Dr. Hackney STA Solutions pg 150

Dr. Hackney STA Solutions pg 150 - 9-8 Solutions Manual for...

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Unformatted text preview: 9-8 Solutions Manual for Statistical Inference 9.25 The confidence interval derived by the method of Section 9.2.3 is C(y) = : y + 1 1 log y + log 1 - n 2 n 2 where y = mini xi . The LRT method derives its interval from the test of H0 : = 0 versus H1 : = 0 . Since Y is sufficient for , we can use fY (y | ). We have (y) = sup=0 L(|y) sup(-,) L(|y) = ne-n (y - 0 )I[0 ,)(y) ne-(y-y) I[,)(y) 0 e-n(y-0 ) if y < 0 if y 0 . = e-n(y-0 ) I[0 ,) (y) = We reject H0 if (y) = e-n(y-0 ) < c , where 0 c 1 is chosen to give the test level . To determine c , set = P { reject H0 | = 0 } = P = P = Y > 0 - 0 - log c n Y > 0 - log c or Y < 0 = 0 n ne-n(y-0 ) dy log c = 0 n = 0 - log c n -e-n(y-0 ) = elog c = c . Therefore, c = and the 1 - confidence interval is C(y) = : y - log n = : y + 1 log y . n To use the pivotal method, note that since is a location parameter, a natural pivotal quantity is Z = Y - . Then, fZ (z) = ne-nz I(0,) (z). Let P {a Z b} = 1 - , where a and b satisfy = 2 a ne-nz dz = -e-nz 0 a 0 = 1 - e-na = 2 ne-nz dz = -e-nz b b = e-nb 2 - log 1 - a= n -nb = log 2 1 b = - log n 2 e-na = 1 - 2 Thus, the pivotal interval is Y + log(/2)/n Y + log(1 - /2), the same interval as from Example 9.2.13. To compare the intervals we compare their lengths. We have Length of LRT interval Length of Pivotal interval = y - (y + = 1 1 log ) = - log n n 1 1 y + log(1 - /2) - (y + log /2) = n n 1 1 - /2 log n /2 Thus, the LRT interval is shorter if - log < log[(1 - /2)/(/2)], but this is always satisfied. 9.27 a. Y = Xi gamma(n, ), and the posterior distribution of is (|y) = (y + 1 )n+a 1 1 1 b e- (y+ b ) , (n + a) n+a+1 ...
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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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