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A Forecasting Example

# A Forecasting Example - var e 2005(1 = var a 2005 = σ 2 a...

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A forecasting example Suppose that annual sales (in millions of dollars) of a company follow the AR(2) model: Z t = 5 + 1 . 1 Z t - 1 - 0 . 5 Z t - 2 + a t with σ 2 a = 2 1. If sales for 2005, 2004, and 2003 were 10 million, 11 million, and 9 million, respec- tively, forecast sales for 2006 and 2007. Solution: ˆ Z 2005 (1) = 5 + 1 . 1 Z 2005 - 0 . 5 Z 2004 = 5 + 1 . 1 × 10 - 0 . 5 × 11 = 10 . 5 ˆ Z 2005 (2) = 5 + 1 . 1 ˆ Z 2005 (1) - 0 . 5 Z 2005 = 5 + 1 . 1 × 10 . 5 - 0 . 5 × 10 = 11 . 55 The forecast sales for 2006 is \$ 10.5 millions. The forecast sales for 2007 is \$ 11.55 millions. 2. Calculate 95% prediction limits for your forecasts in part (a) for 2006. Solution: The error of forcast for sales in 2006 is e 2005 (1) = Z 2006 - ˆ Z 2005 (1) = a 2006 So

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Unformatted text preview: var ( e 2005 (1)) = var ( a 2005 ) = σ 2 a = 2 The 95% prediction limits for the forecast in 2006 is [ ˆ Z 2005 (1)-z α 2 q var ( e 2005 (1)) , ˆ Z 2005 (1) + z α 2 q var ( e 2005 (1))] = [ ˆ Z 2005 (1)-z α 2 σ a , ˆ Z 2005 (1) + z α 2 σ a ] = [10 . 5-1 . 96 √ 2 , 10 . 5 + 1 . 96 √ 2] = [7 . 72814 , 13 . 2719] 1 3. If sales in 2006 are 12 million, show how to update your forecast for 2007. Solution: Using the updating formula ˆ Z 2006 (1) = ˆ Z 2005 (2) + 1 . 1 × [ Z 2006-ˆ Z 2005 (1)] = 11 . 55 + 1 . 1 × [12-10 . 5] = 13 . 2 2...
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A Forecasting Example - var e 2005(1 = var a 2005 = σ 2 a...

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