# Fitness model for scale free networks

In order to generate scale-free networks, we can use this algorithm, derived from Barabási–Albert model:

1) we assign every node a "weight" $\theta_i$ (or two in the direct case).

2) we place $m$ edges between nodes by choosing the ends of the edges with probabilities proportional to the weights.

With the right choice of the coefficients $\{\theta\}$ we can generate a graph with the desired power-law distribution. A very popular method is the so-called Chung-Lu Model and it's, for example, used in the power.law.game function in the igraph library. This model prescribes to number the nodes from 1 to N and then assign to each node a $\theta_i = i^{-\alpha}$, with $\alpha \in [0,1)$, then select nodes as the ends of an edge with probability $p_{ij} \propto \theta_i \theta_j$, and repeating for all the edges to put in. The resulting graph will have a degree distribution $p(k) \propto k^{-\gamma}$, with $\gamma = \frac{\alpha}{\alpha +1}$.

My question is: why can't we simply consider a power-law distribution with a finite extent $p(x) = \frac{1-\gamma}{x_{max}^{1-\gamma} - x_{min}^{1-\gamma}} x^{-\gamma}$, then sample our $\theta_i$ from this distribution? This is more intuitive from my perspective, because if my expected degrees follow a power-law distribution, I expect that the graph degree distribution follows the same power-law.

• Maybe the question is more appropriate for math stack Exchange? ShouldI copy paste this post? – Francesco Di Lauro Sep 30 '17 at 8:26

The algorithm that is used to build the graph is not compatible with a different choice other than $\theta_i = i^{-\alpha}$.
This is because the algo is derived from the Barabási–Albert with some modifies. We start with N nodes, and we add a link between a couple of nodes every time step until we have added $mN$ edges. First we normalized the weights so they will be the probability that we choose the node $i, \;$ $\sum_{i=1}^{N} i^{-\alpha} \sim \frac{1-\alpha}{N^{1-\alpha}}$. We can easily compute the rate at which the node acquires links, since it will be proportional to its degree: $$\frac{d\,k_i}{dt} = \frac{k_i}{2mN} = \frac{1-\alpha}{N^{1-\alpha} i^\alpha},$$ this means that we will select the node $i$ with probability $p(i) = k_i ^{-\frac{1}{\alpha}}$. This is equal to the probability that a node has a degree $k>k_i$, so it is related to the cumulative probability distribution: $$P(k>\bar{k}) \sim k^{-\frac{1}{\alpha}} \rightarrow p(k) \sim k^{-\frac{1}{\alpha} -1} \equiv k^{-\gamma},$$ where $\gamma = \frac{1+\alpha}{\alpha}$. Is this a good answer?