看板 Economics 關於我們 聯絡資訊
希望我沒解錯. let r(t)=x(t), a(t)=y(t), ψ=k and θ=j (only for convenience, it is very hard for me to type those characters...) and suppose |k| < 1 and |j| < 1 therefore, we have x(t)-k(1)x(t-1) = k(0)+y(t)-j(1)y(t-1) by lag operator, we have x(t)-k(1)Lx(t) = k(0)+y(t)-j(1)Ly(t) so, (1-k(1)L)x(t) = k(0)+(1-j(1)L)y(t) [(1-k(1)L)^(-1)](1-k(1)L)x(t) = [(1-k(1)L)^(-1)]k(0)+[(1-k(1)L)^(-1)](1-j(1)L)y(t) x(t) = [(1-k(1)L)^(-1)]k(0)+[(1-k(1)L)^(-1)](1-j(1)L)y(t) if k(1)=j(1), then x(t) = [(1-k(1)L)^(-1)]k(0)+y(t) and x(t) becomes a white noise with intercept term. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 58.39.92.102
bookticket:sincerely thx :) 140.119.232.46 05/04 20:31