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.

• A. Both the statements 1 and 2 are true.
• B. Only 1 is True but 2 is false.
• C. Only 2 is True but 1 is false.
• D. Both the statements 1 and 2 are false.

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.

Answer 1

The Correct Option is B

Step 1: Define $U = X - \mu_1$ and $V = Y - \mu_2$, so $U, V \sim N_3(0,\Sigma)$ independently.
Step 2 (Statement 1): $X+Y = (U+\mu_1)+(V+\mu_2) = (U+V)+(\mu_1+\mu_2)$. Since $\mu_1+\mu_2=(0,0,0)'$, this reduces to $X+Y=U+V$ exactly, a genuinely zero-mean normal vector, so its quadratic form is a central Chi-square with 3 degrees of freedom. Statement 1 is TRUE.
Step 3 (Statement 2): $X-\mathbf{1}$ (subtracting the all-ones vector $(1,1,1)'$) $= U+\mu_1-(1,1,1)' = U+(-1,0,-1)'$, whose mean is $(-1,0,-1)'$, which is NOT the zero vector. Likewise $Y+\mathbf{1} = V+\mu_2+(1,1,1)' = V+(1,0,1)'$, mean $(1,0,1)'$, also not zero.
Step 4: Because these residual means are nonzero (and equal in magnitude, opposite in sign), each quadratic form $(X-1)'\Sigma^{-1}(X-1)$ and $(Y+1)'\Sigma^{-1}(Y+1)$ is a NONcentral Chi-square with the same positive noncentrality parameter $\delta=(1,0,1)\Sigma^{-1}(1,0,1)'>0$, and their independent sum is noncentral Chi-square with noncentrality $2\delta>0$. Statement 2 is FALSE.
\[\boxed{\text{Only statement 1 is true (Option B)}}\]