MATH 668 Midterm: MATH 668 Louisville Exam1Solutions
Document Summary
1. (a) c 1x nn(0, i) since e(c 1x) = c 1e(x) = c 10 = 0 and cov(c 1x) = c 1cov(x)(c 1) = C 1cc c t/2 = et et t/2 which is the mgf of a nn(0, i) random vector x)(cid:17) = e(cid:16)e(c t) x(cid:17) = mx(c t) = ec . (c 1 t0+(c t) (c t)/2 = (b) let z = c 1x. Alternate method 1: we can show that b is idempotent to show that y by 2(rank(b) = r). Since c is upper triangular with positive elements and thus full rank and invertible, we can left multiply both sides by (c ) 1 to obtain (c ) 1c bcc bc = (c ) 1c bc = bcc bc = bc. Then we multiply both sides by c to obtain bcc bcc = Bcc = b b = b = (b )2 = b .
