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by: April Jerde

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# Economic Theory; Macroeconomics Sequence ECON 714

Marketplace > University of Wisconsin - Madison > Economcs > ECON 714 > Economic Theory Macroeconomics Sequence
April Jerde
UW
GPA 3.6

Kenneth West

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COURSE
PROF.
Kenneth West
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Economcs

This 2 page Class Notes was uploaded by April Jerde on Thursday September 17, 2015. The Class Notes belongs to ECON 714 at University of Wisconsin - Madison taught by Kenneth West in Fall. Since its upload, it has received 39 views. For similar materials see /class/205139/econ-714-university-of-wisconsin-madison in Economcs at University of Wisconsin - Madison.

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Date Created: 09/17/15
Econ7l4 Spring 2009 K West Algebra of Solution to Staggered Contract Model The equations are 1 x 5xt1 SEHxM 57E1ytym 2 17 509ml 3 y Wt 17 4 Arquot gAp it ygt0 lglltl it a white noise random variable For future reference observe that 4 can also be written 439 WI gp mt1gp1 Use 4 to eliminate m from 3 and take expectations as of time t l obtaining Emyt 1gE1p mt1gp1 Emit 1gE1p mt1gp1 Use this equation and this equation led one time period to substitute out for EHy and EHyH1 in l x 395xt1 39SEtlxrtl 39SYEt11gpt mtlgptl 39l39gpr1 mtgpt 395xt1 39SEtlxrtl 39SYEt13951gxtxt1 mtlgptl 395139gxr1xt mtgpt 39 5 511g5YEt1x1 11g90113196 511g5YX1 Wm3174 where Et1mtgpt mHng has been used in the last step It is convenient to de ne h 1g5Y and rewrite 5 as 5 5111Et1xM 1hEt1xt 51hxt1 ymt1gpt1 It may be shown that when has 1 the lag polynomial on the left hand side of 5 has two real roots one root strictly greater than one in absolute value one root strictly less than one in absolute value see equation 7a below This suggests a solution of the form 6 x dxtl bmt1gpt17 where d and b are coef cients to be determined ldlltl and the single lag of mHng on the right hand side is motivated by the fact that this variable follows a random walk As is obvious this solution also holds when lh0 so that the terms in Eblx 1 and Eth disappear To solve for d and b use 6 and 6 led once in 5 expressing EHx 1 and Eth in terms of xt1 and mHng The result is 51ha xt1d1bmt1gpt1 1hdxt1bmt1gpt1 51hxt1 Ymt1gl7t1 7a 51ha 1hd 51h 0 7b 51hd1b 1hb 4 Use the formula for roots to a quadratic equation to solve for d It follows that 6 is indeed a solution to 5 with 8 d litZ 7 1 9 b 14 1h7517hd1 17g The second equality for b can be veri ed using the quadratic 7a From 2 3 and 4 9 y 1gp mt1gp1 it 51gxxt1 mtgpt From 6 xt bLmt gptldL xt1 bL2mt gptldL where L is the lag operator After using these eXpressions in 9 rearranging and using the eXpression for b in 8 we get laLy 1 d51gbL 51gbL2mgp 151dL 5ldL2mt gpt 151dL1Lmgpt 151dL t 10 y 0 it 5060

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