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I am trying to show that Pareto distribution with $$f(y;\alpha)=\alpha y^{-\alpha-1} $$ is exponential dispersion (ED) family which means that it can be rewritten as: $$f(y;\theta,\phi)= \exp\left\{\frac {y\theta-b(\theta)} {a(\phi)} + c(y;\phi)\right\}$$ with $E[Y]=b^{\prime}(\theta)$ and $Var[Y]=b^{\prime\prime}(\theta)a(\phi).$

What I found is there are multiple ways to rewrite the PDF in the exponential dispersion family form, but since I know that $$E[Y]=\frac {\alpha} {\alpha-1}$$ and $$Var[Y]=\frac {\alpha} {(\alpha-1)^2(\alpha-2)},$$ I can solve the differential equations to get $$b(\theta)=\alpha + {\rm log}(\alpha-1)$$ and $$a(\phi)=\frac {-\alpha} {\alpha-2}.$$ This suggests that the form of $f(y;\theta,\phi)$ is unique.

So, is the ED family form of the PDF for Pareto distribution unique? If so, what is the correct form?

User1865345
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Sheldon
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    Your question is confusing as (a) you use three different notations for the parameter, ie $\alpha,\theta,\phi$ and (b) you use the same notation for $y$ in the Pareto and $y$ in the exponential family. Since $f(y;\alpha)=\exp{\log(y)(-\alpha-1)+\log(\alpha)}$, you get the natural representation. Note also that you are missing the indicator function in the Pareto density. – Xi'an Nov 10 '16 at 09:05
  • Thanks! (a) I forgot to mention that this is a reparameterization, we can find $\theta$ and $\phi$ in terms of $\alpha$ (b) yes, this is the natural form for exponential family, but not in the form of ED as I illustrated above. Yes, I skip the indicator function $I(Y>1)$. – Sheldon Nov 10 '16 at 15:15
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    You still didn't fix the missing indicator in the density. We need to know the range! – kjetil b halvorsen Oct 18 '17 at 16:14
  • Have you found the solution? Because, me either, I'm interested in showing that the GPD comes from EDM and I can't find a solution – Maria Cristina Apr 01 '19 at 21:05

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You have given a Pareto density with a fixed lower limit of range 1, $$ f(y; \alpha)= \alpha y^{-(\alpha+1)}, \quad y \ge 1 $$ Taking logarithms we can write this in exponential family form $$f(y; \alpha)= \exp\left\{ -(\alpha+1)\log y + \log\alpha \right\} $$ which does not have any dispersion parameter, but introducing $\phi\equiv 1$ we can recognize the exponential dispersion family form as $$f(y; \alpha)= \exp\left\{ -\frac{(\alpha+1)\log y}{\phi} + \log\alpha \right\} $$ so the answer is yes, the Pareto family with known lower limit of 1 is an exponential dispersion family with canonical statistic $\log y$. Note that $\log y$ does have an exponential distribution, see Distribution of the exponential of an exponentially distributed random variable?

And, yes, this form is unique.

kjetil b halvorsen
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