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power law distribution, How could I generate random variables from the truncated power -law distribution and its PDf?

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amberly hadden 2014-10-6

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评论： Image Analyst 2014-10-9

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How could I generate random variables from the truncated power -law distribution and its PDf?

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Image Analyst 2014-10-6

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rayleigh_random_distribution.m

You need to use inverse transform sampling. I attached a demo where I use it to get random numbers from a Rayleigh distribution.

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unable to find CDF and PDF for Power law distribution ?

amberly hadden 2014-10-6

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I mean in terms of normal distribution all we need is sgm and mu but in this case how could I find aplha?

Thanks

Image Analyst 2014-10-6

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Huh? I think you've forgotten to supply some information, such as your equation. I don't know what alpha is. Maybe you'd like to just use RANDRAW: http://www.mathworks.com/matlabcentral/fileexchange/7309-randraw

amberly hadden 2014-10-6

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If I following inverse transform sampling I need to define my probability function for power law distribution and for that I need value of aplha [can be any value] but I' wondering if this parameter is same as let say normal distribuion needs mu and sgma. Is there any formulation to find out alpha?

Thanks

Image Analyst 2014-10-6

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If you have the formula, you can use it. Otherwise generate a distribution with the PDF formula and numerically integrate to get the CDF with cumsum(). You can use the cumsum() output as a lookup table if you don't have the exact formula for the CDF.

amberly hadden 2014-10-6

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what if my PDF is f(x) = (aplpha-1/xmin)(x/xmin)^-alpha

i followed your code but at the end if I plot histogram of resultant distribution im not getting exactly power law dist instead Im getting double exponential.

Image Analyst 2014-10-6

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What is the range of x you're putting in. And what are some typical values for xmin and alpha that you're using?

amberly hadden 2014-10-6

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编辑：Image Analyst 2014-10-7

在 MATLAB Online 中打开

I'm using following :

n=1000;
x=rand(1,n);
for i = 1:1000
    a = 2.5;
    P(i) = ((a - 1)/min(y))*(y(i)/min(y)^(-a))% this PDF
end

Image Analyst 2014-10-7

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InverseTransformSampling_InversePowerLaw_RandomDistribution.m

jay, that's not right. You didn't follow the recipe. See my attached demo (below the plots) for a full illustration.

amberly hadden 2014-10-9

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Hi Thanks,

It turns out if I have to use your inverse transformation then I need to integrate PDF to get CDF and then I can consider CDF as uniform number apply your technique. Thanks for this code it solved my problem.

Regards Jay

Image Analyst 2014-10-9

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You're welcome. Since I solved your problem, could you go ahead and mark the Answer as "Accepted" so I get credit for it? Thanks in advance.

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