Timeline for Space of distributions on $[0,1]^2$: weakly compact or not?
Current License: CC BY-SA 4.0
15 events
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| Nov 15 at 11:53 | comment | added | tom jerry | I mean for a PDF $g$ on $[0,1]^2$, we can calculate its integer $\int_A g$ on the measurable subset $A$--and this value shows the proportion of $g$ distributed on $A$ compared to on $[0,1]^2$. For $f$ with $\int f>0$, the distribution (with respect to $f$) 's integer on $[0,0.5]\cdot [0,0.5]$ is just $\int_[0,0.5]\cdot [0,0.5] f/\int f$. However, when $\int f=0$ (as your example), could we do a samilar thing to know the proportion of $f$ distributed on $[0,0.5]\cdot [0,0.5]$ compared to on $[0,1]^2$? Thanks very much! @NateRiver | |
| Nov 15 at 11:09 | comment | added | Yemon Choi | @tomjerry You have already asked this as a new question, please don't use comment threads to ask new questions. | |
| Nov 15 at 11:04 | comment | added | Nate River | Sorry, what do yoy mean by calculate it’s integer? @tomjerry | |
| Nov 15 at 9:42 | comment | added | tom jerry | May I ask a question: For a distribution which is supposed on a 0-measured subset of $[0,1]^2$, how could we calculate its integer on a measureable subset of $[0,1]^2$? i.e., let $f:[0,1]^2\to [0,1]$ be a measureable function with $\int f=0$. Could we consider the distribution as something like $f(x,y)/(\int f)$ and calculate its integer on $[0,0.5]\cdot [0,0.5]$? Thank you a very much for your time! | |
| Oct 17 at 14:07 | vote | accept | tom jerry | ||
| Oct 11 at 14:55 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 11 at 14:30 | comment | added | Nate River | That is also right I believe. @tomjerry | |
| Oct 11 at 14:22 | comment | added | tom jerry | Thank you, River! And thus, is it wrong to view $\mathcal{X}$ as a subset of $L_1([0,1]^2)$ since there exist the counterexample as you give in your answer? | |
| Oct 11 at 14:01 | comment | added | Nate River | Yes to both! The set of all measurable functions as you wrote has a natural interpretation as the set of distributions admitting a joint density. @tomjerry | |
| Oct 11 at 12:43 | comment | added | tom jerry | Thanks for your answer! So, does the set of all measurable functions on $[0,1]^2$ a subset of $\mathcal{X}$? And does $\mathcal{X}$ equals to the space of all probability distributions on $[0,1]^2$? | |
| Oct 11 at 11:21 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 11 at 11:14 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 11 at 11:06 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 11 at 10:59 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 11 at 10:53 | history | answered | Nate River | CC BY-SA 4.0 |