Timeline for Can we prove the following anti-concentration inequality of polynomials of square Gaussian variables?
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| Nov 30, 2022 at 19:18 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 30, 2022 at 1:51 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 30, 2022 at 1:18 | comment | added | Hermi | @IosifPinelis Does this question make sense now? | |
| Nov 30, 2022 at 1:17 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 30, 2022 at 0:42 | comment | added | Iosif Pinelis | After your last edit, the question seems to have become incomprehensible. I suggest you take time (like a few days) to think about what you actually want and then carefully present your question. | |
| Nov 29, 2022 at 22:38 | comment | added | Hermi | @IosifPinelis So there is no anti-concentration inequality for $N(0,1/n)$ random variables? | |
| Nov 29, 2022 at 21:42 | comment | added | Hermi | @IosifPinelis Thanks. I update my question now. Do you think that right? | |
| Nov 29, 2022 at 21:41 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 29, 2022 at 21:15 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 29, 2022 at 21:12 | comment | added | Iosif Pinelis | What I said previously was assuming that the $X_i$'s were (i) iid $N(0,1)$ as before and (ii) independent of $A$. If (ii) fails to hold, you need to specify the dependence. As for (i), if you replace $N(0,1)$, then it should easy to show that there is no anti-concentration. | |
| Nov 29, 2022 at 21:02 | comment | added | Hermi | @IosifPinelis Oh... I just found a issue. My $X_i\sim N(0, 1/n)$ but not $N(0,1)$. | |
| Nov 29, 2022 at 21:01 | comment | added | Iosif Pinelis | Yes, you can remove the condition. | |
| Nov 29, 2022 at 20:43 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 29, 2022 at 20:36 | comment | added | Hermi | @IosifPinelis Can we just take expectation on the both side? | |
| Nov 29, 2022 at 20:29 | comment | added | Hermi | @IosifPinelis Thanks! Since $\lambda_i$ are random variables following semi-circle law, can we remove this condition probability in that inequality? | |
| Nov 29, 2022 at 20:25 | comment | added | Iosif Pinelis | Yes, you can make this substitution. | |
| Nov 29, 2022 at 19:58 | comment | added | Hermi | @IosifPinelis Also, how about my updated question? Thanks! | |
| Nov 29, 2022 at 19:40 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 29, 2022 at 19:31 | comment | added | Hermi | Ok, I see! So we can take $\epsilon=1/n$, and get my second displayed inequality? | |
| Nov 29, 2022 at 19:26 | comment | added | Iosif Pinelis | The condition $\sum_i a_i=1$ was assumed in that answer without loss of generality. The first displayed inequality in the answer holds for any $a_i>0$. | |
| Nov 29, 2022 at 19:17 | comment | added | Hermi | @IosifPinelis If that concentration inequality is true for my case, I update my thought and questions. Do you think that works? | |
| Nov 29, 2022 at 19:17 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 29, 2022 at 19:09 | comment | added | Hermi | @IosifPinelis I check the proof but he assume that $\sum_i a_i=1$ which is not true in my case... Can we still prove that? | |
| Nov 29, 2022 at 19:09 | history | edited | Hermi | CC BY-SA 4.0 |
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| Nov 29, 2022 at 15:01 | comment | added | Iosif Pinelis | The constant factor in the bound in the linked answer does not depend on the $a_i$'s there. So, the same bound will hold in the present case. | |
| Nov 29, 2022 at 5:17 | history | asked | Hermi | CC BY-SA 4.0 |