Dr. Hackney STA Solutions pg 154

Dr. Hackney STA Solutions pg 154 - 9-12Solutions Manual for...

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Unformatted text preview: 9-12Solutions Manual for Statistical Inference9.39 Letabe such thatRa-f(x)dx=/2. This value is unique for a unimodal pdf if >0. Letbe the point of symmetry and letb= 2-a. Thenf(b) =f(a) andRbf(x)dx=/2.asinceRa-f(x)dx=/21/2 =R-f(x)dx. Similarly,b. And,f(b) =f(a)>0 sincef(a)f(x) for allxaandRa-f(x)dx=/2>f(x)>0 for somex < af(a)>0.So the conditions of Theorem 9.3.2 are satisfied.9.41 a. We show that for any interval [a,b] and>0, the probability content of [a-,b-] isgreater (as long asb-> a). WriteZabf(x)dx-Zb-a-f(x)dx=Zbb-f(x)dx-Zaa-f(x)dxf(b-)[b-(b-)]-f(a)[a-(a-)][f(b-)-f(a)],where all of the inequalities follow becausef(x) is decreasing. So moving the interval towardzero increases the probability, and it is therefore maximized by moving a all the way to zero.b.T=Y-is a pivot with decreasing pdffT(t) =ne-ntI[0,](t). The shortest 1-intervalonTis [0,-1nlog], sinceZbne-ntdt= 1-b=-1nlog....
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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