Timeline for WLLN for bootstrap means of stationary ergodic processes?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2022 at 16:44 | vote | accept | zxmkn | ||
| Sep 27, 2022 at 16:41 | answer | added | Anthony Quas | timeline score: 0 | |
| Sep 27, 2022 at 8:21 | comment | added | Peter O. | @Anthony Quas: If your comment answered the question you should post it as an answer, especially to keep this site from seeing this question as unanswered. | |
| Sep 25, 2022 at 23:00 | comment | added | zxmkn | It works! Thank you :) | |
| Sep 25, 2022 at 19:57 | comment | added | Anthony Quas | It seems as though the answer should be yes. I would suggest writing $X_n$ as $Y_n+Z_n$ where $Y_n$ is $X_n$ if $|X_n|\le m(n)^{1/3}$ and 0 otherwise; similarly $Z_n$ is $X_n$ if $|X_n|>m(n)^{1/3}$ and 0 otherwise. Then the Einmal and Rosalsky result applies to the Bootstrap averages of the $Y_n$, so all that remains is to check that the Bootstrap averages of the $Z_n$ approach 0 in probability. I believe that follows from Markov's inequality once you know that $\mathbb E Z_n\to 0$. | |
| Sep 25, 2022 at 10:24 | history | edited | zxmkn | CC BY-SA 4.0 |
Fixed typo (missing square root symbol)
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| Sep 25, 2022 at 10:16 | history | edited | zxmkn | CC BY-SA 4.0 |
deleted 40 characters in body
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| Sep 25, 2022 at 10:10 | history | asked | zxmkn | CC BY-SA 4.0 |