notes-12-01

# notes-12-01 - M4056 Home Work Answers December 1, 2010...

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Unformatted text preview: M4056 Home Work Answers December 1, 2010 Problem 12 (recalled) In this problem, we are given f ( x | ) = e- x . If vector X = ( X 1 , . . ., X n ) is an i.i.d. sample, then f ( vectorx | ) = n e- ( x 1 + + x n ) = ( e- x ) n = f ( x | ). The log likelihood function is ( ) = n (ln x ). Since d d = n (1 / x ), we see that the MLE of is := 1 / x . The likelihood ratio test for H : = versus H 1 : negationslash = is of the form (see bottom of page 375): reject H if : ( x ) = f ( x | ) f ( x | ) k. Now, f ( x | ) f ( x | ) = ( e- x ) n ( e- x ) n = parenleftBigg e- x (1 / x ) e- 1 parenrightBigg n = ( e x e- x ) n . Thus, the test is of the form: reject H if : x e- x c, where c = k 1 /n e . Problem 13 Continuing Problem 12, suppose = 1, n = 10 and = . 05. We seek the corresponding value of c to define the test. a. A graph of the function y = x e- x appears below. 2 4 6 8 10 0.05 0.10 0.15 0.20 0.25 0.30 0.35 This makes it clear that { x | x e- x c } is a union of two intervals: [0 , x ] [ x 1 , ), where x and x 1 are determined by c . (They are the solutions to x = ce x .) b. We want to choose c so that P ( X e X...
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## notes-12-01 - M4056 Home Work Answers December 1, 2010...

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