Homework 4 Solution - Econ 503 Professor John H Nachbar Fall 2007 Homework 4 Answers 1 By Slutsky Dp(p m = Dp h(p m Dm(p m)x Pre and post multiplying by

Homework 4 Solution - Econ 503 Professor John H Nachbar...

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Econ 503Professor John H. NachbarFall 2007Homework 4Answers1. By Slutsky,Dpφ(p?, m?) =Dph(p?, m?)-Dmφ(p?, m?)x*0. Pre and post mul-tiplying byp?, sinceDph(p?, m?)p?= 0,x?0p?=p?·x?=m?(Walras’s Law),andp?·Dmφ(p?, m?) = 1 (differentiate Walras’s Law with respect tom), theresult follows.Alternatively, you can differentiate Walras’s Law with respect topto getx?0+p?0Dpφ(p?, m?) = 0, which impliesp?0Dpφ(p?, m?) =-x?0; post multiplyingbyp?and applying Walras’s Law again gives the result.2. By the symmetry ofDph,β= 3, henceA=-10α3γ-4δ3ζFrom the fact thatDph p*= 0, we have-10*1 +α*2 + 3*6 = 0, henceα=-4, henceA=-10-43γ-4δ3ζBy symmetry,γ=-4, henceA=-10-43-4-4δ3ζAgain from the fact thatDph p*= 0,-4*1-4*2 +δ*6 = 0 impliesδ= 2,henceA=-10-43-4-423ζOnce again applying symmetry,= 2. Once again from the fact thatDph p*=0, 3*1 + 2*2 +ζ*6 = 0 implies,ζ=-7/6. In summary,A=-10-43-4-4232-761
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3.(a) Recall that for eachn,φn(p, m) =αnpnm.Therefore,Dpφ(p, m) =-α1p21m00-α2p22m.(b)v(p, m) =α1lnα1p1m+α2lnα2p2m=α1[ln(α1)-ln(p1)] +α2[ln(α2)-ln(p2)] + ln(m),where I have used the fact thatα1+α2= 1.
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