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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
22
revised $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
added 28 characters in body
Nov
22
revised $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
added 187 characters in body
Nov
22
asked $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$
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
accepted Independence of Random Variables By Guessing
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!)
Nov
20
comment Is this a proof of $E\int^b_a f dZ = 0$?
@TheBridge details added
Nov
20
revised Is this a proof of $E\int^b_a f dZ = 0$?
added details
Nov
20
comment Independence of Random Variables By Guessing
@MichaelHardy forgive for being a bit slow, but I don't understand what the sentence "is it obvious from the density [...] only if you know the density". Sounds very tautological to me, what am I missing?
Nov
20
comment Independence of Random Variables By Guessing
Two things I don't understand. a) You say that two is common only elmentary probability, what does that mean? I mean, if you have such an elegant approach, why not use it in what you call "advanced probability" (also it's worth noting this course is exactly the course that presupposes measure theory and prerequisite knowledge of mt-based probability theory) b) How does the comment on the third statement contradict it? I don't see that. To me it seems that you just repeated my lecturer's reasoning. Thank you!
Nov
20
revised Independence of Random Variables By Guessing
added 282 characters in body
Nov
19
comment Is this a proof of $E\int^b_a f dZ = 0$?
@TheBridge, thanks for the comment. Once I get home, I will try to provide more details.
Nov
17
asked Is this a proof of $E\int^b_a f dZ = 0$?
Nov
16
comment Why is $Y e^{i\omega t}$ called an “elementary process”
@avid19 Well it certainly doesn't seem to be relevant, this process seems to circle around the circle with diameter $Y$, don't see the step function hidden in there
Nov
16
asked Why is $Y e^{i\omega t}$ called an “elementary process”
Nov
11
awarded  Yearling
Nov
1
comment Independence of Random Variables By Guessing
If you make that an answer, I'll gladly accept it. Of course, if someone writes some extra thoughts that would further clarify this, it'd be even better, but as the question stands, this is sufficient as an answer.