hw8_stat210a_solutions

# hw8_stat210a_solutions - UC Berkeley Department of...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions - Problem Set 8 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006 Graded exercises Problem 8.1 (a) From the definitions given, we can write: S 1 S 2 S 3 . . . S n S n +1 = 1 0 0 ··· 0 0 1 1 0 ··· 0 0 1 1 1 ··· 0 0 . . . 1 1 1 ··· 1 0 1 1 1 ··· 1 1 · E 1 E 2 E 3 . . . E n E n +1 S = M · E It follows that the density of S is given by: f ( S 1 ,S 2 ,...,S n +1 ) = | M |- 1 f E ( E ( S )) = n Y i =1 exp (- ( S i- s i- 1 )) I ( S i- S i- 1 ≥ 0) = exp (- S n- 1 ) n Y i =1 I ( S i- S i- 1 ≥ 0) The characteristic function of an exponential random variable with mean 1 is given by γ E ( t ) = (1- it )- 1 and hence the distribution of S n +1 has characteristic function γ S n +1 = (1- it )- n +1 , that is, S n +1 ∼ Gamma (1 ,n + 1) with density f S n +1 ( s ) = s n exp(- s ) Γ( n +1) = s n exp(- s ) n ! and so: f ( S 1 ,S 2 ,...,S n | S n +1 ) = f S ( S 1 ,S 2 ,...,S n ) I ( S n ≤ S n +1 ) f S n +1 ( S n +1 ) = n ! Q n i =1 I ( S i +1- S i ≥ 0) S n n +1 Noticing that conditional on S n +1 the realized value for S n +1 can be treated as a constant, we get by rescaling: f ( S 1 S n +1 , S 2 S n +1 ,..., S n S n +1 | S n +1 ) = n ! I ( S n S n +1 ≤ 1) n Y i =1 I ( S i S n +1 ≥ S i- 1 S n +1 ) ! I ( S 1 ≥ 0) 1 On the other hand, a sample of sorted uniform variables have density f O given by: f O ( U (1) ,U (2) ,...,U ( n ) ) ∝ I ( U ( n ) ≤ 1 ) Y i = n n I ( U ( i ) ≥ U i- 1 ) I ( U 1 ≥ 0) To determine the constant of the density, notice that there are n ! possible orderings of a sample U 1 ,U 2 ,...,U n . Letting ( σ i ) n ! i =1 denote an enumeration of the possible permutations of n objects, we have that: 1 = Z [0 , 1] n f U ( u 1 ,...,u n ) du = n ! X i =1 Z [0 , 1] n f U ( u 1 ,...,u n ) I ( σ i ( u ) is increasing) du = n ! X i =1 Z [0 , 1] n f U ( u 1 ,...,u n ) I ( σ i ( u ) is increasing) du = n ! Z [0 , 1] n f U ( u 1 ,...,u n ) I ( σ 1 ( u ) is increasing) du By noticing that the last integrand equals f O , it follows that: f O ( U (1) ,U (2) ,...,U ( n ) ) = n ! I ( U ( n ) ≤ 1 ) Y i = n n I ( U ( i ) ≥ U i- 1 ) I ( U 1 ≥ 0) which establishes the result....
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## This note was uploaded on 10/17/2009 for the course STAT 210a taught by Professor Staff during the Fall '08 term at Berkeley.

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hw8_stat210a_solutions - UC Berkeley Department of...

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