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Is a power law distribution right skewed? Is there any relation between concepts of right skewness and power law distribution?

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Is all power-law distributions right skewed? –  user25004 Apr 22 '13 at 22:49

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No, the student-t distribution is a power law distribution and it is always symmetric. The same goes for the Cauchy distribution.

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To answer the question, we need to define mathematically what we mean by a power-law distribution. The usual definition looks like this

$p(x)=\frac{\alpha-1}{x_{\rm min}}\left(\frac{x}{x_{\rm min}}\right)^{-\alpha}$

where $\alpha$ is the scaling parameter or exponent of the distribution, and $x_{\rm min}$ is the smallest value for which the power-law form holds. If this is what you mean by "power law distribution", then absolutely yes, it is right skewed.

Skewness is a measure of the degree to which a distribution is asymmetric, and is thus defined for any probability distribution (that is, it's a very general concept, independent of power-law distributions per se). In the definition above, the support of the distribution is $0< x_{\rm min} \leq x < \infty$, and the power-law is a monotonically decreasing function above $x_{\rm min}$. Therefore, this distribution is highly asymmetric. This idea can be formalized by deriving the equation for the skewness of the distribution, which is related to its third moment. Power-law distributions have a very simply mathematical expression for all their moments. For any distribution, the $k$th moment is defined as

$\langle x^k \rangle = \int_{x_{\rm min}}^\infty x^k p(x) {\rm d}x $

If you plug in the expression for $p(x)$ for the power law and compute the result, you find

$\langle x^k \rangle = \frac{\alpha-1}{\alpha-1-k}x^k_{\rm min}$

For simplicity, consider $x_{\rm min}=1$. In this case, the $k$th moment of the power-law distribution is only defined (meaning less than $\infty$) when $k<\alpha-1$, and whenever it is defined $\langle x^k \rangle>0$ if $\alpha$ takes a finite value.

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