Question:
Given that \(X \sim N_3(\mu_1, \Sigma)\), where \(\mu_1 = (0\ 1\ 0)'\) and \(Y \sim N_3(\mu_2, \Sigma)\), where \(\mu_2 = (0\ -1\ 0)'\), then which of the following statements are true?
Statement 1: \((X+Y)^2\) has a central Chi-square distribution.
Statement 2: \((X-1)^2 + (Y+1)^2\) has a central Chi-square distribution.
Given that \(X \sim N_3(\mu_1, \Sigma)\), where \(\mu_1 = (0\ 1\ 0)'\) and \(Y \sim N_3(\mu_2, \Sigma)\), where \(\mu_2 = (0\ -1\ 0)'\), then which of the following statements are true?
Statement 1: \((X+Y)^2\) has a central Chi-square distribution.
Statement 2: \((X-1)^2 + (Y+1)^2\) has a central Chi-square distribution.
Statement 1: \((X+Y)^2\) has a central Chi-square distribution.
Statement 2: \((X-1)^2 + (Y+1)^2\) has a central Chi-square distribution.
Show Hint
A quadratic form of a normal vector is a central chi-square only when that vector's mean is exactly zero; check carefully whether the mean actually cancels in each expression before concluding centrality.
Updated On: Jul 4, 2026
- Both the statements 1 and 2 are true.
- Only 1 is True but 2 is false.
- Only 2 is True but 1 is false.
- Both the statements 1 and 2 are false.
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Only the second coordinate of each mean vector is non-zero, so in the direction that matters \(X\) behaves like a normal variate with mean 1, and \(Y\) like a normal variate with mean \(-1\), both sharing the same variance \(\sigma^2\) from \(\Sigma\), and independent of each other.
Step 2 (Statement 1): \(X+Y\) has mean \(1+(-1)=0\) and variance \(2\sigma^2\). Since its mean is exactly zero, \(\left(\dfrac{X+Y}{\sigma\sqrt2}\right)^2\) follows a central Chi-square distribution with 1 degree of freedom. So Statement 1 is true.
Step 3 (Statement 2): Consider \(X - \mathbf{1}\), meaning \(X\) shifted by the all-ones vector \((1,1,1)\). Since \(X\) has mean vector \((0,1,0)\), the shifted mean is \((0,1,0)-(1,1,1) = (-1,0,-1)\), which is NOT the zero vector. Similarly \(Y+\mathbf{1}\) has shifted mean \((0,-1,0)+(1,1,1) = (1,0,1)\), also not zero.
Step 4: Because these shifted means are nonzero, the quadratic forms built from \(X-1\) and \(Y+1\) are each NONcentral Chi-square, and their independent sum is also noncentral. So Statement 2 is false.
Final Answer: Only statement 1 is true, statement 2 is false (option B).
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