Ii from part i of the solution we obtain that ? x x n

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(ii) From part (i) of the solution, we obtain that λ ( X ) = X ( n ) θ 0 n X ( n ) θ 0 0 X ( n ) > θ 0 . Then, λ ( X ) < c is equivalent to X ( n ) > θ 0 or X ( n ) < θ 0 c 1 /n . Let c = α . When θ = θ 0 , the type I error rate is P X ( n ) < θ 0 α 1 /n = P ( X 1 < θ 0 α 1 /n ) n = the integer part of θ 0 α 1 /n θ 0 n θ 0 α 1 /n θ 0 n = α. Hence, an LR test of level α rejects H 0 if X ( n ) > θ 0 or X ( n ) < θ 0 α 1 /n . Exercise 39 (#6.87). Let X = ( X 1 , ..., X n ) be a random sample from the exponential distribution on the interval ( a, ) with scale parameter θ . (i) Suppose that θ is known. Find an LR test of size α for testing H 0 : a a 0 versus H 1 : a > a 0 , where a 0 is a known constant. (ii) Suppose that θ is known. Find an LR test of size α for testing H 0 : a =
Chapter 6. Hypothesis Tests 287 a 0 versus H 1 : a = a 0 . (iii) Repeat part (i) for the case where θ is also unknown. (iv) When both θ and a are unknown, find an LR test of size α for testing H 0 : θ = θ 0 versus H 1 : θ = θ 0 . (v) When a > 0 and θ > 0 are unknown, find an LR test of size α for testing H 0 : a = θ versus H 1 : a = θ . Solution. (i) The likelihood function is ( a, θ ) = θ n e na/θ e n ¯ X/θ I ( a, ) ( X (1) ) , where ¯ X is the sample mean and X (1) is the smallest order statistic. When θ is known, the MLE of a is X (1) . When a a 0 , the MLE of a is min { a 0 , X (1) } . Hence, the likelihood ratio is λ ( X ) = 1 X (1) < a 0 e n ( X (1) a 0 ) X (1) a 0 . Then λ ( X ) < c is equivalent to X (1) > d for some d a 0 . To determine d , note that sup a a 0 P ( X (1) > d ) = sup a a 0 ne na/θ θ d e nx/θ dx = ne na 0 θ d e nx/θ dx = e n ( a 0 d ) . Setting this probability to α yields d = a 0 n 1 θ log α . (ii) Note that ( a 0 , θ ) = 0 when X (1) < a 0 . Hence the likelihood ratio is λ ( X ) = 0 X (1) < a 0 e n ( X (1) a 0 ) X (1) a 0 . Therefore, λ ( X ) < c is equivalent to X (1) a 0 or X (1) > d for some d a 0 . From part (i) of the solution, d = a 0 n 1 θ log α leads to an LR test of size α . (iii) The MLE of ( a, θ ) is ( X (1) , ¯ X X (1) ). When a a 0 , the MLE of a 0 is min { a 0 , X (1) } and the MLE of θ is ˆ θ 0 = ¯ X X (1) X (1) < a 0 ¯ X a 0 X (1) a 0 . Therefore, the likelihood ratio is λ ( X ) = [ T ( X )] n X (1) a 0 1 X (1) < a 0 ,
288 Chapter 6. Hypothesis Tests where T ( X ) = ( ¯ X X (1) ) / ( ¯ X a 0 ). Hence the LR test rejects H 0 if and only if T ( X ) < c 1 /n . From the solution to Exercise 33, Y = ( ¯ X X (1) ) / ( ¯ X a ) has the beta distribution with parameter ( n 1 , 1). Then, sup a a 0 P T ( X ) < c 1 /n = sup a a 0 P ¯ X X (1) ¯ X a + a a 0 < c 1 /n = P Y < c 1 /n = ( n 1) c 1 /n 0 x n 2 dx = c ( n 1) /n . Setting this probability to α yields c = α n/ ( n 1) . (iv) Under H 0 , the MLE is ( X (1) , θ 0 ). Let Y = θ 1 0 n ( ¯ X X (1) ). Then the likelihood ratio is λ ( X ) = e n Y n e Y . Thus, λ ( X ) < c is equivalent to Y < c 1 or Y > c 2 . Under H 0 , 2 Y has the chi-square distribution χ 2 2 n 2 . Hence, an LR test of size α rejects H 0 when 2 Y < χ 2 2( n 1) , 1 α/ 2 or 2 Y > χ 2 2( n 1) ,α/ 2 , where χ 2 r,α is the (1 α )th quantile of the chi-square distribution χ 2 r .

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