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stat210a_HW03

# stat210a_HW03 - UC Berkeley Department of Statistics STAT...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 3 - Solutions Fall 2009 Issued: Tuesday, September 15, 2009 Due: Tuesday, September 22, 2009 Problem 3.1 Let X ∼ N ( θ, 1). From, p θ ( x ) = 1 √ 2 π exp(- 1 2 x 2 + θx- θ 2 2 ) , T ( X ) = X is sufficient by the factorization theorem. Notice that N ( θ, 1) is an exponential family with one parameter and one sufficient statistic and thus is of full rank, and T ( X ) = X is linearly independent. Hence T ( X ) = X is complete. From what is given, Z + ∞-∞ f ( x )exp( θx )d x = √ 2 π exp( θ 2 2 ) ⇒ Z + ∞-∞ f ( x )exp( 1 2 x 2 ) 1 √ 2 π exp {- ( x- θ ) 2 2 } d x = 1 Since 1 √ 2 π R + ∞-∞ exp {- ( x- θ ) 2 2 } d x = 1, for any θ ∈ R , we have, Z + ∞-∞ [ f ( x )exp( 1 2 x 2 )- 1] 1 √ 2 π exp {- ( x- θ ) 2 2 } d x = 0 , for any θ ∈ R . From the completeness of T ( X ), we have, f ( x ) e 1 2 x 2- 1 = 0 , w.p.1 Therefore, f ( x ) = e- 1 2 x 2 . Problem 3.2 For θ > 0, let Z 1 , ··· ,Z n ∼ U (0 ,θ ), i.i.d. P θ ( Z 1 , ··· ,Z n ) = n Y i =1 P θ ( Z i ) = n Y i =1 1 θ 1 (0 < Z i < θ ) = 1 θ n 1 ( Z (1) > 0) 1 ( Z ( n ) < θ ) ....
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stat210a_HW03 - UC Berkeley Department of Statistics STAT...

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