Homework9Solutions

Homework9Solutions - Steven Weber Dept. of ECE Drexel...

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Steven Weber Dept. of ECE Drexel University ENGR 361: Statistical Analysis of Engineering Systems (Spring, 2008) Homework 9 Solutions (Monday, May 26) Question 1: #29, § 6.2, p252. Consider a random sample X 1 , . . . , X n from the shifted exponential pdf f ( x ; λ, θ ) = ± λ e - λ ( x - θ ) , x θ 0 , otherwise . (1) Taking θ = 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 θ = 0 . 5 was the minimum possible time headway. a. Obtain the maximum likelihood estimators of θ and λ . The log likelihood is L ( x 1 , . . . , x n ; λ, θ ) = log Q n i =1 f ( x i ; λ, θ ). Note that L = -∞ if any x i < θ , i.e., L = -∞ if min { x i } < θ . When min { x i } ≥ θ : L ( x 1 , . . . , x n ; λ, θ ) = log n Y i =1 λ e - λ ( x i - θ ) = n X i =1 log λ e - λ ( x i - θ ) = n (log λ - λ x - θ )) . (2) For arbitrary x 1 , . . . , x n : L ( x 1 , . . . , x n ; λ, θ ) = ± n (log λ - λ x - θ )) , min { x i } ≥ θ -∞ , else . (3) The derivatives w.r.t. λ, θ are: ∂λ L ( x 1 , . . . , x n ; λ, θ ) = ± n ( 1 λ - x - θ ) ) , θ min { x i } 0 , else ∂θ L ( x 1 , . . . , x n ; λ, θ ) = ± nλ, θ min { x i } 0 , else (4) Setting ∂λ L ( x 1 , . . . , x n ; λ, θ ) = 0, we find ˆ λ = 1 ¯ x - θ . Note that ∂θ L ( x 1 , . . . , x n ; λ, θ ) > 0 for θ min { x i } and = 0 else. It follows that the likelihood is maximized w.r.t. θ for θ = min { x i } . Combining these results: ˆ θ ( x 1 , . . . , x n ) = min { x i } , ˆ λ ( x 1 , . . . , x n ) = 1 ¯ x - min { x i } . (5) b. If n = 10 time headway observations are made, resulting in the values 3.11, 0.64, 2.55, 2.20, 5.44, 3.42,
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Homework9Solutions - Steven Weber Dept. of ECE Drexel...

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