Talk:Rayleigh distribution

Derivation, relationship to Gaussian distribution, and higher dimensions
Some questions that came to mind after reading this article, perhaps appropriate additions: Zolot (talk) 18:10, 21 July 2011 (UTC)
 * How is The Rayleigh distribution related to a normal distribution mathematically? Each of the vector components are supposed to be normally distributed, so how does the Rayleigh parameter (σ) depend upon the normal distribution's parameters (σ and μ)? In particular, how does the R. dist. change when only one of these parameters varies? (What happens when you change either the width or the mean value of the normal distributed vector components generating the R. dist.)
 * Is there a short derivation of the R. dist. from a normal dist. that could be included? (Perhaps this will reveal the answer to my first question.)
 * Is this distribution only valid for two dimensional vectors? Is there a generalization for higher dimensions?

Confusion
This is also a geometry-based distribution in mathematical probabilities. I was disappointed to come here looking for more information on this distribution and significant theorems, only to get redirected to some stuff about radio broadcasting.

Perhaps this could be turned into a disambiguation page? Although I don't know enough about the Rayleigh probability distribution to write a decent article on it myself.

For this distribution and every other probability distribution on Wiki, please include the valid ranges of x. Like for gaussian, x goes from negative infinity to infinity... etc.

5/6/09 - The Rayleigh distribtion is a special case of Weibull, where m (the shape factor) = 2. The Weibull equation is:


 * $$f(t) = \frac{m}{t} \left( \frac{t}{c} \right)^m \exp-\left(\frac{t}{c}\right)^m$$

Setting m = 2 gives:


 * $$f(t) = \frac{2}{t} \left( \frac{t}{c} \right)^2 \exp-\left(\frac{t}{c}\right)^2

= \frac{2t}{c^2} \exp\left(\frac{-t^2}{c^2}\right)$$

Now, let x = 2t (and t = x/2) to get the form on the article page:


 * $$f(x) = \frac{x}{c^2} \exp\left(\frac{-\left(\frac{x}{2}\right)^2}{c^2}\right)

= \frac{x}{c^2} \exp\left(\frac{-\left(\frac{x^2}{2^2}\right)}{c^2}\right)

= \frac{x}{c^2} \exp\left(\frac{-\left(\frac{x^2}{4}\right)}{c^2}\right)$$


 * $$ = \frac{x}{c^2} \exp\left(\frac{-x^2}{4c^2}\right)

$$

This is different than the equation on the article page that has a 2 instead of the 4. So my question is, which is correct? —Preceding unsigned comment added by ChrisHoll (talk • contribs) 05:49, 7 May 2009 (UTC)

Hi Chris: The Matlab documentation has a 2 in the denominator of the exponential - Patrick Tibbits Tibbits (talk) 19:02, 21 September 2009 (UTC)

Example
I thought the wind example addressed the ill-posed nature of the problem of predicting vector components given vector magnitude (i.e. that directional components map onto windspeed in a many:one fashion). Mrdthree (talk) 09:30, 5 October 2010 (UTC)

Sigma notation very misleading if not improper
The sigma character is normally used to represent the standard deviation. For a Rayleigh distribution formula input parameter, it is actually a "scaling" term and is relatable to the standard deviation/variance and mean with the formulas within the Wiki. This scaling term really must be changed to another notation (note Matlab uses the term "parameter B").

Unfortunately lost a day trying to figure out why my standard deviations & means weren't coming out per the stated formulas... — Preceding unsigned comment added by 128.170.224.10 (talk) 02:45, 10 November 2012 (UTC)


 * Yes, it can be confusing, but the sigma is in fact the standard deviation of the bivariate normal distribution to which the Rayleigh distribution is isomorphic -- it comes right through the transformation via polar coordinates between the two. Perhaps best would be to use the parameterization $$\gamma = \sigma^2$$, which is the preferred form of Siddiqui. Dbooksta (talk) 21:03, 19 November 2013 (UTC)

Sigma exponent (obsolete remark)
Integration tests have shown that \frac{x}{\sigma^2} is indeed the correct normalisation (even though it seems strange from a dimension analysis point of view). 150.227.15.253 (talk) 13:14, 3 November 2021 (UTC)

Is the formula:
 * $$f(x;\sigma) = \frac{x}{\sigma^2} e^{-x^2/(2\sigma^2)}, \quad x \geq 0,$$  really correct (YES!), shouldn't it rather be:


 * $$f(x;\sigma) = \frac{x}{\sigma} e^{-x^2/(2\sigma^2)}, \quad x \geq 0,$$  ? (NO!)150.227.15.253 (talk) 12:03, 3 November 2021 (UTC)

Possible source of confusion detected in the opening paragraph example
The opening paragraph states "Assuming that the magnitudes of each component are uncorrelated, normally distributed". I normally understand "magnitude" as a scalar greater than zero. If we take this latter definition, the "magnitudes" of the each component cannot be normally distributed because, by definition, the normal distribution takes values both lower and greater than zero, so that this would be clear contradiction. I am aware that "magnitude", as it is written, might refer to a scalar real value, positive or negative, as vector components may be, so this at least needs clarification. I cannot fix this because I am not certain that given my interpretation the example holds and still produces a Rayleigh distribution, as this needs a proof. This is most probably a semantic problem in the common usage of the words "magnitude" and "component" so, if someone with clear knowledge of both the Rayleigh distribution and this subtleties in mathematical terms can, it would be very helpful.--Fermín MX 05:23, 12 June 2014 (UTC) — Preceding unsigned comment added by Ferminmx (talk • contribs)


 * See how it reads now. PAR (talk) 06:07, 12 June 2014 (UTC)

Generating Random Variates
Formula is incorrect.

Random variate "U" should be "1-U" (non simplified version).

Simple proof:

If random variate U=1 then X should be infinite.

If random variate U=0 then X should be zero.

In the current (simplified) formula this is clearly not the case.

Remember, a random uniform distribution is uniform ONLY if the number of random variables is infinite.

Regards, Rob


 * Note that U and 1-U both have uniform distribution, so it does not matter which one you use.2.240.94.253 (talk) 14:01, 14 September 2015 (UTC)

Linear hazard rate with intercept equal to zero
The Rayleigh distribution has an increasing hazard rate proportional to x. This extends the scope of interpretation.

— Preceding unsigned comment added by 84.83.33.64 (talk) 10:57, 21 April 2015 (UTC)

single dimension?
looking at the pictures it seems like this would represent a random variable in a single dimension that must always be greater than zero. For example, the amount of time something takes must always be greater than zero, but could potentially be much much larger. You could probably model this as a normal too, if the mean wasn't close enough to the zero bound that it would appear skewed? Say, people's heights at a certain age would meet a normal distribution, because there is a negligible probability of your height being near zero, but lifespans seem like they would be a Rayleigh distribution because there are plenty of samples close to zero (birth mortalities)

Would the introduction be much more accessible if it talked about something like this, instead of the vector thing? (it's difficult to think about because it's not obvious why you'd want to model a vector and none of the equations mention vectors, just magnitudes) Dominictarr (talk) 03:00, 12 February 2016 (UTC)