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Unformatted text preview: Lecture 20 20.1 Randomized most powerful test. In theorem in the last lecture we showed how to find the most powerful test with level of significance α (which means that δ ∈ K α ), if we can find c such that 1 f 1 ( X ) f 2 ( X ) < c = α. This condition is not always fulfiled, especially when we deal with discrete distribu tions as will become clear from the examples below. But if we look carefully at the proof of that Theorem, this condition was only necessary to make sure that the like lihood ratio test has error of type 1 exactly equal to α. In our next theorem we will show that the most powerful test in class K α can always be found if one randomly breaks the tie between two hypotheses in a way that ensures that the error of type one is equal to α. Theorem. Given any α ∈ [0 , 1] we can always find c ∈ [0 , ∞ ) and p ∈ [0 , 1] such that 1 f 1 ( X ) f 2 ( X ) < c + (1 p ) 1 f 1 ( X ) f 2 ( X ) = c = α. (20.1) In this case, the most powerful test δ ∈ K α is given by δ = H 1 : f 1 ( X ) f 2 ( X ) > c H 2 : f 1 ( X ) f 2 ( X ) < c H 1 or H 2 : f 1 ( X ) f 2 ( X ) = c where in the last case of equality we break the tie at random by choosing H 1 with probability p and choosing H 2 with probability 1 p. This test δ is called a randomized test since we break a tie at random if necessary. 76 LECTURE 20. 77 Proof. Let us first assume that we can find c and p such that (20.1) holds. Then the error of type 1 for the randomized test δ above can be computed: α 1 = 1 ( δ 6 = H 1 ) = 1 f 1 ( X ) f 2 ( X ) < c + (1 p ) 1 f 1 ( X ) f 2 ( X ) = c = α (20.2) since δ does not pick...
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This note was uploaded on 10/11/2009 for the course STATISTICS 18.443 taught by Professor Dmitrypanchenko during the Spring '09 term at MIT.
 Spring '09
 DmitryPanchenko
 Statistics

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