SOLUTION FOR HOMEWORK 7, STAT np(1 p) (α + β + n) + ( np + α

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1 SOLUTION FOR HOMEWORK 7, STAT Exerc733 Here we just recall that MSE(ˆp B ) = p(1 p) (α + β + ) + ( p + α 2 α + β + p) 2 The you plug i α = β = (/4) 1/2 After simplificatios MSE(ˆp B ) = 4( 1/2 + )2 ad this is costat i p How do we get such α ad β? We take a partial derivative i p ad the set it equal to zero 2 Exerc 737 Let X 1,, X be iid accordig to the pdf f(x θ) = (2θ) 1 I( x < θ), θ Ω = (, ) We discussed i class that X () is the CSS At the same time, (X (1), X () ) is ot CSS because E θ (X (1) = E θ (X () (What do you thik about the SS statistic (X (1), X () ) for the cases of Uif(θ, 2θ) or Uif(θ 1/2, θ + 1/2)?) Now, Y := X () is SS by Factorizatio Theorem because Further, let us fid the pdf of Y Write, f(x θ) = (2θ) I( x () θ) F Y (y) = P(Y y) = P( X () y) = X l y) = θ y I(y (, θ)) Thus, the pdf of Y is f Y (y) = d dz F Y (z) z=y = θ y 1 I(o < y < θ) (1) Let us show that Y is CSS Suppose that E θ (g(y )) = g(y)θ y 1 dy, for all θ > (2) The de θ (g(y ))/dθ for all θ > This yields (remember the Leibitz rule o p69 of how to take the derivative) θ 1 g(y)y 1 dy + g(θ)θ θ 1, θ > 1

2 But the itegral i the above-writte idetity is zero due to (2) This implies that g(θ)θ 1 for all θ >, ad this g(θ for all θ > We proved (directly) that Y is complete Our ext step is to fid a ubiased estimator δ(y ) which is also the UMVUE Let us check, usig (1), that Thus, the UMVUE is E θ (Y ) = yθ y 1 dyθ y dy = θ ( + 1) 1 θ +1 = + 1 θ δ (Y ) = + 1 max X l l 3 Exers 74 Let X 1,, X be a sample from Beroulli(p) We calculate the Fisher iformatio for a sigle observatio: I(p) := E p { 2 p 2 l(px (1 p) 1 X )} = E p { 2 p2[x l(p) + (1 X) l(1 p)]} = E p { p [x p 1 X 1 p ]} = E p{ x p 2 1 X (1 p) 2 } = (1 p)2 p + p 2 (1 p) p 2 (1 p) 2 The Cramér-Rao lower boud tells us that = 1 p + p p(1 p) = 1 p(1 p) Var p (δ(x)) [ E p(δ(x))/ p] 2 I(p) If δ (X) is ubiased, the umerator i the lower boud is 1, ad this yields that Var p (δ (X)) p(1 p), ad because Var p ( X) = p(1 p)/, the sample mea is the best ubiased estimator of p 4 Exerc 741 A sequece of iid RVs X 1,,X is observed It is kow that E µ,σ 2(X) = µ, Var µ,σ 2 = σ 2 (a) If δ := a i X i the m E µ,σ 2(δ) = a i µ = mu i = 1 a i = µiff a i = 1 (b) Let us fid {a i } that miimize the variace Var µ,σ 2(δ) = E µ,σ 2( (a i X i µ) 2 ) = Var µ,σ 2( a i (X i µ)) 2

3 [because observatios are iid we cotiue] = a 2 i Var µ,σ 2(X i) = σ 2 a 2 i Now we should fid {a i } that miimize a 2 i give a i = 1 Let us check that a i 1/ are the extreme (what else ca we try?) Write (i what follows the summatio is over i {1, 2,, }), a 2 i = (a i ) 2 = (a i 1 ) (a i 1 ) + 1 The last sum is zero because a i = 1 As a result, we get that m a 2 i = (a i 1 ) with the equality iff a i 1 5 Exerc 747 We have X = µ + ǫ where ǫ N(, σ 2 ) ad µ is a uderlyig radius A sample of size is observed What is the UMVUE of a = πµ 2? Here X is CSS (due to the expoetial family) so I just ote that E µ,σ 2( X 2 ) 1 σ 2 = µ 2, so â ub = π( X 2 1 σ 2 ) Because it is a fuctio of the CSS, it is UMVUE Here σ 2 is kow, but what if it is also ukow? Cosider the followig estimator: What do you thik about its properties? ã := π( X 1 S 2 )? 6 Exerc 748 Here X 1,,X are iid Beroulli(p) (a) We foud i Exerc 74 that I(p) = 1/[p(1 p)], so Var p ( X) = 1 p(1p) attais the lower boud [I(p)] 1 = 1 p(1 p) (b) Well, we ca write usig iid, 4 E p {X 1 X 2 X 3 X 4} = E p {X l } = p 4 The, because X l is CSS (remember that we are dealig with a expoetial class), the statistic δ( X l ) := E p {X 1 X 2 X 3 X 4 X l } is the UMVUE of p 4 Ca you calculate it? Try by yourself ad the look at this: δ(t) = E p {X 1 X 2 X 3 X 4 X l = t} = P p (X 1 = 1, X 2 = 1, X 3 = 1, X 4 = 1 X l = t) 3

4 = P p(x 1 = 1, X 2 = 1, X 3 = 1, X 4 = 1, X l = t) P p ( X l = t) = p4 [( 4)!/(t 4)!( t)!]p t 4 (1 p) t [!/t!( t)!]p t (1 p) t ] = ( 4)!t!!(t 4)! 7 Exerc 749 Let X 1,, X be a sample from Expo(λ) (a) Fid a UE of λ based o X (1) Well, because f X (x λ) = λ 1 e x/λ I(x > ) we use our techique to fid the desity of the first ordered observatio Remember how we do this: F X(1) (x λ) = P λ (X (1) x) = 1 P(X (1) > x) = 1 P λ (X 1 > x,, X > x) = 1 [λ 1 e z/λ dz] = 1 e x/λ Take derivative ad get f X(1) (x λ) = λ 1 e x/λ As we see, X (1) Expo(λ/) so E(X (1) ) = λ/ ad λ := X (1) is UE (b) Let us fid UMVUE Here Y := X l is CSS (agai due to the expoetial family) Because E λ (Y ) = λ we get ˆλ UMV U = X x 8 Exerc 752 Here X 1,,X are iid from Poisso(λ) (a) Write e λ λ X l f X (x λ) = = e λ λ x l x l! x l! As we see, this is a expoetial family with Y := X l beig the CSS We coclude, usig our theory, that X = Y/ is the UMVUE of λ (b) To aalyze directly E λ (S 2 X) = E λ {( 1) 1 (X l X) X} is possible but rather complicated Let be smart ad use the theory We kow that E λ (S 2 ) = λ because λ is the variace ad S 2 is UE of the variace But λ is also the mea for poisso distributio, so E λ (S 2 X) is the UMVUE of the mea Further, X is also UMVUE of λ ad X is the CSS, so by uiqueess of the UMVUE we have for Poisso distributio! E λ (S 2 X) = X 4

5 The we also ca write that Var λ (S 2 ) = Var λ (E λ (S 2 X)) + E λ {Var λ (S 2 X)} > Var λ (E λ (S 2 X)) = Var λ ( X) (c) If Y is CSS ad Z is ay other statistic such that E θ (Y ) E θ (Z) for all θ Ω the E θ (Z Y ) = Y for all θ Ω Ideed, let E θ (Z Y ) =: g(y ), the E θ (g(y ) Y ) for all θ Ω because E θ g(y ) = E θ (Z) = E θ (Y ) for all θ Ω Because Y is complete this yields that g(y ) = Y as Fially, we kow that coditioig o a CSS reduces the variace, so Var θ (Z) > Var θ (E θ (Z Y )) = Var θ (Y ) 9 Exerc 755 (a) Give that the pdf is f(x θ) = θ 1 I( < x < θ) Here Y := X () is the CSS ad f Y (y θ) = θ y 1 The As a result, E θ (Y r ) = y r θ y 1 dy = θ y +r 1 dy = ˆθ UMV U := + r Xr () + r θ θ +r = + r θr 1 Exerc 759 Here X 1,,X are iid from N(µ, σ 2 ) Fid UMVUE for σ p, p > Well, it is reasoable to try to aalyze (S 2 ) p/2 because S 2 is a good estimate of σ 2 We kow that S 2 D = σ 2 ( 1)χ 2 1, so E µ,σ 2(S 2 ) p/2 = σ p [( 1) p/2 E{(χ 2 1) p/2 }] Deote K p := ( 1) p/2 E{(χ 2 1) p/2 } This is a fuctio i p which ca be calculated for all p (I skip its calculatio but you ca thik about momet geeratig fuctio for chi-squared RV), ad the ˆδ p := (S2 ) p/2 K p (3) is the UMVUE Ideed, ( X, S 2 ) is the CSS for the ormal distributio ad the mea of the estimator is σ p 5