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Jan
25
comment Conditional probability and almost sure equality confusion
Well I follow your reasoning, which means I can now prove my original statement (with the added measurability), yet I feel I am no closer to the understanding. Is there anything more you could say on this? I suppose the intuition is what I largely lack and what makes me feel like I don't actually understand any of it.
Jan
25
comment Conditional probability and almost sure equality confusion
@sinbadh I agree that if $X$ is $\mathcal G$-measurable, we get that equality. But I'm talking about $Y$ - if $Y$ is $\mathcal G$-measurable, I don't see how we get the equality $Y\stackrel {a.s.} = E^{\mathcal G} X$directly from the definition of conditional expectance.
Jan
25
comment Conditional probability and almost sure equality confusion
Ah! We might be gettting somehwere. Or at least I am, you already are "somewhere". In an hour I'll sit down with this and see if I can now clear the mess in my mind.
Jan
25
comment Conditional probability and almost sure equality confusion
@BCLC: Well I didn't, by "is such that" I only wanted to include the (what seemed to me as) relevant part of the definition, keeping in mind that we all know that it has to be measurable. I'll add that to the question.
Jan
25
comment Conditional probability and almost sure equality confusion
Thanks. This might be the problem. When you say $Y=Z$ in the $P$-a.s. sense in relation to $\mathcal G$, does this mean that in the context a.s. means with relation to $\mathcal G$ and not $\mathcal F$? I don't see why.
Jan
25
comment Conditional probability and almost sure equality confusion
Thanks, that clears some of the confusion (and possibly answers this question). Could you elaborate on why the measurability is sufficient?
Jan
10
comment Limit of a random walk with zero mean
Ah, dammit. Well, thank you :)
Dec
25
comment Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
@ZevChonoles True, I will think about that from now on. Any suggestions for a better title?
Dec
25
comment Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
@AsafKaragila Ah, I didn't think of seeing it as an problem with the associative property! Thanks
Dec
25
comment Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
Which is another reason why this will be an embarrassement. The problem is, how do you search for things like this? I tried a couple of searches but couldn't find anything. Perhaps improving my searching skills rather than set theoretic knowledge is what I am after :)
Dec
24
comment Past open problems with sudden and easy-to-understand solutions
This is great! Showing this in an integral calculus course would improve the students' self confidence for sure.
Dec
23
comment Importance of Locally Compact Hausdorff Spaces
Haha, you're right about my guess, if math doesn't work out, I'll go into politics. Thank you, will read the answer(s) properly once this christmas madness passes.
Dec
9
comment Is $(f \circ g)(x) = g(f(x))$ Common in Group Theory?
Thank you, I added my own answer, because I sadly completely missed that the author comments on this just below the definition. Not sure what that says about my attention, since I read the text twice already.
Nov
22
comment $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
Well $E[X|Y]$ where $X,Y$ are random variables is usually defined as $E[X|\sigma (Y)]$, more on en.wikipedia.org/wiki/…
Nov
22
comment $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
@robjohn as BCLC correctly said, it's the random variable $\max \{X-Y,0\}$
Nov
22
comment $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
Well $U=X-Y$, I used that to calculate the auxiliary conditional expectation and used the tower rule (/law of total expectation) to use that to calculate the original.
Nov
20
comment Markov chains: Condtitional independence implies independence?
Oh, right! I will re-read the theorem and then your comment and see if it will then make sense.
Nov
20
comment Markov chains: Condtitional independence implies independence?
@Did well that's the thing, the text was proving independence that way. Or at least that's what I thought months ago. I do not have the text at hand right now, but this is still an unresolved question in my mind, will revisit it as soon as I get home. Thanks for the comment.
Nov
20
comment Independence of Random Variables By Guessing
Yes, that is exactly what was meant, glad it's resolved (and glad you don't have to argue with Did)
Nov
20
comment Independence of Random Variables By Guessing
a) Ah, yes. In my mind you painted a world where, if you're advanced enough, you wouldn't use these cavemen techniques :) b) Ah, this clear things up! You and Did agree, it's just that you misunderstood the question - I am saying you know them immediately up to constants and those constants can be calculated exactly the way you said (please check my post again to see if you agree that we agree!)