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100BHWS6

# 100BHWS6 - STAT 100B Homework VI Solution Problem 1 Suppose...

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STAT 100B Homework VI Solution Problem 1: Suppose we want to test hypotheses H 0 : X p 0 ( x ) versus H 1 : X p 1 ( x ). Suppose our decision rule is to reject H 0 if p 1 ( X ) /p 0 ( X ) > C . (1) Show that this decision rule is the most powerful in the sense that among all the decision rules with the same type I error, it has the minimum type II error. A: Please see the picture in the lecture note. (2) Suppose the prior probability that H 1 is true is λ . Suppose that the loss due to type I error is l 1 , and the loss due to type II error is l 2 . Show that the optimal decision rule is to reject H 0 if p 1 ( X ) /p 0 ( X ) > C , where C is determined by λ , l 1 and l 2 . A: The posterior probability of H 1 is p ( H 1 | X ) = λp 1 ( X ) / ( λp 1 ( X ) + (1 - λ ) p 0 ( X )) . The posterior probability of H 0 is p ( H 0 | X ) = (1 - λ ) p 0 ( X ) / ( λp 1 ( X ) + (1 - λ ) p 0 ( X )) . Given X , if we reject H 0 , the risk or the expected loss is p ( H 0 | X ) l 1 . If we accept H 0 , the risk is p ( H 1 | X ) l 2 . So we reject H 0 if p ( H 1 | X ) l 2 > p ( H 0 | X ) l 1 , i.e., λp 1 ( X ) l 2 > (1 - λ ) p 0 ( X ) l 1 , i.e., if p 1 ( X ) /p 0 ( X ) > (1 - λ ) l 1 / ( λl 2 ).

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100BHWS6 - STAT 100B Homework VI Solution Problem 1 Suppose...

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