1 Slutsky Matrix og Negative deniteness

Transcription

1 Slutsky Matrix og Negative deniteness This is exercise 2.F. from the book. Given the demand function x(p,) from the book page 23, here β = and =, e shall :. Calculate the Slutsky matrix S = D p x(p, ) + D x(p, )x(p, ) T evaluate S at p = (,,) 2. Sho that x(p,) does not fullll the eak axiom. Since e calculate the Slutsky Matrix and therefore changes in consumer income, e shall ait by inserting = assumption.. Solution We start out by calculating D p x(p, ): D p x(p, ) = p2 P p2 p3 P 2 p p 3P P 2 p p2 P p3 2 P p3 P 2 p P 2 p2 p p3 2P P 2 p P p P 2 () Where P (p + + ). No e have the ra eect on demand from changing prices. This eect includes income eects, and since e only ant to consider substitution eects e have to compensate the consumer. Firstly e calculate the ra eects from changing income D x(p, ): D x(p, ) = ( p P Then e calculate the full income eect as D x(p, )x(p, ) T, hich is a 3 3 matrix. D x(p, )x(p, ) T = p P p3 P p P D x(p, )x(p, ) T = P ( p P 2 P 2 P 2 p P ) P 2 P 2 p P 2 p ) P P 2 p P 2 P 2 (2) (3) (4)

2 So the isolated substitution eects is: S = D p x(p, ) + D x(p, )x(p, ) T = p2(2p+p3) (p +2) (p ) P 2 ( p ) (2+) P 2 p3(p+2p2) (2p +) 2 P 2 P 2 p(p2 p3) P 2 p(p2+2p3) P 2 (5) When evaluating the Slutsky matrix in p = (,, ) and =, one gets: S(,, ) = 3 (6) This Matrix does not have full rank. Take as an example (- column vector - column vector 3) = column vector 2. Also p Sp =, hen p = (,,). But this actually applies for all p. We no examine if S is negative semidenite v, v Sv. This is a necessary condition for x(p,) fulllling the eak axiom. Let p = (,, ɛ), ɛ >, and insert this subset of price vectors in S: S(,, ɛ) = 2+ɛ +2ɛ ɛ 3ɛ 3 (2+ɛ)ɛ ɛ +2ɛ 2 (7) We search for ɛ s and vectors v here S(,, ɛ) is not negativ semidenite (that is positive denite). We no apply 2.F.3, hich says that if x(p,) fulllls WL and is homogenous of degree of, then p S(p) = and S(p)p = (p, ). From exercise, set 2 e kno that x(p,) fulllls WL and is homogenous of degree of. Therefore e can apply M.D.4 page 939. Theorem M.D.4: If Sp = p S = and a reduced matrix S is negative denit, then S is negative denit for all vectors in the subspace T z = {z z p = }. We therefore examine the reduced matrix S given by removing one ro and one 2

3 column: S(,, ɛ) = 2+ɛ +2ɛ 3ɛ (8) We choose an arbitrary vector ṽ R 2 and calculate: ṽ Sṽ = + 2ɛ (2 + ɛ) 2 ṽṽ (2 + ɛ) 2 (2 + ɛ) 2 ṽ2 (3ɛ) (2 + ɛ) 2 ṽ2 2 (9) Especially e search for vectors hich ensures ṽ Sṽ >. Since ɛ > e can remove (2 + ɛ) 2 and reduce the expression: ṽ Sṽ > (ṽ ṽ 2 2ṽ 2 ) > ɛ(ṽ 2 + 3ṽ 2 2 2ṽ ṽ 2 ) () The expression on the right hand side next to ɛ is allays positive, since: (ṽ 2 + 3ṽ 2 2 2ṽ ṽ 2 ) = (ṽ ṽ 2 ) 2 + 2ṽ 2 2 > () Therefore e can convert the expression to: (ṽ ṽ 2 2ṽ 2 ) (ṽ 2 + 3ṽ2 2 2ṽ ṽ 2 ) > ɛ > (2) Basically e achieve our goal if ṽ 2 > 2 ṽ. Thus if e choose ṽ = (, 4) e ill ensure that ṽ Sṽ > for some ɛ. By insertion of ṽ e get: ( ) ( ) = 2 4 > ɛ > (3) Thus S is not negative denite for all vectors, thus S cannot be either. Renaming our v vector to p, e have found that: p R 3 ; p = (, 4, ) p S p >, hen p = (,, 2/4 ρ) (4) Thus there exists price changes p and prices p hich makes S postive denite, and x(p,) cannot fullll the eak axiom. Q.E.D. 2 Strange demand changes In execise 2.F.6 from the book e are given the folloing demand function: x(p, ) = 3 p (5)

4 That is, the consumer demands positively good and 3, but delivers good 2. Also the demand for the rst to good does not vary ith its on price or the consumers overall income. We could think of good as special consumption good hich our consumer basically demands according to his real age salery, namely. Good 2 is then labour supply hich is supplied according to consumer price p. Only good 3 seems to be a normal good, and could be thought of as some kind of investment good here bought from savings. The demand could be explained like this: Assume the orker kno, at hich prices he can buy good for in the next period, but not hat his orking salary is. It could be that the payo from ork is uncertain. Hoever, he chooses his labour supply, so that if prices on good are high, then he ill ork more. When the ork is completed, the project pays of at some price, and all ork income is spend. Basically this story involve prices being a signal to the consumer, thus changing the prices, changes the the signal and the behaivior even if the consumer are compensated through. In the exercise e are asked to:. sho that x(p, ) is homogenous of degree of. and fulllls Walras La. 2. sho that x(p, ) does not fullll the eak axiom 3. sho that the Slutsky matrix S fulllls v Sv = v R Proof - Homogenity of degree. and Walras La. From insertion of the real number λ one gets: x(λp, λ) = λ λ λp λ λ λ = x(p, ) (6) Thus x(p,) is homogenous of degree. Also x(p,) satises Walras lo since: Both properties follos directly. p x(p, ) = p p + = (7) Q.E.D. 2.2 Proof - Violation of the eak axiom Next e sho that x(p,) does not satisfy the eak axiom. This is due to fact that is not used in good and 2. Let: p = (,, ) = x = (,, ) (8) No lets change price of good, and compensate our consumer, so he still can aord x in this ne situation: p = (2,, ) = p x = 2 x = (, 2, 2) (9) 4

5 So x is revealed preferred to x, and x should not be revealed preferred x, hich is basically the same as x is not aordable in situation. Unfortunately that is indeed the case: p = (,, ) p x = = (2) Thus the eak axiom is not fulllled. The problem is, that e use savings for compensating price changes, hich cannot be remedied through increased savings. Q.E.D. 2.3 The Slutsky Matrix Finally e deduct the Slutsky matrix. First demand changes from price changes. p2 D p x(p, ) = p p3 (2) Next the compensated income eect D x(p, ) x = (22) p Note that only good 3 is aected by our income changes. Finally e calculate the substitution eects shon in the Slutsky matrix S: S = D p x(p, ) + D x(p, )x(x, p) T = p3 Choose a vector v R 3 and calculat: v Sv = v v 2 p3 p2 v 3 v p3 + p v 3 v 2 v v 2 p3 p3 + p2 v 3 v p3 p p2 p (23) p v 3 v 2 = (24) p3 So each pair in the sum negates each other. So. term negates 4. term and so on. So v Sv = for arbitrary vectors v. Q.E.D. Both exercises demonstrates an application of the negative deniteness of S. Firstly, S being negative semidenite, is an necessary condition for the eak axiom. On the other hand, S can be negative semidenite, even if the demand does not fullll the eak axiom. Thus there is no biimplication in proposition 2.F.2. 5