Discussion
Let \(X = (X_{1}, X_{2}, X_{3}, X_{4}) \sim N_{4}(\underline{0}, \Sigma)\) where
\[ \Sigma = \begin{bmatrix} 1 & \rho & \rho & \rho \\ \rho & 1 & \rho & \rho \\ \rho & \rho & 1 & \rho \\ \rho & \rho & \rho & 1 \end{bmatrix} \]
is positive definite.
Then which of the following statements is true ?
Answer: A
The true statement is \(X_{1} + X_{2}, X_{2} + X_{3}, X_{3} + X_{4}\) all have the same distribution.
Each sum of two different coordinates is normal with mean 0 and variance \(1 + 1 + 2\rho = 2(1 + \rho)\), because every pair has covariance \(\rho\). So all three are \(N(0, 2 + 2\rho)\).
The others fail. \(X_{1} - X_{2}\) and \(X_{1} - X_{3}\) are correlated (covariance \(1 - \rho\), correlation 1/2), so their squared ratio is not \(F_{1,1}\). \(X_{1} - X_{3}\) and \(X_{2} - X_{4}\) are independent with variance \(2(1 - \rho)\) each, so the sum of squares over 2 is \((1 - \rho)\chi_{2}^{2}\), equal to \(\chi_{2}^{2}\) only if \(\rho = 0\). And \(\sum X_{i}\) has variance \(4 + 12\rho\), not 4.
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