APPM 5720 Solutions to Problem Set Five

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1 APPM 5720 Solutios to Problem Set Five 1. The pdf is f(x; α, β) 1 Γ(α) βα x α 1 e βx I (0, ) (x). The joit pdf is f( x; α, β) 1 [Γ(α)] β α ( i1 x i ) α 1 e β i1 x i i1 I (0, ) (x i ) 1 [Γ(α)] βα } {{ } a(θ) So, by two-parameter expoetial family, is complete ad sufficiet for θ (α, β). I (0, ) (x i ) exp (α 1) l x i β x i i1 i1 i1 c 1 (θ) c 2 (θ) d 1 ( x) d 2 ( x) S (d 1 ( X), d 2 ( X)) ( l X i, X i ) i1 i1 (We also say that l X i ad X i are joitly complete ad sufficiet for θ (α, β). ) 2. First, we will use the expoetial family factorizatio to fid a complete ad sufficiet statistic for this model. The pdf is The joit pdf is f( x; λ) e λ λ f(x; λ) e λ λ x I x! {0,1,2,...} (x). x i (xi!) I{0,1,2,...} (x i ) e} λ I{0,1,2,...} (x i ) {{} exp[(l λ) ( x i )] (xi!) a(λ) c(λ) d( x) So, by oe-parameter expoetial family, S d( X) i1 X i is complete ad sufficiet for this model. (a) To fid the UMVUE for λ, we eed to fid a fuctio of S X i that is ubiased for λ. Let s look at E[S]. E[S] E[ X i ] E[X i ] idet E[X 1 ] P oisso λ

2 Thus, the UMVUE for λ is λ S Xi X. (b) To fid the UMVUE for τ(λ) e λ, we eed to fid a fuctio of S X i that is ubiased for λ If you ca guess oe ad verify that it is ubiased for e λ, great. If ot, try the Rao-Blackwell Theorem. The Rao-Blackwell Theorem says (amog other thigs) that if T is ubiased for τ(λ) ad S is sufficiet, the T : E[T S] is still ubiased for τ(λ). Furthermore, T is a fuctio of S. While we did t have completeess back the ad were t talkg about UMVUEs, if we also have that S is complete the we have the UMVUE sce we have a ubiased estimator of τ(λ) that is a fuctio of the complete ad sufficiet statistic S! e λ, hmmm... where have we see this i relato to the Poisso distributio... Is t it part of the pdf? P (X 1 x) e λ λ x I x! {0,1,2,...} (x) We ca see that e λ P (X 1 0). Also, we ca always get a ubiased estimator for a probability out of a idicator: E[I X1 0] P (X 1 0) so we will take I I X1 0. Our UMVUE will be E[I X1 0 S]. For cocreteess we will compute this by first fixig S s. E[ τ 1 (λ) S s] E[I {X1 0} S s] P (X 1 0 S s) P (X 10,Ss) Ss P (X 10, i1 X is) P ( i1 X is) We would like to be able to break apart the probability i the umerator but, curretly, there is a X 1 i both terms so they are ot idepedet. Sice we already kow that X 1 0, we ca write the other evet, which says that the sum of all of the X i is s as just that the sum of the remaiig X 2, X 3,..., X is s 0 s. Thus, we have E[ τ 1 (λ) S s] P (X 1 0, i2 X is) P ( i1 X is) idep P (X 1 0) P ( i2 X is) P ( i1 X is) The sum of the Poissos i the deomiator has agai the Poisso distributio with rate λ. The sum of the Poissos i the umerator has the Poisso distributio with rate

3 ( 1)λ. Thus, E[ τ 1 (λ) S s] P (X 10) P ( i2 X is) P ( i1 X is) e λ λ 0 e ( 1)λ [( 1)λ] s 0! s! e λ [λ] s s! ( ) s 1. Removig the cocrete s we have that the UMVUE for τ(λ) e λ is τ(λ) E[ τ 1 (λ) S] ( ) 1 S ( ) 1 X i. 3. The pdf is The joit pdf is f(x; θ) θx θ 1 I (0,1) (x). θ [ ] i1 x θ 1 i1 i I (0,1) (x i ) θ [ i1 x i ] θ 1 i1 I (0,1) (x i ) So, by oe-parameter expoetial family, θ I (0,1) (x i ) exp[(θ 1) l x i ] a(θ) i1 i1 c(θ) d( x) is complete ad sufficiet for θ. S l X i i1 (a) We eed to fid a fuctio of S that is ubiased for 1/θ. Lettig y l x g(x), a simple g-iverse trasformatio shows us that f Y (y) θe θy I (,0) (x). This is similar to a expoetial distributio. To make it a actual expoetial distributio (easier to work with), we will istead take Y l X. Now Y exp(rate θ). Thus, S lx i W where W Γ(, θ). So, i1 E[S] E[W ] θ

4 which implies that is the UMVUE for θ. θ 1 S 1 l X i (b) We ow eed to fid a fuctio of S whose expected value is τ(θ) (θ/(θ + 1)). Note that this looks like the momet geeratig fuctio of a Γ(, θ), evaluated at t 1. Ideed, ( ) θ E[e S ] E[e W ] M W ( 1). θ + 1 Thus, is the UMVUE for τ(θ). τ(θ) e S e l Xi X i 4. First, we will show that T is sufficiet. Fix ay x. Let t t( x). Let y t be ay vector that maps to t, uder t( ). Note the that t( y t ) t( x). Thus, by the give property, we must have that f( y t ; θ) f( y t( x) ; θ) is θ-free. We will write f( y t( x) ; θ) h( x, y t ) for some fuctio h(, ). Note that y t( x) is a fuctio of t( x), which is a fuctio of x, so we ca actually say that f( y t( x) ; θ) h( x) for some fuctio h( ). So, we have h( x) f( y t( x) ; θ) g(t( x);θ) ad, by the Factorizatio Criterio for sufficiecy, we have that T t( X) is sufficiet for the model. Now let s show that T t( X) is miimal sufficiet. Suppose that S s( X is sufficiet for the model. The we ca write for some fuctios h ad g. h( x)g(s( x); θ)

5 So, we have, for ay x ad y, that h( x)g(s( x); θ) f( y; θ) h( y)g(s( y); θ). (1) We wat to show that T is a fuctio of S. Suppose that x ad y are such that s( x) s( y). By (1), we the have f( y; θ) h( x) h( y), which is θ-free. Thus, by the assumptio of the problem, we have that t( x) t( y). Now s( x) s( y) t( x) t( y) t( ) is a oe-to-oe fuctio of s( ). Sice S s( X) was a arbitrary sufficiet statistic ad T is a fuctio of S, we have that T is miimal sufficiet! 5. Note that this distributio ca t be a two-parameter expoetial family because it does ot have two parameters! (a) Use the Factorizatio Criterio for sufficiecy. (b) Show that E[2( X i ) 2 ( + 1) Xi 2] 0. Sice 2( X i ) 2 ( + 1) Xi 2 0, we have exhibited a g for which E[g(S)] 0 g(s) 0. Thus, S is ot complete. (c) This is a easy applicatio of Problem 4.