Timeline for Is there monotonicity of measure concentration?
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| Oct 8, 2013 at 12:32 | comment | added | R W | @Dustin G. Mixon: I can just repeat my previous comment. The way you formulated the problem it only asks about distributions of $\overline X$ and $\overline Y$ separately, so that their joint distribution is irrelevant. | |
| Oct 8, 2013 at 12:13 | comment | added | Dustin G. Mixon | @R.W. - If we realize $X$ and $Y$ in the same probability space, can't we draw the $X_i$'s and $Y_i$'s with $2n$ independent outcomes of the probability space? I think the difficulty would not be independence between the $X_i$'s and $Y_i$'s, but rather a more exotic joint distribution. | |
| Oct 8, 2013 at 9:05 | comment | added | R W | @Joris Bierkens: No. Let $\overline X= (X_i)$ and $\overline Y = (Y_i)$ be the corresponding sequence valued random variables. The question involves just the distributions of $\overline X$ and $\overline Y$, but not their joint distribution. | |
| Oct 8, 2013 at 6:44 | comment | added | Joris Bierkens | Can something be said when all (X_i) and (Y_i) are assumed to be independent? | |
| Oct 8, 2013 at 4:42 | vote | accept | Dustin G. Mixon | ||
| Oct 8, 2013 at 4:34 | comment | added | R W | No, they don't. By the way, if your $X$ and $Y$ are compactly supported, one can always make them positive just by adding an appropriate constant. | |
| Oct 8, 2013 at 4:20 | comment | added | Dustin G. Mixon | So $X$ and $Y$ don't need to be nonnegative? | |
| Oct 8, 2013 at 3:59 | history | edited | R W | CC BY-SA 3.0 |
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| Oct 8, 2013 at 3:54 | history | answered | R W | CC BY-SA 3.0 |