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 Sep 3 comment Are the converses of the following special cases of conditional expectation also true? Thanks. what is about X that is equivalent to E(X|N)=EX a.e., if independence is too much? Sep 3 asked Are the converses of the following special cases of conditional expectation also true? Sep 2 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? Let me put it in another way: If Y1 and Y2 are both in E(Y|N), are they equal a.e. (wrt N)? Sep 1 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? "It's conceivable that you can specify criterion to partition E(X|N) into disjoint classes" doesn't make sense. I guess you know that the a.e. is a equivalent relation on the set of random variables, so the a.e. relation partitions the set of r.v.s. into equivalent classes. it isn't what I can change. My last comment is a well defined question Sep 1 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? In that new measure space, can E(X|N) be several a.e. equivalent classes of N measurable functions? Sep 1 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? In my last comment, by a.e.equivalent class, I meant for the new measure space with N as sigma algebra. so measurable functions are N measurable. In that new measure space, Is E(X|N) exactly an entire a.e. equivalent class of N measurable functions? Or several such classes? Or part of such a class? Sep 1 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? Is E(X|N) exactly an entire a.e. equivalent class of N measurable functions? Or several such classes? Or part of such a class? Sep 1 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? Can you show "there can exist a Y′ that equals Y a.e. but Y′ is not a member of E(X|N)"? I don't think so, because their integrals on any measurable set are same Sep 1 comment Does $X ⊥ Y \leftrightarrow X ⊥ Y | Z$ implies $(X,Y) ⊥ Z$? When is reverse used and when is converse? Sep 1 revised How to formulate the requirements that a counterexample must satisfy? added 486 characters in body Sep 1 comment How to formulate the requirements that a counterexample must satisfy? @Brian. In the linked post, to construct an example, we have to choose what the three random variables $X$ $Y$ and $Z$ are, though the three statements about the random variables are given and not changeable. Sep 1 revised How to formulate the requirements that a counterexample must satisfy? added 486 characters in body Sep 1 comment How to formulate the requirements that a counterexample must satisfy? @Brian: I should have added that $p_1, p_2$ and $p_3$ are given statements, which can't be changed. For an example, math.stackexchange.com/questions/1416468/… Sep 1 comment How to formulate the requirements that a counterexample must satisfy? @Brian: what do you mean by " take p2 and p3 to be the same statement"? p1, p2 and p3 are supposed to be given statements which can't be changed. Sep 1 comment How to formulate the requirements that a counterexample must satisfy? @Brian: Thanks. Is this a counterexample: $p_2$ and $p_3$ both are true, but $p_1$ isn't? How shall we find a counterexample that satisfies the requirement that $p_2$ and $p_3$ imply each other, although I know it means either both $p_2$ and $p_3$ are true, or neither is true. Sep 1 revised Does $X ⊥ Y \leftrightarrow X ⊥ Y | Z$ implies $(X,Y) ⊥ Z$? added 39 characters in body Sep 1 comment Is conditional expectation E(X|N) an a.e. equivalence class wrt N or underlying sigma algebra? E(X|N) is a set of r.v.s. If a random variable $Y$ is in $E(X|N)$, then any random variable that is equal to $Y$ a.e. must also be in $E(X|N)$. That is, the a.e. equivalent class where $Y$ is in must be a subset of $E(X|N)$. My first comment is asking if $E(X|N)$ is just that one single a.e. equivalent class, or can it contain two or more different a.e. equivalent class(es) where some other class(es) don't contain $Y$? (I guess you know that the a.e. is a equivalent relation on the set of random variables, so the a.e. equivalent classes form a partition of the set of r.v.s.) Sep 1 comment How to formulate the requirements that a counterexample must satisfy? Thanks. I am not looking for a concrete example, but a general formulation of a counterexample in terms of $p_1$, $p_2$ and $p_3$. (For a concrete example, see math.stackexchange.com/questions/1416468/…) Sep 1 asked Does $X ⊥ Y \leftrightarrow X ⊥ Y | Z$ implies $(X,Y) ⊥ Z$? Sep 1 comment How to formulate the requirements that a counterexample must satisfy? what is to look for when searching for a counterexample?