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: A is rank-1 projection.
\(A=\frac{1}{3}\mathbf{1}\mathbf{1}^T\), idempotent, rank 1.
Step 2: (A) TRUE.
\(Y^TAY\sim\chi_1^2\).
Step 3: (B) FALSE.
\(B\) rank 2, \(Y^TBY\sim\chi_2^2\).
Step 4: (C).
\(v=(1,-2,1)\), \(v\cdot\mathbf{1}=0\), so independent. TRUE.
Step 5: (D).
\(u=(1,2,1)\), \(u\cdot\mathbf{1}=4\ne0\), not independent. FALSE.
Final Answer: \[ \boxed{\text{(A) and (C)}} \]