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Economics Dynamics Problems 142

# Economics Dynamics Problems 142 - − 2 = then the...

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126 Economic Dynamics the autonomous component of investment is I 0 (1 + g ) t . Hicks’ model can then be expressed C t = bY t 1 I t = I 0 (1 + g ) t + v ( Y t 1 Y t 2 ) Y t = C t + I t (3.24) Substituting, we get Y t = bY t 1 + I 0 (1 + g ) t + v ( Y t 1 Y t 2 ) = ( b + v ) Y t 1 vY t 2 + I 0 (1 + g ) t Since the model involves a moving equilibrium, then assume equilibrium income at time t is Y (1 + g ) t and at time t 1 it is Y (1 + g ) t 1 , etc. Then in equilibrium Y (1 + g ) t ( b + v ) Y (1 + g ) t 1 + v Y (1 + g ) t 2 = I 0 (1 + g ) t (3.25) Dividing throughout by (1 + g ) t 2 , then Y (1 + g ) 2 ( b + v ) Y (1 + g ) + v Y = I 0 (1 + g ) 2 i.e. Y = I 0 (1 + g ) 2 (1 + g ) 2 ( b + v )(1 + g ) + v (3.26) Note that in the static case where g = 0, that this reduces down to the simple result Y = I 0 / (1 b ). The particular solution to equation (3.25) is then Y p = Y (1 + g ) t = I 0 (1 + g ) t + 2 (1 + g ) 2 ( b + v )(1 + g ) + v Since the homogeneous component is Y (1 + g ) t ( b + v ) Y (1 + g ) t 1 + v Y (1
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Unformatted text preview: − 2 = then the complementary function, Y c is Y c = c 1 r t + c 2 s t where r , s = − ( b + v ) ± ± ( b + v ) 2 − 4 v 2 The complete solution to equation (3.25) is then Y t = c 1 r t + c 2 s t + I (1 + g ) t + 2 (1 + g ) 2 − ( b + v )(1 + g ) + v r = − ( b + v ) + ± ( b + v ) 2 − 4 v 2 s = − ( b + v ) − ± ( b + v ) 2 − 4 v 2 (3.27) Once again the stability of (3.27) depends on the sign of ( b + v ) 2 − 4 v , and the various possibilities we have already investigated....
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