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hw12soln - ORIE 3500/5500 Fall'10 HW 12 Solution Assignment...

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ORIE 3500/5500, Fall ’10 HW 12 Solution Assignment 12 Solution Problem 1 The joint pdf of X 1 , . . . , X n is given by f X 1 ,...,X n ( x 1 , . . . , x n ; θ ) = n Y i =1 f X i ( x i ; θ ) = n Y i =1 1 2 e -| x i - θ | = 1 2 n e - n i =1 | x i - θ | . Taking the logarithm of this expression, we obtain log f X 1 ,...,X n ( x 1 , . . . , x n ; θ ) = - n log 2 - n X i =1 | x i - θ | . Let us call this function h ( θ ) for notational convenience. Note that h is not differentiable at θ = x 1 , . . . , x n , so we cannot simply set the derivative of h equal to zero to find out where h is maximized. Instead, we note that the derivative of h , when it exists, is equal to h 0 ( θ ) = n X i =1 sgn( x i - θ ) , where sgn denotes the sign function sgn( x ) = - 1 if x < 0 , 0 if x = 0 , 1 if x > 0 . Let x (1) < x (2) < . . . < x ( n ) denote the order statistics of the sample. Observe that for θ < x (1) , we have sgn( x i - θ ) = 1 for all i , so h 0 ( θ ) = n , for x (1) < θ < x (2) , we have sgn( x i - θ ) = ( - 1 if x i = x (1) , 1 otherwise , so h 0 ( θ ) = - 1 + ( n - 1)(1) = n - 2, for x (2) < θ < x (3) , h 0 ( θ ) = n - 4, ... for θ > x ( n ) , h 0 ( θ ) = - n . Thus we see that the graph of h is a continuous polygonal curve with vertices at θ = x 1 , . . . , x n . If n is odd, h is strictly increasing for θ x ( n +1 2 ) and strictly decreasing for θ x ( n +1 2 ) . So h is maximized at θ = x ( n +1 2 ) . If n is even, h is 1
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ORIE 3500/5500, Fall ’10 HW 12 Solution strictly increasing for θ x ( n 2 ) , constant for x ( n 2 ) θ x ( n 2 +1 ) , and strictly decreasing for θ x ( n 2 +1 ) . So h is maximized at any θ h x ( n 2 ) , x ( n 2 +1 ) i . In either case, setting θ equal to ˜ x , the sample median, maximizes h . So we conclude that the MLE of θ is ˆ θ = e X = median { X 1 , . . . , X n } .
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