STAT 400 Lecture AL1 Answers for 6.4, Part 2 Spring 2015 Dalpiaz 4. Let X 1 , X 2 , … , X n be a random sample of size n from the distribution with probability density function f ( x ; ) = otherwise010θ1 θ θ1xx 0 < < . Recall: Maximum likelihood estimator of is niin1 X 1 lnθˆ. Method of moments estimator of is 1X1 XX1 θ~. E ( X ) = θ11. Def An estimator θˆ is said to be unbiased for if E(θˆ) = for all . unbiased, large variance biased, small variance unbiased, small varianced) Is θˆ unbiased for ? That is, does E(θˆ) equal ? 10θ θ1X1 θ1ln θ ln Xln E ; dxxxdxxfx. Integration by parts: babaduvabvudvu Choice of u: L ogarithmic A lgebraic T rigonometric E xponential u = ln x, dv = dxxdxx 1 θ1θ θ1θ1 θ1 , du = dxx 1, v = θ 1x. 10θ 1θ 110θ θ11 1 01 ln θ1ln Xln E dxxxxxdxxx = θ 01 θ 11 1 θ 1101 θ110θ 1 xdxxdxxx. Therefore, niniinn11 1 Xln E 1 θˆ E = , that is, θˆ is an unbiased estimator for .= = = = = = = = = = = = = = = = = = = = = = = = = = = = = Jensen’s Inequality: If g is convex on an open interval I and X is a random variable whose support is contained in I and has finite expectation, then E [ g ( X ) ] g [ E ( X ) ]. If g is strictly convex then the inequality is strict, unless X is a constant random variable. E ( X 2 ) [ E ( X ) ] 2 Var ( X ) 0 E ( e t X ) e t E ( X ) M X ( t ) e t XE1 X1E for a positive random variable X E [ X 3 ] [ E ( X ) ] 3 for a non-negative random variable X E [ ln X ] ln E ( X ) for a positive random variable X XE XE for a non-negative random variable X = = = = = = = = = = = = = = = = = = = = = = = = = = = = = e) Is θ~ unbiased for ? That is, does E(θ~) equal ? Since g ( x ) = 111xxx, 0< x < 1, is strictly convex, and X is not a constant random variable, by Jensen’s Inequality, E (θ~) = E ( g (X) ) > g ( E (X) ) = . θ~ is NOT an unbiased estimator for .6. Let X 1 , X 2 , … , X n be a random sample of size n from a population with mean and variance 2. Show that the sample mean X and the sample variance S 2 are unbiased for and 2, respectively. nn X...XXX21 E ( X 1 + X 2 + … + X n ) = n E (X) = E ( X 2 ) = Var ( X ) + [ E ( X ) ] 2 = 2 + 2. Var ( X 1 + X 2 + … + X n ) = n 2 Var (X) = 2/ n 2 XE = Var (X) + [ E (X) ] 2 = n22 σμ . S 2 = 2 X X11 in = 22 XX 11 nni E ( S 2 ) = XEXE 1122 nni = 112222 σμσμnnnn = 2 For an estimator θˆ of , define the Mean Squared Error of θˆ by MSE ( θˆ ) = E [ ( θˆ – ) 2 ]. E [ ( θˆ – ) 2 ] = ( E ( θˆ ) – ) 2 + Var ( θˆ ) = ( bias ( θˆ ) ) 2 + Var ( θˆ ).7. Let X 1 , X 2 , … , X n be a random sample of size n from a distribution with probability density function 21 θ XXθ ;xxfxf, – 1 < x < 1, – 1 < < 1. a) Obtain the method of moments estimator of , θ~. = E ( X ) = 11 21 θdxxx = 11 3 2 64 θxx = 3θ. 3 Xθ~ θ~ = 3 X. b) Is θ~ an unbiased estimator for ? Justify your answer. E (θ~) = E ( 3 X) = 3 E (X) = 3 = 3 3θ = . θ~ an unbiased estimator for . c) Find Var ( θ~ ). E ( X 2 ) = 11 2 21 θ dxxx = 11 4 3 86 θxx = 31. 2 = Var ( X ) = 2 331 θ = 932 θ. Var ( θ~ ) = 9 Var (X) = 9 n2 σ = n2 θ3 . MSE ( θ~ ) = n2 θ3.8. Let X 1 , X 2 be a random sample of size n = 2 from a distribution with probability density function 21 θ XXθ ;xxfxf, – 1 < x < 1, – 1 < < 1. Find the maximum likelihood estimator θˆ of . L ( ) = 21212 1 θθ xx = 4121221 θθ xxxx L ( ) is a parabola with vertex at ab 2 = 2121 2 xxxx . Case 1: a = x 1 x 2 > 0. Parabola has its “antlers” up. The vertex is the minimum. Subcase 1: x 1 > 0, x 2 > 0. Vertex = 2121 2 xxxx < 0. Maximum of L ( ) on – 1 < < 1 is at θˆ = 1. Subcase 2: x 1 < 0, x 2 < 0. Vertex = 2121 2 xxxx > 0. Maximum of L ( ) on – 1 < < 1 is at θˆ = – 1. Case 2: a = x 1 x 2 < 0. Parabola has its “antlers” down. The vertex is the maximum. Vertex is at 2121 2 xxxx .Subcase 1: 2121 2 xxxx > 1. That is, 12112 xxx. Maximum of L ( ) on – 1 < < 1 is at θˆ = 1. Subcase 2: 2121 2 xxxx < – 1. That is, 12112 xxx. Maximum of L ( ) on – 1 < < 1 is at θˆ = – 1. Subcase 3: – 1 < 2121 2 xxxx < 1. Maximum of L ( …
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