# HW5 - 9.35 L(y1 yn | = n-1 yi n = h(y1 yn)g n Yi By theorem...

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9.35 L ( y 1 , ··· ,y n | α,θ ) = α n Q y α - 1 i θ = h ( y 1 , ··· ,y n ) g ( Q Y i ) By theorem 9.4, where h ( y 1 , ··· ,y n ) = 1 , Q n Y i is suﬃcient for α 9.38 The exponential distribution is given by f ( y ) = ( 1 θ ) e y/θ . L ( y 1 , ··· ,y n | θ ) = 1 θ e y 1 implying by the factorization theorem that Y i is suﬃcient for θ . Then also ¯ Y is also suﬃcient for θ . Notice this problem is a special case of 9.37. 9.68 For a single observation y , the ﬁrst sample moment is m 0 1 = Y , and since Y has a geometric distribution, E ( Y ) = μ 0 1 = 1 p . Hence the moment estimator of p is ˆ p = 1 Y . 9.72 a L = n Y i =1 λ y i e - λ y ! = λ y i e - λ Q n i =1 y i ! ln L = ( X y i ) ln λ - - X ln y i ! ∂λ [ln L ] = ± X y i λ ² - n = 0 We obtain ˆ λ = (∑ y i n ) = ¯ Y b Since E ( Y i ) = λ, V ( Y i ) = λ , E ( ˆ λ ) = λ, V ( ˆ λ ) = λ n c Since E ( Y i ) = λ, V ( Y i ) = λ
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