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A linear operator T on a flnite dimensional vector space V is diagonalizable if and only if there is an ordered basis fl for V consisting of eigenvectors of T. Furthermore, (¡1)n. In this case, the entries on the diagonal of B are of the form b i;i ¡t. ... Read Here

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Case 1, b= n=2: In this case, from equation 2, the whole sample space is covered so p e = 1. Also F(x) = F (x) >1=2 so p c = 1. Because the unique peak of the f(x) function happens at n=2 when nis even (nmust be even when b= n=2), we can see that p m = 1. Also because of that peak, min n F(x);F (x) o is maximized at b= n=2 and p b = 1. Case 2 ... View This Document

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