Suppose we have X_1,…,X_n i.i.d, X_i∼Exp(1,μ) (pdf is f_μ(x)=e^(−(x−μ)) for x≥μ and 0 for x<μ). Is there any one dimensional (i.e. T:R^n→R) sufficient statistic for parameter μ? Obvious two dimensional sufficient statistic is T(x_1,…,x_n)=(∑_(xi),min x _i)

zaviknuogg

zaviknuogg

Answered question

2022-09-23

Suppose we have X 1 , , X n i.i.d, X i Exp ( 1 , μ ) (pdf is f μ ( x ) = e ( x μ ) for x μ and 0 for x < μ). Is there any one dimensional (i.e. T : R n R ) sufficient statistic for parameter μ? Obvious two dimensional sufficient statistic is T ( x 1 , , x n ) = ( x i , min x i )

Answer & Explanation

Davian Nguyen

Davian Nguyen

Beginner2022-09-24Added 9 answers

The likelihood of an i.i.d. sample x = ( x i ) 1 i n of distribution E ( 1 , μ ) is
L ( μ , x ) = exp ( s ( x ) + n μ ) [ t ( x ) μ ] ,
where s ( x ) = i = 1 n x i and t ( x ) = min 1 i n x i . This is
L ( μ , x ) = h ( x ) g ( μ , t ( x ) ) ,
with h ( x ) = exp ( s ( x ) ) and g ( m , u ) = exp ( n m ) [ u m ] hence t : x t ( x ) = min 1 i n x i is a sufficient statistic for μ.
Elias Heath

Elias Heath

Beginner2022-09-25Added 2 answers

We have
L ( x 1 , , x n , μ ) = j = 1 n e ( x j μ ) 1 [ μ , + [ ( x j ) = e n x ¯ e n μ 1 [ μ , + [ ( min 1 j n x k ) ,
so by Fischer-Neyman's theorems, φ ( x 1 , , x n ) = min 1 j n x j is a sufficient statistic.

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