Dahn Jahn
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 Dec 21 accepted When does continuity with probability one imply mean-square continuity Dec 21 accepted Is $(f \circ g)(x) = g(f(x))$ Common in Group Theory? Dec 21 asked Importance of Locally Compact Hausdorff Spaces Dec 9 revised Is $(f \circ g)(x) = g(f(x))$ Common in Group Theory? more accurate 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. Dec 9 answered Is $(f \circ g)(x) = g(f(x))$ Common in Group Theory? Dec 9 asked Is $(f \circ g)(x) = g(f(x))$ Common in Group Theory? Nov 28 answered Chebyshev’s Theorem and Lightbulbs Nov 28 awarded Informed Nov 28 awarded Announcer 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 revised $E[Y^2| (X-Y)^+]$ for $X,Y\stackrel{iid}\sim Unif(0,1)$ edited body 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)