent of \(X_1\) and \(X_2\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Read the covariances directly from \(\Sigma\): \(\text{Cov}(X_1,X_2)=0\), \(\text{Cov}(X_1,X_3)=0\) and \(\text{Cov}(X_2,X_3)=-1\), with \(\text{Var}(X_1)=1\), \(\text{Var}(X_2)=4\), \(\text{Var}(X_3)=1\).
Step 2: For a jointly (multivariate) normal vector, zero covariance between two sub-vectors is equivalent to independence between them. Since \(X_1\) has zero covariance with both \(X_2\) and \(X_3\), \(X_1\) is independent of the pair \((X_2,X_3)\).
Step 3: However \(\text{Cov}(X_2,X_3)=-1\neq 0\), so \(X_2\) and \(X_3\) are NOT independent of each other. This rules out option (A), and also rules out (C) and (D) since neither \(X_2\) nor \(X_3\) is independent of the other.
Step 4: Only \(X_1\) is independent of \(X_2\) and \(X_3\).
\(\boxed{\text{Option (B)}}\)
Step 2: For a jointly (multivariate) normal vector, zero covariance between two sub-vectors is equivalent to independence between them. Since \(X_1\) has zero covariance with both \(X_2\) and \(X_3\), \(X_1\) is independent of the pair \((X_2,X_3)\).
Step 3: However \(\text{Cov}(X_2,X_3)=-1\neq 0\), so \(X_2\) and \(X_3\) are NOT independent of each other. This rules out option (A), and also rules out (C) and (D) since neither \(X_2\) nor \(X_3\) is independent of the other.
Step 4: Only \(X_1\) is independent of \(X_2\) and \(X_3\).
\(\boxed{\text{Option (B)}}\)
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