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Macroeconomics Exam Review 88

# Macroeconomics Exam Review 88 - Solutions for Foundations...

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2.123 By the Banach fixed point theorem, for every 𝜃 Θ, there exists 𝑥 𝜃 𝑋 such that 𝑓 𝜃 ( 𝑥 𝜃 ) = 𝑥 𝜃 . Choose any 𝜃 0 Θ. 𝜌 ( 𝑥 𝜃 , 𝑥 𝜃 0 ) = 𝜌 ( 𝑓 𝜃 ( 𝑥 𝜃 ) , 𝑓 𝜃 0 ( 𝑥 𝜃 0 )) 𝜌 ( 𝑓 𝜃 ( 𝑥 𝜃 ) , 𝑓 𝜃 ( 𝑥 𝜃 0 )) + 𝜌 ( 𝑓 𝜃 ( 𝑥 𝜃 0 ) , 𝑓 𝜃 0 ( 𝑥 𝜃 0 )) 𝛽𝜌 ( 𝑥 𝜃 , 𝑥 𝜃 0 ) + 𝜌 ( 𝑓 𝜃 ( 𝑥 𝜃 0 ) , 𝑓 𝜃 0 ( 𝑥 𝜃 0 )) (1 𝛽 ) 𝜌 ( 𝑥 𝜃 , 𝑥 𝜃 0 ) 𝜌 ( 𝑓 𝜃 ( 𝑥 𝜃 0 ) , 𝑓 𝜃 0 ( 𝑥 𝜃 0 )) 𝜌 ( 𝑥 𝜃 , 𝑥 𝜃 0 ) 𝜌 ( 𝑓 𝜃 ( 𝑥 𝜃 0 ) , 𝑓 𝜃 0 ( 𝑥 𝜃 0 )) (1 𝛽 ) 0 as 𝜃 𝜃 0 . Therefore 𝑥 𝜃 𝑥 𝜃 0 . 2.124 1. Let x be a fixed point of 𝑓 . Then x satisfies x = ( 𝐼 𝐴 ) x + c = x 𝐴 x + 𝑐 which implies that 𝐴 x = 𝑐 . 2. For any x 1 , x 2 𝑋 𝑓 ( x 1 ) 𝑓 ( x 2 ) = ( 𝐼 𝐴 )( x 1 x 2 ) ≤ ∥ 𝐼 𝐴 x 1 x 2 Since 𝑎 𝑖𝑖 = 1, the norm of 𝐼 𝐴 is 𝐼 𝐴 = max 𝑖 𝑗 = 𝑖 𝑎 𝑖𝑗 = 𝑘 and 𝑓 ( x 1 )
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