Question

Suppose that \(X\) and \(Y\) are independent and identically distributed \(N_p(\mu, \Sigma)\) random vectors, where \(\mu \in \mathbb{R}^p\) and \(\Sigma\) is a positive definite matrix. Let \(\chi^2_m\) denote chi-square distribution with \(m\)-degrees of freedom. Then which of the following statements is correct?

• A. \(\dfrac{1}{2}(X-Y)^T\Sigma^{-1}(X-Y)\) follows \(\chi^2_p\)
• B. \(2(X-Y)^T\Sigma(X-Y)\) follows \(\chi^2_p\)
• C. \(\dfrac{1}{2}(X-Y)^T\Sigma^{-1}(X-Y)\) follows \(\chi^2_{2p}\)
• D. \(2(X-Y)^T\Sigma(X-Y)\) follows \(\chi^2_{2p}\)

Hint:

Find the distribution of X-Y first, then use that a zero-mean normal vector times the inverse of its own covariance gives a chi-square with degrees of freedom equal to the vector's dimension.

Answer 1

The Correct Option is A

Step 1: Reduce to a simpler vector.
$D=X-Y\sim N_p(0,2\Sigma)$.

Step 2: Whiten using square root of covariance.
$W=\frac{1}{\sqrt2}\Sigma^{-1/2}D\sim N_p(0,I_p)$.

Step 3: Sum of squares is chi-square.
$W^TW\sim\chi^2_p$.

Step 4: Write back in terms of D.
$W^TW=\frac{1}{2}D^T\Sigma^{-1}D$. So $\frac{1}{2}(X-Y)^T\Sigma^{-1}(X-Y)\sim\chi^2_p$.

Step 5: Rule out remaining options.
Plain $\Sigma$ wrong; $2p$ degrees wrong.
\[ \boxed{\tfrac{1}{2}(X-Y)^T\Sigma^{-1}(X-Y)\sim\chi^2_p} \]