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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T10:10:50+00:00 2026-06-15T10:10:50+00:00

I am trying to model shadowing and fast fading for mobile wireless networks. For

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I am trying to model shadowing and fast fading for mobile wireless networks. For fast fading, Rayleigh fading is a reasonable model to use. The envelope of the channel response will be Rayleigh distributed. Calling this random variable R, it will have a probability density function (PDF) of

PR(r) = ((2r)/Ω)*exp(-r^2/Ω), r >= 0, Ω = 2σ^2

http://en.wikipedia.org/wiki/Rayleigh_fading to see the equation written nicely.

So, I have the PDF, now I am just wondering how to get the random variable from it?

I have looked at these questions:

Generate Array of Numbers that fit to a Probability Distribution in Ruby?

Generate Random Numbers with Probabilistic Distribution

but I am still not sure how to do it. It has taken me forever to understand PDF’s so if anyone knows a way in Java to get a random variable with a specific PDF, that would be much appreciated!

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  1. Editorial Team
    Editorial Team
    2026-06-15T10:10:51+00:00Added an answer on June 15, 2026 at 10:10 am

    Rayleigh distribution is a special case of the Weibull distribution. If you google around, there are lots of Weibull generators written in Java, for example:

    • http://commons.apache.org/math/apidocs/org/apache/commons/math3/distribution/WeibullDistribution.html
    • http://www.iro.umontreal.ca/~simardr/ssj/doc/html/umontreal/iro/lecuyer/randvar/WeibullGen.html
    • http://www.icsa.inf.ed.ac.uk/research/groups/hase/simjava/distributions/doc/eduni/distributions/Weibull.html
    • http://www.ee.ucl.ac.uk/~mflanaga/java/PsRandom.html

    One way to generate a random number from a given distribution is to generate a random number uniformly distributed between zero and one, and apply the target distribution’s inverse CDF to that random number. See Wikipedia.

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