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UIUC STAT 400 - 400Ex6_4_3

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STAT 400 Lecture AL1 Examples for 6.4, Part 3 Spring 2015 Dalpiaz 1. Let θ > 0 and let X 1 , X 2 , … , X n be a random sample from the distribution with the probability density function ( ) ( ) XXθ 2 θ θ;xexxfxf−==, x > 0. a) Find E ( X k ), k > – 21. “Hint”: Consider u = θ x. E ( X k ) = ∫∞−⋅0 θ 2 θdxxxxke u = θ x du = dxx 2θ = ∫∞−0 2 θduuuek = ∫∞−0 22 θ1 duuuekk = ( ) 12 θ1 2+Γkk. b) Find the method of moments estimator θ~ of θ. Suppose n = 4, and x 1 = 0.01, x 2 = 0.04, x 3 = 0.09, x 4 = 0.36. Find the method of moments estimate of θ. E ( X ) = E ( X 1 ) = ( ) 3 θ1 2Γ = 2 θ2! = 2 θ2. 21 θ~2X1X ==∑=⋅niin ⇒ ∑=⋅==niin1 X2X2θ~ ∑=nii1 X = 0.50. x = 0.125. θ~ = 0.1252 = 4.OR E ( X ) = ( )∫∞⋅0 dxxfx = ∫∞−⋅0 θ 2θdxxxxe y = x 2 xdxdy = E ( X ) = ∫∞−⋅⋅0 2θ θ dyyye Integration by parts: ∫∫−=babaduvabvudvu Choice of u: L ogarithmic A lgebraic T rigonometric E xponential u = y 2, dv = dyye θ θ−⋅, du = 2 y dy, v = ye θ −−. E ( X ) = ∫∞−−−⋅⋅∞⋅+0 2θθ 20 dyyyyeey = ∫∞−⋅⋅0 θ 2dyyye u = y, dv = dyye θ −, du = dy, v = ye θ θ1−⋅−. E ( X ) = ∞−∫−−⋅∞⋅⋅⋅ +0 θθ θθ10 1 2 dyyyeey = ∫∞−⋅0 θ θ2dyye = −∞⋅⋅−0 1 2 θ θθye = 2 θ2. ORE ( X ) = ( )∫∫∞∞−⋅⋅ =0 0 θ 2 θdxxxdxxfxxe y = x 2 xdxdy = E ( X ) = ∫∞−⋅0 2θ θ dyyye = E ( Y 2 ), where Y has Exponential distribution with mean θ1. E ( X ) = E ( Y 2 ) = Var ( Y ) + [ E ( Y ) ] 2 = 2 θ1 + 2 θ1 = 2 θ2. c) Find the maximum likelihood estimator θˆ of θ. Suppose n = 4, and x 1 = 0.01, x 2 = 0.04, x 3 = 0.09, x 4 = 0.36. Find the maximum likelihood estimate of θ. L ( θ ) = ∏=−nixiiex1 θ 2θ ln L ( θ ) = ( )∑∑==⋅⋅ −−niinixnix1 1 2 θ lnθln ( ln L ( θ ) )' = ∑=−niixn1 θ = 0 ⇒ θˆ = ∑=niin1 X ∑=niix1 = 1.2. θˆ = 214. = 310 ≈ 3.3333.2. Let λ > 0 and let X 1 , X 2 , … , X n be a random sample of size n from a double exponential distribution. That is, f ( x ; λ ) = λ 2 λxe−, – ∞ < x < ∞. a) Find E ( X k ) for positive integer k. E ( X k ) = ∫∞∞−−⋅ dxxxke λ 2 λ = ∫∫∞+∞−−⋅⋅00 λ 2 λ 2 λλdxxdxxxkxkee = … k odd … = 0. k even … = ∫∞−⋅⋅0 λ 2 2 λdxxxke = ∫∞−⋅0 λ λ dxxxke = ( )( )∫∞+Γ+Γ−−++⋅0111 λ 1 1 λλdxxkkxkkke = ( )kk λ 1+Γ = kk λ! . b) Obtain the maximum likelihood estimator of λ, λˆ. L ( λ ) = −∑=⋅ 1 λλexp 2niinnx. ln L ( λ ) = ∑=⋅−−niixnn1 λ λ 2lnln. ( )∑=−=niiddxn1 λλλLln = 0. ⇒ ∑==niin1 Xλˆ.3. Let λ > 0 and let X 1 , X 2 , … , X n be a random sample from the distribution with the probability density function ( )2 53λ ; λλ xef xx−=, x > 0. a) Find E ( X k ), k > – 6. “Hint”: Consider u = λ x 2. E ( X k ) = ∫∞−⋅0 532 λ λ dxxxxek u = λ x 2 du = 2 λ x = ∫∞−+⋅0 222 2 λλduuuek = ∫∞−+−⋅0 222 21 λ duuuekk = +Γ− 23 21 2 λkk. b) Obtain the method of moments estimator of λ, λ~. Suppose n = 4, and x 1 = 4, x 2 = 2, x 3 = 4, x 4 = 3. Find the method of moments estimate of λ. E ( X ) = +Γ− 213 21 21 λ = +Γ⋅⋅− 21225 21 21 λ = +Γ⋅⋅⋅− 2112325 21 21 λ = Γ⋅⋅⋅⋅− 21212325 21 21 λ = πλ 212325 2121 ⋅⋅⋅⋅− = λπ 1615⋅. λπ 1615⋅ = X ⇒ λ~ = ( )2 X 256 225 π. x 1 = 4, x 2 = 2, x 3 = 4, x 4 = 3. x = 413 = 3.25. λ~ = ( )2 X 256 225π = 2704225π ≈ 0.2614.OR E ( X 2 ) = +Γ− 223 21 22 λ = ( ) 4 21 1λ Γ⋅− = λ! 23 = λ3. λ3 = 2 X = ∑⋅=niin12 X 1 ⇒ 2 λ~ = X 32 = ∑=niin12 X3 . x 1 = 4, x 2 = 2, x 3 = 4, x 4 = 3. ∑=niix12 = 45. 2 λ~ = ∑=niixn12 3 = 4512 = 154 ≈ 0.2667. c) Obtain the maximum likelihood estimator of λ, λˆ. Suppose n = 4, and x 1 = 4, x 2 = 2, x 3 = 4, x 4 = 3. Find the maximum likelihood estimate of λ. L ( λ ) = ∏=−niiixex1 53 2 λ λ. ln L ( λ ) = ()∑∑==⋅⋅−+niiniixn x12 15 3 λλ lnln. ( ln L ( λ ) )' = ∑=−niixn12 3λ = 0. ⇒ λˆ = ∑=niin12 X3 . x 1 = 4, x 2 = 2, x 3 = 4, x 4 = 3. ∑=niix12 = 45. λˆ = ∑=niixn12 3 = 4512 = 154 ≈ 0.2667.Useful facts: Def ( )∫∞Γ−−=01 duuxuxe, x > 0. Γ ( 1 ) = 1. Γ ( x ) = ( x – 1 ) Γ ( x – 1 ). ⇒ ( ) ( )! 1 −Γ =nn if n is an integer. π


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UIUC STAT 400 - 400Ex6_4_3

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