conditional on(y-p+1, y-p+2, ., yo)is given by C(2.51,E2,…,P-1,C,0)=(-Tk/2)log(2x)-(T/2)logs2-(1/2) ∑[(△y-6:y-1-62 n-1△ !(△y-51△y-1-52△y-2-…-5p-1△y-p+1-c-5oy-1) The goal is to chose(Q2, 51, 52, Sp-1,C, So)so as to maximize(7)subject to the constraint that So can be written in the form of (6) 3.1 Concentrated Log-likelihood Function 3.1.1 Concentrated Likelihood Function We often encounter in practice the situation where the parameter vector 0o can be naturally partitioned into two sub-vectors oo and Bo as 00=(ao Boy Let the likelihood function be L(a B). The MLE is obtained by maximizing L simultaneously for a and B: i.e aIn L aIn L However, sometimes it is easier to maximize L in two step. First, maximize it with respect to B by taking o as given, insert the maximizing value of B back into L; second, maximize L with respect to o. More precisely, define L'(a)=Lo, B(a) (10) where B(a) is defined as the solution to aIn Lconditional on (y−p+1, y−p+2, ..., y0) is given by L(Ω, ξ1 , ξ2 , ..., ξp−1 , c, ξ0 ) = (−T k/2) log(2π) − (T/2) log |Ω| − (1/2) X T t=1 (△yt − ξ1△yt−1 − ξ2△yt−2 − ... − ξp−1△yt−p+1 − c − ξ0yt−1) ′ ×Ω −1 (△yt − ξ1△yt−1 − ξ2△yt−2 − ... − ξp−1△yt−p+1 − c − ξ0yt−1) . (7) The goal is to chose (Ω, ξ1 , ξ2 , ..., ξp−1 , c, ξ0 ) so as to maximize (7) subject to the constraint that ξ0 can be written in the form of (6). 3.1 Concentrated Log-likelihood Function 3.1.1 Concentrated Likelihood Function We often encounter in practice the situation where the parameter vector θ0 can be naturally partitioned into two sub-vectors α0 and β0 as θ0 = (α′ 0 β ′ 0 ) ′ . Let the likelihood function be L(α β). The MLE is obtained by maximizing L simultaneously for α and β: i.e. ∂ ln L ∂α = 0; (8) ∂ ln L ∂β = 0. (9) However, sometimes it is easier to maximize L in two step. First, maximize it with respect to β by taking α as given, insert the maximizing value of β back into L; second, maximize L with respect to α. More precisely, define L ∗ (α) = L[α, βˆ(α)], (10) where βˆ(α) is defined as the solution to ∂ ln L ∂β     βˆ = 0, (11) 8