Macroeconomics Exam Review 150

# Macroeconomics Exam Review 150 - Solutions for Foundations...

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3.148 Since 𝑓 is quasiconcave 𝑓 ( 𝛼 x 1 + (1 𝛼 ) x 2 ) min { 𝑓 ( x 1 ) , 𝑓 ( x 2 ) } for every x 1 , x 2 𝑆 and 0 𝛼 1 Since 𝑔 is increasing 𝑔 ( 𝑓 ( 𝛼 x 1 + (1 𝛼 ) x 2 ) ) 𝑔 ( ( min { 𝑓 ( x 1 ) , 𝑓 ( x 2 ) } ) ) min { 𝑔 ( 𝑓 ( x 1 ) ) , 𝑔 ( 𝑓 ( x 2 ) ) } 𝑔 𝑓 is quasiconcave. 3.149 When 𝜌 1, the function ( x ) = 𝛼 1 𝑥 𝜌 1 + 𝛼 2 𝑥 𝜌 2 + . . . 𝛼 𝑛 𝑥 𝜌 𝑛 is convex (Example 3.58) as is 𝑦 1 /𝜌 . Therefore 𝑓 ( x ) = ( ( x )) 1 /𝜌 is an increasing convex function of a convex function and is therefore convex (Exercise 3.133). 3.150 𝑓 is a monotonic transformation of the concave function ( x ) = x . 3.151 By Exercise 3.39, there exist linear functionals ˆ 𝑓 and ˆ 𝑔 and scalars 𝑏 and 𝑐 such that 𝑓 ( x ) = ˆ 𝑓 ( x ) + 𝑏 and 𝑔 ( x ) = ˆ 𝑔 ( x ) + 𝑐 The upper contour set ( 𝑎 ) = { 𝑥 𝑆 : ( x ) 𝑎 } = { 𝑥 𝑆 : ˆ 𝑓 ( 𝑥 ) + 𝑏 ˆ 𝑔 ( 𝑥 ) + 𝑐 𝑎 } = { 𝑥 ∈ ℜ 𝑛 + : ˆ 𝑓 ( x ) + 𝑏 𝑎 ˆ 𝑔 ( x ) + 𝑎𝑐 } = { 𝑥 ∈ ℜ 𝑛 + : ˆ 𝑓 ( x ) 𝑎 ˆ 𝑔 ( x ) 𝑏 𝑎𝑐 } which is a halfspace in
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