I am confused as to the relationship between a non-stationary ARIMA(1,1,0) [integrated autoregressive moving average (p,d,q)] and an ARMA.

1st off this question says the process is an ARIMA(1,1,0).  Is the zero necessary.  Isn't this just an ARI(1,1)?

[NEAS: Correct; ARI is an abbreviation for an ARIMA process with q = 0]

An ARIMA(p,d,q) process can be rewritten as am ARMA(p+1,q) process (Cryer Chan page 92).

[NEAS: No, and Cryer Chan do not say this. The first differences of an ARIMA process with d = 1 are an ARMA process of the same order.]

So how can this non-stationary ARIMA(1,1,0) be put into terms of an AR(1)?  Shouldn't it be an AR(2)?

[NEAS: The first differences of ARIMA(1,1,0) are an AR(1) process.]

Also, no where in the Module 15 reading does it mention 1st differences.  I have come across modules like this before where the reading does not match up with the homework assignments very well.  I came across this with the QQ plots in module 14.  I am growing frustrated with doing readings where the corresponding homeworks come out of left field.

[NEAS: Cryer Chan discuss first differences throughout their text, beginning in chapter 5.]

 

 

Tom McNamara III

Hi Tom,

First off, ARIMA(1,1,0) and ARI(1,1) are exactly the same process. Either notation should be acceptable to write in the course of a problem. The ARIMA(1,1,0) process can indeed be written in terms of a nonstationary AR(2) process. To achieve stationary, you would need to take first differences of the ARIMA(1,1,0) process which would result in an AR(1) process. The concept of first differences came up in an earlier module. I've found that writing up the explicit equation for both processes helps here in seeing how to derive the one and two period ahead forecasts.

Can NEAS can explain this problem?

[NEAS: Here is the outline of the solution. Many exam problems are similar. Once you get the intuition, the problems are straight-forward.

For forecasting, why don't you use equation 9.3.6 Δyt = 5 + φ1 [Δyt-1 - 5] .

Step 3 will be 0 = 5+ φ1 [10 - 5], φ1 = -1, not .5???

[NEAS: 5 is the constant term (theta-0), not the mean.]

So basically, the ARIMA(1,1,0) could have been any type of process - we don't care about what type of process it is because all we really know about is the type of process the first difference is. Is that correct?

[NEAS: An ARIMA(1,1,0) process is the integration (summation) of an AR(1) process. If the first difference is AR(1), the process is ARIMA(1,1,0).]

The first differences are an AR(1) model: Δyt = 5 + φ1 Δyt-1 + εt-1

why the last term is et-1 but not et?

Can someone clarify how the first difference was derived? I think I'm getting confused with the subscripts.

It's given that the first difference is AR(1) --> Yt = 5 + φΔΥt-1 + et-1

Here's how I began to derive the first difference:
ΔΥt = Yt – Yt-1
ΔΥt = φΥt-1 + et – θet-1 – [φΥt-2 + et-1 – θet-2]


Also, for Part B: I believe the formula being used is 9.3.26 --> Y_hat(L) = Yt + (θ-zero)*L
Does "L" = 0 in this case, then? That way Y41 = Y40.

AJB, pages 96-97 explain your dilema here.  Also, the AR(1) process is the first difference of the ARIMA(1,1,0) process, so that

ARIMA(1,1,0): unknown, and AR(1): deltaY_t = 5 + (phi)deltaY_t-1 + e_t-1.  Where the 5 is accommidating for a nonzero constant mean by theta = mu(1 - phi), also explained on page 97.

Part B wants to know the AR(1) one period ahead forecast, which is the ARIMA(1,1,0) one period ahead forecast (given) less the ARIMA(1,1,0) actual (given).

RDH

So is that what the d means in the ARIMA(p,d,q)? Does that just mean that the d-th difference of the time series is an ARMA(p,q) process?

Also, is a second difference the difference between Y_t and Y_{t-2}?

[NEAS: No, the second difference is the first difference of the first differences.]