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Unformatted text preview: 92 Solutions Manual for Statistical Inference therefore 0 is the MLE. The LRT statistic is ^2 X ^2 0
m/2 n/2 Y ^2 m/2 (^0 ) 2 (m+n)/2 , and the test is: Reject H0 if (x, y) < k, where k is chosen to give the test size . 2 2 2 b. Under H0 , Yi2 /(0 X ) 2 and Xi /X 2 , independent. Also, we can write m n n/2 m/2 1 1 (X, Y ) = 2 2 (Yi2 /0 2 )/m (Xi /X )/n n m m n + (X 2 /2X m+n + (Y 2 /0 2 )/m m+n m+n m+n )/n
i X i X = where F = n n+m 1 m + m+n F n/2 m m+n 1 +
n 1 m+n F m/2 Yi2 /0 m 2 Xi /n Fm,n under H0 . The rejection region is 1
n n+m (x, y) : + m m+n F n/2 m m+n 1 +
n 1 m+n F m/2 < c where c is chosen to satisfy P n m + F n+m m+n
n/2 m n + F 1 n+m m+n m/2 < c = . 2 c. To ease notation, let a = m/(n + m) and b = a yi / x2 . From the duality of hypothesis i tests and confidence sets, the set m/2 n/2 1 1 c() = : c a + b/ (1  a)+ a(1a) b is a 1 confidence set for . We now must establish that this set is indeed an interval. To do this, we establish that the function on the left hand side of the inequality has only an interior maximum. That is, it looks like an upsidedown bowl. Furthermore, it is straightforward to establish that the function is zero at both = 0 and = . These facts imply that the set of values for which the function is greater than or equal to c must be an interval. We make some further simplifications. If we multiply both sides of the inequality by [(1  a)/b]m/2 , we need be concerned with only the behavior of the function h() = 1 a + b/
n/2 1 b + a m/2 . Moreover, since we are most interested in the sign of the derivative of h, this is the same as the sign of the derivative of log h, which is much easier to work with. We have d log h() d = = = d n m  log(a + b/)  log(b + a) d 2 2 n b/2 m a  2 a + b/ 2 b + a 1 a2 m2 +ab(n  m)+nb2 . 22 (a + b/)(b + a) ...
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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.
 Spring '12
 Dr.Hackney
 Statistics

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