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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)