Variance and Cauchy Distribution

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SUMMARY

This discussion focuses on the challenges of analyzing random signals, specifically noise, characterized by Cauchy distribution and Lorentzian shapes. The second-order moment, or variance, is infinite for Cauchy-distributed data, making traditional comparisons of noise statistics impractical. Alternatives such as the interquartile range and characteristic functions are suggested for quantifying spread in non-Gaussian distributions. The conversation highlights the complexity of comparing different noise types and the limitations of conventional statistical measures.

PREREQUISITES
  • Understanding of Cauchy distribution and its properties
  • Familiarity with statistical measures such as variance and interquartile range
  • Basic knowledge of Fourier transforms and characteristic functions
  • Experience with analyzing random signals in physics
NEXT STEPS
  • Research the properties of the Cauchy distribution and its implications in signal processing
  • Learn about the interquartile range and its application in comparing non-Gaussian distributions
  • Explore the use of characteristic functions in statistical analysis
  • Investigate Voigt functions and their significance in convolution of distributions
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Physicists, data analysts, and statisticians working with random signals, particularly those dealing with non-Gaussian noise distributions and seeking alternative methods for statistical comparison.

ZeMike
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Dear all,
I'm not a mathematician so please excuse me for a certain lack of strictness ...

I work on random signals in physics, these signals are most of the time called "noise" for us. For example, we can speak about x(t), a time domain random signal.
Very usually, the statistics for noise is gaussian or poissonian, and so we use to speak of the histogram of this signal, in terms of second order moment or variance : it's practical because this single number permits to compare various noises.

Unfortunately, recently I started to work on signals which histogram is a lorentzian shape. If I understood, this corresponds to the Cauchy distribution. And again, if I understood, the second order momentum is inifinite : when I try to solve the second order moment integral, I obtain the infinite ... not practical. And indeed, a Lorentzian width is usually known thanks to its full with half maximum and not by its variance (before calculating the variance, I thought this choice was arbitrary ;-) ).
Moreover, things can become more complex as some histograms can be voigt-function shaped (= lorentzian-gaussian convolution).

So here is my question : I would like to compare the histograms of all my noises, whatever their statistics, by speaking of their "range" (= variance for a gaussian shape for example). I understand it is not possible to use the second order momentum for that purpose. Is there some other way to compare the range of these statistics ?

Perhaps my question does not makes sense for some fundamental reason, if this is the case I would like to understand why.

Best Regards,
Mike
 
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If you're just looking for quantities of spread, then perhaps you could check the "interquartile range". You might also be able to do something with the characteristic function.

By the way, if your data was really from the Cauchy distribution, then you would find that the first moment doesn't exist either.
 
Thanks a lot for this answer ! the interquartile range seems to be an interesting indicator.

However, concerning the chareteristic function, I do not understand what you suggest. If I understood, the charateristic function computes like the Fourier transform of the histogram. Then there are two problems form me :
- what kind of width calculatin may I use on the characteristic function ?
- if I use for example the variance for this, it is possible to obtain a Lorentzian in the Fourier space, so the problem remains unsolved

For the first moment of the Cauchy distribution, it's a result I have seen, however the integral calculs seems to give good values. I think I did not really understand the problem.

Best Regards,
Mike
 

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