Analytic solution for preferential attachment probabilities scheme

TL;DR

This paper presents an explicit analytical solution for the preferential attachment probabilities scheme, modeling it as a sequence of dependent Bernoulli variables to better understand scale-free network formation.

Contribution

It introduces a novel explicit expression for transition probabilities in the preferential attachment scheme, framing it as dependent Bernoulli variables.

Findings

Explicit transition probability formula derived

Modeling as dependent Bernoulli variables enhances understanding

Applicable to scale-free network analysis

Abstract

Preferential attachment probabilities scheme appear in the context of scale free random graphs [1],[2]. In this work we present preferential attachment probabilities scheme as a sequence of dependent Bernoulli random variables and we give an explicit expression of the transition probabilities.

Equations

$$P(X_m(n)=k)=$$

$$+P(X_m(n)=k|X_m(n-1)=k)P(X_m(n-1)=k)=$$

$$=P(X_m(n-1)+Y_m(n)=k|X_m(n-1)=k-1)P(X_m(n-1)=k-1)+$$

$$+P(X_m(n-1)+Y_m(n)=k|X_m(n-1)=k)P(X_m(n-1)=k)=$$

$$=P(Y_m(n)=1|X_m(n-1)=k-1)P(X_m(n-1)=k-1)+$$

$$+P(Y_m(n)=0|X_m(n-1)=k)P(X_m(n-1)=k)=$$

$$=\frac{k-1}{2n-1}P(X_m(n-1)=k-1)+\frac{2n-1-k}{2n-1}P(X_m(n-1)=k)$$

$$p_{n,k}^m=p_{n-1,k-1}^m\frac{k-1}{2n-1}+p_{n-1,k}^m\frac{2n-1-k}{2n-1},0\le m<n,0\le k\le n-m$$

$$\displaystyle\left\{\begin{array}[]{ll}p_{1,1}=1,\quad p_{n,k}=0,\quad\forall k>n\\ p_{n,k}=\frac{k-1}{2n-1}p_{n-1,k-1}+\frac{2n-1-k}{2n-1}p_{n-1,k},\quad\forall k\leq n\end{array}\right.$$

$$p_{n,1}=\frac{2(n-1)}{2n-1}{p}_{n-1,1}\forall n\ge 2$$

$$p_{n,1}=\frac{2^{n-1}(n-1)!}{{\prod }_{i=1}^n(2i-1)}\forall n\ge 1$$

$$p_{n,n}=\frac{n-1}{2n-1}{p}_{n-1,n-1}\forall n\ge 2$$

$$p_{n,n}=\frac{(n-1)!}{{\prod }_{i=1}^n(2i-1)}\forall n\ge 1$$

$$a_{n,k}=\frac{2^{n-k}{\prod }_{i=1}^n(2i-1)}{(n-1)!}{p}_{n,k}$$

$$\displaystyle\left\{\begin{array}[]{ll}a_{n,1}=2^{2(n-1)},\quad a_{n,n}=1,\quad a_{n,k}=0\quad\forall k>n\\ a_{n,k}=\frac{1}{n-1}\left[(k-1)a_{n-1,k-1}+2(2n-1-k)a_{n-1,k}\right]\quad\forall k=1,...,n-1\quad n>1\end{array}\right.$$

$$\displaystyle\left\{\begin{array}[]{ll}a_{n,n}=1\\ a_{n,k}=2^{2(n-k)}\left[1+(k-1)\sum_{j=0}^{n-k-1}\frac{2^{-2(j+1)}(k+2j)!}{(j+1)!(k+j)!}\right]\quad\forall k=1,...,n-1\quad n>1\end{array}\right.$$

$$a_{n,n-1}=\frac{1}{n-1}\left[(n-2)a_{n-1,n-2}+2n{a}_{n-1,n-1}\right]$$

$$2^2\left[1+(n-2)\sum _{j=0}^0\frac{2^{-2(j+1)}(n-1+2j)!}{(j+1)!(n-1+j)!}\right]=$$

$$=\frac{n-2}{n-1}{2}^{2(n-1-n+2)}\left[1+(n-3)\sum _{j=0}^{n-1-n+2-1}\frac{2^{-2(j+1)}(n-2+2j)!}{(j+1)!(n-2+j)!}\right]+\frac{2n}{n-1}$$

$$2^2+(n-2)=\frac{n-2}{n-1}{2}^2\left[1+(n-3)2^{-2}\right]+\frac{2n}{n-1}$$

$$2^2(n-1)-2n=2n-4$$

$$-4=-4$$

$$(n-1)\left[1+(k-1)\sum _{j=0}^{n-k-1}\frac{2^{-2(j+1)}(k+2j)!}{(j+1)!(k+j)!}\right]=(k-1)\left[1+(k-2)\sum _{j=0}^{n-k-1}\frac{2^{-2(j+1)}(k-1+2j)!}{(j+1)!(k-1+j)!}\right]$$

$$(n-1)\left[1+(k-1)\sum _{j=0}^{n-k-1}\frac{2^{-2(j+1)}(k+2j)!}{(j+1)!(k+j)!}\right]=\frac{1}{2}(2n-1-k)\left[1+(k-1)\sum _{j=0}^{n-k-2}\frac{2^{-2(j+1)}(k+2j)!}{(j+1)!(k+j)!}\right]$$

$$(n-1)\left[1+(k-1)\sum _{j=0}^{n-k-1}\frac{2^{-2(j+1)}(k+2j)!}{(j+1)!(k+j)!}\right]=\left[(n-1)-\frac{1}{2}(k-1)\right]\left[1+(k-1)\sum _{j=0}^{n-k-2}\frac{2^{-2(j+1)}(k+2j)!}{(j+1)!(k+j)!}\right]$$

$$\sum _{j=0}^{n-k-2}\frac{2^{-2(j+1)}}{(j+1)!}\left[2(k-2)\frac{(k-1+2j)!}{(k-1+j)!}-(k-1)\frac{(k+2j)!}{(k+j)!}\right]=2^{-2(n-k-1)}\frac{(2n-k-3)!}{(n-k-1)!(n-2)!}-1$$

$$\sum _{j=0}^r\frac{2^{-2(j+1)}}{(j+1)!}\left[2(k-2)\frac{(k-1+2j)!}{(k-1+j)!}-(k-1)\frac{(k+2j)!}{(k+j)!}\right]=2^{-2(r+1)}\frac{(k+1+2r)!}{(r+1)!(k+r)!}-1$$

$$\frac{2^{-2}}{1!}[2(k-2)-(k-1)]=\frac{2^{-2}(k+1)!}{1!k!}-1$$

$$2^{-2(r+1)}\frac{(k+1+2r)!}{(r+1)!(k+r)!}+\frac{2^{-2(r+2)}}{(r+2)!}\left[2(k-2)\frac{(k+1+2r)!}{(k+r)!}-(k-1)\frac{(k+2+2r)!}{(k+1+r)!}\right]=2^{-2(r+2)}\frac{(k+3+2r)!}{(r+2)!(k+r+1)!}$$

$$2^{2(r+2)}\frac{(k+1+2r)!}{(k+r)!}+2(k-2)\frac{(k+1+2r)!}{(k+r)!}-(k-1)\frac{(k+2+2r)!}{(k+1+r)!}=\frac{(k+3+2r)!}{(k+r+1)!}$$

$$2(r+2)(k+r+1)+2(k-2)(k+r+1)-(k-1)(k+2+2r)=(k+3+2r)(k+2+2r)$$

$$p_{n,k}$$

$$p_{n,k}=\frac{n!(n-1)!}{(2n)!}{2}^{2n-k}\left[1+(k-1)\sum _{j=0}^{n-k-1}\frac{2^{-2(j+1)}}{(j+1)!}\frac{(k+2j)!}{(k+j)!}\right]$$

Full text

1 Introduction

Preferential attachment probabilities schemes arise in the context of scale free random graphs, following the description used in [2]:

The preferential linking arises in this simple model not because of some special rule including a function of vertex degree as in [1],[3] but quite naturally. Indeed, in the model that we consider here, the probability that a vertex has a randomly chosen edge attached to it is equal to the ratio of the degree $k$ of the vertex and the total number of edges, $2t-1$.

we want to mimic this dynamic using a sequence of dependent Bernoulli random variables.

Let $\{Y_{m}(n)\}_{n>0}$ a sequence of r.v. such that $Y_{m}(n)\sim Ber\left(\frac{\sum_{i=m}^{n-1}Y_{m}(i)}{2n-1}\right)$, with $Y_{m}(1)=1$. The sequence $\{Y_{m}(n)\}_{n>0}$, represent the trails to attach a new arcs at node $m$ when a new node $n$ is generated and jointed to the network, obviously $n>m$. The degree of the node $m$ will be $X_{m}(n)=\sum_{i=m}^{n}Y_{m}(i)$.

We recover transition probabilities using elementary probability arguments, then

[TABLE]

Now set $p^{m}_{n,k}=P(X_{m}(n)=k)$, we have that

[TABLE]

2 Recovering explicit expression of transition probabilities

Choosing $m=0$ in (1), i.e. we consider the probabilities transition equations for the first node, we have

[TABLE]

Observe that (1) is time invariant for all nodes scaling properly the quantity $n-m$.

For (4) holds that:

[TABLE]

i.e.

[TABLE]

also

[TABLE]

i.e.

[TABLE]

Let

[TABLE]

from (4),(5) and (6) we have that

[TABLE]

Now we give an analytic solution of (9).

Theorem 1**.**

Let $\{a_{n,k}\}$ as in (9), then

[TABLE]

**Proof

**

If $k=1$ obvious.

If $k=n-1$ and $\forall n\geq 2$, then (9) will be

[TABLE]

now substituting (12) in (13) we have that

[TABLE]

Let $k=2,...,n-2$ substituting we have that

[TABLE]

we recover that (2) is equivalent to

[TABLE]

Now let $n=k+2+r$, $r\geq 0$ the equality (15) become

[TABLE]

By induction we prove equality (16):

For $r=0$ we have

[TABLE]

i.e. $2^{-2}(k-3)=2^{-2}(k-3)$.

By induction hypothesis we prove equality (16) for $r+1$, supposing it holds for generic $r$, then:

[TABLE]

i.e.

[TABLE]

from obvious arrangement w.r.t. $k$ the thesis follow.

Now consider that $\prod_{i=1}^{n}(2i-1)=\frac{(2n)!}{2^{n}n!}$, for $k=1,...,n-1$ holds that

[TABLE]

i.e.

[TABLE]