Dr. Hackney STA Solutions pg 88

Dr. Hackney STA Solutions pg 88 - 6-2Solutions Manual for...

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Unformatted text preview: 6-2Solutions Manual for Statistical Inference6.6 The joint pdf is given byf(x1,...,xn|,) =nYi=11()xi-1e-xi/=1()nnYi=1xi!-1e-ixi/.By the Factorization Theorem, (Qni=1Xi,ni=1Xi) is sufficient for (,).6.7 Letx(1)= mini{x1,...,xn},x(n)= maxi{x1,...,xn},y(1)= mini{y1,...,yn}andy(n)=maxi{y1,...,yn}. Then the joint pdf isf(x,y|)=nYi=11(3-1)(4-2)I(1,3)(xi)I(2,4)(yi)=1(3-1)(4-2)nI(1,)(x(1))I(-,3)(x(n))I(2,)(y(1))I(-,4)(y(n))|{z}g(T(x)|)1|{z}h(x).By the Factorization Theorem,(X(1),X(n),Y(1),Y(n))is sufficient for (1,2,3,4).6.9 Use Theorem 6.2.13.a.f(x|)f(y|)=(2)-n/2e-i(xi-)2/2(2)-n/2e-i(yi-)2/2= exp-12"nXi=1x2i-nXi=1y2i!+2n(y-x)#.This is constant as a function ofif and only if y= x; thereforeXis a minimal sufficientstatistic for.b. Note, forXlocation exponential(), the range depends on the parameter. Nowf(x|)f(y|)=Qni=1(e-(xi-)I(,)(xi))Qni=1(e-(yi-)I(,...
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