Exercise Solution 9.9
Yes. Here is why: Adopting the notation of [9.72], 1R is a linear polynomial of 1D. Because 1R is nonsingular, the relationship is invertible, and 1D is a linear polynomial of 1R. As indicated in Section 3.10.4, a linear polynomial of a joint-normal random vector is joint-normal. 1R is joint normal, so 1D is joint-normal. comprises components of 1D. As discussed in Section 3.10.4, this means that is also joint-normal.