Question

Let \(Y = (Y_1, Y_2, Y_3)^T \sim N_3(0, I_3)\), where \(I_3\) denotes the identity matrix of order \(3\). Let
\[ A = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{pmatrix} \quad \text{and} \quad B = I_3 - A. \] Let \(\chi_1^2\) denote chi-square distribution with \(1\) degree of freedom. Then which of the following statements is/are correct?

• A. \(Y^T A Y \sim \chi_1^2\)
• B. \(Y^T B Y \sim \chi_1^2\)
• C. \(Y^T A Y\) and \(Y_1 - 2Y_2 + Y_3\) are independently distributed
• D. \(Y^T A Y\) and \(Y_1 + 2Y_2 + Y_3\) are independently distributed

Hint:

\(A\) is the rank-1 projection onto \((1,1,1)^T\); \(Y^TAY\) is a function of \(Y_1+Y_2+Y_3\) alone, so it is independent of any linear form whose coefficients dot to zero with \((1,1,1)\).

Answer 1

The Correct Option is A, C

Step 1: Simplify Y^TAY.
\(Y^TAY=(Y_1+Y_2+Y_3)^2/3=S^2/3\), \(S\sim N(0,3)\), so \(S^2/3\sim\chi_1^2\). (A) TRUE.
Step 2: (B) by subtraction.
\(Y^TBY=Y^TY-Y^TAY\sim\chi_2^2\). FALSE.
Step 3: (C) covariance.
\(\text{Cov}(S,Y_1-2Y_2+Y_3)=1-2+1=0\), independent. TRUE.
Step 4: (D) covariance.
\(\text{Cov}(S,Y_1+2Y_2+Y_3)=4\ne0\), not independent. FALSE.
\[ \boxed{\text{(A) and (C)}} \]