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Macroeconomics Exam Review 34

Macroeconomics Exam Review 34 - c 2001 Michael Carter All...

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Note that 0 𝛾 𝑖 1, so that either 𝛾 𝑚 < 1 or 𝛾 𝑚 = 1. We show that both cases lead to a contradiction. 𝛾 𝑚 < 1. Then x 𝑚 = 1 1 𝛾 𝑚 𝑚 1 𝑖 =1 ( 𝛼𝛼 𝑖 + (1 𝛼 ) 𝛽 𝑖 ) x 𝑖 which contradicts the minimality of the set { x 1 , x 2 , . . . , x 𝑚 } . 𝛾 𝑚 = 1. Then 𝛾 𝑖 = 0 for every 𝑖 = 𝑚 . That is 𝛼𝛼 𝑖 + (1 𝛼 ) 𝛽 𝑖 = 0 for every 𝑖 = 𝑚 which implies that 𝛼 𝑖 = 𝛽 𝑖 for every 𝑖 = 𝑚 and therefore x = y . Therefore, if { x 1 , x 2 , . . . , x 𝑚 } is a minimal spanning set, every point must be an extreme point. 1.181 Assume to the contrary that one of the vertices is not an extreme point of the simplex. Without loss of generality, assume this is x 1 . Then, there exist distinct y , z 𝑆 and 0 < 𝛼 < 1 such that x 1 = 𝛼 y + (1 𝛼 ) z (1.25) Now, since y 𝑆 , there exist 𝛽 1 , 𝛽 2 , . . . , 𝛽 𝑛 such that y = 𝑛 𝑖 =1 𝛽 𝑖 x 𝑖 , 𝑛 𝑖 =1 𝛽 𝑖 = 1 Similarly, there exist 𝛿 1 , 𝛿 2 , . . . , 𝛿 𝑛 such that z = 𝑛 𝑖 =1 𝛿 𝑖 x 𝑖
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