Solutions_to_PS7_Fall2011 - Solutions to PS7 Econ 202A -...

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Unformatted text preview: Solutions to PS7 Econ 202A - Second Half Fall 2011 Prof. David Romer GSI: Victoria Vanasco 1 Saddle-Path of the q-Theory Model. Consider the equations q ( t ) = rq ( t )- ( K ( t )) K ( t ) = f ( q ( t )) (a) De ne the steady state of the model, ( q, K ) . Show that the model's linear (Taylor) approximation in the neighborhood of the steady state takes the form: bracketleftBigg q ( t ) K ( t ) bracketrightBigg = bracketleftBigg A B C bracketrightBiggbracketleftBigg q ( t )- q K ( t )- K bracketrightBigg Remember that the Taylor approximation around the steady state is given by: bracketleftBigg q ( t ) K ( t ) bracketrightBigg = bracketleftBigg q ( t ) q ( t ) q ( t ) K ( t ) K ( t ) q ( t ) K ( t ) K ( t ) bracketrightBigg | SS bracketleftBigg q ( t )- q K ( t )- K bracketrightBigg bracketleftBigg q ( t ) K ( t ) bracketrightBigg = bracketleftBigg r- ( K ) f (1) bracketrightBiggbracketleftBigg q ( t )- q K ( t )- K bracketrightBigg where- ( K ) > because we assumed ( K ( t )) < , and f ( q ( t )) > since the higher the inside value of capital, the higher the incentives to invest. This implies that A,B,C > . From now on, G bracketleftBigg r- ( K ) f (1) bracketrightBigg (b) Characteristic Roots. To nd the roots of our matrix G , we need to nd the conditions under which the matrix G- I is not invertible, i.e.: det ( G- I ) = 0 det bracketleftBigg r- - ( K ) f (1)- bracketrightBigg = 0 1- ( r- ) + ( K ) f (1) = 0 2- r + ( K ) f (1) = 0 and using the usual formula for roots of a quadratic function: 1 , 2 = r radicalBig r 2- 4 ( K ) f (1) 2 where 1 = r + radicalBig r 2- 4 ( K ) f (1) 2 > and 2 = r- radicalBig r 2- 4 ( K ) f (1) 2 < Note that r = r 2 and- 4 ( K ) f (1) > . Since . is an increasing function, we know that r < radicalBig r 2- 4 ( K ) f (1) , and thus 2 < . (c) Show that the eigenvectors of the matrix are proportional to X = bracketleftBigg 1 f (1) 2 f (1) 1 1 bracketrightBigg . To nd the eigenvectors, recall that there exists a matrix of eigenvectors X that satis es the following property: X 1 GX = where X = bracketleftBigg v 11 v 21 v 12 v 22 bracketrightBigg and = bracketleftBigg 1 2 bracketrightBigg Let's normalize v 12 = v 22 = 1 . Now we solve for v 11 and v 21 : GX = X bracketleftBigg r- ( K ) f (1) bracketrightBiggbracketleftBigg v 11 v 21 1 1 bracketrightBigg = bracketleftBigg v 11 v 21 1 1 bracketrightBiggbracketleftBigg 1 2 bracketrightBigg bracketleftBigg rv 11- ( K ) rv 21- ( K ) f (1) v 11 f (1) v 21 bracketrightBigg = bracketleftBigg 1 v 11 2 v 21 1 2 bracketrightBigg This is a system of four equations and two unknowns, you can check that two of these equations are redundant.This is a system of four equations and two unknowns, you can check that two of these equations are redundant....
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This note was uploaded on 02/28/2012 for the course ECON 202A taught by Professor Akerlof during the Fall '07 term at University of California, Berkeley.

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Solutions_to_PS7_Fall2011 - Solutions to PS7 Econ 202A -...

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