HW 10 Solutions

HW 10 Solutions - Steven Weber Dept. of ECE Drexel...

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Unformatted text preview: Steven Weber Dept. of ECE Drexel University ENGR 361: Statistical Analysis of Engineering Systems (Spring, 2010) Homework 10 Solutions Question 1: #29, 6.2, p252. Consider a random sample X 1 ,...,X n from the shifted exponential pdf f ( x ; , ) = e- ( x- ) , x , otherwise . (1) Taking = 0 gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example 4.5 in which the variable of interest was time headway in traffic flow and = 0 . 5 was the minimum possible time headway. a. Obtain the maximum likelihood estimators of and . The log likelihood is L ( x 1 ,...,x n ; , ) = log Q n i =1 f ( x i ; , ). Note that L =- if any x i < , i.e., L =- if min { x i } < . When min { x i } : L ( x 1 ,...,x n ; , ) = log n Y i =1 e- ( x i- ) = n X i =1 log e- ( x i- ) = n (log - ( x- )) . (2) For arbitrary x 1 ,...,x n : L ( x 1 ,...,x n ; , ) = n (log - ( x- )) , min { x i } - , else . (3) The derivatives w.r.t. , are: L ( x 1 ,...,x n ; , ) = n ( 1 - ( x- ) ) , min { x i } , else L ( x 1 ,...,x n ; , ) = n, min { x i } , else (4) Setting L ( x 1 ,...,x n ; , ) = 0, we find = 1 x- . Note that L ( x 1 ,...,x n ; , ) > 0 for min { x i } and = 0 else. It follows that the likelihood is maximized w.r.t. for = min { x...
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HW 10 Solutions - Steven Weber Dept. of ECE Drexel...

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