Hw3-ORIE3510_S12_solns - ORIE 3510 Homework 3 Solutions...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ORIE 3510 – Homework 3 Solutions Instructor: Mark E. Lewis due 2PM, Wednesday February 15, 2012 (ORIE Hallway drop box) 1. (a) { X n } is not a Markov chain. To see this, it sufces to check that P ( X 4 = 1 | X 3 = 0 , X 2 = 1) n = P ( X 4 = 1 | X 3 = 0 , X 2 = 1) . Indeed, P ( X 4 = 1 | X 3 = 0 , X 2 = 1) = P (min(0 , 1) + ǫ 4 = 1 , X 3 = 0 , X 2 = 1) P ( X 3 = 0 , X 2 = 1) = P ( ǫ 4 = 1) = 1 2 , but P ( X 4 = 1 | X 3 = 0 , X 2 = 1) = P (min(0 , 1) + ǫ 4 = 1 , X 3 = 0 , X 2 = 1) P ( X 3 = 0 , X 2 = 1) = P ( ǫ 4 = 2) = 0 . (b) De±ne Y n = ( X n , X n - 1 ), n = 1 , 2 , . . . , then the sequence Y n is a Markov chain. To prove this, we see that P ( Y n = ( i n , i n - 1 ) | Y n - 1 = ( i n - 1 , i n - 2 ) , . . . , Y 1 = (0 , 0)) = P ( X n = i n , X n - 1 = i n - 1 | X n - 1 = i n - 1 , X n - 2 = i n - 2 , . . . , X 1 = 0 , X 0 = 0) = P (min( i n - 1 , i n - 2 ) + ǫ n = i n , X n - 1 = i n - 1 , . . . , X 0 = 0) P ( X n - 1 = i n - 1 , . . . , X 0 = 0) = P ( ǫ n = i n min( i n - 1 , i n - 2 )) = P (min( i n - 1 , i n - 2 ) + ǫ n = i n , X n - 1 = i n - 1 , X n - 2 = i n - 2 ) P ( X n - 1 = i n - 1 , X n - 2 = i n - 2 ) = P ( Y n = ( i n , i n - 1 ) | Y n - 1 = ( i n - 1 , i n - 2 )) . 2. We will show that P (( X n +1 , Y n +1 ) = ( i n +1 , j n +1 ) | ( X n , Y n ) = ( i n , j n ) , . . . , ( X 0 , Y 0 ) = ( i 0 , j 0 )) = P (( X n +1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

Hw3-ORIE3510_S12_solns - ORIE 3510 Homework 3 Solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online