Talk:Exponential distribution

The graphs

I don't understand how to read the two graphs at the top. Whats on the axis? —Preceding unsigned comment added by 87.48.41.14 (talk) 12:28, 20 January 2010 (UTC)

In both graphs, the thing on the horizontal axis is the values of the random variable that is exponentially distributed. In the second graph, the thing on the vertical axis is probability—that of being less than the corresponding value on the horizontal axis. In the first graph, the unit is probability divided by the units of measurement of the thing on the horizontal axis. See probability density function. Michael Hardy (talk) 19:01, 20 January 2010 (UTC)

moment-generating functions

I noticed that the Gaussian and Uniform distributions have a section on the mgf, but this page lacks such a section. 75.142.209.115 (talk) 07:21, 30 October 2008 (UTC)

Is there a reason the moment generating function of the exponential is written as

$\left(1-{\frac {t}{\lambda }}\right)^{-1}$

instead of the more familiar, and easy to understand

${\frac {\lambda }{\lambda -t}}$

in the sidebox? --129.34.20.19 (talk) 20:48, 11 August 2010 (UTC)

Exponential

I understand that the infobox text reads "Name: Exponential" and that makes sense, but what is displayed is just "Exponential" and that makes no sense, its an adjective thats missing a noun to modify. Alternatively, we could change the infobox to display "{{{name}}} distribution"? PAR 10:31, 1 Apr 2005 (UTC)

Take the discussion here please: Template talk:Probability distribution Cburnett 15:04, 1 Apr 2005 (UTC)

Does this page need to be cited more for the technical parts? I ask specifically because I am unable to explain the memoryless property to somebody, and would like to cite more resources. —Preceding unsigned comment added by 151.190.254.108 (talk) 12:58, 2 June 2008 (UTC)

Rayleigh & Expo fishyness

Something's fishy about the formula relating the Raleigh and the exponential distribution. The parameters λ and β have to be related somehow. 141.140.6.60 22:36, 7 May 2005 (UTC)

That's right. I will fix it right now. Also, you seem to know this subject. How about some help with the other Category:probability distributions? We could sure use it. We are trying to put full infoboxes in all of them. PAR 23:57, 7 May 2005 (UTC)

Yup, I'll continue to do little things here and there. I forgot to log in for the edits to this article. AxelBoldt 00:53, 8 May 2005 (UTC)

I got the relationship from Statistical Inference by Casella & Berger (ISBN 0534243126) but it I don't think it gave the parameter relationship (it might be lambda = beta) but I'd have to pull out some paper and confirm by transforming it. Unfortunately, I'm in the middle of moving and it's boxed up. Cburnett 06:36, May 8, 2005 (UTC)

Email query

From an email to the Foundation:

"Could [you] explain why the Exponential distribution can be generated by taking the natural log of the uniform distribution? Is there any analytical proof? The author cited the quantile function, but it is not clear how. Thanks.

The part of the article quoted was:

"Given a random variate U drawn from the uniform distribution in the interval (0,1], the variate has an exponential distribution with parameter £f. This follows from the form of the quantile function given above and yields a convenient way to produce exponentially distributed values using a random number generator on a computer, for instance to conduct simulation experiments."

I've directed the correspondent here - can anyone help? Thanks -- sannse (talk) 19:33, 15 January 2006 (UTC)

This may be the wrong place. I've added a pointer to inverse transform sampling method here, and the (rather trivial) formal proof for the general case there. --MarkSweep (call me collect) 23:37, 15 January 2006 (UTC)

Comment

Sorry if I'm doing this wrong, I'm a wikipedia newbe. I just wanted to mention that I got confused for a while in the Bayesian inference section because the Gamma distribution is not parameterized the same way as on the Gamma distribution page. I'm not an expert, there might be a good reason for this which is over my head. If not, it would probably be nice to change it to make it more consistant with the gamma distribution page so that the resulting likelyhood is

$p(\lambda )=\mathrm {Gamma} (\lambda \,;\,\alpha +n,(\beta +n{\overline {x}})^{-1}).$

--BenE 19:20, 14 July 2006 (UTC)

Confusion with alternative spec.

Well, the text takes non-alternative spec., but I see at least the last two formulas of Related distributions are using alter. spec. So, I have modified them to non-alter. spec. Ping my talk page if you need a discussion. Thanks. --Amr 12:04, 1 September 2006 (UTC)

Occurrence and applications - Gamma distribution

"In queueing theory, the inter-arrival times (i.e. the times between customers entering the system) are often modeled as exponentially distributed variables. The length of a process that can be thought of as a sequence of several independent tasks is better modeled by a variable following the gamma distribution (which is a sum of several independent exponentially distributed variables)."

This is quoted from the current page. Isn't the exponential distribution just a special case of the gamma distribution (alpha=1, beta=1/lambda)? If so, it doesn't make sense to me to say that gamma distributions are better when the exponential is itself a gamma distribution. Should this section of text be changed or clarified? --JRavn ^talk 22:09, 7 February 2007 (UTC)

Sum of exponential distributions

It currently says that sum of exponential distributions is distributed as Gamma(n;lambda). Looks like it should be Gamma(n;1/lambda) in order to be consistent with the page about the Gamma distribution. 128.84.154.137 20:39, 28 October 2007 (UTC)

Where does it say that? I can't find it. Michael Hardy 22:57, 28 October 2007 (UTC)

Related distributions, the one before the last. 128.84.154.13 23:46, 11 November 2007 (UTC)

I was just about to write a similiar comment. It sure looks like an inconsistency to me. —Preceding unsigned comment added by 83.130.133.199 (talk) 18:03, 28 December 2007 (UTC)

Yes, I think this is false too. Should be 1/lambda —Preceding unsigned comment added by 129.67.134.154 (talk) 15:42, 29 January 2008 (UTC)

I think the sum of n variables i.i.d. exponential distributions with mean lambda should give a gamma distribution with parameter n and lambda rather than 1/lambda. Would appreciate a proof if otherwise.

Language: Model? Or something else?

"used to model" is used three times on this page. I think that this wording is a bit weak and could be strengthened. "Used to model" implies that the distribution is used out of convenience or an approximate fit. In all three cases where the phrase is used (such as with a Poisson process), the distribution can be derived mathematically from the basic assumptions. Thus, those assumptions are the modeling assumption and the exponential distribution is a result of the model, not really part of the model itself. Contrast this with, for example, the use of a beta distribution either as a Bayes prior or as the family for a parametric model. If these distributions are used mainly because of convenience or goodness of fit, then they are "used to model". I think the exponential distribution is a bit of a different animal, however. I'm going to change this but you may want to check over my changes because I'm not exactly sure what words/phrases to use. Cazort 14:55, 4 November 2007 (UTC)

Shift parameter?

Why not discussing the exponential distribution with a rate parameter $\lambda$ and a shift parameter a, especially, focusing on the use of finding unbiased estimators for these parameters? —Preceding unsigned comment added by 84.83.33.64 (talk) 15:26, 11 February 2009 (UTC)

index of the minimum

I am thinking to add the following to the section on Distribution of the minimum of exponential random variables

The index of the variable which achieves the minimum is distributed according to the law.

$\Pr(X_{k}=\min\{\,X_{1},\dots ,X_{n}\,\})={\frac {\lambda _{k}}{\lambda _{1}+\ldots +\lambda _{n}}}$

Can someone verify my calculations or find a reference to this fact? (Igny (talk) 18:47, 1 May 2009 (UTC))

First find the conditional probability that X₁ ≤ X₂ and ... and X₁ ≤ X_n given the value of X₁.

That is a random variable that is just a function of X₁. Its expected value is the unconditional probability that you seek (see law of total probability). The expected value is a simple integral. That proves the result.

But I'd rather write

${}+\cdots +{}\,$

than

${}+\ldots +{}\,$

Michael Hardy (talk) 00:25, 2 May 2009 (UTC)

OK, I can comment in a more leisurely way now. First we have

$\Pr(X_{k}>x)=e^{-\lambda _{k}x}.\,$

So (using independence)

$\Pr(X_{k}>X_{1}\mid X_{1})=e^{-\lambda _{k}X_{1}}{\text{ for }}k\in \{\,2,\dots ,n\,\}.\,$

Then, again using independence,

${\begin{aligned}\Pr(X_{2}>X_{1}{\text{ and }}\dots {\text{ and }}X_{n}>X_{1}\mid X_{1})&=e^{-\lambda _{2}X_{1}}\cdots e^{-\lambda _{n}X_{1}}\\&=e^{-(\lambda _{2}+\cdots +\lambda _{n})X_{1}}.\end{aligned}}$

Therefore, by the law of total probability,

${\begin{aligned}&{}\quad \Pr(X_{2}>X_{1}{\text{ and }}\dots {\text{ and }}X_{n}>X_{1})=E\left(e^{-(\lambda _{2}+\cdots +\lambda _{n})X_{1}}\right)\\&=\int _{0}^{\infty }e^{-(\lambda _{2}+\cdots +\lambda _{n})x}f_{X_{1}}(x)\,dx=\int _{0}^{\infty }e^{-(\lambda _{2}+\cdots +\lambda _{n})x}\lambda _{1}e^{-\lambda _{1}x}\,dx\\&=\int _{0}^{\infty }e^{-(\lambda _{1}+\cdots +\lambda _{n})x}\lambda _{1}\,dx={\frac {\lambda _{1}}{\lambda _{1}+\cdots +\lambda _{n}}}.\end{aligned}}$

Michael Hardy (talk) 03:35, 2 May 2009 (UTC)

Ok I have added that. My reasoning was to prove (by integration) that

$\Pr(X_{1}<X_{2})={\frac {\lambda _{1}}{\lambda _{1}+\lambda _{2}}}$

and then use induction over $\min\{\min\{X_{1},\ldots ,X_{k}\},X_{k+1}\}$ (Igny (talk) 18:28, 2 May 2009 (UTC))

confusion between gamma and phase-type distributions

The article currently says "Both an exponential distribution and a gamma distribution are special cases of the phase-type distribution"

but gamma distributions are only phase-type when the shape parameter (k) is an integer. —Preceding unsigned comment added by 66.184.77.15 (talk) 18:16, 4 March 2011 (UTC)

About the confidence interval section

I am not a mathematician. And I was shocked to see that the section about the confidence interval includes a link to a geology journal. Can any expert comment and/or revert the change? --Jbarcelo (talk) 10:28, 15 April 2011 (UTC)