Dahn Jahn
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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.