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Given a probability measure space $(\Omega, \mathcal{F}, P)$, if a random variable X and a sub $\sigma$-algebra $\mathcal{A}$ are independent, I was wondering why:

  1. $$E (X|\mathcal{A}) = (EX)I_Ω;$$
  2. $$E(I_A \times X) = P (A)EX, \, \forall A \in \mathcal{A}.$$

Thanks and regards!

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Tim: Either my hints allowed you to answer the question or they did not, in both cases you have ways of signaling it. –  Did May 7 '11 at 13:12
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1 Answer

up vote 2 down vote accepted

Hint for 1.: Recall the definition of the random variable $E(X|\mathcal{A})$. Hint for 2.: Recall the definition of $X$ being independent from $\mathcal{A}$. (And add the condition that $X$ is integrable.)

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Thanks! For 2, a random variable and a sigma algebra are independent, if the sigma algebra and the sigma algebra generated by the random variable are independent. Two sigma algebras are independent, if any two subsets, each from each sigma algebra, are independent. But I still can see how 2 can be derived from the definitions. –  Tim May 4 '11 at 13:50
    
Fix A and try to find a class of random variables X for which assertion 2. reduces to a trivial statement. If X is a constant x, of course 2. holds because both sides are xP(A). Just a little bit more elaborate would be to choose X as... –  Did May 4 '11 at 14:12
    
Thanks! For 2, I would like to use the following, but not sure if it is correct. A random variable can be approximated pointwise by a sequence of simple functions, and each simple functions can be represented as a sum of finitely many step functions. For each random variable, is the subset corresponding to each such step functions in the sigma algebra generated by the random variable? –  Tim May 7 '11 at 15:26
    
It is true. For a simple construction and for the answers to many of your other questions, see the small book Probability with martingales by David Williams goodreads.com/book/show/615497.Probability_with_Martingales. –  Did May 7 '11 at 16:58
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