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Unformatted text preview: Econ 387L:MacroII Spring 2008 University of Texas at Austin Solution to Problem Set#8 Ex1ac. We want to show that under the stated assumptions, the hypotheses of Theorem 4.24.5, which establishes the equvilance between sequential problem, (SP), given in (1), and the function equation, (FE), given in (2), are satisfied. Then solving (2), which is a relatively simpler problem, provides characterization for the solution to (1). This amounts to verifying that assumptions 4.1 through 4.8 are satisfied. We show each in turn. A4.1: Here ( x t ) = { x t +1 : 0 x t +1 f ( x t ) } . For any x t R + , f ( x t ) 0 implies that x t +1 = 0 is feasible, so that 0 ( x t ) . It follows that feasible set ( x ) is nonempty for all x. A4.2: Here F ( x t ,x t +1 ) = U ( f ( x t ) x t +1 ). Pick any x X and consider an arbitrary feasible sequence, x ( x ) generated given this initial condition. There are two cases to consider. case 1: if x x then x t +1 f ( x t ) x so that x t +1 x . By an inductive argument x t x for all t. It follows that U ( f ( x t ) x t +1 ) U ( x ) U < . Since (0 , 1), by assumption U 1, n U converges to 0 as n increases, and hence the sequence { n t =0 t U ( f ( x t ) x t +1 ) } n is bounded above by 1 1 U . Moreover, since f is strictly increasing the sequence bounded is below by 1 1 U (0). Then from the BolzanoWeierstrass theorem the sequence has a limit, so that lim n n t =0 t U ( f ( x t ) x t +1 ) exists. case 2: x x . Since x t +1 xt , either x t +1 x for some t or x t is a decreasing sequence. In either case, from the similar argument as in above follows boundedness results and hence convergence. Since x is chosen arbitrary the desired result follows. A4.3: To show the convexity of constraint set, X t = { x t +1 : x t +1 [0 ,f ( x t )] } (1) for all t , pick two feasible sequences, { x t } t and { x t } t , and a constant, [0 , 1]. Since these1]....
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 Spring '07
 CORBAE

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