Talk:Chebyshev's inequality

message: Let Χ be the mathematician in question until you sort out his name =p Then there are at least two other inequality-type Χ's theorems.
One is aka Bertrand's postulate.
Another was a step toward prime number theorem: if π(n) is the number of primes not exceeding n, then 0.92\frac n{\ln n}<\pi(n)<1.11\frac n{\ln n} [Yaglom and Yaglom problem 170; vaguely mentioned at Mathworld]. 142.177.19.171 19:12, 13 Sep 2004 (UTC)
thread_title: Other inequalities

message: I felt that the intoductory paragraph was excessively dense and technical:
:'Chebyshev's inequality (also known as Tchebysheff's inequality, Chebyshev's theorem, or the Bienaymé-Chebyshev inequality) is a theorem of probability theory named in honor of Pafnuty Chebyshev. Chebyshev's inequality gives a lower bound for the probability that a value of a random variable of any distribution with finite variance lies within a certain distance from the variable's mean; equivalently, the theorem provides an upper bound for the probability that values lie outside the same distance from the mean.
It's quite correct, but I was afraid it would be difficult for a general audience to understand.  This seemed a shame, since the importance of the theorem can be readily understood by a general audience.  So I inserted a nontechnical paraphrase:
:Chebyshev's inequality (also known as Tchebysheff's inequality, Chebyshev's theorem, or the Bienaymé-Chebyshev inequality) is a theorem of probability theory. Simply put, it states that in any data sample, nearly all the values are close to the mean value, and provides a quantitiative description of "nearly all" and "close to".
:The theorem gives a lower bound for the probability that a value of a random variable of any distribution with finite variance lies within a certain distance from the variable's mean; equivalently, the theorem provides an upper bound for the probability that values lie outside the same distance from the mean.
I thought that people would be more likely to appreciate the meaning of the second paragraph if the idea was set up in less technical language first. 
-- Dominus 14:41, 11 August 2005 (UTC)
thread_title: Introductory Paragraph

message: The second statement is not true for arbitrary nondecreasing nonnegative extended valued measurable g.  This is because g composed with f need not be measurable if g and f are measurable and therefore the integral on the right does not make sense.  If g is continuous the theorem holds since the inverse image of g is open and the inverse image of this under g of this open set will by definition by measurable.    — Preceding unsigned comment added by 73.36.169.23 (talk) 00:58, 24 June 2014 (UTC)
replies: text: Except that that is called the Markov inequality.  The Chebyshev inquality is about second moment about the mean. There is a Gauss inequality for the second moment about the mode. —The preceding unsigned comment was added by 81.151.197.86 (talk) 11:37, 14 April 2007 (UTC)
thread_title:  measure theoretic statement 

message: eh?  —Preceding unsigned comment added by 68.161.204.136 (talk) 18:04, 17 December 2007 (UTC)
thread_title:  whats the difference between chebyshev's inequality and chebyshev sum inequality 

message: I removed the statement that Chebyshev was the first person to prove the inequality. Bienayme proved it (as I recall) some 20 years earlier. For those who care, here is a little more history about the inequality, and about who the inequality "should" be named after. Markov, who was Chebyshev's student, wrote a letter saying that Chebyshev deserved the credit because he understood its purpose, which was to lay the groundwork for stronger probabilistic inequalities (that culminated in the central limit theorem).
Markov also wrote that Bienayme had just proved the bound to refute something or other that Cauchy had claimed. --AS314 (talk) 15:46, 22 February 2008 (UTC)
thread_title:  Who first proved it? 

message: The page An inequality on location and scale parameters refers to a simple corollary of the one-sided version of Chebyshev's inequality and has a misleading (or at least non-standard) name.  Searching around I can't find a "real" name for the theorem proven on that page; rather, everyone simply proves it as a natural result of Chebyshev's inequality.  Romanempire (talk) 12:09, 11 June 2008 (UTC)
replies: text: Keep as separate article. The inequality is about different quantities than involved in Chebyshev's inequality ... median as opposed to probabilities of typical values. By "everyone simply proves it as a natural result of Chebyshev's inequality", I guess you mean everyone except whoever put the article up. It is good to have a direct proof. Are there any elaborations of that would allow tighter bounds to be formulated? Melcombe (talk) 13:24, 11 June 2008 (UTC)
replies: text: ...and there isn't a proof given of the "One-sided Chebyshev inequality". Melcombe (talk) 13:37, 11 June 2008 (UTC)
replies: text:  shows that tighter bounds are impossible, ie \frac{|\mu - M|}{\sigma} can be made as close as possible.  That reference and  seem to call the result something like "a median-mean inequality." Romanempire (talk) 11:34, 12 June 2008 (UTC)

text: The problem here is that the other page seems to be violating WP:NOR.  Wikipedia is not a probability textbook.  We can't just have pages called "a theorem on ..." with non-standard results and original proofs. Romanempire (talk) 11:18, 12 June 2008 (UTC)
thread_title:  Merger proposal 

message: I noticed this is not the form of Chebyshev's inequality I am familiar with.  The inequality that occurs most often in analysis would is what Wikipedia refers to as Markov's inequality.  The terminology seems to vary a bit in probability as well (for example Krylov's book on Introduction to Diffusion processes for a different usage then here.) Should we make a comment about different usages of the phrase.
It might also make a lot of sense to merge the articles since the two concepts are so closely related.  Thenub314 (talk) 11:51, 4 November 2008 (UTC)
thread_title:  Chebyshev's inequality in mathematics. 

message: In the measure theory versions,
for the choice of g(t) so that the extended version implies the first version might need a little tweaking.
g(t)=\begin{cases}t^2&\text{if }t\geq0\\
0&\text{otherwise,}\end{cases}
is decreasing on the range (0,1), which violates the precondition that g(t) be nondecreasing on the range of easily chosen f, such as f(t)=t. Did I miss something?
: Sorry, but the function g(t) you wrote is definitely non-decreasing on (0,1). --Quantrillo (talk) 15:10, 15 December 2020 (UTC)
thread_title:  g(t) decreasing on some intervals? 

message: If (X, Σ, μ) is an arbitrary probability space, then there is no guarantee that the event { x∈X : |f(x)| ≥ t } is even measurable. I think we need Σ to be Borel.  // stpasha »  01:53, 5 November 2010 (UTC)
replies: text: The set \textstyle \{x:|f(x)|\ge t\} is measurable, because h:x\mapsto |f(x)| is the composition of measurable functions, thus measurable, and our set is just h^{-1}([t,\infty]), i.e., the pre-image of a measurable set via a measurable function. --Quantrillo (talk) 15:18, 15 December 2020 (UTC)
thread_title:  Error in measure-theoretic formulation 

message: The following quote is not correct: "the proportion which are between 600 and 1400 words (...) must be more than 3/4".  One can say that on average over many samples, or for an infinitely high pile of articles (!) the proportion between 600 and 1400 words must be more than 3/4, but for any particular finite pile Chebyshev's inequality cannot guarantee anything about the contents.  The inequality is a statement about probability distributions, not about finite statistical samples.  —Preceding unsigned comment added by 62.141.176.1 (talk) 16:44, 4 February 2011 (UTC)
thread_title: Probabilistic statement interpretation wrong

message: More exact inequality for random values with normal distribution (average m and variance s^2)
P(|X-m| > ks) < sqrt(2/pi) * exp(-(k^2)/2) / k — Preceding unsigned comment added by 81.162.76.85 (talk) 11:54, 21 June 2012 (UTC)
thread_title:  Exact inequality for Probabilistic statement 

message: Sitting comparing the Wiki page with reference [30], there seem to be several errors.
  - Bhattacharyya uses normed central moments, not normed raw moments.
  - The second term in the denominator is squared.  — Preceding unsigned comment added by 203.110.235.129 (talk) 05:43, 7 February 2013 (UTC)
replies: text: I think Wiki has the first correct: the inequality as stated requires the mean to be zero. DrMicro (talk) 11:37, 18 March 2013 (UTC)
thread_title:  Error in Bhattacharyya's inequality? 

message: The inequality in this section is the same as given higher up under the heading "higher moments".  As such it is rather trivial and very old and I don't think it deserves its own section and modern attribution.  Anyone disagree? McKay (talk) 07:18, 8 March 2013 (UTC)
replies: text: Pardon me if I have missed this but I don't think this inequality is given under the higher moments section. The inequality is somewhat tighter than the one given under higher moments.DrMicro (talk) 11:26, 18 March 2013 (UTC)
thread_title:  Mitzenmacher and Upfal's inequality 

message: Either some previous editors or me has had some confusion when it comes to the effects of a finite population size. As I see it, if you have a finite sample and calculate the sample mean and sample standard deviation, you can choose to see this finite sample as a distribution in of itself. And thus 99% of the samples must lie within 10 standard deviations of the sample mean. 
The article by Saw et. al. which is discussed in the section on finite samples seems to have been misunderstood. The situation it deals with is when you have taken a sample of n elements and wants to use the sample mean and sample deviation to get some bounds for what you can expect of a new element drawn from the same distribution. It is not an alternative to Samuelson's inequality. MathHisSci (talk) 11:02, 30 July 2015 (UTC)
thread_title:  Finite sample size 

message: The subsection "Chernoff bounds" says: "If the random variables may also be assumed to be independently distributed ..."
I do not understand what "variables" are there. Chebyshev's inequality speaks of a single variable, not many variables. --Erel Segal (talk) 13:03, 14 January 2016 (UTC)
replies: text: You are right; it makes no sense.  It has been like that since it was first inserted and I don't see that it corresponds immediately to anything in the source added later.  It is something about a variable which is the sum of independent variables, but exactly what I don't know.  So I'm deleting it.  If anyone figures it out, please feel free to put it back with the missing details and source. McKay (talk) 03:47, 15 January 2016 (UTC)
thread_title:  Subsection on Chernoff bounds unclear 

message: The section Chebychev's inequality#Notes says 
:For any symmetrical unimodal distribution:
* approximately 5.784% of the distribution lies outside 1.96 standard deviations of the mode
* 5% of the distribution lies outside 2√10/3 (approximately 2.11) standard deviations of the mode
Such a blanket statement about all symmetric uniform distributions cannot possibly be right, and indeed it is not even close to right for a rectangular distribution modified slightly to make it slightly higher at the mean, in which case the percentages should be 0% and 0%. So I'm removing this passage. Loraof (talk) 17:50, 1 February 2016 (UTC)
thread_title:  Removing nonsense assertion 

message: The section Chebyshev's inequality#Vector_version says that Ferentinos showed the inequality for an arbitrary norm, but Ferentinos (1982) only shows the inequality for the Euclidean norm.  I don't think it's true for the max-norm, for example (there's a number of inequalities on the max-norm that are like the Chebyshev inequality, but I've never seen this one).  I've never edited Wikipedia before so I'm not sure what to do.  Just noticed the error (or missing citation of the proof) and thought I'd pass the info along. Best,
Zffff (talk) 23:07, 18 February 2016 (UTC)
replies: text: You can just go right ahead and edit the page. Thanks for pointing out the error, I made an edit. MathHisSci (talk) 15:18, 28 February 2016 (UTC)
thread_title:  Ferentinos inequality 

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thread_title:  External links modified 

message: Currently the article opens with
:In probability theory, Chebyshev's inequality ...  guarantees that, for a wide class of probability distributions, "nearly all" values are close to the mean—the precise statement being ....
This is not true for any reasonable concept of "close to". For example, if P(X= –10100) = 1/2 = P(X=10100), which is in the wide class because the mean and variance exist, no value is close to the mean, nor does the inequality claim otherwise. So I'm going to reword this. Loraof (talk) 15:36, 30 June 2017 (UTC)
thread_title:  Bad opening sentence 

message: Shouldn't Ferentinos' vector version read 
 \Pr(\| X - \mu \| \ge k \| \sigma \|) \le \frac{ 1 } { k^2 }. 
(instead of ||\sigma^2||) ? See page 8 of his paper and the dimension 1 case. Wurtemberg (talk) 14:27, 18 October 2018 (UTC)
replies: text: Looking at the paper, the statement actually uses σ for the variance, which is probably why it's like that here.  However, based on the rest of the paper, and the simpler case, that's probably just a typo, so I'll go ahead and change it.  –Deacon Vorbis (carbon • videos) 20:24, 20 October 2018 (UTC)
thread_title:  Vector version 

message: The formula after the sentence "Olkin and Pratt derived an inequality for n correlated variables" has capital "U" in it, but it is never defined, as far as I see. — Preceding unsigned comment added by 194.166.191.94 (talk) 19:40, 29 October 2018 (UTC)
thread_title:  Error in "Multivariate case"? 

message: In one line P, in the other Pr ... WTF?  — Preceding unsigned comment added by 31.183.229.101 (talk) 22:33, 4 January 2020 (UTC)
replies: text: WP:SOFIXIT. –Deacon Vorbis (carbon • videos) 22:43, 4 January 2020 (UTC)
thread_title:  This article is trash 

message: {\displaystyle K(t)=\log \left(\operatorname {E} \left(e^{tx}\right)\right).}
should be
{\displaystyle K(t)=\log \left(\operatorname {E} \left(e^{tX}\right)\right).}
2A01:CB0C:CD:D800:2D20:85BA:B0A9:5626 (talk) 08:49, 13 February 2021 (UTC)
thread_title:  {\displaystyle K(t)=\log \left(\operatorname {E} \left(e^{tx}\right)\right).} 

message: In the measure theoretic statement  the proposition that the measure of the inverse image of the set of real numbers superior or equal to the real number t is inferior or equal to
to the division of the integral g(f) and the function g is false , the function g the square of identity is a counter example  because there is R integrable functions that converge slower to 0 at infinity.  — Preceding unsigned comment added by Boutarfa Nafia (talk • contribs) 12:34, 23 March 2022 (UTC)
thread_title:  Discussion 

message: In fact the proposition can be true under the condition that g is monotonous and do not take the value of 0 on R and it should not be g(f) inside the integral but bur g multiplied by f.  — Preceding unsigned comment added by Boutarfa Nafia (talk • contribs) 12:38, 23 March 2022 (UTC)
thread_title:  Notification