Dr. Hackney STA Solutions pg 41

# Dr. Hackney STA Solutions pg 41 - 3-14Solutions Manual for...

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Unformatted text preview: 3-14Solutions Manual for Statistical InferenceThe inequality is becausex-θ2> x-θ1, andFis nondecreasing. To get strict inequalityfor somex, let (a,b] be an interval of lengthθ1-θ2withP(a < Z≤b) =F(b)-F(a)>0.Letx=a+θ1. ThenF(x|θ1)=F(x-θ1) =F(a+θ1-θ1) =F(a)< F(b) =F(a+θ1-θ2) =F(x-θ2) =F(x|θ2).b. Letσ1> σ2. LetX1∼f(x/σ1) andX2∼f(x/σ2). LetF(z) be the cdf corresponding tof(z) and letZ∼f(z). Then, forx >0,F(x|σ1)=P(X1≤x) =P(σ1Z≤x) =P(Z≤x/σ1) =F(x/σ1)≤F(x/σ2) =P(Z≤x/σ2) =P(σ2Z≤x) =P(X2≤x)=F(x|σ2).The inequality is becausex/σ2> x/σ1(becausex >0 andσ1> σ2>0), andFisnondecreasing. Forx≤0,F(x|σ1) =P(X1≤x) = 0 =P(X2≤x) =F(x|σ2). Toget strict inequality for somex, let (a,b] be an interval such thata >0,b/a=σ1/σ2andP(a < Z≤b) =F(b)-F(a)>0. Letx=aσ1. ThenF(x|σ1)=F(x/σ1) =F(aσ1/σ1) =F(a)< F(b) =F(aσ1/σ2) =F(x/σ2)=F(x|σ2).3.43 a.FY(y|θ) = 1-FX(1y|θ)y >0, by Theorem 2.1.3. Forθ1> θ2,FY(y|θ1) = 1-FX1yθ1≤1-FX1yθ2=FY(y|θ2)for all...
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