Compassion
08-15-2003, 10:42 AM
I am using the Fifth Edition of the ASM study manual and I have two questions regarding Lesson 42.
1. In the last paragraph on page 221, it mentions that y = psi(B) * epsilon by definition, and then plugging this into the main equation:
phi(B) * (1-B)^d * y = theta(B) * epsilon
to get
phi(B) * (1-B)^d * psi(B) = theta(B) * epsilon
Why didn't the epsilon cancel out?
2. On page 223, in the secondary solution to Example 42.2 (using all of the B's) we were told that it was an ARMA(1,1) process, but then we are spending time calculating (1-0.4B)^-1. Shouldn't the (1-B)^-d portion of the equation just drop out because d=0? The notation ARMA(1,1) is like saying ARIMA(1,0,1) in short form, meaning that d=0. Then the (1-B)^-d portion should turn into 1 if d=0.
Or am I just getting confused by the way the question is written and mistaking (1-0.4B)^-1 for the (1-B)^-d portion of the equation, when really the (1-0.4B)^-1 is really the phi(B) part?
1. In the last paragraph on page 221, it mentions that y = psi(B) * epsilon by definition, and then plugging this into the main equation:
phi(B) * (1-B)^d * y = theta(B) * epsilon
to get
phi(B) * (1-B)^d * psi(B) = theta(B) * epsilon
Why didn't the epsilon cancel out?
2. On page 223, in the secondary solution to Example 42.2 (using all of the B's) we were told that it was an ARMA(1,1) process, but then we are spending time calculating (1-0.4B)^-1. Shouldn't the (1-B)^-d portion of the equation just drop out because d=0? The notation ARMA(1,1) is like saying ARIMA(1,0,1) in short form, meaning that d=0. Then the (1-B)^-d portion should turn into 1 if d=0.
Or am I just getting confused by the way the question is written and mistaking (1-0.4B)^-1 for the (1-B)^-d portion of the equation, when really the (1-0.4B)^-1 is really the phi(B) part?