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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 ∈ σ ( Y ) } . Note that L is a λ-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 π-system generates F s = σ ( X s 1 , X s 2- X s 1 , ··· , X s m- 1- X s m ), by Dynkin’s π- λ 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 = σ ( 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 π-system, appealing once more to the π- λ theorem you conclude that F X t = σ ( 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→ μ ( 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( μ ( 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( μ ( 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 π : { 1 , . . . , n } → { 1 , . . . , n } . Note that for index sets { A π ( k ) } the preceding constructs mutually independent N ( b π (1) ,...,b π ( n ) ) of Poisson( μ ( A b )) laws, hence the resulting law of ( N A π (1) , . . . , N A π ( n ) ) is merely the image of the law of ( N A 1 , . . . , N A n ) under the permutation π 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 μ ( · ) that μ ( A b ) = μ ( A b , ) + μ ( 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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