# what family of distributions is this?

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.

• some sort of power law distribution maybe? – Alex Feb 4 '15 at 11:15
• 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 $\alpha$. – wolfies Feb 4 '15 at 14:09
• H is a constant. Not really relevant to the problem. Added to the question. – luchonacho Feb 4 '15 at 15:13
• This is a good source of distributions but I have not checked it for yours. causascientia.org/math_stat/Dists/Compendium.pdf – user121049 Feb 4 '15 at 15:37
• 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? – luchonacho Feb 4 '15 at 16:04

## 1 Answer

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.