This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PSTAT 120C: Solutions to Assignment # 2 April 20, 2009 1. Suppose that we have n independent observations from a normal Y 1 ,...,Y n where each Y i N ( , ) and we want to test H : = versus H a : > L ( , ) = ( 2 2 ) n/ 2 exp " 1 2 2 n X i =1 ( y i ) 2 # The estimate of the mean is always y whatever the is. Thus, max , L ( , ) = ( 2 2 ) n/ 2 exp " 1 2 2 n X i =1 ( y i y ) 2 # The MLE of is = q 1 n ( y i y ) 2 . Therefore, max L ( , ) = ( 2 2 ) n/ 2 exp h n 2 i Thus, the GLRT test statistic is = max L ( , ) max , L ( , ) = 2 2 n/ 2 exp n 2 1 2 2 We can write this likelihood ratio in terms of the test statistic that we are interested in X = S 2 ( n 1) / 2 = n 2 / , = ( X/n ) n/ 2 exp n 2 X 2 . To see if this is an increasing or decreasing function, differentiate with respect to X d dx = n 2 X n/ 2 1 n n/ 2 exp n 2 X 2 1 2 ( X/n ) n/ 2 exp n 2 X 2 which is negative when n X < 0 = X > n. For a onesided test, when 2 2 then the likelihood ratio is going to be 1. Therefore, we only need to explore the case where 2 / 2 > . These are the same cases where X > n and the derivative is negative. For these values of X , the likelihood ratio is a decreasing function of X which implies that the GLRT is of the form X > c . The last step is to observe that under the null hypothesis, X has a 2 n 1 distribution and c can be obtained from the table of 2 distributions....
View Full
Document
 Spring '08
 EduardoMontoya

Click to edit the document details