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# lec22 - Lecture 22 22.1 One sided hypotheses continued It...

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Lecture 22 22.1 One sided hypotheses continued. It remains to prove the second part of the Theorem from last lecture. Namely, we have to show that for any δ K α Π( δ * , θ ) Π( δ, θ ) for θ > θ 0 . Let us take θ > θ 0 and consider two simple hypotheses h 1 : = θ 0 and h 2 : = θ . Let us find the most powerful test with error of type one equal to α. We know that if we can find a threshold b such that θ 0 f ( X | θ 0 ) f ( X | θ ) < b = α then the following test will be the most powerful test with error of type 1 equal to α : δ θ = ( h 1 : f ( X | θ 0 ) f ( X | θ ) b h 2 : f ( X | θ 0 ) f ( X | θ ) < b But the monotone likelihood ratio implies that f ( X | θ 0 ) f ( X | θ ) < b f ( X | θ ) f ( X | θ 0 ) > 1 b V ( T, θ, θ 0 ) > 1 b and, since now θ > θ 0 , the function V ( T, θ, θ 0 ) is strictly increasing in T. Therefore, we can solve this inequality for T and get that T > c b for some c b . This means that the error of type 1 for the test δ θ can be written as θ 0 f ( X | θ 0 ) f ( X | θ ) < b = θ 0 ( T > c b ) . 86

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LECTURE 22. 87 But we chose this error to be equal to α = θ 0 ( T > c ) which means that c b should
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