conditional-expectation0

conditional-expectation0 - Conditional expectation Jason...

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Unformatted text preview: Conditional expectation Jason Swanson April 17, 2009 1 Conditioning on -algebras Let ( , F ,P ) be a probability space and let A F with P ( A ) > 0. Define Q ( B ) = P ( B | A ) = P ( B A ) P ( A ) , for all B F . It is easy to to check that Q is a probability measure on ( , F ). If X is a random variable, we define the conditional expectation of X given A as E [ X | A ] = Z X dQ, (1.1) whenever this integral is well-defined. Note that E [1 B | A ] = P ( B | A ). Theorem 1.1. If E [ | X | 1 A ] < , then X is Q-integrable. If X or E [ | X | 1 A ] < , then E [ X | A ] = E [ X 1 A ] P ( A ) . (1.2) Remark 1.2. Note that (1.2) may be written as E [ X | A ] = ( A ) P ( A ) , (1.3) where d = X dP . Also note that (1.2) gives us the formula E [ X 1 A ] = P ( A ) E [ X | A ]. If X = 1 B , then this reduces to the familiar multiplication rule, P ( A B ) = P ( A ) P ( B | A ). Proof of Theorem 1.1. Note that if P ( B ) = 0, then Q ( B ) = 0. Hence Q P . Also note that Q ( B ) = Z B 1 A P ( A ) dP, for all B F . Thus, dQ/dP = 1 A /P ( A ). It follows that if X 0, then E [ X | A ] = Z X dQ = Z X dQ dP dP = E X 1 A P ( A ) = E [ X 1 A ] P ( A ) . Therefore, if E [ | X | 1 A ] < , then X is Q-integrable, and the same formula holds. 2 1 Lemma 1.3. Any finite -algebra F on can be written as F = ( { A j } n j =1 ) , where { A j } n j =1 is a partition of , that is, = S n j =1 A j . Moreover, the partition { A j } n j =1 is unique. Proof. For each , let A be the smallest measurable set containing . That is, A = T F , where F = { A F : A } . Since this is a finite intersection, A F . In particular, E = { A : } is a finite set. We claim that E is a partition of and that F = ( E ). Clearly, = S A , so to show that E is a partition, it suffices to show that this is a disjoint union. More specifically, we wish to show that if , , then either A = A or A A = . Let , . Note that for any A F , if A , then A F , which implies A A . Hence, if A , then A A ; and if A c , then A A c . That is, either A A or A A c . By symmetry, either A A or A A c . Taken together, this shows that either A = A or A A = . To see that F = ( E ), simply note that any A F can be written as A = S A A , and that this is a finite union. For uniqueness, suppose that F = ( { B j } n j =1 ), where = S n j =1 B j . If B j , then A = B j . Therefore, E = { B j } n j =1 . 2 Exercise 1.4. Show that every infinite -algebra is uncountable....
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conditional-expectation0 - Conditional expectation Jason...

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