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Unformatted text preview: Stochastic Calculus in Finance Assignment 1  SOLUTIONS (February 2007) 1. (a) Ω = { ( H,H,H ) , ( H,H,T ) , ( H,T,H ) , ( H,T,T ) , ( T,H,H ) , ( T,H,T ), ( T,T,H ) , ( T,T,T ) } , F consists of all subsets of Ω, and P ( A ) = #( A ) / 8. In other words, P makes each outcome equally likely. (b) G is a σfield with 4 atoms, namely { ( T,H,H ) , ( H,T,H ) , ( H,H,T ) } , { ( H,H,H ) } , { ( H,T,T ) , ( T,H,T ) , ( T,T,H ) } , and { ( T,T,T ) } . The σ field G itself consists of all possible unions of these four atoms, so the events in G consist of all possible unions of 0, 1, 2, 3, or 4 of the atoms. That is, # G = 2 4 = 16. Put another way, each element of G is deter mined by making 4 choices: whether to include the first atom, whether to include the second atom, etc. and there are 16 ways to make all these choices. To see that G actually is the desired σfield, observe that X is G measurable, since any event of the form { ω  X ( ω ) ≤ x } can be written as a union of atoms, so belongs to G . Conversely, if H is any σfield with respect to which X is measurable, then { ω  X ( ω ) = x } ∈ H for every x . In particular, all the atoms of G belong to H , so G ⊂ H too. In other words, G really is the smallest such σfield. (c) We use the formula from class, that E [1 A  G ]( ω ) = P ( A  B ), which is valid whenever B is the atom of G containing ω . Thus E [1 A  G ]( ω ) = , if ω = ( T,T,T ) 1 / 3 , if ω = ( H,T,T ) , ( T,H,T ), or ( T,T,H ) 2 / 3 , if ω = ( T,H,H ) , ( H,T,H ), or ( H,H,T ) 1 , if ω = ( H,H,H ) . 2. G has 4 atoms, namely A ∩ B , A \ B , B \ A , and A c ∩ B c . Our assumptions imply that none of these events is empty. Putting them together in all possible ways yields the following 16 events: G = {∅ ,A ∩ B,A \ B,B \ A,A c ∩ B c ,A,B,A c ,B c , ( A ∩ B ) ∪ ( A c ∩ B c ) , ( A \ B ) ∪ ( B \ A ) ,A ∪ B,A ∪ B c ,B ∪ A c ,A c ∪ B c , Ω } 3. To show that max( X,Y ) (which I denote X ∨ Y ) is Gmeasurable, we need to show that { X ∨ Y ≤ z } ∈ G for every any real number z . But { X ∨ Y ≤ z } = { X ≤ z } ∩ { Y ≤ z } ∈ G 1 2 since the latter two sets belong to G (as X and Y are Gmeasurable), and G is closed under intersections (as it is a σfield). Similarly, { X ∧ Y ≤ z } = { X ≤ z } ∪ { Y ≤ z } ∈ G , so X ∧ Y is Gmeasurable. 4. Let X be the number of days with trades, and let A i denote the event that there is a trade on the i ’th day. Then E [ X ] = E [ 10 X i =1 1 A i ] = 10 X i =1 E [1 A i ] = 10 P ( A 1 ) . Let B j be the event that the j ’th manager doesn’t trade on the first day. Therefore P ( A 1 ) = 1 P ( A c 1 ) = 1 P ( B 1 ∩···∩ B 12 ) = 1 P ( B 1 ) 12 = 1 ( . 9) 12 ....
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 Brownian Motion, Xt, zt, /2T dw, Bs Bt

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