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Let $X$ be a random variable on the measurable space $ (\Omega, \mathcal{A})$ and $\mathcal{B} $ be a sub-$\sigma$-field of $\mathcal{A}$.

Question 1: how to prove that $ \mathbb{E}(X |\mathcal{B})\in L^2(\Omega,\mathcal{B})$ is solution of the variational problem $$\min\{\mathbb{E}(X-Y)^2 : Y\quad \mathcal{B}-\text{measurable}\}$$ is $ \mathbb {E}(X | \mathcal {B}) $ ?

Question 2: Is that solution unique in $L^2(\Omega,\mathcal{B})$ ?

Question 3: What is the best characterization for $\mathbb {E}(X | \mathcal {B})\,$?

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    You want to apply to the general case of $L^1$ random variables a characterization of conditional expectation as a projection, which is valid for $L^2$ random variables only. For example, if $X$ is in $L^1$, there is no guarantee that $X-E(X\mid B)$ is even in $L^2$.2011-12-18
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    Didier Piau understood! Thank you for watching. I made the necessary corrections.2011-12-18
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    Now, the answer to Q2 is YES, as explained in every good textbook on the subject, the answer to Q1 depends on your definition of conditional expectation, and the meaning of Q3 is unclear.2011-12-18
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    Hello, Didier Piau. In q1 question the definition is the same exercise 17 on page 92 of the book of Folland (Real Analysis: Modern Techniques and Their Applications).2011-12-19
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    The best characterization refers to that which provides the best geometric interpretation of the conditional expectation. That is, ownership similar to a Hilbert space.2011-12-19
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    1. Questions on MSE should be as self-contained as possible. I recommend that you copy here Folland's relevant passage. 2. The Hilbert setting is nice but definitely not the only one nor does it provide *the best characterization*, whatever that means. One flaw is that it ccan only encompass the $L^1$ case by approximation. 3. Since your background seems to be more in analysis than in probability, you could try an accessible reference on conditioning such as the small and wonderful book *Probability with martingales* by David Williams.2011-12-19

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I think you need $X \in L^2$.

Anyway you can write, for Q1, $$E[(X-Y)^2]=E[\{(X-Z)+(Z-Y)\}^2]$$ where $Z=E[X|\mathcal{B}]$.