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10.3. INFINITESIMAL MARGINAL INTERVENTION 315Solution under Krugman intervention. To solve (10.30), replace y(t) in(10.32) with G(f), set a 1 = 2η , aσ 2 2 = −2 , and b = aασ 2 2 .TheresultiswhereG[f(t)] = ηα + f(t)+Ae λ 1f(t) + Be λ 2f(t) , (10.33)λ 1 = −ηsησ + 22 σ + 2 > 0, (10.34)4 ασ2 λ 2 = −ηsησ − 22 σ + 2 < 0. (10.35)4 ασ2 To solve for the constants A and B, you need two additional pieces of information.These are provided by the intervention rules. 4 From (10.33),you can see that the function mapping f(t) intos(t) is one-to-one. Thismeans that there is a lower and upper band on the fundamentals, [f, ¯f]that corresponds to the lower and upper bands for the exchange rate[s, ¯s]. When s(t) hits the upper band ¯s, the authorities intervene toprevent s(t) from moving out of the band. Only inÞnitesimally smallinterventions are required. During instants of intervention, ds = 0 fromwhich it follows thatG 0 ( ¯f) =1+λ 1 Ae λ 1 ¯f + λ 2 Be λ 2 ¯f =0. (10.36)Similarly, at the instant that s touches the lower band s, ds =0andG 0 (f) =1+λ 1 Ae λ 1f + λ 2 Be λ 2f =0. (10.37)(10.36) and (10.37) are 2 equations in the 2 unknowns A and B, whichyou can solve to getA =B =e λ 2 ¯f − e λ 2fλ 1 [e (λ 1 ¯f+λ 2 f) − e (λ 1f+λ 2 ¯f)]e λ1f − e λ 1 ¯fλ 2 [e (λ 1 ¯f+λ 2 f) − e (λ 1f+λ 2 ¯f)]< 0, (10.38)> 0. (10.39)4 In the case of a pure ßoat and in the absence of bubbles, you know thatA = B =0.