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Give an example satisfying the following conditions: give out a sequence of random variables defined on a probability space, and a sub sigma algebra: the sequence converges almost surely to a limit and it is also uniformly integrable, but the sequence of the conditional expectation of the random variable sequence on the sub sigma algebra does not converges almost surely to the limit, which is the conditional expectation of the limit of the random variable sequence on the sub sigma algebra.

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closed as too localized by Nate Eldredge, Andres Caicedo, George Lowther, Zev Chonoles, Mark Sapir Feb 23 '11 at 4:07

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This isn't really a research level question (see the faq: mathoverflow.net/faq#whatquestions). Actually, it's not a question, rather, its a command. –  George Lowther Feb 23 '11 at 3:30
    
math.stackexchange is probably a more suitable site although, if you post it there, it should still be written as a question. –  George Lowther Feb 23 '11 at 3:31

1 Answer 1

Consider independent random variables $X_n, Y_n, n\in\mathbb N$, such that $\mathbb P(X_n = 1) = 1-\mathbb P(X_n = 0) = 1/n$, $\mathbb P(Y_n = n) = 1-\mathbb P(Y_n = 0) = 1/n$. Set $Z_n = X_nY_n$. Then, $Z_n$ is uniformly integrable and $Z_n\to 0$ as $n\to\infty$ almost surely, by Borel-Cantelli Lemma. However, set ${\cal A} = \sigma(X_n:n\in\mathbb N)$. Then $\mathbb E(Z_n\mid{\cal A}) = X_n\mathbb E(Y_n\mid {\cal A}) = X_n$, which converges to 0 in $L^1$ but not almost surely.

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