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HO1-sol-2009 - Homework 1 Solutions Problem 1 First we...

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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 [ ξ ( - 1) N t 1 = a 1 , ξ ( - 1) N t 2 = a 2 , . . . , ξ ( - 1) N tn = a n ] = = P [ ξ ( - 1) N t 1 = a 1 , ξ ( - 1) N t 1 + N t 2 - N t 1 = a 2 , . . . , ξ ( - 1) N t n - 1 + N tn - N t n - 1 = a n ] = = P [ ξ ( - 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 [ ξ ( - 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 ξ is independent of it. In the same way the right hand side has the form P [ ξ ( - 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 ξ and N t are independent P [ ξ ( - 1) N s = 1] = P [ ξ = 1] P [( - 1) N s = 1] + P [ ξ = - 1] P [( - 1) N s = - 1] = = 1 / 2 ( P [( - 1) N s = 1] + P [( - 1) N s = - 1] ) = 1 / 2 and similarly P [ ξ ( - 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 [ ξ ] = 0 implies E [ X ( t )] = E [ ξ ( - 1) N t ] = E [ ξ ] 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 [ | ξ 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 P ( N t - N s = odd) To compute the latter observe that N t - N s is Poisson with parameter t - s and so P ( N t - N s = odd) = summationdisplay k =0 e - λ λ 2 k +1 (2 k + 1)! = e - λ e λ - e - λ 2 = 1 - e - 2 λ 2 .
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