# Talk:Berry–Esseen theorem

WikiProject Mathematics (Rated Start-class, Low-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 Start Class
 Low Importance
Field:  Probability and statistics
WikiProject Statistics (Rated Start-class, Mid-importance)

This article is within the scope of the WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page or join the discussion.

Start  This article has been rated as Start-Class on the quality scale.
Mid  This article has been rated as Mid-importance on the importance scale.

The formula Fn is INCORRECT.

Pls check against the German's page (http://de.wikipedia.org/wiki/Satz_von_Berry-Ess%C3%A9en) or the source already listed in "External links" (Chen, Po-Ning (2002). Asymptotic Refinement of the Berry-Esseen Constant)

The correct one is:

```  Fn = Yn / (standard deviation/sqrt(n))
```

[Someone else] You are wrong: Fn is cdf of (X1+...+Xn)/(stdev*sqrt(n)) = Yn*sqrt(n)/stdev That's what on the German page you mention, but you read too quickly. —Preceding unsigned comment added by 86.212.95.141 (talk) 09:22, 26 January 2008 (UTC)

All three versions of the formula given above are equivalent. (The discussion above has been changed by several anon editors, which is why the "disagreement" no longer makes any sense.) I originally gave the formula in the first form above because that's closer to the way the standardized sample mean is written in introductory statistics textbooks currently used in the U.S. (where I'm from). Some books place the sqrt(n) in the numerator, as in the third form; the second form is talking about the sum, not the mean. - dcljr (talk) 07:52, 16 April 2009 (UTC)

## Spelling of Esseen

I claim that the correct spelling of Gustav Esseen's name has no accent on the 'e'. This is a tricky point because it is written with an accent in many sources. He's definitely Swedish; I don't know if this helps determine the truth. Credible citations with no accent:

[[1]]

[[2]]

[[3]] —Preceding unsigned comment added by 67.186.15.220 (talk) 00:39, 26 November 2007 (UTC)

I asked to Svante Janson, a former student of Esseen, and a very reliable (among many other qualities) mathematician, about the correct spelling. Here is the answer
"Carl-Gustav Esseen" is correct.
We also had a professor Matts Essén at the department. (Actually, they were distantly related; different parts of the family adopted different spellings at some time long ago. Esséen is incorrect for both of them.)
so we are as close to 100% credible as possible, there. The only question remaining is "is the incorrect spelling so frequent in the literature that the correct spelling would prevent interested readers from finding the article in Wikipedia ?" I think the answer is "no", and that we should rename the article according to the correct spelling very soon.--193.50.42.138 (talk) 16:24, 16 January 2009 (UTC)
I agree with this (the original research part notwithstanding ;) and have moved the page. Note that the German article has already been moved to the spelling "Esseen" and their original with the accent deleted. We should keep the accented version as a redirect, however. - dcljr (talk) 08:02, 16 April 2009 (UTC)

## X_1 =? X_j

Hi everybody,

Under "Identically distributed summands" it says E(X_1) = mu and E(|X_1|^3) = rho, should that in fact be X_j for j in {1, ..., n}?

Right now it only defines moments for the _first_ variable being summed. Surely this should be for every variable?

Salutacions 88.128.80.6 (talk) 07:29, 17 May 2013 (UTC)

If they are "identically distributed", then every Xj has the same distribution (as X1). - dcljr (talk) 08:54, 20 October 2015 (UTC)

## External links modified

Hello fellow Wikipedians,

I have just modified one external link on Berry–Esseen theorem. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:

When you have finished reviewing my changes, please set the checked parameter below to true or failed to let others know (documentation at `{{Sourcecheck}}`).

An editor has reviewed this edit and fixed any errors that were found.

• If you have discovered URLs which were erroneously considered dead by the bot, you can report them with this tool.
• If you found an error with any archives or the URLs themselves, you can fix them with this tool.

If you are unable to use these tools, you may set `|needhelp=<your help request>` on this template to request help from an experienced user. Please include details about your problem, to help other editors.

Cheers.—InternetArchiveBot 18:02, 31 October 2016 (UTC)

Seems fine. - dcljr (talk) 23:51, 1 November 2016 (UTC)