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: Find the distribution of \(X-Y\).
\(E(X-Y)=0\), \(\text{Cov}(X-Y)=\Sigma+\Sigma=2\Sigma\). So \(X-Y\sim N_p(0,2\Sigma)\).

Step 2: Standard quadratic form result.
If \(Z\sim N_p(0,V)\), \(Z^TV^{-1}Z\sim\chi^2_p\).

Step 3: Apply with \(Z=X-Y\), \(V=2\Sigma\).
\[ (X-Y)^T(2\Sigma)^{-1}(X-Y)=\frac{1}{2}(X-Y)^T\Sigma^{-1}(X-Y) \sim \chi^2_p. \]

Step 4-6: Rule out others.
(B),(D) use \(\Sigma\) instead of \(\Sigma^{-1}\); (C),(D) claim wrong degrees of freedom \(2p\) instead of \(p\).

Final Answer: \[ \boxed{\tfrac{1}{2}(X-Y)^T\Sigma^{-1}(X-Y)\sim\chi^2_p} \]