For x 1 0 substituting ˆ θ 0 and maximizing with

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Forx(1)>0, substitutingˆθ0= 0 and maximizing with respect toλ, as above, yieldsˆλ0= ¯x.Therefore,λ(x) =supΘ0L(θ, λ|x)supΘL(θ, λ|x)=sup{(λ,θ):θ0}L(λ, θ|x)L(ˆθ,ˆλ|x)=1ifx(1)0Lx,0|¯x)L(ˆλ,ˆθ)ifx(1)>0,whereLx,0|x)L(ˆλ,ˆθ|x)=(1/¯x)ne-n¯x/¯x1/ˆλ·ne-nx-x(1))/x-x(1))=ˆˆλ¯x!n=¯x-x(1)¯xn=1-x(1)¯xn.So rejecting ifλ(x)cis equivalent to rejecting ifx(1)/¯xc*, wherec*is some constant.b. The LRT statistic isλ(x) =supβ(1n)e-ixisupβ,γ(βnn)(Qixi)γ-1e-ixγiThe numerator is maximized atˆβ0= ¯x.For fixedγ, the denominator is maximized atˆβγ=ixγi/n. Thusλ(x) =¯x-ne-nsupγ(γn/ˆβnγ)(Qixi)γ-1e-ixγi/ˆβγ=¯x-nsupγ(γn/ˆβnγ)(Qixi)γ-1.The denominator cannot be maximized in closed form. Numeric maximization could be used tocompute the statistic for observed datax.2
8.12 a. ForH0:μ0 vs.H1:μ >0 the LRT is to rejectH0if ¯x > cσ/n(Example 8.3.3). Forα=.05 takec= 1.645. The power function isβ(μ) =Pˆ¯X-μσ/n>1.645-μσ/n!=PˆZ >1.645-σ!.Note that the power will equal .5 whenμ= 1.645σ/n.b.ForH0:μ= 0 vs.HA:μ6= 0 the LRT is to rejectH0if|¯x|> cσ/n(Example 8.2.2). Forα=.05 takec= 1.645. The power function isβ(μ) =P(-1.96-nμ/σZ1.96 +nμ/σ).In this case,μ= +-1.96σ/ngives power of approximately .5.8.14 The CLT tells us thatZ= (iXi-np)/pnp(1-p) is approximatellyn(0,1). For a test thatrejectsH0wheniXi> c, we need to findcandnto satisfyPˆZ >c-n(.49)pn(.49)(.51)!=.01andPˆZ >c-n(.51)pn(.51)(.49)!=.99.We thus wantc-n(.49)pn(.49)(.51)= 2.33andc-n(.51)pn(.51)(.49)=-2.33.Solving these equations givesn= 13,567 andc= 6,783.5.8.15 From the Neyman-Pearson lemma the UMP test rejectsH0iff(x|σ1)f(x|σ0)=(2πσ21)-n/2e-ix2i/(2σ21)(2πσ20)-n/2e-ix2i/(2σ20)=σ0σ1nexp(12Xix2i1σ20-1σ21)> kfor somek0. After some algebra, this is equivalent to rejecting ifXix2i>2 log(k(σ10)n)1σ20-1σ21·=c(because1σ20-1σ21>0)This is the UMP test of sizeα, whereα=Pσ0(iX2i> c). To determinecto obtain a specifiedα, use the fact thatiX2i20χ2n. Thusα=Pσ0ˆXiX2i20> c/σ20!

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