→ bookticket:sincerely thx :) 140.119.232.46 05/04 20:31
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.
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