... becomesµ(j+1)σ 2 (j+1)=µ(j)σ 2 (j)−∂ ¯m1∂µ∂ ¯m1∂σ 2 ∂ ¯m 2 ∂µ∂ ¯m 2 ∂σ 2 ¯m1(µ(j), σ 2 (j)) − ¯z¯m 2 (µ(j), σ 2 (j)) − ¯y,Copyrightc 1999, Russell Davidson ... as¯m1(µ, σ 2 ) = ¯z and ¯m 2 (µ, σ 2 ) = ¯y, (9.111)where ¯z and ¯y are the sample averages of the zt and the yt, respectively, and ¯mi(µ, σ 2 ) ≡1−nnt=1mti(µ, σ 2 ), i = 1, 2. Equations ... scalar factors, we obtainσ 2 y− aσyzaσ 2 z− σyz−1 2 aσyz1 2 aσ 2 zι00 ιf1(µ, σ 2 )f 2 (µ, σ 2 )= 0.Since the leftmost factor above is a 2 2 nonsingular matrix, we...