I want to know what type of family this distribution belongs to:

  • The variable is $x$.
  • There is a parameter $\alpha$ with $\alpha>0$
  • There is a constant $H > 0$.
  • The pdf is

$$ f(x,\alpha) = \frac{\alpha x^{\alpha-1}}{\left[H\bigg(\frac{\alpha+1}{\alpha}\bigg)\right]^{\alpha}} $$

  • The distribution support is $[0,H\big(\frac{\alpha+1}{\alpha}\big)]$.

For example, if $\alpha=1$, $f(x,1)$ becomes the Uniform distribution. If $\alpha \to \infty$, $f(x,\infty)$ degenerates at $H$.

I want to know the name of the family (if it exist) so I can learn more about its moments, etc. I have not found it online so far.

  • $\begingroup$ some sort of power law distribution maybe? $\endgroup$ – Alex Feb 4 '15 at 11:15
  • $\begingroup$ Without specifying what $H$ is, or its properties, $H(\frac{\alpha+1}{\alpha})$ appears to say nothing. If you want a finite domain of support, set the upper bound to $c$, and solve $P(X<c) == 1$ for $c$ as a function of $\alpha$. $\endgroup$ – wolfies Feb 4 '15 at 14:09
  • $\begingroup$ H is a constant. Not really relevant to the problem. Added to the question. $\endgroup$ – luchonacho Feb 4 '15 at 15:13
  • $\begingroup$ This is a good source of distributions but I have not checked it for yours. causascientia.org/math_stat/Dists/Compendium.pdf $\endgroup$ – user121049 Feb 4 '15 at 15:37
  • $\begingroup$ My bad, I made a terrible mistake. Formula updated. Well, if you take away all the constant/parameters stuff it is basically $f(x,k)=c(k)x^{k}$ and so I guess it is indeed a power law distribution. However, does the parameter $\alpha$ mean that this is a sub-family of the power law distribution? $\endgroup$ – luchonacho Feb 4 '15 at 16:04

After some analysis based on @alex's reply it seems the distribution is in fact a power law.

The power law distribution can be characterized as $f(x)=cx^{k}$ defined over the $[0,A]$ interval, where $A=\big(\frac{k+1}{c} \big)^{\frac{1}{k+1}}$

The function in my question is indeed a power law under the following parameterization:

  • $k=\alpha+1$
  • $c=\bigg(\frac{\alpha}{\big[H\big(\frac{\alpha+1}{\alpha}\big)\big]^{\alpha}}\bigg)$
  • and so $A=H\big(\frac{\alpha+1}{\alpha}\big)$

If you replace these three expressions into the power law distribution given in this answer then you get the pdf from the original problem.


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