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Dr. Hackney STA Solutions pg 37

Dr. Hackney STA Solutions pg 37 - 3-10Solutions Manual for...

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Unformatted text preview: 3-10Solutions Manual for Statistical Inferenceb.fT(t) =γβtγ-1e-tγ/β,t≥0 andFT(t) =Rtγβxγ-1e-xγ/βdx=Rtγ/βe-udu=-e-u|tγ/β=1-e-tγ/β, whereu=xγ/β. Thus,hT(t) =(γ/β)tγ-1e-tγ/βe-tγ/β=γβtγ-1.c.FT(t) =11+e-(t-μ)/βandfT(t) =e-(t-μ)/β(1+e-(t-μ)/β)2. Thus,hT(t) =1βe-(t-μ)/β(1+e-(t-μ)/β)21e-(t-μ)/β1+e-(t-μ)/β=1βFT(t).3.27 a. The uniform pdf satisfies the inequalities of Exercise 2.27, hence is unimodal.b. For the gamma(α,β) pdff(x), ignoring constants,ddxf(x) =xα-2e-x/ββ[β(α-1)-x], whichonly has one sign change. Hence the pdf is unimodal with modeβ(α-1).c. For the n(μ,σ2) pdff(x), ignoring constants,ddxf(x) =x-μσ2e-(-x/β)2/2σ2, which only hasone sign change. Hence the pdf is unimodal with modeμ.d. For the beta(α,β) pdff(x), ignoring constants,ddxf(x) =xα-2(1-x)β-2[(α-1)-x(α+β-2)],which only has one sign change. Hence the pdf is unimodal with modeα-1α+β-2....
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