# Hw5sol - APPM 4/5520 Solutions to Problem Set Five 1(a ∑ n i =1 x i − μ 2 = ∑ n i =1 x i − x x − μ 2 = ∑ n i =1 bracketleftbig x i

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Unformatted text preview: APPM 4/5520 Solutions to Problem Set Five 1. (a) ∑ n i =1 ( x i − μ ) 2 = ∑ n i =1 ( x i − x + x − μ ) 2 = ∑ n i =1 bracketleftbig ( x i − x ) 2 + 2( x i − x )( x − μ ) + ( x − μ ) 2 bracketrightbig = ∑ n i =1 ( x i − x ) 2 + 2( x − μ ) ∑ n i =1 ( x i − x ) + ∑ n i =1 ( x − μ ) 2 For the middle term, note that n summationdisplay i =1 ( x i − x ) = n summationdisplay i =1 x i − n x = n summationdisplay i =1 x i − n summationdisplay i =1 x i = 0 . The last term is n summationdisplay i =1 ( x − μ ) 2 = n ( x − μ ) 2 . So, we can conclude that n summationdisplay i =1 ( x i − μ ) 2 = n summationdisplay i =1 ( x i − x ) 2 + n ( x − μ ) 2 , as desired. (b) f X 1 ,...,X n ( x 1 ,... ,x n ) iid = producttext n i =1 f ( x i ) = producttext n i =1 1 √ 2 πσ 2 e − 1 2 σ 2 ( x i − μ ) 2 = (2 πσ 2 ) − n/ 2 e − 1 2 σ 2 ∑ n i =1 ( x i − μ ) 2 Part ( a ) = (2 πσ 2 ) − n/ 2 e − 1 2 σ 2 [ ∑ n i =1 ( x i − x ) 2 + n ( x − μ ) 2 ] (c) Y 1 = g 1 ( X 1 ,... ,X n ) = X Y 2 = g 2 ( X 1 ,... ,X n ) = X 1 − X . . . . . . . . . Y n = g ( X 1 ,... ,X n ) = X n − X ⇓ X 1 = g − 1 1 ( Y 1 ,... ,Y n ) = Y 1 − ∑ n i =2 Y i X 2 = g − 1 2 ( Y 1 ,... ,Y n ) = Y 1 + Y 2 X n = g − 1 n ( Y 1 ,... ,Y n ) = Y 1 + Y n So, the Jacobian is J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ∂x 1 ∂y 1 ∂x 1 ∂y 2 ··· ∂x 1 ∂y n ∂x 2 ∂y 1 ∂x 2 ∂y 2 ··· ∂x 2 ∂y n . . . . . . . . . . . . ∂x n ∂y 1 ∂x n ∂y 2 ··· ∂x n ∂y n vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 − 1 − 1 ··· − 1 − 1 1 1 ··· ....
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## This note was uploaded on 02/07/2012 for the course APPM 4520 taught by Professor Manuel during the Fall '11 term at University of Colorado Denver.

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Hw5sol - APPM 4/5520 Solutions to Problem Set Five 1(a ∑ n i =1 x i − μ 2 = ∑ n i =1 x i − x x − μ 2 = ∑ n i =1 bracketleftbig x i

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