STAT 400 Homework 09 Spring 2018 Dalpiaz UIUC Due: Friday, April 6, 2:00 PM

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1 STAT Homework 9 Sprg 28 Dalpaz UIUC Due: Fray, Aprl 6, 2: PM Exercse f(x, θ) = θ e x/θ, x >, θ > Note that, the momets of ths strbuto are gve by E[X k ] = Ths wll be a useful fact for Exercses 2 a 3. x k θ e x/θ = k! θ k. (a) Obta the maxmum lkelhoo estmator of θ, ˆθ. (Ths shoul be a fucto of the uobserve x a the sample sze.) Calculate the estmate whe (Ths shoul be a sgle umber, for ths ataset.) x =.5, x 2 =.5, x 3 =., x = 3.. L(θ) = f(x ; θ) = ( θ e x/θ = θ = exp x ) θ Istea of rectly maxmzg the lkelhoo, we stea maxmze the log-lkelhoo. = log L(θ) = log θ x θ To maxmze ths fucto, we take a ervatve wth respect to θ. log L(θ) = θ θ + = x θ 2 We set ths ervatve equal to zero, the solve for θ. θ + = x θ 2 = Solvg gves our estmator, whch we eote wth a hat.

2 = ˆθ = x = x Usg the gve ata, we obta a estmate. ˆθ = = 2.25 (b) Calculate the bas of the maxmum lkelhoo estmator of θ, ˆθ. (Ths wll be a umber.) Note that we have a expoetal strbuto. E[X ] = θ Var[X ] = θ 2 Bas(ˆθ) = E[ˆθ] θ [ = = E X ] θ = E[X ] θ = = θ θ = θ θ = (c) F the mea square error of the maxmum lkelhoo estmator of θ, ˆθ. (Ths wll be a expresso base o the parameter θ a the sample sze. Be aware of your aswer to the prevous part, as well as the strbuto gve.) MSE(ˆθ) = [Bas(ˆθ)] 2 + Var(ˆθ) ( = = + Var X ) = 2 Var(X ) = = 2 θ2 = θ2 () Prove a estmate for P [X > ] whe x =.5, x 2 =.5, x 3 =., x = 3.. 2

3 P [X > ] = e /θ ˆP [X > ] = e /ˆθ = e /2.25 =.69 Exercse 2 f(x, α) = α 2 xe x/α, x >, α > (a) Obta the maxmum lkelhoo estmator of α, ˆα. Calculate the estmate whe x =.25, x 2 =.75, x 3 =.5, x = 2.5, x 5 = 2.. L(α) = f(x ; α) = ( ) ( α 2 x e x/α = α 2 = x exp x ) α Istea of rectly maxmzg the lkelhoo, we stea maxmze the log-lkelhoo. log L(α) = 2 log α + = log x x α To maxmze ths fucto, we take a ervatve wth respect to α. = 2 log L(α) = α α + = x α 2 We set ths ervatve equal to zero, the solve for α. 2 α + = x α 2 = Solvg gves our estmator, whch we eote wth a hat. = ˆα = x 2 = x 2 Usg the gve ata, we obta a estmate. ˆα = =.7 (b) Obta the metho of momets estmator of α, α. Calculate the estmate whe 3

4 x =.25, x 2 =.75, x 3 =.5, x = 2.5, x 5 = 2.. We frst obta the frst populato momet. Notce the tegrato s oe by etfyg the form of the tegral s that of the seco momet of a expoetal strbuto. E[X] = x α 2 xe x/α x = α x 2 α e x/α x = α (2α2 ) = 2α We the set the frst populato momet, whch s a fucto of α, equal to the frst sample momet. = 2α = x Solvg for α, we obta the metho of momets estmator. = α = x 2 = x 2 Usg the gve ata, we obta a estmate. α = Note that, ths case, the MLE a MoM estmators are the same. =.7 Exercse 3 f(x, β) = 2β 3 x2 e x/β, x >, β > (a) Obta the maxmum lkelhoo estmator of β, ˆβ. Calculate the estmate whe x = 2., x 2 =., x 3 = 7.5, x = 3.. L(β) = f(x ; β) = ( ) ( 2β 3 x2 e x/β = 2 β 3 = x exp x ) β Istea of rectly maxmzg the lkelhoo, we stea maxmze the log-lkelhoo. log L(β) = log 2 3 log β + = log x x β =

5 To maxmze ths fucto, we take a ervatve wth respect to β. 3 log L(β) = β β + = x β 2 We set ths ervatve equal to zero, the solve for β. 3 β + = x β 2 = Solvg gves our estmator, whch we eote wth a hat. ˆβ = = x 3 = x 3 Usg the gve ata, we obta a estmate. ˆβ = =.375 (b) Obta the metho of momets estmator of β, β. Calculate the estmate whe x = 2., x 2 =., x 3 = 7.5, x = 3.. We frst obta the frst populato momet. Notce the tegrato s oe by etfyg the form of the tegral s that of the thr momet of a expoetal strbuto. E[X] = x 2β 3 x2 e x/β x = x 3 2β 2 β e x/β x = 2β 2 (6β3 ) = 3β We the set the frst populato momet, whch s a fucto of β, equal to the frst sample momet. = 3β = x Solvg for β, we obta the metho of momets estmator. β = = x 3 = x 3 Usg the gve ata, we obta a estmate. β = Note aga, the MLE a MoM estmators are the same. =.375 5

6 Exercse f(x, λ) = λx λ, < x <, λ > (a) Obta the maxmum lkelhoo estmator of λ, ˆλ. Calculate the estmate whe x =., x 2 =.2, x 3 =.3, x =.. L(λ) = f(x ; λ) = λx λ ( ) λ = λ x Istea of rectly maxmzg the lkelhoo, we stea maxmze the log-lkelhoo. log L(λ) = log λ + (λ ) To maxmze ths fucto, we take a ervatve wth respect to λ. log x = λ log L(λ) = λ + log x We set ths ervatve equal to zero, the solve for β. = λ + log x = Solvg gves our estmator, whch we eote wth a hat. = ˆλ = = log x Usg the gve ata, we obta a estmate. ˆλ = = log x = log(..2.3.) =.663 Note that ths s actually a reparameterzato of a example see class where λ = θ. Ha you realze ths, you coul have smply fou the aswer va varace. (b) Obta the metho of momets estmator of λ, λ. Calculate the estmate whe x =., x 2 =.2, x 3 =.3, x =.. 6

7 We frst obta the frst populato momet. E[X] = x λx λ x = λ λ + We the set the frst populato momet, whch s a fucto of β, equal to the frst sample momet. λ λ + = = x = x Solvg for λ, we obta the metho of momets estmator. Usg the gve ata, we obta a estmate. x = λ = x x λ = = 3 Note that the MLE a MoM estmators are fferet. =.25 7