Timeline for Scaling in Mehta's integral
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 24, 2020 at 4:38 | comment | added | Iosif Pinelis | I have now combined the cases $\gamma=o(1/n^2)$ and $\gamma\sim a/n^2$ for real $a>0$ into one case: $\gamma n^2\to a\ge0$. | |
| Feb 24, 2020 at 4:36 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
deleted 600 characters in body
|
| Feb 24, 2020 at 4:29 | comment | added | Pritam Bemis | thank you very much. | |
| Feb 24, 2020 at 4:29 | vote | accept | Pritam Bemis | ||
| Feb 24, 2020 at 4:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 884 characters in body
|
| Feb 24, 2020 at 4:27 | comment | added | Iosif Pinelis | @SolidStatePhysicist : I have added the modification you requested. | |
| Feb 24, 2020 at 4:27 | comment | added | Iosif Pinelis | @XinWang : I have added the modification you requested. | |
| Feb 24, 2020 at 3:18 | comment | added | Pritam Bemis | @IosifPinelis Do you see a way to extend your argument to $\gamma=\mathcal O(1/n^2)$ rather than $o(1/n^2)$? | |
| Feb 24, 2020 at 3:18 | comment | added | Pritam Bemis | @XinWang good point, I guess this point would require some clarification. I think I looked through the proof too fast when I read it and was satisfied to see that it was essentially uniform integrab. + convergence in measure | |
| Feb 24, 2020 at 1:39 | comment | added | Xin Wang | @IosifPinelis I think there is an issue with this answer. The OP claims and asking for a proof that this property holds for $\gamma=1/n^2$. In your answer however, you assume that $\gamma=o(1/n^2).$ This seems to be irrelevant for the uniform integrability, although the property is mentioned there, but it is clearly used to show convergence in probability. I'd be curious to hear what you think? | |
| Jan 23, 2020 at 4:08 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 14 characters in body
|
| Jan 23, 2020 at 4:01 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
deleted 8 characters in body
|
| Jan 23, 2020 at 4:00 | comment | added | Iosif Pinelis | @SolidStatePhysicist : You were right. I had indeed overlooked that the dimension of the integral, $n$, goes to infinity. This is now fixed. Instead of the integration over $\mathbb R^n$, with the variable $n$, we now use the expectation, which is the integration over a fixed background probability space. Instead of dominated convergence, we now use uniform integrability, which complicates the reasoning just a bit. | |
| Jan 23, 2020 at 3:53 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 20 characters in body
|
| Jan 23, 2020 at 3:47 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 253 characters in body
|
| Jan 23, 2020 at 2:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 599 characters in body
|
| Jan 23, 2020 at 1:57 | vote | accept | Pritam Bemis | ||
| Feb 24, 2020 at 3:19 | |||||
| Jan 23, 2020 at 1:56 | comment | added | Iosif Pinelis | @SolidStatePhysicist : I have added the requested details. | |
| Jan 23, 2020 at 1:56 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 338 characters in body
|
| Jan 23, 2020 at 1:47 | comment | added | Pritam Bemis | interesting, but how did you estimate this product? Also, the dimension of the integral changes, which is perhaps a bit different from the usual DCT. | |
| Jan 23, 2020 at 1:44 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |