Timeline for estimate the error term in CLT
Current License: CC BY-SA 2.5
20 events
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| Aug 4, 2016 at 13:22 | vote | accept | gondolier | ||
| Jul 28, 2016 at 16:29 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Jul 28, 2016 at 15:45 | history | edited | Iosif Pinelis |
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| Sep 20, 2010 at 20:51 | history | edited | gondolier | CC BY-SA 2.5 |
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| Sep 20, 2010 at 19:14 | comment | added | Yaroslav Bulatov | I take it back, if Stein considers this, it must be interesting :) BTW, equation 37 in his book also requires f' to be bounded. | |
| Sep 20, 2010 at 15:42 | answer | added | gondolier | timeline score: 0 | |
| Sep 20, 2010 at 14:54 | answer | added | Mark Meckes | timeline score: 3 | |
| Sep 20, 2010 at 1:07 | answer | added | user6096 | timeline score: 2 | |
| Sep 19, 2010 at 22:54 | comment | added | gondolier | sure. let's focus on the case where the sum is properly centralized and normalized. | |
| Sep 19, 2010 at 20:54 | comment | added | Yaroslav Bulatov | OK, maybe "uninteresting" behavior rather than unusual. If mean is not 0, limiting distribution is a delta distribution. Also, CLT-type theorems for (properly scaled) distribution of $f(\bar{n})$ don't require f to be bounded. | |
| Sep 19, 2010 at 20:24 | comment | added | R Hahn | Is $f$ bounded over all of $\mathbb{R}$, because the two polynomial examples discussed so far aren't, right? Are the $Z_k$ uniform on $\lbrace -1, 1 \rbrace$ or do you want to consider more general cases, as the parenthetical at the end of para two suggests? | |
| Sep 19, 2010 at 20:13 | comment | added | gondolier | why is the behavior unusual? $f(\sqrt{m} \bar{X})$ converges in distribution to the image measure of standard normal under $f$. This only requires continuity of $f$. | |
| Sep 19, 2010 at 20:04 | comment | added | Yaroslav Bulatov | Counterpart of Central Limit Theorem gives the distribution of $\sqrt{n}f(\bar{X})$. Distribution of $f(\sqrt{n}\bar{X})$ seems to have unusual behavior, for instance if $Z_i$'s are uniform on {0,1}, mean of $X_m$ goes to infinity, but because $f$ is bounded, distribution of $f(X_m)$ gets squished into a delta function | |
| Sep 19, 2010 at 19:25 | comment | added | gondolier | well if something converges to Gaussian weakly, then all its moments must converge. | |
| Sep 19, 2010 at 19:18 | comment | added | Yaroslav Bulatov | oh...Z's are symmetric...so the error term is 0 regardless of m, right? | |
| Sep 19, 2010 at 19:12 | comment | added | gondolier | No. I meant $f(X_m)$. For your $f$ the error term is zero. | |
| Sep 19, 2010 at 19:02 | comment | added | Yaroslav Bulatov | Do you perhaps mean to look at $f(\bar{X})$ instead of $f(X_m)$? For $f(x)=x^2$ your error term doesn't converge to 0. | |
| Sep 19, 2010 at 16:36 | answer | added | Nate Eldredge | timeline score: 1 | |
| Sep 19, 2010 at 10:28 | history | edited | gondolier | CC BY-SA 2.5 |
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| Sep 19, 2010 at 10:02 | history | asked | gondolier | CC BY-SA 2.5 |