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I use the poweRlaw package in R to fit a power law to my data. I am trying to figure out what is the value of the Pareto exponent.

Assume the probability mass function is defined by:

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

and the complementary cumulative density function is defined by:

$$ P(x) = \int_x ^\infty p(x’) dx’ = \left(\frac{x}{x_{min}}\right)^{-\alpha + 1} $$

Is the Pareto exponent $\alpha$ or $- \alpha + 1$ or $\alpha - 1$? In most literature, the CCDF is used to describe the income/wealth distribution, and $-(\alpha-1)$ is the slope of the CCDF on a log-log plot, so $\alpha-1$ seems the most intuitive. I'm pretty sure the R library poweRlaw returns $\alpha$ as defined above. I am using Newman as a reference.

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A random variable $X$ has a Pareto distribution with Pareto exponent $\theta$ if $$\text{P}(X>x)=\begin{cases} \left(\frac{x}{x_m}\right)^\theta \quad \text{ if } x\geq x_{min}\\ \ \ 1 \quad \quad \quad \text{ if } x<x_{min} \end{cases}$$

In this case, the Pareto exponent is $\theta = \alpha - 1$. Remember that $P(X>x)=\int_x^\infty p(x')\mathrm{d}x'$.

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  • $\begingroup$ Is $\theta$ defined correctly? $\endgroup$ – 0x90 Sep 1 at 23:22

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