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.