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Logistic?

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Thatsme314 asserted in an edit https://en.wikipedia.org/w/index.php?title=Cumulative_distribution_function&oldid=904343161...

"In the case of a continuous distribution, it gives the area under the probability density function from minus infinity to , and is a logistic distribution. Cumulative distribution functions are also used to specify the distribution of multivariate random variables."

Am I misunderstanding what is meant here? The obvious reading is clearly wrong, but perhaps there is a less obvious reading that I am missing. Best clase, it still needed clarification so I deleted it.

--Livingthingdan (talk) 07:36, 7 August 2019 (UTC)[reply]

Serious Error

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There is written that if the cumulative distribution function is continuos then X is absolutely continuos. This is just false, you need F continuos and with continuous derivative!

cadlag

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while the distribution is required to be cadlag? a discussion section on this will be valuable. Jackzhp (talk) 18:46, 15 August 2009 (UTC)[reply]

Moreover there is a tradition here (I suppose because of Kolmogorov's original notation, but I'm not sure) that the CDF should be left continous... Drkazmer Just tell me... 23:01, 2 January 2012 (UTC)[reply]

EDIT: My sincere apologies, but I don't know where else to report this. Unlike other Wikipedia pages, when this page is googled, its title shows up with the first letter of the first word uncapitalized. Try it. Cheers. -- Anonymous user

Complementary Comulative Distribution function

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I assume there is an error after "Proof: Assuming X has density function f, we have for any c > 0", regarding integration limits for E(X) ? —Preceding unsigned comment added by Amir bike (talkcontribs) 05:54, 19 May 2011 (UTC)[reply]

It is said that Markov's inequality states that: However it is only correct in continuous case, as in discrete case Although the Inequality still holds, the current version is weaker than the proper Markov's inequality — Preceding unsigned comment added by Colinfang (talkcontribs) 18:59, 4 March 2012 (UTC)[reply]

The current version is the standard statement of Markov's inequality found in reference books. If there is a stronger result, it could be stated with a citation. If the stronger resuly is still generally known as Markov's inequality, then the Markov's inequality article could be updated as well. But the version in the article (now) states valid conditions under which the results hold. Melcombe (talk) 16:58, 15 April 2012 (UTC)[reply]

Utility

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It would be helpful if in the entry there was a discussion of the utility performing a CDF plot. This would include when to perform one, and what information is learned from performing the CDF plot. What real world applications would this include? Maybe an example would be helpful — Preceding unsigned comment added by 209.252.149.162 (talk) 14:10, 1 August 2011 (UTC)[reply]

Table of cdfs

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I have moved the recently added table of cdfs to here for discussion/revision. The version added was ...

Distribution Cumulative Density function
Binomial B(n, p)  
Negative binomial NB(r, p)  
Poisson Pois()  
Uniform U(a, b)   for
Normal N(µ, )  
Chi-squared  
Cauchy Cauchy(µ, )  
Gamma G(k, )  
Exponential Exp()  

There are several problems here, particularly with inconsistent notations. But there are structural problems in defining the cdfs of the discrete distributions, as the formulae given are only valid at the integer points (within the range of the distribution) and would give incorrect values of the cdf at non-integer values. Also several of the functions involved require definitions/wikilinks. So, if the table is to be included, thought needs to be given to possibly dividing it into discrete/continuous tables and/or adding extra columns. Melcombe (talk) 09:25, 21 October 2011 (UTC)[reply]

@Melcombe: I certainly hope we won't see tables that say "cumulative density function" instead of "cumulative distribution function". Michael Hardy (talk) 18:40, 23 February 2019 (UTC)[reply]

citation needed.... really?

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I think it is a little bit ridiculous to expect a citation that a CDF is càdlàg. It is an almost trivial observation that follows directly from the probability space axioms and the definition of a càdlàg function. Surely this is a routine calculation. --217.84.60.220 (talk) 11:32, 3 November 2012 (UTC)[reply]

Please, for didactic, show area relation

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EXAMPLE

http://beyondbitsandatomsblog.stanford.edu/spring2010/files/2010/04/CdfAndPdf.gif — Preceding unsigned comment added by 187.66.187.183 (talk) 07:25, 3 February 2013 (UTC)[reply]

CDF is definitely LEFT-continuous.

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CDF must be left-continuous, not right as stated on the wiki page. Source: current ongoing University studies, 3 separate professors, books from 4 different authors. — Preceding unsigned comment added by 213.181.200.159 (talk) 07:46, 6 March 2013 (UTC)[reply]

This article follows the convention reached via the link right-continuous. This is "continuous from the right". Perhaps you are thinking of "continuous to the left". 81.98.35.149 (talk) 11:34, 6 March 2013 (UTC)[reply]

I'd like to mention that some textbooks define the CDF as the probability that the random variable X at a given point, F_X(a), that X is strictly less the constant, rather than less than or equal to a. This is where the confusion of this talk section is coming from. --128.42.66.81 (talk) 20:23, 25 August 2016 (UTC)[reply]

Not that redundant?

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The passage that I deleted but which was restored said

Point probability
The "point probability" that X is exactly b can be found as
This equals zero if F is continuous at x.

However, at the end of the section "Definition" it says

In the case of a random variable X which has distribution having a discrete component at a value x0,
where F(x0-) denotes the limit from the left of F at x0: i.e. lim F(y) as y increases towards x0.

What I deleted looks identical to that, except that it doesn't include the sentence This equals zero if F is continuous at x.

I propose that we re-delete it but put the last-mentioned sentence into the existing section.

Okay, no problem. Nijdam (talk) 07:08, 19 April 2013 (UTC)[reply]

cdf notation

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I went through and changed the notation to everywhere in the definition section to try to obtain notational consistency through the article, but the change was reverted by Nijdam with edit summary "Difference between cdf of X and just a cdf". But that conflicts with much notation in the article that uses F(x) for the cdf of X. In the Properties section:

the CDF of X will be discontinuous at the points xi and constant in between:
If the CDF F of X is continuous, then X is a continuous random variable; if furthermore F is absolutely continuous, then there exists a Lebesgue-integrable function f(x) such that
for all real numbers a and b. The function f is equal to the derivative of F almost everywhere, and it is called the probability density function of the distribution of X.

In the Examples section:

As an example, suppose X is uniformly distributed on the unit interval [0, 1]. Then the CDF of X is given by

In the Derived functions section:

In the multivariate case section:

for a pair of random variables X,Y, the joint CDF is given by

So we need to establish consistency of notation -- either use FX every time we mention a cdf "of X", or else never. Your thoughts? Duoduoduo (talk) 15:08, 17 May 2013 (UTC)[reply]

In the literature both and are used, the latter for ease of notation, and only if there is no confusion about the random variable. In the examples you mention there is not always an inconsistency. In the first one you're right, but if it reads: If the CDF F of X ..., it merely states , where is some specified function. Nijdam (talk) 06:11, 18 May 2013 (UTC)[reply]
But there's no confusion anywhere in the article regardless of which is used. So why not use the same one everywhere? Duoduoduo (talk) 12:30, 18 May 2013 (UTC)[reply]

Continuous Random Variables

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Elsewhere on Wikipedia, and in many published books, a continuous random variable has an absolutely continuous c.d.f., not merely continuous as stated in the properties section. I suggest that this page should also state that the c.d.f. is absolutely continuous so that there is a p.d.f. Paulruud (talk) 17:26, 30 March 2015 (UTC)[reply]

Definition as expectation value

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I found this in the introduction of Characteristic function:

The characteristic function provides an alternative way for describing a random variable. Similarly to the cumulative distribution function

( where 1{X ≤ x} is the indicator function — it is equal to 1 when X ≤ x, and zero otherwise), which completely determines behavior and properties of the probability distribution of the random variable X, the characteristic function

also completely determines behavior and properties of the probability distribution of the random variable X. The two approaches are equivalent in the sense that knowing one of the functions it is always possible to find the other, yet they both provide different insight for understanding the features of the random variable.

The notation is so confusing I ask the community a clarification in this page (Characteristic function).

Kind of reciprocity

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There is a theorem stating that for each F nondecreasing right-continuous with limits 0 in -∞ and 1 in +∞, there exists a unique probability mesure μ such that F is a the cdf of a random variable X distributed according to μ. I think this important theorem should be displayed in this page.

GizTwelve (talk) 12:28, 5 October 2017 (UTC)[reply]

Colors in the first graph are too similar

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Please use clearly distinguishable colors, that work also with color vision deficiency. — Preceding unsigned comment added by 88.219.233.164 (talk) 12:31, 11 December 2018 (UTC)[reply]

Connection to measure theory

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I would like to know the full connection to measure theory. It seems if I have a probabilistic measure space and a measurable function it might be the pushforward measure which I guess would be a mapping which for some reason is then restricted to half open sets . But this is original research and feels a bit patchy (unclear to me how to generalize to random variables with values in ) so please if someone who has connected the dots could add the connection. — Preceding unsigned comment added by Rostspik (talkcontribs) 05:58, 27 April 2019 (UTC)[reply]

h — Preceding unsigned comment added by 2.183.8.104 (talk) 05:39, 16 November 2019 (UTC)[reply]

Check Required

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Hello! I am new editor of Wikipedia and I have tried to do some edits and give some citations I hope I have done things right if not pls let me know my mistakes. This would help me a lot Thank you. Stobene45 (talk) 09:49, 2 July 2021 (UTC)[reply]

Hello! ALong with your addition to my talk page I have taken a look at some of your edits. I would like to direct you to the teahouse to actually ask a question about what to add to the article and what not to add to articles (most of which can be found in the Manual of Style). I have reverted your edits however please note that I do see your edits as being in good faith considering you contacted me first saying that you are a new user wanting to learn. Otherwise, I probably would've given you warnings. Blaze The Wolf | Proud Furry and Wikipedia Editor (talk) 18:17, 2 July 2021 (UTC)[reply]
@Stobene45: this is mostly a matter of Wikipedia's accepted style conventions. We don't artificially underline subheadings, but rather use the pre-defined styles generated by the staggered classes of headings; and while italics can be used for emphasis, this should be done very sparingly (see Wikipedia:Manual_of_Style/Text_formatting#Emphasis). As for the citations, references to Google results are not suitable anywhere on Wikipedia - the cited source must be the specific work used, not just a search result containing the work's title. Cheers! --Elmidae (talk · contribs) 16:26, 3 July 2021 (UTC)[reply]