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TS module 16 ARMA forecasts practice problems (The attached PDF file has better formatting.) ** Exercise 16.1: ARMA (1,1) Process A time series of 45 observations yt, t = 1, 2, …, 45, is represented by an ARMA(1,1) model, with ì = 22, y45 = 21, å45 = 1, ö = –0.8, è = 0.5 What is è0 (the constant term in the ARMA process)? What is the one period ahead forecast (for Period 46)? What is the expected residual in Period 46? What is the two periods ahead forecast (for Period 47)? Solution 16.1: We solve this exercise two equivalent ways. Use the method that is clearer for you. Part A: From the mean and ö parameter, we solve for è0, the constant term in the time series: ì = è0 / (1 – ö) = 22 ö = –0.8, so è0 = 22 × (1 – (–0.8)) = 39.600 Part B: The ARMA(1,1) forecast is è0 + ö × Yt – è × åt. The residual in Period 45 is 1, so the one period ahead forecast is 39.6 – 0.8 × 21 – 0.5 × 1 = 22.300 Part C: The forecast is the best estimate, so the expected residual in all future periods is zero. Part D: The two periods ahead forecast is 39.6 – 0.8 × 22.3 – 0.5 × 0 = 21.760 Alternatively, we subtract the mean from the observed value, generate the forecasts, and add the mean to the forecasts. Part A: Subtracting the mean gives y45 = 21 – 22 = –1. Part B: The one period ahead forecast (with a mean of zero) is ö × Yt – è × åt. The residual in Period 45 remains 1, so the one period ahead forecast is –0.8 × –1 – 0.5 × 1 = 0.300. Adding the mean of 22 gives a forecast of 22.300. Part C: The expected residual in future periods remains zero. Part D: The two periods ahead forecast (with a mean of zero) is ö × Yt+1 – è × åt+1. The time series value residual in Period 46 is 0.300 and the residual in Period 46 is zero, two periods ahead forecast is –0.8 × 0.300 – 0.5 × 0 = -0.240. Adding the mean of 22 gives 22 – 0.240 = 21.760. ** Exercise 16.2: ARMA (1,1) Process A time series of 45 observations yt, t = 1, 2, …, 45, is represented by an ARMA(1,1) model, with ì = 22, y45 = 21, and å45 = 1. The one period ahead forecast is 22.300. The two periods ahead forecast is 21.760. What is ö for this time series? What is è for this time series? Solution 16.2: This exercise uses the same values as the previous exercise, but it derives the time series parameters from the forecasts. Final exam problems may ask either derivation. We show both solution methods. From the mean and ö parameter, we solve for è0: ì = è0 / (1 – ö) = 22 ö = –0.8, so è0 = 22 × (1 – (–0.8)) = 39.600 We have two linear equations for the two forecasts: One period ahead forecast: 39.6 + ö × 21 – è × 1 = 22.300 Two periods ahead forecast: 39.6 + ö × 22.3 – è × 0 = 21.760 Part A: From the two periods ahead forecast, we obtain 39.6 + ö × 22.3 = 21.760 ö = (21.760 – 39.6) / 22.3 = -0.800 Part B: From the one period ahead forecast, we obtain 39.6 + –0.8 × 21 – è × 1 = 22.300 è = (22.300 – 39.6 + 0.8 × 21) / 1 = -0.500 Alternatively, we subtract the mean from the observed values and forecasts and derive the parameters. Subtracting the mean gives y45 = 21 – 22 = –1 and forecasts of 0.300 and –0.240. We have two linear equations for the two forecasts: One period ahead forecast: ö × –1 – è × 1 = 0.300 Two periods ahead forecast: ö × 0.300 – è × 0 = -0.240 Part A: From the two periods ahead forecast, we obtain ö × 0.300 = -0.240 ö = –0.240 / 0.300 = -0.800 Part B: From the one period ahead forecast, we obtain –0.8 × –1 – è × 1 = 0.300 – è = (0.300 + 0.8 × –1) / 1 = -0.500 è = 0.500
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