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Unformatted text preview: IE 4521 Midterm #2 Prof. John Gunnar Carlsson April 20, 2010 Before you begin: This exam has 11 numbered pages and a total of 8 problems. Make sure that all pages are present. To obtain credit for a problem, you must show all your work. if you use a formula to answer a problem, write the formula down. Do not open this exam until instructed to do so. 1 1. (12 points) Consider the probability density function f ( x ) = ( c (1 + θx ) 1 ≤ x ≤ 1 otherwise (a) Compute the value of the constant c (since f ( x ) must de ne a valid p.d.f.) Solution We must have ˆ 1 1 c (1 + θx ) dx = 1 c x + θ x 2 2 1 1 = 1 c = 1 / 2 (b) Use the method of moments to estimate θ . Solution We have E ( X ) = ˆ 1 1 1 2 (1 + θx ) xdx = θ/ 3 and therefore θ = 3 E ( X ) , so ˆ θ = 3¯ x is the methodofmoments estimate. (c) Show that ˆ θ = 3¯ x is an unbiased estimator for θ . Solution The bias of ˆ θ is E ˆ θ θ = E ( 3 ¯ X θ ) = 3 ( E ( ¯ X ) θ ) and since ¯ x is always an unbiased estimator, the bias is zero. (d) Suppose we have two samples x 1 and x 2 . Find the maximum likelihood estimator for θ . For the time being, do not worry about any additional constraints on θ that may be necessary. Solution We have L ( x 1 ,x 2 ; θ ) = 1 2 (1 + θx 1 ) · 1 2 (1 + θx 2 ) ‘ ( x 1 ,x 2 ; θ ) = log L ( x 1 ,x 2 ; θ ) = log (1 / 4) + log (1 + θx 1 ) + log (1 + θx 2 ) ∂‘ ∂θ = x 1 1 + θx 1 + x 2 1 + θx 2 = 0 solving this we nd that ˆ θ = x 1 x 2 2 x 1 x 2 . (e) (Bonus; 2 points) Suppose we sample x 1 = 0 . 2 , x 2 = 0 . 8 , x 3 = 0 . 3 . What is wrong with the moment estimator in (c)? (f) (Bonus; 2 points) Other than having x 1 = 0 or x 2 = 0 , what else could go wrong in (d), and how might you resolve it? 2 2. (12 points) Consider the probability density function f ( x ) = ( 1 θ 2 xe x/θ x ≥ otherwise Find the maximum likelihood estimator of θ , given a collection of samples x 1 ,...,x n . Solution We have L ( x 1 ,...,x n ; θ ) = 1 θ 2 x 1 e x 1 /θ ··· 1 θ 2 x n e x n /θ ‘ ( x 1 ,...,x n ; θ ) = 2 n log θ + (log x 1 + ··· + log x n ) x 1 + ··· + x n θ ∂‘ ∂θ = 2 n θ + x 1 + ··· + x n 2 θ 2 = 0 Solving for θ we nd that ˆ θ = x 1 + ··· + x n 4 n . 3 3. (12 points) The Rayleigh distribution has a probability density function given by f ( x ) = ( x θ e x 2 / (2 θ ) x ≥ otherwise (a) Find the maximum likelihood estimator of θ , given a collection of samples x 1 ,...,x n . Solution We have L ( x 1 ,...,x n ; θ ) = h x 1 θ e x 2 1 / (2 θ ) i ··· h x n θ e x 2 n / (2 θ ) i ‘ ( x 1 ,...,x n ; θ ) = (log x 1 + ··· + log x n ) n log θ x 2 1 + ··· + x 2 n 2 θ ∂‘ ∂θ = x 2 1 + ··· + x 2 n 4 θ 2 n θ = 0 and therefore ˆ θ = x 2 1 + ··· + x 2 n 4 n ....
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 Spring '08
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 Normal Distribution, Maximum likelihood, Estimation theory, Bias of an estimator, θ

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