Question

Let \( X_1, X_2, \dots, X_{20} \) be a random sample of size 20 from \( N_6(\mu, \Sigma) \), with det(\(\Sigma\)) \(\neq 0\), and suppose both \(\mu\) and \(\Sigma\) are unknown. Let \[ \bar{X} = \frac{1}{20} \sum_{i=1}^{20} X_i \quad \text{and} \quad S = \frac{1}{19} \sum_{i=1}^{20} (X_i - \bar{X})(X_i - \bar{X})^T. \] Consider the following two statements: \begin{enumerate} \item The distribution of \( 19S \) is \( W_6(19, \Sigma) \) (Wishart distribution of order 6 with 19 degrees of freedom). \item The distribution of \( (X_3 - \mu)^T S^{-1} (X_3 - \mu) \) is \( \chi^2_6 \) (Chi-square distribution with 6 degrees of freedom). \end{enumerate} Then which of the above statements is/are true?

• A. (I) only
• B. (II) only
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

In a multivariate normal distribution, the sample covariance matrix \( S \) follows a Wishart distribution. For a normal sample, linear combinations like \( (X - \mu)^T S^{-1} (X - \mu) \) follow a Chi-square distribution with degrees of freedom equal to the dimension of the data.

Answer 1

The Correct Option is C

Step 1: Analyzing Statement (I).
The Wishart distribution \( W_p(n, \Sigma) \) is the distribution of the sample covariance matrix \( S \) for a random sample from a multivariate normal distribution. Since \( X_1, X_2, \dots, X_{20} \) are from a \( N_6(\mu, \Sigma) \) distribution, the distribution of \( 19S \) will indeed be \( W_6(19, \Sigma) \), where the order is 6 and the degrees of freedom are 19. This makes statement (I) true.

Step 2: Analyzing Statement (II).
The quantity \( (X_3 - \mu)^T S^{-1} (X_3 - \mu) \) follows a \( \chi^2 \) distribution with degrees of freedom equal to the number of variables, which is 6 in this case. Therefore, statement (II) is also true.

Step 3: Conclusion.
Both statements (I) and (II) are true, so the correct answer is (C).