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conditional-expectation0

# conditional-expectation0 - Conditional expectation Jason...

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Conditional expectation Jason Swanson April 17, 2009 1 Conditioning on σ -algebras Let (Ω , 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 (Ω , F ). If X is a random variable, we define the conditional expectation of X given A as E [ X | A ] = 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 0 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 ] = α ( A ) P ( A ) , (1.3) where = 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 ) = 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 ] = X dQ = 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. 1

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Lemma 1.3. Any finite σ -algebra F on Ω can be written as F = σ ( { A j } n j =1 ) , where { A j } n j =1 is a partition of Ω , that is, Ω = ˙ n j =1 A j . Moreover, the partition { A j } n j =1 is unique. Proof. For each ω Ω, let A ω be the smallest measurable set containing ω . That is, A ω = F ω , where F ω = { A ∈ F : ω A } . Since this is a finite intersection, A ω ∈ F . In particular, E = { A ω : ω Ω } is a finite set. We claim that E is a partition of Ω and that F = σ ( E ). Clearly, Ω = ω Ω A ω , 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 ω, ω Ω, then either A ω = A ω or A ω A ω = . Let ω, ω Ω. Note that for any A ∈ F , if ω A , then A ∈ F ω , which implies A ω A . Hence, if ω A ω , then A ω A ω ; and if ω A c ω , then A ω A c ω . That is, either A ω A ω or A ω A c ω . By symmetry, either A ω A ω or A ω A c ω . Taken together, this shows that either A ω = A ω or A ω A ω = . To see that F = σ ( E ), simply note that any A ∈ F can be written as A = ω A A ω , and that this is a finite union. For uniqueness, suppose that F = σ ( { B j } n j =1 ), where Ω = ˙ n j =1 B j . If ω B j , then A ω = B j . Therefore, E = { B j } n j =1 . Exercise 1.4. Show that every infinite σ -algebra is uncountable. Let (Ω , F , P ) be a probability space and X an integrable random variable. Let G ⊂ F be a finite σ -algebra. Write G = σ ( { A j } n j =1 ), where { A j } n j =1 is a partition of Ω. The conditional expectation of X given G , written E [ X | G ], is a random variable defined by E [ X | G ]( ω ) = E [ X | A 1 ] if ω A 1 , E [ X | A 2 ] if ω A 2 , . . . E [ X | A n ] if ω A n . Note that we may write E [ X | G ] = n j =1 E [ X | A j ]1 A j . (1.4) We also define the conditional probability of A given G as P ( A | G ) = E [1 A | G ]. Note that P ( A | G )( ω ) = P ( A | A j ) if ω A j .
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