Talk:Bessel's correction

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The Source of the Bias (Intuitive Explanation)[edit]

The text is written in first person and unnecessarily wordy. Not written like an encyclopedia entry. — Preceding unsigned comment added by 2A02:908:1087:60:E0EC:9DBF:195B:76D4 (talk) 16:34, 11 November 2019 (UTC)[reply]

At least one sentence here is wrong[edit]

In the middle of the article, it says "The answer is "yes" except when the sample mean happens to be the same as the population mean."

Say I have the population {1,1,1,1,1,3,3,3,3,3,5,5,5,5,6}. My population mean is 3.06667. If I happen to choose a sample {1,6}, I'll have a sample mean of 3.5, but my sample variance estimated by dividing by N will obviously be larger than the population variance.

Mglerner (talk) 19:14, 1 September 2011 (UTC)[reply]

Yeah, I think it should say that the expected value of the variance of a sample is always smaller than the actual variance. But the sample variance does not always have to be smaller. Only the expected value of the sample variance. Could someone fix this? 193.110.36.16 (talk) 09:46, 19 January 2018 (UTC)[reply]

Tim

There seems to be no problem with the statement in the article. It still holds in your population/sample example

the sample variance is 6.25 which is smaller than 6.437777778, obtained if the variance of the sample is calculated by using the population mean (3.06667), instead of the sample mean (3.5). The proof is sound.

OK, I'm confused...[edit]

How did they get from here:

${\begin{aligned}&={\frac {1}{n-1}}\operatorname {E} \left(\sum _{i=1}^{n}(x_{i}-\mu )^{2}-2({\overline {x}}-\mu )\sum _{i=1}^{n}(x_{i}-\mu )+\sum _{i=1}^{n}({\overline {x}}-\mu )^{2}\right)\\\end{aligned}}$

to here:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \begin{align} & = \frac{1}{n-1}\operatorname{E}\left(\sum_{i=1}^n(x_i-\mu)^2 - 2(\overline{x}-\mu)n (\frac{\sum_{i=1}^nx_i}{n}-\mu) + n(\overline{x}-\mu)^2\right) \\ \end{align} }

??? - 114.76.235.170 (talk) 15:20, 6 August 2010 (UTC)[reply]

Two things were done. One was this:

\sum _{i=1}^{n}({\overline {x}}-\mu )^{2}=n({\overline {x}}-\mu )^{2}.

The index i does not appear within the scope of the summation sign, i.e. the term

({\overline {x}}-\mu )^{2}\,

does not change as i goes from 1 to n. In other words the sum

\sum _{i=1}^{n}({\overline {x}}-\mu )^{2}

is just

\underbrace {({\overline {x}}-\mu )^{2}+\cdots +({\overline {x}}-\mu )^{2}} _{n{\text{ terms}}}\,

hence it is

n({\overline {x}}-\mu )^{2}.\,

The other thing that was done is as follows:

{\begin{aligned}\sum _{i=1}^{n}(x_{i}-\mu )&=\left(\sum _{i=1}^{n}x_{i}\right)-\left(\sum _{i=1}^{n}\mu \right)\\[10pt]&=n\left({\frac {\sum _{i=1}^{n}x_{i}}{n}}\right)-\left(n\mu \right)\\[10pt]&=n\left({\frac {\sum _{i=1}^{n}x_{i}}{n}}-\mu \right).\end{aligned}}:~~~~

I know its not very Wiki-like, but I have to say: Brilliant Article...! Mmick66 (talk) 07:15, 6 May 2011 (UTC)[reply]

And I'm going to have to go even further in non-Wikiness and agree; brilliant! The subtle point about the estimator becoming unbiased for the variance but not for the standard deviation via Jensen's inequality -- absolutely brilliant! Superpronker (talk) 12:45, 2 June 2011 (UTC)[reply]

Bessel usage for standard deviation[edit]

Why is the formula with Bessel's correction used as a default (ie in Matlab) to estimate standard deviation, instead of the unbiased version (such as Unbiased_estimation_of_standard_deviation) to estimate standard deviation? Frankmeulenaar (talk) 06:43, 12 July 2011 (UTC)[reply]

Isn't n*(a/n) = a ?[edit]

In the final formula:

{\frac {1}{n-1}}\left[\sum _{i=1}^{n}\sigma ^{2}-n(\sigma ^{2}/n)\right]

Isn't:

n(\sigma ^{2}/n)=\sigma ^{2}?

And therefore the sum is zero?

Or is that factor outside the summation?

Even if it's outside, why not just put $\sigma ^{2}$ ? — Preceding unsigned comment added by 71.107.55.85 (talk) 08:12, 28 June 2012 (UTC)[reply]

The first term reduces to

n\sigma ^{2}

while second is

\sigma ^{2}

, so the result is not zero. The formula is presumably there because it relates to a formula above, with corresponding terms. Melcombe (talk) 12:46, 28 June 2012 (UTC)[reply]

Ah ok, so it's outside the summation then. A few more steps would make it more clear. — Preceding unsigned comment added by 71.107.55.85 (talk) 07:10, 2 July 2012 (UTC)[reply]

Formula doesn't make sense in context[edit]

In the final section, when dicussing a particular random sampling, the article throws out the formula:

\operatorname {Var} ({\overline {x}})=\operatorname {E} (({\overline {x}}-\mu )^{2})=\sigma ^{2}/n.\,

However, in the context of a single random sampling of the data, this formula doesn't make sense. It seems to want to measure the variance of the sample mean with respect to the real mean. However, in the case of a single random sampling, you just have one sample mean, and one number. Therefore the expected value is just that number.

Or is this formula stating the variance you'd get if you took a number of random samplings, took the sample mean of each of those samplings, and then calculated the variance of the the sample means of several random samplings to the true mean?

If the second is true (and it's the only explanation that makes sense to me), this formula has been thrown a bit of context, given the text above it, that dicusses a single random sampling, and then says "Also", then seems to throw the formula up on the wall, and see if it sticks...

Furthermore, it's unexplained nature makes it very difficult to make the leap to the final formula below it, which essentially justifies the results of the whole page. — Preceding unsigned comment added by 71.107.55.85 (talk) 08:31, 28 June 2012 (UTC)[reply]

This relates to the question above. The formula complained of is set down because it is the thing that is substituted in the earlier formula to get the final result. Melcombe (talk) 12:50, 28 June 2012 (UTC)[reply]

I'm not complaining about the formula - I understand it's trying to link the two parts. I'm complaining about the context - reading from top to bottom (as is common), you're dicussing one random sampling from the population, and then introducting a formula that *appears* to be talking about results from a set of random samplings. But it's unclear whether this is so, or I just misunderstand the formula. Or both. — Preceding unsigned comment added by 71.107.55.85 (talk) 07:12, 2 July 2012 (UTC)[reply]

Putting the -1 in n-1.[edit]

Do I understand that the "real" answer in the example would be 36/5 = 7.2, the observed answer is 16/5 = 3.2 and using Bessel's correction would give: 16/(5-1) = 4? Four being closer to the correct answer of 7.2 than 3.2 which is considered a low value per the article text? Thank you. Fotoguzzi (talk) 11:55, 1 February 2013 (UTC)[reply]

Comment as a very occasional editor.[edit]

I had need to read this article. For me it read well overall. I found the article well set out in its flow of explanation for a person yet to understand this topic. Better than a number of other basic stats articles. Providing this feedback to those who have worked on this article. With my thanks.

CitizenofEarth001 (talk) 10:15, 17 February 2014 (UTC)[reply]

Proof Alternate #3 moved to #1 with intuition unhidden[edit]

The intuitive explanation in proof alternate #3 is very clear for even a lay person to understand, I think this makes it an ideal candidate for being in the main text. MATThematical (talk) 05:00, 26 May 2014 (UTC)[reply]

Why is this factor called Bessel's correction?[edit]

In Whittaker, E.T.; Robinson, G. (1924). Calculus of Observations. p. 206. {{cite book}}: External link in |title= (help) a reference is made to Bessel's formulae. No work or date of publication is given for Bessel. The factor for π observations and ρ unknowns appears is Gauss' Theory of the Combination of Observations Least Subject to Error published in 1823. --Jbergquist (talk) 06:00, 13 December 2014 (UTC)[reply]

Bessel is known for the introduction of the term "probable error" in 1815. The translation of the passage cited is,

"The probable error of a single observation, to judge by the real errors occurring = 1s.067, and therefore the probable error of the final result = 0s.154. The basis of this estimate of the probable error is based on the development given by Gauss of the probability of committing an error of a given size; their communication I must save for another occasion."

This statement is in Bessel, Ueber den Ort des Polarsterns. --Jbergquist (talk) 23:47, 13 December 2014 (UTC)[reply]

In Bessel, 1818, Fundamenta astronomiae pro anno MDCCLV it is stated that the probable error is greater than expected. --Jbergquist (talk) 11:02, 16 December 2014 (UTC)[reply]

The answer is yes except when the sample mean happens to be the same as the population mean.[edit]

jump in the function? 1/n-1 is chosen arbitrarily to be different from 1/n (the whole/full set), it could be defined as 1/n+1, but it is always better to estimate the larger error rather than the smaller. — Preceding unsigned comment added by 82.140.130.86 (talk) 10:58, 27 January 2019 (UTC)[reply]

Identity[edit]

s_{n}^{2}={\frac {1}{n}}\sum _{i=1}^{n}\left(x_{i}-{\overline {x}}\right)^{2}={\frac {\sum _{i=1}^{n}\left(x_{i}^{2}\right)}{n}}-{\frac {\left(\sum _{i=1}^{n}x_{i}\right)^{2}}{n^{2}}}

That seems false to me :

(a-b)^{2}\neq a^{2}-b^{2}

Perhaps would it be nice to include some more steps to better understand ? — Preceding unsigned comment added by Neo 13 (talk • contribs) 18:31, 8 January 2020 (UTC)[reply]

Proof of correctness[edit]

More concisely: Let ${\rm {Var\,}}X=\sigma ^{2}$ . Then, since the uncorrected variance is $s_{n}^{2}={\overline {x^{2}}}-({\overline {x}})^{2}$ we have that

${\rm {E\,}}s_{n}^{2}={\rm {E\,}}X^{2}-{\big (}{\rm {Var\,{\overline {X}}}}+({\rm {E\,}}{\overline {X}})^{2}{\big )}.$

But ${\rm {E\,{\overline {X}}={\rm {E\,X\,}}}}$ , ${\rm {Var\,{\overline {X}}=\sigma ^{2}/n\,}}$ and $\,{\rm {E\,}}X^{2}-({\rm {E\,}}X)^{2}=\sigma ^{2}$ , so

${\rm {E\,}}s_{n}^{2}=\,{\frac {n-1}{n}}\,\sigma ^{2}\,.$

Therefore we have to multiply $s_{n}^{2}$ by a factor ${\frac {n}{n-1}}$ in order to get an unbiased estimation, which yields the usual variance $s^{2}$ . José Luis Arregui (talk) 21:40, 2 May 2023 (UTC)[reply]

Alternate proof #3 is dubious[edit]

I believe alternate proof #3 substitutes an incorrect expression for the population standard deviation σ. It sums over on the same number of elements as the sample but should sum over all the elements in the population (not the same as n) Antoine1619 (talk) 06:50, 12 December 2023 (UTC)[reply]

Reversion request: Restore alternative proof #3 present before 21 Jan 2024[edit]

The proof of correctness alternative #3 was better accessible for those with highschool backgrounds, and I hence request restoring it to before its deletion on 21 Jan 2024 please. CalebSohSweeKai (talk) 06:01, 8 February 2024 (UTC)[reply]