exponent of the powerlaw distribution and Barabasi-Albert-Model; value of the exponent

Question

I am new to studying the scale-free networks and applying it to Microbial communities for research work which include Barabasi-Albert-Model. In the course of the study, I came across the degree distribution which follows power-law from study on the https://arxiv.org/abs/cond-mat/9910332 by Albert and Barabsi(1999). I don't know how to calculate or measure and compare the value of $\alpha$ from the equation $p(x) $~$ cx^{-\alpha}$.

Secondly, $$ \alpha=1+n\left[\sum_{i=1}^n \ln\frac{x_i}{x_{min}}\right]^{-1}.$$

1. I tried using the equation above to obtain the value of $\alpha$ with respect to $p(x) $~$ cx^{-\alpha}$, but still couldn't obtain the value of $\alpha$ for an estimation against the degree of distribution on the graph.

2. From Barabasi article (1999), for the citation and power grid date, how are the values of gamma obtained to be 2.3 and 4? Note, '$\gamma$' can be used in the place of '$\alpha$' from the equation.

I would appreciate any help with this.

Thanks

Answer 1

Let's start from your definition of the power law distribution: $p(x)=cx^{-\alpha}$, that is valid from some $x_{min}$. First we want to calculate $c$. We know that the integrated probability is $1$. $$1=c\int_{x_{min}}^\infty x^{-\alpha}dx=c\frac{x_{min}^{1-\alpha}}{\alpha-1}$$ $$c=\frac{\alpha-1}{x_{min}^{1-\alpha}}$$ We can therefore write $$p(x)=\frac{\alpha-1}{x_{min}}\left(\frac{x}{x_{min}}\right)^{-\alpha}$$ There is a nice article that explains the concept of log likelihood. It boils down to this: the probability of getting $x_1, x_2, ..., x_n$ is$$\cal L(\alpha,x_1,x_2,...,x_n)=p(x_1)p(x_2)...p(x_n)$$ You want to take the derivative with respect to $\alpha$, such that you maximize the likelihood. This is a little tedious, so you can consider $\ln \cal L$ instead, You can do that since the logarithmic function is monotonic, so a maximum of $\cal L$ occurs at the same position $\ln\cal L$. $$\begin{align}\frac{d}{d\alpha}\ln\cal L&=\sum_{i=1}^n\frac{d}{d\alpha}\ln\frac{\alpha-1}{x_{min}}\left(\frac{x_i}{x_{min}}\right)^{-\alpha}\\&=\sum_{i=1}^n\left(\frac 1{\alpha-1}-\ln\frac{x_i}{x_{min}}\right)\\&=0\end{align}$$ From here $$\frac n{\alpha-1}=\sum_{i=1}^n\ln\frac{x_i}{x_{min}}$$ A simple re-arrangement of terms give you the initial formula.

Note a simple way to get $\alpha$, without requiring that the function is normalized, is to take the logarithm of the initial form of the distribution. $$\ln p(x)=c-\alpha \ln x$$ You can see that in the log-log coordinates this is the equation of a straight line with slope $-\alpha$. Just do some linear fit to get $\alpha$.