sol2 - PSTAT 120C Solutions to Assignment 2 1 Suppose that...

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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 ( μ,σ ) and we want to test H : σ = σ versus H a : σ > σ L ( μ,σ ) = ( 2 πσ 2 )- n/ 2 exp "- 1 2 σ 2 n X i =1 ( y i- μ ) 2 # • The estimate of the mean is always ¯ y whatever the σ is. • Thus, max μ,σ ∈ Ω L ( μ,σ ) = ( 2 πσ 2 )- n/ 2 exp "- 1 2 σ 2 n X i =1 ( y i- ¯ y ) 2 # • The MLE of σ is ˆ σ = q 1 n ∑ ( y i- ¯ y ) 2 . • Therefore, max L ( μ,σ ) = ( 2 π ˆ σ 2 )- n/ 2 exp h- n 2 i • Thus, the GLRT test statistic is λ = max L ( μ,σ ) max μ,σ ∈ Ω L ( μ,σ ) = ˆ σ 2 σ 2 n/ 2 exp n 2 1- ˆ σ 2 σ 2 We can write this likelihood ratio in terms of the test statistic that we are interested in X = S 2 ( n- 1) /σ 2 = n ˆ σ 2 /σ , λ = ( 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 λ = 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 ˆ σ 2 ≤ σ 2 then the likelihood ratio is going to be 1. Therefore, we only need to explore the case where ˆ σ 2 /σ 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 α . The last step is to observe that under the null hypothesis, X has a χ 2 n- 1 distribution and c α can be obtained from the table of χ 2 distributions....
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This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.

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sol2 - PSTAT 120C Solutions to Assignment 2 1 Suppose that...

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