# Conditional expectation and independence

1. Consider conditional expectation of a real-valued r.v. $X$ given a sub sigma algebra $\mathcal{B}$ of the probability space $(\Omega, \mathcal{F}, P)$.

Will independence between $\sigma(X)$, i.e. the sigma algebra of the r.v., and the given sub sigma algebra $\mathcal{B}$ make $E(X \mid \mathcal{B}) \equiv E(X)$ ?

Is independence between $\sigma(X)$ and the given sub sigma algebra $\mathcal{B}$ is the only way to define independence between $X$ and $\mathcal{B}$?

If there are other ways, will they make $E(X \mid \mathcal{B}) \equiv E(X)$ ?

2. Consider conditional expectation of a real-valued r.v. $X$ given another $S$-valued r.v. $Y$ on the same probability space $(\Omega, \mathcal{F}, P)$.

Will independence between $X$ and $Y$ make $E(X \mid Y) \equiv E(X)$ ?

Why? References are also appreciated! Thanks and regards!

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The answer to the first question in 1. is yes.

As a special case, the answer to 2. is also yes.

Concerning the second question in 1.: It's the only way I can think of. One definition is enough, isn't it? You can of course spell out what indepence actaully means in this case and take what you get as your definition.

Concerning the third question in 1.: They should, otherwise they cannot be equivalent defitions.

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@Rasmus: Thanks! Why the answer to the first question is yes? Can you explain? –  Tim Feb 25 '11 at 15:11
For (1): Without loss of generality, suppose $EX=0$. Let $Y = E[X|\mathcal{B}]$. Consider the event $A = \{Y > 0\} \in \mathcal{B}$. We have $A \in \mathcal{B}$ so $E[Y;A] = E[X;A]$. But $X$ and $A$ are independent... Also, in general, it is nice if you show some evidence that you have thought about a question before you ask it. –  Nate Eldredge Feb 25 '11 at 15:18
@Nate: Thanks! I will try to. I have limited knowledge about the topic, and I don't know how to prove it. I have searched my books, and don't find my books that tell me about this. I just want to figure out if the measure theoretical view is consistent with my elementary knowledge. –  Tim Feb 25 '11 at 15:31
@Nate: AWhat does E[Y;A] mean? If it means E[Y|A], what does E[Y|A] mean? –  Tim Feb 25 '11 at 15:35
@Tim: Just to prevent a confusion which might creep in, the fact that $E(X \mid \mathcal{B}) \equiv E(X)$ does not imply that $\sigma(X)$ and $\mathcal{B}$ are independent. (And +1 to Nate's remark.) –  Did Feb 25 '11 at 16:54