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Thank you for your enlightenment and patience, and for giving me the best memory in my life.


2d
comment When will functions of two uncorrelated random variables be still uncorrelated?
Thanks! The uniform example seems carefully constructed, but it is also simple enough.
Mar
29
comment Difference and relation between dependency graph and graphical model?
Thanks, Bitwise! (1) Can a graphical model capture independence besides conditional independence? Also see here (2)Can a dependency graph capture conditional independence besides independence?
Mar
20
comment What is the name of this sequence/progression?
E.g., arithmetic progression, geometric progression, hypergeometric progression, and more.
Mar
19
comment Name for such special functions?
This situation stats.stackexchange.com/a/88738/1005
Mar
19
comment Name for such special functions?
@StellaBiderman: The second.
Mar
19
comment Name for such special functions?
Sorry, I updated my post.
Mar
11
comment Measurability from taking out property of conditional expectation
For 1, (1) the alternative and equivalent definition of $E(X|\mathcal G)$ requires $Y$ to be $\mathcal G$-measurable. Here, however, I ask if $Y$ is $\mathcal G$-measurable, if it satisfies the equation in the alternative and equivalent definition of $E(X|\mathcal G)$. So I don't quite understand "if it is true for any bounded r.v. this is correct". (2) Do you have non-trivial counterexamples?
Mar
11
comment Measurability from taking out property of conditional expectation
For 2, I am sometimes confused for some unclear reason, so I can't explain.
Mar
10
comment Conditional expectation as an orthogonal projection to what subspace?
Thanks, Davide! I confused this subspace with the set of random variables which generate the same sigma algebra $N$. could you take a look at it here math.stackexchange.com/questions/706924/…
Mar
10
comment The subspace which $X-E(X|\mathcal G)$ is orthogonal to, and the set of r.v.s generating the same $\mathcal G$.
I realized I made some error in my post. I changed to conditioning on sigma algebra instead of on random variable, so that we don't have to be bothered by which is the domain of $(X|Y)$. Do you have some thought about my posted question? Your reply in the link is a nice one. Thanks!
Mar
10
comment Formal definition of conditional probability
Thanks! Nice! +1
Mar
10
comment The subspace which $X-E(X|\mathcal G)$ is orthogonal to, and the set of r.v.s generating the same $\mathcal G$.
@StefanHansen: Your comment was also part of my original confusion. Now, if I understand correctly, here $E(X|Y)$ is not defined on $\Omega$ but on the codomain of $Y$, (in the process going from $E(X|\sigma(Y))$ to $E(X|Y)$, we shift the domain $\Omega$ of $E(X|\sigma(Y))$ to the codomain of $Y$, by some factorization theorem such as Doob-Dinkyns theorem?) , see here en.wikipedia.org/wiki/…
Mar
10
comment A case when uncorrelatedness imply mean independence.
@Did: I edited my post, trying to pin down why I don't understand the post.
Mar
10
comment What structure does this set of mapping have?
In the general settings, they are bijective. In the example they are Borel measurable bijections.
Mar
9
comment A case when uncorrelatedness imply mean independence.
The "not very coherent" comes from the present tense using in "why are you editing your comments", and from not understanding which questions "were answered". Still bothered by something else, and my mind can't focus, but ironically can deal with the word game which I will now stop.
Mar
9
comment A case when uncorrelatedness imply mean independence.
It was pure fun and imitating your comment (not only you can do that), not actually shouting.
Mar
9
comment A case when uncorrelatedness imply mean independence.
I am not doing anything, and haven't had activity in the past half an hour for having to deal with something else, and hoping will have time to come back to find out why I avoided your query, and address that. What "comments"? Sometimes you assume me (and maybe others) to know something or what you are referring to, but actually I (and others) don't. Sorry.
Mar
9
comment A case when uncorrelatedness imply mean independence.
Those points in the bulletted questions. I studied something before, which doesn't mean I understand and am able to use it completely and correctly.
Mar
9
comment A case when uncorrelatedness imply mean independence.
Here YOU are. :)
Mar
9
comment When $E(X|Y)E(Y) = E(XY)$?
@Feanor: Thanks! I updated the questions in my post. Could you also try to answer them?