Question:
Given that \(X=(X_1\ X_2\ X_3)' \sim N_3(\mu,\Sigma)\), where \(\mu=(0\ 0\ 0)'\) and \(\Sigma=\begin{pmatrix}1 & 1 & 0\\1 & 4 & 1\\0 & 1 & 4\end{pmatrix}\). The correlation coefficient between \(X_2\) and \(X_3\) is given by:
Given that \(X=(X_1\ X_2\ X_3)' \sim N_3(\mu,\Sigma)\), where \(\mu=(0\ 0\ 0)'\) and \(\Sigma=\begin{pmatrix}1 & 1 & 0\\1 & 4 & 1\\0 & 1 & 4\end{pmatrix}\). The correlation coefficient between \(X_2\) and \(X_3\) is given by:
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Correlation = Cov(X2,X3) divided by the square root of the product of their variances, all read straight off \(\Sigma\).
Updated On: Jul 4, 2026
- 0.25
- 0.5
- 1
- -1
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The Correct Option is A
Solution and Explanation
Step 1: From \(\Sigma\), read off \(\text{Var}(X_2) = 4\), \(\text{Var}(X_3)=4\) and \(\text{Cov}(X_2,X_3)=1\) (the (2,3) entry of \(\Sigma\)).
Step 2: The correlation coefficient is \[\rho_{X_2X_3} = \frac{\text{Cov}(X_2,X_3)}{\sqrt{\text{Var}(X_2)}\sqrt{\text{Var}(X_3)}} = \frac{1}{\sqrt{4}\sqrt{4}} = \frac{1}{2\times 2} = \frac{1}{4}.\]
Step 3: So \(\rho_{X_2X_3} = 0.25\).
\(\boxed{0.25}\)
Step 2: The correlation coefficient is \[\rho_{X_2X_3} = \frac{\text{Cov}(X_2,X_3)}{\sqrt{\text{Var}(X_2)}\sqrt{\text{Var}(X_3)}} = \frac{1}{\sqrt{4}\sqrt{4}} = \frac{1}{2\times 2} = \frac{1}{4}.\]
Step 3: So \(\rho_{X_2X_3} = 0.25\).
\(\boxed{0.25}\)
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