Timeline for Where does this coupling result use independence when bounding total variational distance?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2023 at 14:40 | comment | added | AspiringMat | Thanks I accepted it (I previously upvoted it) | |
| Sep 12, 2023 at 14:40 | vote | accept | AspiringMat | ||
| Sep 12, 2023 at 3:34 | comment | added | AspiringMat | I see I misunderstood their proof. They were actually doing what you're doing (but it's not as clear as your writing). Thank you! | |
| Sep 12, 2023 at 2:00 | comment | added | Iosif Pinelis | @AspiringMat : The $X_i$'s are independent and the $X_i^*$'s are independent and each $X_i^*$ equals $X_i$ in distribution. So, $(X_1^*,\dots,X_n^*)$ equals $(X_1,\dots,X_n)$ in distribution. So, $S_n^*=X_i^*+\dots+X_n^*$ equals $S_n=X_i+\dots+X_n$ in distribution. | |
| Sep 12, 2023 at 1:55 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 71 characters in body
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| Sep 12, 2023 at 1:54 | comment | added | AspiringMat | Thanks for the answer. I'm still a bit confused how independence implies $S_n^\ast$ equals $S_n$ in distribution? (I just learned about coupling recently, so this might be an obvious question). | |
| Sep 12, 2023 at 1:47 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |