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5.1. CALIBRATING THE ONE-SECTOR GROWTH MODEL 147The roots of the polynomial (1 − b 1 z − b 2 z 2 )=(1− ω 1 L)(1 − ω 2 L)satisfy b 1 = ω 1 + ω 2 and b 2 = −ω 1 ω 2 . Using the quadratic formulaand evaluating at the parameter values thatqwe used to calibrate themodel, the roots are, z 1 =(1/ω 1 )=[−b 1 − b 2 1 +4b 2 ]/(2b 2 ) ' 1.23, andqz 2 =(1/ω 2 )=[−b 1 + b 2 1 +4b 2 ]/(2b 2 ) ' 0.81. There is a stable root,|z 1 | > 1 which lies outside the unit circle, and an unstable root, |z 2 | < 1which lies inside the unit circle. The presence of an unstable root meansthat the solution is a saddle path. If you try to simulate (5.24) directly,the capital stock will diverge.To solve the difference equation, exploit the certainty equivalenceproperty of quadratic optimization problems. That is, you Þrst getthe perfect foresight solution to the problem by solving the stable rootbackwards and the unstable root forwards. Then, replace future randomvariables with their expected values conditional upon the time-tinformation set. Begin by rewriting (5.24) asW t = (1− ω 1 L)(1 − ω 2 L)ˆk t+1= (−ω 2 L)(−ω −12 L −1 )(1 − ω 2 L)(1 − ω 1 L)ˆk t+1= (−ω 2 L)(1 − ω −12 L−1 )(1 − ω 1 L)ˆk t+1 ,and rearrange to get(1 − ω 1 L)ˆk t+1 = −ω−1 2 L −11 − ω2 −1 L W −1 tµ 1= − L −1 X ∞ω 2= −∞Xj=1j=0µ 1ω 2 jW t+jµ 1ω 2 jW t+j . (5.25)The autoregressive speciÞcation (5.18) implies the prediction formulaej=1E t W t+j = b 3 E t  t+j+1 + b 4 E t  t+j =[b 3 ρ + b 4 ]ρ j  t .Use this forecasting rule in (5.25) to get∞Xµ 1 j ∞Xµ " #ρjρE t W t+j =[b 3 ρ + b 4 ]Ât = (b 3 ρ + b 4 )Ât.ω 2 ω 2 ω 2 − ρj=1