sol-310c - 310c: Homework Solutions 2010 Homework 1...

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Unformatted text preview: 310c: Homework Solutions 2010 Homework 1 Solution. 7.1.12 Fixing h > 0 and t 0, let Y = X t + h- X t and L = { A F : P ( A C ) = P ( A ) P ( C ) for all C ( Y ) } . Note that L is a -system. Also, for s = ( s 1 , . . . , s m ) such that 0 s 1 < < s m = t considering our assumption that X s 1 , X s 2- X s 1 , , X s n- X s n- 1 are P-mutually independent, with and without s m +1 = t + h , we deduce from Definition 1.4.3 that { X s 1 B 1 } T m k =2 { X s i- X s k- 1 B k } L for any ( B 1 , B 2 , , B m ) B m . Since this -system generates F s = ( X s 1 , X s 2- X s 1 , , X s m- 1- X s m ), by Dynkins - theorem we have that F s L . The measurable map from ( X s 1 , X s 2 , , X s m ) to ( X s 1 , X s 2- X s 1 , , X s m- X s m- 1 ) is invertible, hence F s = ( X s 1 , X s 2 , , X s m ) (see Exercise 1.2.30). Let P = s F s where the union is over all finite sets s [0 , t ]. Since P is a -system, appealing once more to the - theorem you conclude that F X t = ( P ) L . That is, as claimed X t + h- X t is independent of F X t . Solution. 7.1.13 Given A 1 , . . . , A n T , we define the corresponding f.d.d. as follows. First set B k 1 = A k = B c k for k = 1 , . . . , n and note that by monotonicity of A 7 ( A ), for each non-zero b = ( b 1 , . . . , b n ) { , 1 } n the set A b = T k B kb k is in T . With N b a Poisson( ( A b )) random variable independent of { N b , b 6 = b } , let the random vector ( N A 1 , . . . , N A n ) have the same law as ( { b : b k =1 } N b , k = 1 , . . . , n ). Then, considering n = 1 we see that N A has the Poisson( ( A )) law for each A T . Moreover, if A k , k = 1 , . . . , n are disjoint sets then k b k > 1 yields A b = and hence N b 0. So, in this case our construction results with mutually independent N A k , k = 1 , . . . , n . Turning to check that these f.d.d. are consistent, fix n and a permutation : { 1 , . . . , n } { 1 , . . . , n } . Note that for index sets { A ( k ) } the preceding constructs mutually independent N ( b (1) ,...,b ( n ) ) of Poisson( ( A b )) laws, hence the resulting law of ( N A (1) , . . . , N A ( n ) ) is merely the image of the law of ( N A 1 , . . . , N A n ) under the permutation of the coordinates. That is, the identity (7.1.1) holds (where A k T serve as the index points t k there). Next, fixing { A k , k n } and non- zero b { , 1 } n- 1 , since A b is the disjoint union of A b , and A b , 1 we have by finite additivity of ( ) that ( A b ) = ( A b , ) + ( A b , 1 ). This applies for any non- zero b , so by the thinning property of Poisson variables we have the representation N b = N b , + N b , 1 (see Exercise 3.4.16, part (a)), and as a result the identity (7.1.2) also holds....
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sol-310c - 310c: Homework Solutions 2010 Homework 1...

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