International macroe.. - Free

198CHAPTER 6. FOREIGN EXCHANGE MARKET EFFICIENCY(6.57) displays the familiar property of constant absolute risk aversionutility in which portfolio positions are proportional to the expectedasset payoff. The factor of proportionality is inversely related to theindividual’s absolute risk aversion coefficient. Recall that individualsundertake zero-net investment strategies. The portfolio position in oursetup does not depend on wealth because traders are endowed with zeroinitial wealth.The foreign fundamental trader faces an analogous problem. Thesecond period euro-wealth of fundamentalist foreign agents is the payofffrom portfolio investments plus an exogenous euro payment of ‘labor’income c ∗ , W f ∗t+1 = [∆s t+1 − x t ] λf ∗tS t+ c ∗ . The solution is to chooseλ f ∗t = S t λ f t . Because individuals at home and abroad have identicaltastes but evaluate wealth in national currency units, they will pursueidentical investment strategies by taking positions of the same size asmeasured in monetary units of the country of residence.These portfolios combined with the market clearing condition (6.49)imply the difference equation 15E t ∆s t+1 − x t = Γ t (E t−1 ∆s t − x t−1 ), (6.58)where Γ t ≡ [(S t /S t−1 )+S t−1 R t−1 ]/(1 + S t ). The level of the exchangerate is indeterminate but it is easily seen that a solution for the rate ofdepreciation is∆s t = 1 ρ x t = x t−1 + 1 ρ v t. (6.59)The independence of v t and x t−1 implies E t (∆s t+1 )=x t and the fundamentalssolution therefore does not generate a forward premium biasbecause uncovered interest parity holds in the fundamentals equilibriumeven when agents are risk averse. The reason is that under homogeneousexpectations and common knowledge, you demand the same riskpremium as I do, and we want to do the same transaction. Since wecannot Þnd a counterparty to take the opposite side of the transaction,no trades take place. The only way that no trades will occur inequilibrium is for uncovered interest parity to hold.15 The left side of the market clearing condition (6.49) is λ t + λ ∗t =(1 + S t )λ t =(1 + S t )/(γσ s )[E t ∆s t+1 − x t ]. The right side is, (S t /S t−1 )R ∗ λ t−1 + R t−1 S t−1 λ t−1=[(S t /S t−1 )+(1 + x t−1 )S t−1 ]λ t−1 . Finally, using λ t−1 =[E t−1 ∆s t − x t−1 ]/(γσs 2),we get (6.58).