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204Lecture132008Web

204Lecture132008Web - Economics 204 Lecture 13Wednesday...

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Economics 204 Lecture 13–Wednesday, August 13, 2008 Section 5.5 (Cont.) Transversality Theorem The Transversality Theorem is a particularly convenient formulation of Sard’s Theorem for our purposes: Theorem 1 (2.5’, Transversality Theorem) Let X × Ω R n + p be open F : X × Ω R m C r with r 1+max { 0 ,n m } If F ( x, ω )=0 DF ( x, ω ) has rank m then for all ω except for a set of Lebesgue measure zero, F ( x, ω D x F ( x, ω ) has rank m In particular, if m = n , there is a local implicit function x ( ω ) characterized by F ( x ( ω ) x is a C r function of ω , and the correspondence ω x ( ω ) is lower hemicontinuous. Interpretation of Tranversality Theorem Ω: a set of parameters (agents’ endowments and preferences, or players’ payoF functions). 1

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X : a set of variables (price vectors, or strategies). R m is the range of F (excess demand, or best-response strategies). F ( x, ω ) = 0 is equilibrium condition, given parameter ω . Rank DF ( x, ω )= m says that, by adjusting either the variables or parameters, it is possible to move F in any direction. When m = n ,Rank D x F ( x, ω m says det D x F ( x, ω ) 6 = 0, which says the economy is regular and is the hypothesis of the Implicit Function Theorem. This will tell us that the equilibrium prices are given by a ±nite number of implicit functions of the parameters (endowments), and the equilibrium correspondence is thus lower hemicontinuous. Parameters of any given economy are ±xed. However, we want to study the set of parameters for which the resulting economy is well-behaved. Theorem says the following: “If, whenever F ( x, ω ) = 0, it is possible by perturbing the parameters and variables to move F in any direction, then for almost all parameter values, all equilibria are regular, and hence there are ±nitely many equilibria, the equilibria are implicitly de±ned C r functions of the parameters, and the equilibrium correspondence is lower hemicontinuous.” If n<m D x F ( x, ω ) min { m, n } = . Therefore, ( F ( x, ω )=0 DF ( x, ω )hasrankm) 2
for all ω except for a set of Lebesgue measure zero F ( x, ω ) = 0 has no solution Why is it true? Sard’s Theorem says the set of critical values of F is a set of Lebesgue measure zero.

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204Lecture132008Web - Economics 204 Lecture 13Wednesday...

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