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I'm wondering what distribution results in adding two (or more) type-one Pareto distributions of the form $x^{-\alpha}$. Experimentally, it looks like a two-mode power-law, asymptotic to the difference of alphas.

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    $\begingroup$ The last remark makes it sound like you contemplate the alphas differing among the distributions. Are you going to fix the domains (aka "scales") of the distributions or not? A quick Mathematica calculation indicates the PDF includes, as one of its terms, the product of $x^{-\alpha-\beta}$ and the difference of a Beta$(-\alpha,1-\beta)$ distribution in $1-1/x$ and a Beta distribution in $1/x$. It is unimodal for $0\lt\alpha\lt\beta\lt 1$. This result would not hold for larger $\alpha$ and $\beta$, so are there any limits on the possible values of the parameters in which you are interested? $\endgroup$
    – whuber
    Commented Sep 12, 2014 at 20:22
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    $\begingroup$ The following paper proposes an expansion of the CDF and a way to approximate it : docs.isfa.fr/labo/2012.16.pdf $\endgroup$
    – RUser4512
    Commented Sep 14, 2015 at 12:31

1 Answer 1

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Edited to be a bit more readable. Distributions add by convolution. The Pareto distribution is piece wise defined as a $k^a x^{-a-1}$ for $x\geq k$ and 0 for $x<k$. The convolution of two Pareto functions $k^a x^{-a-1}$ and $j^b x^{-b-1}$ is:

$$ a (-1)^{-b} b k^a j^b \Gamma (a+b+1) \times \\ \left(\left(\frac{1}{t-j}\right)^{a+b+1} \, _2\tilde{F}_1\left(b+1,a+b+1;a+b+2;\frac{t}{t-j}\right)- \\ \left(\frac{1}{k}\right)^{a+b+1} \, _2\tilde{F}_1\left(b+1,a+b+1;a+b+2;\frac{t}{k}\right)\right), $$

where $j+k<x$ and 0 for $x\leq j+k$, which although complex field within the that term, is real valued outside of it. $\, _2\tilde{F}_1(w,x;y;z)$ is Hypergeometric2F1Regularized here in Mathematica code. Not all choices for the parameters will yield positive valued density functions. Here is an example of when they are positive. For the two Pareto distributions let a = 2, b = 3, j = 0.1 and k = 0.3.
where
and their plots are in blue for the {k, a} function and in orange for the {j, b} function. Their convolution is then graphically
enter image description here
which, when the tails are examined looks like
enter image description here
where the green is the convolution.

From your question, you may be asking about the ordinary addition of two Pareto distributions. In that case, the area under the curve is two, so the sum is not a density function, which needs to have an area under the curve of one. However, if that is the question then $\frac{a k^a t^{-a-1}+b j^b t^{-b-1}}{t^{a-b-1}}$ for $b>a>0$ simplifies to $t^{-2 a} \left(b t^a j^b+a k^a t^b\right)$, which has a limit of $a k^a$ only if $b=2a$, and is 0 or infinity in all other cases. In other words, the arithmetic sum of two Pareto distributions only has tails that are the difference between $a$ and $b$ when $b=2a$, and the arithmetic sum is not a density function, and the sum would have to be scaled for two probabilities, $1=p+q$ in order to be a density function. Although arithmetic addition of density functions to define another density function does occur, it is unusual. One example of this occurs in pharmacokinetics, where the sum of two or more exponential distributions is used to define a density function. To make a long story short, that is not something I would recommend.

Hopefully this answers your question. If it does not, please object to my answer or please add some more information.

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    $\begingroup$ @gung Thanks for the cleanup. Is there some etiquette required of me for this? Does one get reputation allotted for cleanup, or just good will? $\endgroup$
    – Carl
    Commented Oct 17, 2016 at 21:17
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    $\begingroup$ You're welcome, @Carl. If your reputation is <2k(?), when you suggest an edit & it is approved, you get +2. After that, edits get you nothing. I don't need the rep, so it's no problem. Your answer here is good (+1), I just edited it to make it a little easier to read. $\endgroup$ Commented Oct 17, 2016 at 21:20
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    $\begingroup$ Can you tell me how this might work for three or more pareto distributions? Is it still a ${}_{2}F_{1}$, or does it change to some other generalised hypergeometric function? Also, does this distribution have a name, like Erlang is to Exponential? $\endgroup$
    – ck1987pd
    Commented Apr 15, 2022 at 15:53
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    $\begingroup$ To get a convolution of three or more, one would have to convolve the answer above by one or more additional Pareto distributions, which I have not attempted to do, and which given how messy the convolution of only two Pareto type I distributions is, might be quite tedious to derive and rather long to write out. As for hypergeometric and/or beta functions, such answers generally contain them, but which ones exactly is difficult to predict and there are a lot of identities such that.... $\endgroup$
    – Carl
    Commented Apr 15, 2022 at 17:20
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    $\begingroup$ con't... no one answer can be thought of as definitely being uniquely expressed. As for a name I would call the convolution answer above a Pareto-Pareto convolution distribution, PPD (types I). $\endgroup$
    – Carl
    Commented Apr 15, 2022 at 17:25

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