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hw3sol

# hw3sol - Math 280A Fall 2011 Homework 3 Solutions 1 Problem...

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Math 280A, Fall 2011 Homework 3 Solutions 1. Problem 14: As suggested, we assume (without loss of generality) that E [ X ] = 0. (a) Let M denote the matrix in question, and observe that det M = μ 2 μ 4 - μ 3 2 - μ 2 3 . To see that this is non-negative, consider the function f ( t ) := Var bracketleftbig ( X + t ) 2 bracketrightbig , t R . Explicitly, f ( t ) = E bracketleftbig ( X + t ) 4 bracketrightbig - ( E bracketleftbig ( X + t ) 2 bracketrightbig) 2 = μ 4 + 4 3 + 6 t 2 μ 2 + t 4 - ( μ 2 + t 2 ) 2 = μ 4 + 4 3 + 4 t 2 μ 2 - μ 2 2 Of course f ( t ), being a variance, is non-negative. Now a quadratic (in t ) like f ( t ) is non-negative for all real t if and only if its discriminant is non-positive. That is, we may conclude that 0 (4 μ 3 ) 2 - 4(4 μ 2 )( μ 4 - μ 2 2 ) . Simplifying, this inequality becomes 0 16 · bracketleftbig μ 2 3 - μ 2 μ 4 + μ 3 2 bracketrightbig = - 16 det M, so det M 0 as desired. (b) The (symmetric) matrix M is non-negative definite if and only if Q ( a ) := a t M a 0 for all real vectors a = ( a 0 , a 1 , a 2 ) t . (The superscript t indicates transpose.) Define a random vector X := (1 , X, X 2 ) t , and observe that XX t = 1 X X 2 X X 2 X 3 X 2 X 3 X 4 . Thus M is the (entry-by entry) expected value of XX t . Finally, Q ( a ) = a t M a = E bracketleftbig a t XX t a bracketrightbig = E bracketleftbig ( a t X ) 2 bracketrightbig = E bracketleftbig ( a 0 + a 1 X + a 2 X 2 ) 2 bracketrightbig 0 .

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hw3sol - Math 280A Fall 2011 Homework 3 Solutions 1 Problem...

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