Advanced Macroeconomics Solution Manual by David Romer

By David Romer

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CASE 1 : x  Pt Pt 1 She will consume half of her endowment, store the rest and not hold any money since the rate of return on money is less than the rate of return on storage. Thus C1,t  A 2 M dt Pt  0 Ft  A 2 C 2 ,t 1  xA 2 . CASE 2 : x  Pt Pt 1 Now storage is dominated by holding money. She will consume half of her endowment and then sell the rest for money: C1,t  A 2 C 2 ,t 1   Pt Pt 1   A 2 . M dt Pt  A 2 Ft  0 CASE 3 : x  Pt Pt 1 Money and storage pay the same rate of return.

Thus selling a unit of the good in period t allows the individual to buy Qt /Qt+1 units of the good in period t + 1. Thus the gross rate of return on trading is Qt /Qt+1 . Now, Qt+1 = Qt /x for all t > 0 is equivalent to x = Qt / Qt+1 for all t > 0. In other words, the rate of return on storage is equal to the rate of return on trading and hence the individual is indifferent as to the amount to store and the amount to trade. Let t  [0, 1] represent the fraction of "saving", A/2, that the individual sells in period t.

Thus (4) Kt+1 = St Lt , where St is the amount of saving done by a young person in period t. Note that St  s(rt+1 )Atwt ; the amount of saving done is equal to the fraction of income saved times the amount of income. Thus equation (4) can be rewritten as (5) Kt+1 = Lts(rt+1 )Atwt . To get this into units of time t + 1 effective labor, divide both sides of equation (5) by At+1Lt+1 : K t 1 A t Lt  (6)  s( rt1 ) w t  . A t 1 L t 1 A t 1L t 1 Since At+1 = (1 + g)At , we have At /At+1 = 1/(1 + g).

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