Problem set 3

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PROBLEM SET 3 - JOSE (el) > LET AND SUPPOSE is THE OPTIMAL CONSUMPTION BUNDLE WHEN PRICES ARE P! SINCE P FOR ALL x THEREFORE AND so FEASIBLE WHEN PRICES ARE P, WHICH MEANS V(P,W) = HENCE V is NONINCREASING. NOW SUPPOSE U AND U LOCALLY NON-SATIATED IMPLIES THAT = W CAW) so SINCE NOW AND IMPLIES AND SINCE A CONTINUOUS FUNCTION IN FOR = SUCH THAT AND so ALL x E HAVE THAT = WHICH IMPLIES x E BECAUSE U SUCH THAT > AND SINCE =D WE HAVE THAT V(P,W) = VCP', w), V 15 SRICTLY DECREASING IN P. (P2) SINCE U CONTINUOUS AND SRICTLY EXISTS AND UNIQUE FOR ALL (P,W) So > U(x) FOR x E WITH SUPROSE AND SINCE E Bew WE HAVE THAT > AND BECAUSE THE MARSHALIAN DEMAND WHEN PRICES ARE p' AND WEALTH IS w', > U(x) FOR ALL SUCH THAT THEREFORE > IMPLIES & WHICH IMPLIES > w'. So SATISFIES THE Axion of REVEALED (P3) CONSIDER THE SET M= SINCE BECONGS TO IMAGE u, 7 R", SUCH THAT = M 15 ALSO BECAUSE FOR THEREFORE A LOWER BOUND FOR M, M A NON-EMPTY BOUNDED FROM BELOW SUBSET of R, WHICH IMPLIES INF M EXISTS (SINCE R NOW CONSIDER MN SUCH THAT y" BECAUSE E M N, SUCH THAT > AND y" = THIS DEFINGS SEQUENCE SINCE N SUCH THAT noSUCH THAT THE EXPENDITURE HAS A SOLUTION SUPPOSE THAT x' x" ARE BOTH SOLUTIONS to (E) WITH CONSIDE = (1-t) FOR SINCE AND x" x">0 ALSO, SINCE U STRICTLY AND > MIN SINCE 15 > SUCH THAT x U(x) SINCE FOR LEAST ONE COORDINATE so SUCH THAT LET MIN AND CONSIDER - WHERE IS THE VECTOR SUCH THAT FOR AND e; =1 FOR THEN = 0 BECAUSE BUT = P. = = e.x' (SINCE AND x" ARE SOLUTIONS TO (E) AND = WHICH CONTRADICTS THE FACT THAT x' SOLUTION TO SINCE z' is AN ECEMENT or THE CONSTRAINT SET. HENCE, IT MUST BE THAT THE SOLUTION to (E) UNIQUE. (P4) FIRST NOTICE = IS A LINEAR OF CONCAVE FUNCTIONS WITH POSITIVE ENTS AND THEREPORE STRICTLY CONCAVE (AND ALSO, THE CONSTRAINTS OF THE CONSUMER ARE ALL THEREFORE THEY ARE (AND SINCE THE WALRASIAN SET Be,w = ALWAYS HAVE AN INTERIOR POINT IF AND w>o, IF WE ASSUME AND w>o BOTH SLATER'S CONDITIONS AND THE CONDITIONS FROM THEOREM ARE 80 THE K-T CONDITIONS ARE NECESSARY AND SUFFICIENT FOR THE SOLUTION to THIS PROBLEM. SINCE THE ONLY HAVE A SOLUTION ALL we MIGHT ASSUME THIS AND SINCE V IS CONTINUOUS, THE HAVE A SOLUTION). ALSO ONLY 50 = If AND THEN THE SOLUTION to THE CONSUMER PROBLEM, WHICH UNIQUE SINCE U 15 STRICTLY IS CHARACTERIZED BY THE K-T (1) xᵢ" FOR ALL ns (2) = = 0 (3) xizo SINCE ARE by (1) WHICH IMPLIES FOR ALL WE CAN DROP Mi FROM THE NOW (1) IMPLIES = so = THEREPORE FOR REAL SOLUTION, WHICH IMPLIES by (2). SINCE AND WE WRITE xᵢ AS A FUNCTION of x; ALL AND FLXED je nl, FOR CONDITIC (1) IMPLIES: = FOR ALL WHERE j SUBSTITUTING THOSE IN GET " W = = + NOTICE THIS Pi SOLUTION AGREES WITH x*=0 WHEN FOR AND THE MARSHALLIAN DEMAND = w), WHERE = FOR(PS) SINCE U THE SUM of CONCAVE FUNCTIONS, CONCAVE AND AS IN THE PREVIOUS EXERCISE, IF AND THE SOCUTION TO THE CONSUMER'S PROBLEM BY THE K-T CONDITIONS, 50 is SUCH 1/2 - + = (18) t = (2.) = 0, AND (3) + FIRST NOTICE THAT OTHERWISE (I.A) IMPLIES THE POSITIVE NUMBER THE NON- POSITIVE NUMBER 50 BY (2) + P2 = W. WE WILL NOW DECOMPOSE THE PROBLEM IW THREE CASES: FIRST SUPPOSE SUBSTITUTING THIS CASE IN (1.B) > 2P, AND = 1/2 WHICH IS NEGATIVE SINCE BUT THIS CONTRADICTS SUBSTITUTING Mi=0. GET THEN By (2) IMPLIES AND SINCE So THIS IN AND WE Now GET ASSUME = By (2) AND = WHICH IS So = P2 SATISFIES THE K-T WHICH ARE AN (2) WE GET = (1.B) so JUPPOSE NOW By BUT AND IF THEN > W By AND WE GET THAT A IF SATISFIES K-T, x', = BUT SINCE 2P, IF SATISFIES (2), WHICH MEANS 2P, CANNOT SATISFY AND (1.8) AT THE SAME TIME, WHICH MEANS IS NOT A HENCE, SINCE THE K-T CONDITIONS ARG THE UNIQUE SOLUTION IN THE FIRST CASE. SECOND CASE > 2P, AND AS IN THE FIRST CASE, > IMPLIES AND SUPPOSE M; = THEN AS ABOVE By (1.A) AND (1.8) WE GET = BUT SINCE > W, by (2) AND 2>0 WE GET WHICH CONTRADICTS > WHICH IMPLIES by (2) THAT AND AGAIN by (2) AND 2>0 WE GET THAT = WHICH IS POSITIVE AND = 1, WHICH POSITIVE SINCE IMPLIES SINCE THE K-T CONDITIONS ARE NECESSARY = W IS THE UNIQUE IN THIS CASE. THIRD CASE SUPPOSE THEN AND By (1.A) AND (1.B) WE = BUT AND IMPLIES W ) If P2 2P,(i) = AND THEN Bw THE SET THE CONSUMER'S IF P SEQUENCE IN Bw THAT CONVERGES TO y, THEN ≥ y." y,t SINCE gcy) IS CONTINUOUS IN WHICH CONTRA THE FACT THAT AND K' ARE BUTH TO THE CONSUMER'S (ii) SINCE DIFFERENTIABLY it AND SINCE THE CONSTRAINT FUNCTIONS THE CONSUMER'S PROBLEM ARE ALL THEY ARE AND QUASI LET WHERE AND THEN, FOR E SUFFICIENTLY SMALL, = AND (SINCE >0). OBVIOUSLY AND so THE CONSTRANT SET HAS AN POINT. HENCE BOTH CONDITION AND THE CONDITIONS FROM ARE SATISFIED, THE K-T CHARAC THE To me SINCE SATISFIES THE CONDITION, we THAT IF THE THEN ALSO ALL IMPLIES THAT THE BUDGET CONSTRAINT 15 BINDING AT THE HENCE WE MAY WRITE THE K-T CONDITIONS AS; (1.A) Эх. (1.5) (2) = = (4) By AND (1.B) AND THE FACT THAT AND THAT = 2 SUPPOSE THEN By (2) = P AND By (1.6) = = >,0 AND SUCH THAT SATISFIES THE K-T CONDITIONS. SINCE THE K-T CONDITIONS ARE s THE PROBLEM, AND SINCE THIS SOCUTION UNIQUE, THERE No WITH so > s NELESSA RY SINCE IF So DOES NOT SATISFY THE K-T CONDITION, AND SINCE THEY ARE NECESSARY, NOT A SOLUTION & THE CONSUMER'S PROBLEM SINCE WE KNOW THAT EXISTS, AND THE CONDITION > 0, IT MUST BE THAT > A SUFFICIENT CONDITION FORBy THE PREVIOUS EXERCISE we KNOW THAT IF wa (0,1) AND IN THE K-T THEREFORE if WE DEFINE SUCH THAT = 1, THEN DIFINES THE CONSUMER'S DEMAND AND A FUNCTION of AND P. NOTICE THAT WHERE : CONSIDER CINEAR SYSTEM (x.)) -6' = 0 0 -6" 0 WHERE AND LET BE SUCH THAT SINCE WE MAY WRITE 16 = 0 by THE K-T CONDITIONS IF = WE HAVE THAT = = by THE CONDITIONS MPOSED ON U ALSO > 1. THE SUBSTITUTION WHICK ALWAYS - THE INCOME EFFECT, WHICH WILL BE POSITIVE if NE ASSUME THAT IF THE GOOD DIMINISHING MARGINAL UTILITY AND if AN INCREASE IN ONE NOT THE UTILITY DERIVED FROM OTHER GOOD, THEN THE INCOME 15 POSITIVE, AND IF THIS EFFECT LARGER IN THEN THE VALUE OF THE TUTION EFFECT, WILL BE POSITIVE. THE INTUITION THAT EVEN THOU IS NOW THE CONSUMER IS GETTING MORE INCOME A GIVEN LABOR SUPPLY of