Talk:Pareto distribution/Archive 1

Archive 1

Existence of the mean and the variance

Technically, when the exponent is < 1, the mean "does not exist"; "is infinity" is slightly off. Similar comment for < 2, variance.

I don't think "is infinity" is "slightly off". There are distributions like the Cauchy distribution for which the expected value does not exist even if one allows ∞ as the expected value, and there are those like the Pareto distribution for which "is ∞" makes sense. Michael Hardy (talk) 00:21, 22 July 2009 (UTC)

Error in CDF formula for Pareto

I believe that the expression for the cumulative distribution function has an error. It currently reads

cdf =$1-\left({\frac {x_{\mathrm {m} }}{x_{\mathrm {m} }+x}}\right)^{k}\!$

and should read

cdf =$1-\left({\frac {x_{\mathrm {m} }}{x}}\right)^{k}\!$

This is perhaps part of the confusion arising out of not shifting the origin to x_m. There should also be a reference to the excellent (highly technical) article in mathworld: http://mathworld.wolfram.com/ParetoDistribution.html

Unless I get, within a short period of time, some indication that I am wrong, I will change it in the main article.

69.15.90.194 (talk) 21:03, 22 July 2009 (UTC) Can anyone expand on this^ Is the xm in the denominator a shifting term for the origin? I can't find any references to this anywhere...

It's not clear to mean what changes are proposed. What is written above says the article "now" gives the version in which the lower bound is zero. That may have been the case in the past, but it's not now. Michael Hardy (talk) 21:23, 22 July 2009 (UTC)

I got the wrong PDF?

I Changed the

cdf =$1-\left({\frac {x_{\mathrm {m} }}{x}}\right)^{k}\!$

For this one

cdf =$1-\left({\frac {x_{\mathrm {m} }}{x_{\mathrm {m} }+x}}\right)^{k}\!$

I got the result from integrating a pdf...which is a bit different from the one given; it is essentially the same one but mine did not shift the origin to x_m ... this is not a fake result or anything. Something should indicate this "kinda" conflict between 2 version of the same probability function. but definitively...i will remove my mistake...only because the current cdp does not reflect the shifting nature of the pdf. I'll specify in the generating topic that it will generate a random sample from a non shifted pareto distribution.
Cyberyder 04:24, 6 April 2006 (UTC)

Many types of Pareto

Hi :) I'm studying My Actuarial Exam 4/C along with loss models and i can't help but to notice that the pareto distribution listed in wikipedia is the 1 parameter for of the distribution. In fact this distribution is pretty much the same as the 2 parameter instead, Xm. I was wondering if it was possible to change the name of the article for "Single parameter pareto distribution" and i will eventually add a subtopic for the 2 parameter distribution. 64.18.166.93 18:16, 7 April 2007 (UTC)

Distribution Example: size of sand particles?

Is it true that the size of sand particles (probably from the same part of the same beach) are pareto-distributed? I would have thought they'd be gaussian-normal-distributed...

I never looked at sand particles under a microscope, though, I just thought they were all the same size...

<--- Feels Educated now. :)

Archive 1

Assessment comment

The comment(s) below were originally left at Talk:Pareto distribution/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Comment(s)

Press [show] to view →

I just tried to understand the passage: A characterization theorem Suppose Xi, i = 1, 2, 3, ... are independent identically distributed random variables whose probability distribution is supported on the interval [k, ∞) for some k > 0. Suppose that for all n, the two random variables min{ X1, ..., Xn } and (X1 + ... + Xn)/min{ X1, ..., Xn } are independent. Then the common distribution is a Pareto distribution. Now the following questions have come up to my mind: 1.) The statement "min{ X1, ..., Xn } and (X1 + ... + Xn)/min{ X1, ..., Xn } are independent" is very counterintuitive to me. I would expect that two random variables which are somehow functions of the same set of underlying random variables to be somehow dependent. My notion is probably naive but it would be nice to have a supporting hint in the explanation... 2.) What is ment by "the common distribution"? Do both min{ X1, ..., Xn } and (X1 + ... + Xn)/min{ X1, ..., Xn } have the same distribution and this is the pareto distribution? It is also quiet counterintutitive to me and I am asking myself if my interpretation is realy adequate...

Last edited at 09:11, 22 June 2009 (UTC). Substituted at 15:31, 1 May 2016 (UTC)