chap-10 - Chapter 10 1(a After stage k the algorithm has...

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Chapter 10 141 Chapter 10 1. (a) After stage k the algorithm has generated a random permutation of 1, 2, …, k . It then puts element k + 1 in position k + 1; randomly chooses one of the positions 1, …, k + 1 and interchanges the element in that position with element k + 1. (b) The first equality in the hint follows since the permutation given will be the permutation after insertion of element k if the previous permutation is i 1 , …, i j 1 , i , i j , …, i k 2 and the random choice of one of the k positions of this permutation results in the choice of position j . 2. Integrating the density function yields that that distribution function is F ( x ) = 0 , 2 / 1 0 , 2 / 2 2 > > x e x e s x which yields that the inverse function is given by F 1 ( u ) = 2 / 1 if 2 / ]) 1 [ 2 log( 12 if 2 / ) 2 log( > < u u u u Hence, we can simulate X from F by simulating a random number U and setting X = F 1 ( U ). 3. The distribution function is given by F ( x ) = 6 3 , 2 12 / , 3 2 , 1 4 / 2 2 + x x x x x x Hence, for u 1/4, F 1 ( u
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This note was uploaded on 03/31/2008 for the course STAT 418 taught by Professor G.jogeshbabu during the Spring '08 term at Penn State.

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chap-10 - Chapter 10 1(a After stage k the algorithm has...

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