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 one-sided 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