Talk:Weibull distribution: Difference between revisions

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I believe the graph needs replacing. the weibull reaches its max at t = a (b-1)^1/b / b^1/b, not at t=a.

Pdf graph may be wrong since it exceeds probability of one (i.e:1.5) for one parameter set —Preceding unsigned comment added by 76.226.118.244 (talk) 02:17, 5 February 2008 (UTC)[reply]

and

may be better, but not as pretty.

Noodle snacks (talk) 00:00, 24 March 2008 (UTC)[reply]

in the current graph, k = 1 is not displayed ... it would be nice to also include the example values which are mentioned:

• Under certain parameterizations, the Weibull distribution reduces to several other familiar distributions:
• When k = 1, it is the exponential distribution.
• When k = 2, it becomes equivalent to the Rayleigh distribution, which models the modulus of a two-dimensional uncorrelated bivariate normal vector. In this case, the Rayleigh variance σ2 is equal to λ2 / 2. => included
• When k = 3.4, it appears similar to the normal distribution.
• As k goes to infinity, the Weibull distribution asymptotically approaches the Dirac delta function. —Preceding unsigned comment added by 134.2.190.254 (talk) 09:37, 31 October 2008 (UTC)[reply]

Do I understand this correctly?: The exponential distribution and the Rayleigh distribution are contained in the definition of the Weibull distribution as special cases? If this is so, then I feel the article should state this more clearly (i.e. human-readable :-) ) --BjKa 14:58, 16 Jun 2005 (UTC)

The Weibull distribution is essentially an exponential distribution with an extra parameter 'k', that describes the time-dependance of the distribution. Remember that the exponential distribution describes a Poisson process: it describes the time T until a sample fails. The Weibull distribution also depicts the time T until a sample fails, but it doesn't require the sample failure chance to be independant of the time elapsed.

${\displaystyle P(T<t_{1}+t_{2}|T>t1)!=P(T<t_{2})}$

The notation here could use some work. The shape parameter is more commonly denoted as "beta" and the characteristic life (or scale parameter) as "eta" (sometimes "alpha"). I would fix it, but I don't know how to fix the graphics to match.

Can we get some consistency in the parameter usage? I see lamdas, gammas, betas, k, and mu all floating around out there. I have read that standard usage involves parameters alpha and beta, not lamda and k.

This has always annoyed me about statistical distributions; We are not all professional statisticians.

Please see Template talk:probability distribution for template standards. As far as article standards go - We have been trying to use subscript ${\displaystyle f_{k}(...)}$ for discrete probability density, where k is the random variate, and ... are the parameters. ${\displaystyle F_{k}(...)}$ represents the CMF. For continuous, ${\displaystyle f(x|...)}$ is the PDF and ${\displaystyle F(x|...)}$ is the CDF, with x being the continuous random variate. The parameters are named according to whatever is "common" usage. I don't think setting them all to alphas and betas would be helpful. Generally ${\displaystyle \mu }$ stands for the location parameter (usually the mean), ${\displaystyle \sigma }$ for the scale parameter (usually std. dev.) These "standards" are by no means universal and 80% of the time it just needs to be fixed, other times the non-standard usage is so widespread that it would be inappropriate to change it. Maybe we should write this down somewhere. PAR 14:39, 11 July 2005 (UTC)[reply]

The plots of the Weibull function in this article are not correct. For example, the curve for lambda=1 and k=2 has a mode of ~0.7 at x=~1 whereas it should be 0.86 at x=1/sqrt(2)=0.707. Likewise for the other plots. [8 April 2006]

It would be useful to coordinate the notation with that used on pages for other distributions, especially those that are related. For example, if I am not mistaken, the term used for lambda on this page is the reciprocal of that used for lambda on the exponential distribution page. Perhaps this could be checked by someone who is more expert than myself [23 October 2008] —Preceding unsigned comment added by Pboettcher (talk • contribs) 12:12, 23 October 2008 (UTC)[reply]

I think the mode is simpler to calculate when written this way:

${\displaystyle \lambda \left({\frac {k-1}{k}}\right)^{\frac {1}{k}}\,}$ if ${\displaystyle k>1}$

It is currently

${\displaystyle {\frac {\lambda (k-1)^{\frac {1}{k}}}{k^{\frac {1}{k}}}}\,}$ if ${\displaystyle k>1}$

This is the first time I have made a submission - so I don't know what the process is here.

Shape of distribution can change a great deal (at least when viewed as pdf) as shape parameter goes to 1 and then below 1. Would be very helpful to get graphics that show that. Would also be neat to show graphics of hazard rate with range of parameters. —The preceding unsigned comment was added by AndrewRA (talk • contribs) 09:50, 13 December 2006 (UTC).[reply]

The article could be usefully expanded by inserting a section about the reversed Weibull distribution. DFH 21:55, 27 January 2007 (UTC)[reply]

Currently, the 2 parameter formula contains ln(U) with a subsequent warning about watching out for zeros when U=[0,1).

Of course, if one were to derive the formula from the CDF, the result would be ln(1-U) for which ln(U) is a common substitution. Of course, in this case it might be better to use the ln(1-U) formulation when u=[0,1). —Preceding unsigned comment added by Tusharm (talk • contribs) 13:05, 11 September 2007 (UTC)[reply]

This has been done, please find and correct any inadvertent mistakes made.Noodle snacks (talk) 00:01, 24 March 2008 (UTC)[reply]

The Weibull distribution "collapses" into several other useful and common distributions. Specifically, at k=2, it is the Rayleigh distribution, which describes the net displacement of a pure two-dimensional random walk. As k approaches infinity, the distribution approaches a Dirac delta function at ${\displaystyle \lambda }$. I made some of these points in the lead, since there was some talk about that there already. I see now that some of them are made (in a strangely formatted and slightly confusing way) in the Related Distributions section.

I think it might be worth noting that the skewness of the Weibull distribution switches from negative to positive as k increases, though I'm not sure where. I'm assuming that the statement "When k=3.4, the Weibull distribution appears normal" is referring to this, though it doesn't seem very rigorously stated. In any case, I would suggest that in illustrating this distribution, a greater range of k values should be selected to display the great variety of shapes it can take. Varying ${\displaystyle \lambda }$ doesn't seem as important or interesting, since it is just equivalent of squishing or stretching the x axis.

Cheers, Eliezg (talk) 16:30, 29 October 2008 (UTC)[reply]

I just read in an article that this-and-that behavior follows Weibull distributions -- is this really of any information worth, if the distribution can mimic other distribution like exponential and normal? In the end you have to look at the graph or the parameters to grasp which form of distribution it follows -- or am I missing something? —Preceding unsigned comment added by 134.2.190.254 (talk) 09:07, 3 November 2008 (UTC)[reply]

Characteristic Function of Weibull Distribution

The Characteristic Function (φ(t)) of Weibull distribution given by Muraleedharan et.al(2007), Coastal Engineering,54(8),630-638 is correct. It can generate all the moments of Weibull distribution and hence characteristic functions are also known as moment generating functions(mgf).It also gives all the moments of Rayleigh and exponential distributions for b (shape parameter)equal to 2 and 1 respectively. Also it gives the characteristic and moment generating functions of Rayleigh and exponential distributions for b equal to 2 and 1 respectively. It satisfied all the conditions for a function to be a characteristic function (See Reply to Saralees Nadarajah (2008),Coastal Engineering 55(2),191-193).

Also by Taylor's Theorem for complex functions,the characteristic function of any random variable, X, with finite mean μ can be written as

φ(t)=1+itμ+0(t); t→0

This is also true for the characteristic function of Weibull,Rayleigh (b=2) and exponential (b=1) distributions as

μ= aГ(1+1/b)

Hence the characteristic function given by Muraleedharan et.al(2007) is correct. —Preceding unsigned comment added by 220.227.156.129 (talk) 09:34, 10 February 2009 (UTC)[reply]

According to the article as of 2009-06-09, the distribution "changes character sharply when k=3/2". The only explanation offered for this claim is that the distribution looks like an exponential distribution for smaller k, and like a Rayleigh distribution for larger k. That's rather vague, to say the least. There isn't any sudden change in the shape of the pdf curve when k crosses 3/2, at least not to my eye; the point at which the skewness changes sign is nearby at about k=1.39, and that's about all I can find. Would someone who knows more about the Weibull distribution than I do care to comment? Does anything special really happen at k=3/2? Gareth McCaughan (talk) 12:29, 9 June 2009 (UTC)[reply]

Now improved a little I hope.Melcombe (talk) 10:31, 10 June 2009 (UTC)[reply]

Seems much better now. Gareth McCaughan (talk) 13:20, 11 June 2009 (UTC)[reply]

I have been visiting this page every few months for a few years now because I always forget exactly how to generate Weibull random variables from uniform random variables. This time, it took me much longer to find the description of how to do this as it is no longer in its own section as of May 11, 2009. I suggest reinstating the "Generating Weibull-distributed random variates" section because surely the method of practically generating Weibull random variables is more important/useful to more people than say, Information Entropy, which has its own section. Bjp716 (talk) 18:51, 26 August 2009 (UTC)[reply]

The parameter domains appear to me to be incompletely defined. Parameters are stated to be ${\displaystyle \lambda >0}$ and ${\displaystyle k>0}$. However, ${\displaystyle f(x=0,\lambda >0,0<k<1)}$ results in the term ${\displaystyle 0^{k-1}}$ where ${\displaystyle k-1<0}$. In other words, 1/0. As far as I can see there is no rescue from L'Hopital, perhaps others can prove me wrong. --Phays (talk) 03:31, 3 November 2009 (UTC)[reply]

This behaviour is allowed and is exemplified in a plot in the article. Densities are allowed to be infinite within the range of the distribution, provided that the integral exists. Melcombe (talk) 10:42, 3 November 2009 (UTC)[reply]

Hmm, I'll reveal my full ignorance by saying that I thought the value in this case was undefined, not infinite. I wonder if you know any resources that talk about actually determining the integral in this condition. Your comment, by the way, applies to one of the comments at the top as well. Thanks for moving this too. --134.253.26.12 (talk) 14:23, 3 November 2009 (UTC)[reply]

I was writing loosely ... f(x)→ ∞ as x→ ∞. The integral can be dealt with in many ways I think, one to is leave out a small part of the range around the problem point and then let the size of the part left out approach zero. Otherwise, you could start from the cumulative distribuition function as stated in the article, show that is essentially fully defined on [0,1] for the range of parameters staed, and show that the density is as stated by taking the derivative. Melcombe (talk) 17:36, 3 November 2009 (UTC)[reply]

The three bullets in the introductory section that describes the meaning of different k values hint at possible applications but probably would not help someone new to the Weibull distribution to understand its uses. I can see, for example, where the reference to "infant mortality" is just enough to leave someone hanging. I would suggest that if these references are left in, they are further described or, better yet, another section might be added to explain applications in more detail.Hubbardaie (talk) 00:57, 15 February 2010 (UTC)[reply]