HO1-sol-2009

HO1-sol-2009 - Homework 1 Solutions January 24, 2009...

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Unformatted text preview: Homework 1 Solutions January 24, 2009 Problem 1 First we check that X ( t ) is stationary. Let ( a 1 ,...,a n ) be any n-tuple with a i = 1. We need to verify that P ( X ( t 1 ) = a 1 ,X ( t 2 ) = a 2 ,...,X ( t n ) = a n ) = P ( X ( t 1 + h ) = a 1 ,X ( t 2 + h ) = a 2 ,...,X ( t n + h ) = a n ) The left hand side has the form P [ X ( t 1 ) = a 1 ,X ( t 2 ) = a 2 ,...,X ( t n ) = a n ] = = P [ (- 1) N t 1 = a 1 , (- 1) N t 2 = a 2 ,..., (- 1) N tn = a n ] = = P [ (- 1) N t 1 = a 1 , (- 1) N t 1 + N t 2- N t 1 = a 2 ,..., (- 1) N t n- 1 + N tn- N t n- 1 = a n ] = = P [ (- 1) N t 1 = a 1 , (- 1) N t 2- N t 1 = a 2 a 1 ,..., (- 1) N tn- N t n- 1 = a n a n- 1 ] = = P [ (- 1) N t 1 = a 1 ] P [(- 1) N t 2- N t 1 = a 2 a 1 ] ...P [(- 1) N tn- N t n- 1 = a n a n- 1 ] , where in the last step we used that the Poisson process N t has independent increments and is independent of it. In the same way the right hand side has the form P [ (- 1) N t 1 = a 1 ] P [(- 1) N t 2 + h- N t 1 + h = a 2 a 1 ] ...P [(- 1) N tn + h- N t n- 1 + h = a n a n- 1 ] Now noting that the random variables N t i + h- N t i- 1 + h and N t i- N t i- 1 have the same (Poisson, parameter t i- t i +1 ) distribution we see that the above expressions are identical except for the first terms. But since and N t are independent P [ (- 1) N s = 1] = P [ = 1] P [(- 1) N s = 1] + P [ =- 1] P [(- 1) N s =- 1] = = 1 / 2 ( P [(- 1) N s = 1] + P [(- 1) N s =- 1] ) = 1 / 2 and similarly P [ (- 1) N s =- 1] = 1 / 2, so they do not depend on s . This means that the first terms in the products are also the same. We have thus verified directly that X ( t ) is stationary. Now we compute the mean and covariance (autocorrelation). The mean is easy, since E [ ] = 0 implies E [ X ( t )] = E [ (- 1) N t ] = E [ ] E [(- 1) N t ] = 0. To find the covariance, suppose that s t , so that R ( X t ,X s ) = E [ X t X s ] = E [ | 2 (- 1) N t + N s | ] = E [(- 1) N t- N s +2 N s ] = E [(- 1) N t- N s ] = = (- 1) P ( N t- N s = odd) + (1) P ( N t- N s = even) = 1- 2...
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This note was uploaded on 11/08/2009 for the course MATH 236 taught by Professor Papanicolaou during the Winter '08 term at Stanford.

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HO1-sol-2009 - Homework 1 Solutions January 24, 2009...

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