Laplace distribution

#1
Hi all.

I have a query regarding the Laplace (double exponential) distribution.

I have conducted some experiments measuring the orientation of fibres within a material. The orientations are biased towards 0 degrees. I have summarised the orientations in a histogram between (Pi/2) and (-Pi/2) and then fitted a curve to the data to give me a continuous function. I used NLREG non-linear regression software to do this.

Upon first inspection, the experimental data appeared to follow a normal distribution, but this gave a poor fit for intermediate orientations between 0.35-1.22 radians. A Laplace distribution (double exponential distribution) was selected as the most appropriate model, based on the coefficient of determination values (R2). It was assumed that the fibre orientations were symmetrical about the origin (i.e. the probability of 0.5 radians is the same as for -0.5 radians).

All was fine up to this point.

I then submitted the results in a scientific journal and I got the following response form one of the reviewers:

"It is troublesome to see the sharp tip on the density curves – implying that the change rate of the probability density is infinite at the center, something first impossible and also a reflection of ignorance about the probability distributions."

I presume what he is talking about is that when you differentiate the Laplace distribution there is no solution at zero.

In the current context why does this matter? All I am doing is plotting probability against orientation, between (Pi/2) and (-Pi/2). Personally, I am not interested in the differential. I have successfully integrated the distribution to find the percentage of fibres between +10degrees and -10degrees, but why would I ever need to differentiate it? Am I missing something vital?

I am happy to challenge the reviewer on this point, but I first want to check that I am correct, especially since he has already implied that I am ignorant :mad: .
 

TheEcologist

Global Moderator
#2
Hi all.

I have a query regarding the Laplace (double exponential) distribution.

I have conducted some experiments measuring the orientation of fibres within a material. The orientations are biased towards 0 degrees. I have summarised the orientations in a histogram between (Pi/2) and (-Pi/2) and then fitted a curve to the data to give me a continuous function. I used NLREG non-linear regression software to do this.

......is plotting probability against orientation, between (Pi/2) and (-Pi/2). Personally, I am not interested in the differential. I have successfully integrated the distribution to find the percentage of fibres between +10degrees and -10degrees, but why would I ever need to differentiate it? Am I missing something vital?

I am happy to challenge the reviewer on this point, but I first want to check that I am correct, especially since he has already implied that I am ignorant :mad: .

My first point would be: why did you use OLS to fit the distribution? And could you also post your histogram?

Your distribution can be approximated by n x pdf(x) were n is your sample size. However the data is discrete, namely counts at certain discrete radial values, therefore I would use a Maximum likelihood method to fit a pdf to your observed distribution.

First determine what kind of error distribution would fit your count data. By using OLS you automatically assume a continuous normally distributed error. This yields the same results as running a ML and calculating the likelihoods with the normal pdf. However your data does not have normal errors, counts lower than 0 do not exist.

I would guess that probably your data would have Poisson or binomial errors (which would seem perfectly acceptable to me as reviewer).
Using one of these error distributions you can then simply optimize the shape and location parameters of the pdf's by maximizing the likelihood of each pdf given your data. You maximize P(pdf(sigma,mu)|DATA), or more practically you minimize the log-likelihood.

Fit a range of pdf's using ML to your observed dataset also try the Student's T if you haven’t already. Then compare those pdf’s with each other using AIC, BIC or log-likelihood ratio tests. I think there is a possibility that the best fit then would be different.

Another thing at what level do you discretizise the radials? At full degrees (1 - 360)? I guess this would depend on how accurate you can measure the angle a fiber is orientated at.

If you someway transformed the data to use OLS, how and why didn't you just use ML on the original distribution?

If this tickle's your fancy there should be ample examples in the literature, I can help you with some ref's (but they will be from the field of Mathematical Ecology).

Hope this helps,
 
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#3
Thanks for the reply.
I don't fully understand your answer and I am not sure that you have actually answered my initial question. What is wrong with using a Laplace distribution to represent this data?
The reason I wanted to turn my discrete data into continuous data is that I have written a program to model the mechanical properties of this material. The model generates a random number and then returns an orientation according to the ditribution.

Here are the histograms:



Here is an example of the fitted data:

 
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TheEcologist

Global Moderator
#4
My point is that I don’t believe using Ordinary Least Squares to fit a set of discrete observations to a model is the right way. In the case of more general models (with discrete data points as in your case) maximum likelihood is the criterion of fit. One can only guess if laplace would still yield the "best" fit when using ML (although it would probably still be one of the top ranking candidates, seeing your 115mm data).

Is his response trivial? The negative exponential is a popular PDF used in a variety of fields, the fact that rate of change is infinite at it's centre is of course known and easily overcome. If this reviewer is attacking the credibility of the Laplace distribution I would see this as a reflection of his ignorance about the probability density functions. You can simply add a few lines on how this "problem" is overcome (use peer-reviewed sources) and maybe give some examples of it's uses in your and other field(s) (again overwhelm him with credible sources). Or you can just choose to ignore him and leave it up to the editor (however this is a more political move, you know the situation and your chances better).

My only problem is that I believe that the correct way to fit your data to a certain PDF is through ML not OLS.

Cheers,
 
#6
I think I know what you are getting at now regarding OSL and the ML approach.

For this current scenario I have 6 different sets of data. Using the ML approach, how do I ensure that the same type of distribution is used for each? Sorry if this is a stupid question, but I have had no previous experience using the ML method.

Without sounding lazy, in general, how is the problem overcome regarding the rate of change at zero when using the Laplace distribution? Just a brief description will suffice. I have done a google search and not found anything.
 

mp83

TS Contributor
#7
Hi!

I'd say that it's fair fit the Laplace.You could check for a chi-square too, to relax the symmetry assumption.

For this current scenario I have 6 different sets of data. Using the ML approach, how do I ensure that the same type of distribution is used for each? Sorry if this is a stupid question, but I have had no previous experience using the ML method.
Welcome to the world of model uncertainty! It's possible one of your dataset not to fit well to the Laplace,for a model must be an approximation to the true (?) model but fit the observations as well...
 

TheEcologist

Global Moderator
#8
I think I know what you are getting at now regarding OSL and the ML approach.

For this current scenario I have 6 different sets of data. Using the ML approach, how do I ensure that the same type of distribution is used for each? Sorry if this is a stupid question, but I have had no previous experience using the ML method.

Without sounding lazy, in general, how is the problem overcome regarding the rate of change at zero when using the Laplace distribution? Just a brief description will suffice. I have done a google search and not found anything.
I was hoping you'd find this yourself but the problem is that there is no problem, you can integrate the Laplace from -Infinity to Infinity, and the area will be 1. Just like any other PDF. How can the area be 1 when apparantly the rate of change is infinite at the centre? It cant, so one of the two statements is wrong. So just find out for yourself if the area under the curve is one, if so.. you've got your answer. This is the one I'm talking about:

pdf(x) = 1/(2*b)*exp(-abs(x-a)/b)

or

pdf(x) = 1/(2*b)*exp(-sqrt((x-a)^2)/b)

It is used for modeling in signal processing, various biological processes (eddy-covariance CO2 flux, seed shadow fitting) and economics.

This is were you can start learing more about it:

The Laplace Distribution and Generalizations
A Revisit with Applications to Communications, Economics, Engineering, and Finance

By

Kotz, Samuel, Kozubowski, Tomasz, Podgorski, Krzystof


the book review:


The Laplace Distribution and Generalizations: A Revisit With Applications to Communications, Economics, Engineering, and Finance

Author: McNeil A.J.1

Source: Journal of the American Statistical Association, Volume 97, Number 460, 1 December 2002 , pp. 1210-1211(2)


Articles that you might like:

*A discrete analogue of the Laplace distribution

Author's

INUSAH Seidu (1) ; KOZUBOWSKI Tomasz J. (2) ;


*A time-series model using asymmetric Laplace distribution

K. Jayakumara, Corresponding Author Contact Information, E-mail The Corresponding Author and A.P. Kuttykrishnan


Ensuring that only the Laplace fits all you data using ML should not be done in my opinion. Why? Because that is not fair, you should fit a few PDF's to your data and then use their ML values to determine which is best. It could be the Laplace.......
 
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#9
Just to give a bit of an update.

I downloaded a trial version of EasyFit software and performed a Maximum Likelihood Estimate. For the six data sets I have, the Laplace distribution was ranked the highest three times, and was ranked in the top three for the other three data sets.

Because I just want to reproduce the data for modelling purposes I am happy to use the Laplace distribution for all six data sets and accept the small associated error. Because the experimental orientations were only measure to ±1º then this is not such a problem.

I am pleased to say that I have learned something useful from all of this. I guess it was down to luck that I chose the Laplace distribution over any of the others.
 

TheEcologist

Global Moderator
#10
Just to give a bit of an update.

I downloaded a trial version of EasyFit software and performed a Maximum Likelihood Estimate. For the six data sets I have, the Laplace distribution was ranked the highest three times, and was ranked in the top three for the other three data sets.

Because I just want to reproduce the data for modelling purposes I am happy to use the Laplace distribution for all six data sets and accept the small associated error. Because the experimental orientations were only measure to ±1º then this is not such a problem.

I am pleased to say that I have learned something useful from all of this. I guess it was down to luck that I chose the Laplace distribution over any of the others.
Great!

Your motivation concerning the use of the Laplace seems acceptable to me.

Hope it all works out for you.
 
#11
So just find out for yourself if the area under the curve is one, if so.. you've got your answer. This is the one I'm talking about:

pdf(x) = 1/(2*b)*exp(-abs(x-a)/b)

or

pdf(x) = 1/(2*b)*exp(-sqrt((x-a)^2)/b)

I have just differentiated the two distributions above, and sure enough there is no solution at theta=0 for the first pdf. However, there is for the second pdf and the area under the curve equals 1. Great.

Unfortunately, I have yet to source copies of any of the recommended books. Does this modified version of the Laplace Distribution have a common name?
 

mp83

TS Contributor
#12
The second version simply relies on the fact that

|x-a|=sqrt{(x-a)^2}

It is frequently used to make it easier to differentiate an expression containing the abs() function.
 

TheEcologist

Global Moderator
#13
The second version simply relies on the fact that

|x-a|=sqrt{(x-a)^2}

It is frequently used to make it easier to differentiate an expression containing the abs() function.
Thats the reason I added it, so it could be intregrated. The two models are identical.