Advanced Macroeconomics Solution Manual by David Romer

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By David Romer

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Again, use equation (4) to put everything in terms of k and (18) becomes (19) y  ((k * )1  e t (k (0)1  (k * )1 ))  (1 ) . Now we would like to see whether the speed of convergence to the balanced growth path is constant. Using equation (19) and subtracting the balanced growth path y*, where     y  (k )      g  * *   (1  ) from equation (2), and taking logs results in © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.

Thus aggregate demand will be less than aggregate supply and the market for the good will not clear. Thus the proposed price path cannot be an equilibrium. © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Solutions to Chapter 2 2-33 Suppose instead that the auctioneer announces Qt+1 > Qt /x or equivalently x > (Qt /Qt+1 ) for some date t.

Isolating for z z yields a complementary solution of (13) z c  A1e  t , where A1 is a constant of integration. To solve for the particular solution, we consider the non-homogeneous case, where z  z  (1  ) . Using an integrating factor, we find the solution to be (14) z p  (1  )   A 2 e  t , where A 2 is a constant of integration. Therefore, (15) z  z p  z c  z  (1  )   (A1  A 2 )e t . Using the initial condition z(0), we can substitute for A 1  A 2 and (15) becomes (16) z  (1   )   e  t (z(0)  (1   )  ) .

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