NEAS


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TS Module 15: Forecasting basics HW (The attached PDF file has better formatting.) Homework assignment: ARIMA(1,1,0) forecasts An ARIMA(1,1,0) process has 40 observations y_{t}, t = 1, 2, …, 40, with y_{40} = 60 and y_{39} = 50. This time series is not stationary, but its first differences are a stationary AR(1) process. The parameter è_{0} of the stationary AR(1) time series of first differences is 5. The 1 period ahead forecast ŷ_{40}(1) is 60. We determine the 2 period ahead forecast ŷ_{40}(2).
A. What is the most recent value of the autoregressive model of first differences? Derive this value from the most recent two values of the ARIMA(1,1,0) process. B. What is the one period ahead forecast of the first differences? Derive this value from the the one period ahead forecast of the ARIMA(1,1,0) process. C. What is the parameter ö_{1} of the AR(1) process of first differences? Derive this parameter from the 1 period ahead forecast. D. What is the two periods ahead forecast of the AR(1) process of first differences? Use the parameter of the AR(1) process. E. What is the two periods ahead forecast of the ARIMA(1,1,0) process? Derive this from the two periods ahead forecast of the AR(1) process.



Tom McNamara III


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I am confused as to the relationship between a nonstationary 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 nonstationary 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



sbenidt


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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.



Lidong Wang


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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 straightforward. The first differences are an AR(1) model: Δy_{t} = 5 + φ_{1} Δy_{t1} + ε_{t}.
Step 1: Determine the most recent value of the autoregressive model from the most recent two values of the original time series: Äy_{40} = y_{40} – y_{39} = 60 – 50 = 10 Step 2: Convert forecasts of the time series to forecasts of the first differences for the one period ahead forecast. The forecast for period 41 is 60, so the forecasted first difference is 60 – 60 = 0. Step 3: Find the parameter ö_{1} from the 1 period ahead forecast. Äy_{41} = 5 + ö_{1} × 10 = 0 ö_{1} = –0.5. è_{0} = 5 is not the mean of the AR(1) process; it is the constant term. The AR(1) process can be written two ways: Y_{t} = è_{0} + ö_{1} × Y_{t1} or Y_{t} – ì = ö_{1} × (Y_{t1} – ì). The textbook uses both formats, but è_{0} and ì are different values. Step 4: Solve for the two period ahead forecast from the autoregressive equation. Äy_{42} = 5 – 0.5 × 0 = 5 Step 5: Convert the forecast of the first differences for two periods to the initial time series. y_{42} = y_{41} + 5 = 65
The textbook has formulas for forecasts and variances of ARIMA processes.
If you understand the intuition, the formulas are easy to recall and provide a good check on your work. If you do not understand the intuition, you will mix up the formulas for the various ARIMA processes. Focus on the intuition. After a few problems, it is easy.



Woody


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For forecasting, why don't you use equation 9.3.6 Δyt = 5 + φ1 [Δyt1  5] .
Step 3 will be 0 = 5+ φ1 [10  5], φ1 = 1, not .5???
[NEAS: 5 is the constant term (theta0), not the mean.]



Luke Grady


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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).]



benjaminttp


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The first differences are an AR(1) model: Δyt = 5 + φ1 Δyt1 + εt1
why the last term is et1 but not et?



AJB1011


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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 + φΔΥt1 + et1
Here's how I began to derive the first difference: ΔΥt = Yt – Yt1 ΔΥt = φΥt1 + et – θet1 – [φΥt2 + et1 – θet2]
Also, for Part B: I believe the formula being used is 9.3.26 > Y_hat(L) = Yt + (θzero)*L Does "L" = 0 in this case, then? That way Y41 = Y40.



RayDHIII


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AJB, pages 9697 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_{t1} + e_{t1}. 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



Luke Grady


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So is that what the d means in the ARIMA(p,d,q)? Does that just mean that the dth difference of the time series is an ARMA(p,q) process?
Also, is a second difference the difference between Y_t and Y_{t2}? [NEAS: No, the second difference is the first difference of the first differences.]


