Surprising Behavior of the Average Degree for a Node’s Neighbors in Growth Networks

Abstract

We study the variation of the stochastic process that describes the temporal behavior of the average degree of the neighbors for a fixed node in the Barabási-Albert networks. It was previously known that the expected value of this random quantity grows logarithmically with the number of iterations. In this paper, we use the mean-field approach to derive difference stochastic equations, as well as their corresponding approximate differential equations, in order to find the dynamics of its variation in time. The noteworthy fact proved in this paper is that the variation of this process is bounded by a constant. This behavior is fundamentally different from the dynamics of variation in most known stochastic processes (e.g., the Wiener process), in which its value tends to infinity over time.

This work was supported by the Ministry of Science and Higher Education of the Russian Federation in the framework of the basic part of the scientific research state task, project FSRR-2020-0006.

Appendices

A The Proof of Lemma 1

Proof

Denote $\beta _i(t):=\frac{s_i(t)}{d_i^2(t)}$. We have

$$\begin{aligned}&\varDelta \beta _i(t+1):=\beta _i(t+1)-\beta _i(t) \nonumber \\&\qquad \qquad \quad =\,\left( \frac{s_i(t)+m}{(d_i(t)+1)^2}-\frac{s_i(t)}{d_i^2(t)}\right) \xi _i^{t+1}+\left( \frac{s_i(t)+1}{(d_i(t)+1)^2}-\frac{s_i(t)}{d_i^2(t)}\right) \eta _i^{t+1}. \end{aligned}$$

(5)

Since

$$ \mathbb {E}(\xi _i^{t+1})=\frac{d_i(t)}{2t},\ \mathbb {E}(\eta _i^{t+1})=\frac{s_i(t)}{2t}, $$

we get the difference equation

$$\begin{aligned}&\mathbb {E}\left( \varDelta \beta _1(t+1)|G_t\right) \nonumber \\&\qquad =\,\left( \frac{s_i(t)+m}{(d_i(t)+1)^2}-\frac{s_i(t)}{d_i^2(t)}\right) \frac{d_i(t)}{2t}+\left( \frac{s_i(t)+1}{(d_i(t)+1)^2}-\frac{s_i(t)}{d_i^2(t)}\right) \frac{s_i(t)}{2t} \nonumber \\&\qquad =\,\frac{md_i(t)}{2t(d_i(t)+1)^2}-\frac{s_i(t)}{t(d_i(t)+1)^2}-\frac{s_i(t)}{2td_i(t)(d_i(t)+1)^2}+\frac{s_i(t)}{2td_i^2(t)} \nonumber \\&\qquad \qquad =\, -\frac{s_i(t)}{2td_i^2(t)}+\frac{m}{2td_i(t)}+\frac{3s_i(t)}{2td_i^3(t)}-\frac{m}{td_i^2(t)}-\frac{4s_i(t)}{td_i^4(t)}+O\left( \frac{s_i(t)}{td_i^5t)}\right) . \end{aligned}$$

(6)

Using $\mathbb {E}\left( \frac{1}{d_i(t)}\right) \sim \frac{i^{\frac{1}{2}}}{mt^{\frac{1}{2}}}$, we get the following approximate first order differential equation:

$$ \frac{df(t)}{dt}=-\frac{f(t)}{2t} + \frac{i^{\frac{1}{2}}}{2t^{\frac{3}{2}}}, $$

the solution of which is $f(t)=\frac{1}{2} \left( \frac{i}{t}\right) ^{\frac{1}{2}}(\log t+b)$, where b is a constant.

   $\square$

B The Proof of Lemma 2

To prove Lemma 2 we need two auxiliary Lemmas 3 and 4.

The next lemma complements the results of papers [9, 16].

Lemma 3

$$ \mathbb {E}\left( \alpha _i(t)\right) \sim \frac{m}{2}\log t+a-\frac{1}{2} \left( \frac{i}{t}\right) ^{\frac{1}{2}}\log t+\frac{1}{2} b\left( \frac{i}{t}\right) ^{\frac{1}{2}}, $$

where a is a constant.

Proof

We have

$$\begin{aligned}&\varDelta \alpha _i(t+1):=\alpha _i(t+1)-\alpha _i(t) \nonumber \\&\qquad \qquad \qquad \qquad =\,\left( \frac{s_i(t)+m}{d_i(t)+1}-\frac{s_i(t)}{d_i(t)}\right) \xi _i^{t+1}+\left( \frac{s_i(t)+1}{d_i(t)+1}-\frac{s_i(t)}{d_i(t)}\right) \eta _i^{t+1}. \end{aligned}$$

(7)

Since

$$ \mathbb {E}(\xi _i^{t+1})=\frac{d_i(t)}{2t},\ \mathbb {E}(\eta _i^{t+1})=\frac{s_i(t)}{2t}, $$

we get the difference equation

$$\begin{aligned}&\mathbb {E}\left( \varDelta \beta _1(t+1)|G_t\right) \nonumber \\&\qquad \qquad =\,\left( \frac{s_i(t)+m}{d_i(t)+1}-\frac{s_i(t)}{d_i(t)}\right) \frac{d_i(t)}{2t}+\left( \frac{s_i(t)+1}{d_i(t)+1}-\frac{s_i(t)}{d_i(t)}\right) \frac{s_i(t)}{2t} \nonumber \\&\qquad \qquad \qquad \qquad =\,\frac{m}{2t}+\frac{s_i(t)}{2t(d_i(t)+1)}-\frac{m}{2td_i(t)(d_i(t)+1)} \nonumber \\&\qquad \qquad \qquad =\,\frac{m}{2t}+\frac{s_i(t)}{2td_i^2(t)}-\frac{m}{2td_i(t)}-\frac{s_i(t)}{td_i^3(t)}+\frac{m}{2td_i^2(t)}+O\left( \frac{s_i(t)}{td_i^4(t)}\right) . \end{aligned}$$

(8)

Using Lemma 1, we get the following approximate first order differential equation:

$$ \frac{df(t)}{dt}=\frac{m}{2t} + \frac{i^{\frac{1}{2}}}{4t^{\frac{3}{2}}}\log t+ \frac{bi^{\frac{1}{2}}}{4t^{\frac{3}{2}}}-\frac{i^{\frac{1}{2}}}{2t^{\frac{3}{2}}}, $$

the solution of which is $f(t)=\frac{m}{2}\log t-\frac{1}{2} \left( \frac{i}{t}\right) ^{\frac{1}{2}}\log t+\frac{1}{2} b\left( \frac{i}{t}\right) ^{\frac{1}{2}}+a$, where a is a constant.

   $\square$

Lemma 4

$$ \mathbb {E}\left( \frac{s_i^2(t)}{d_i^3(t)}\right) \sim m \left( \frac{i}{t}\right) ^{\frac{1}{2}}\left( \frac{1}{4}\log ^2 t+\frac{b}{2}\log t+c\right) , $$

where c is a constant.

Proof

We have

$$\begin{aligned}&\varDelta _i(t+1):=\frac{s_i^2(t)(t+1)}{d_i^3(t)(t+1)}-\frac{s_i^2(t)}{d_i^3(t)} \nonumber \\&\qquad \qquad \quad =\,\left( \frac{(s_i(t)+m)^2}{(d_i(t)+1)^3}-\frac{s_i^2(t)}{d_i^3(t)}\right) \xi _i^{t+1}+\left( \frac{(s_i(t)+1)^2}{(d_i(t)+1)^3}-\frac{s_i^2(t)}{d_i^3(t)}\right) \eta _i^{t+1}. \end{aligned}$$

(9)

Since

$$ \mathbb {E}(\xi _i^{t+1})=\frac{d_i(t)}{2t},\ \mathbb {E}(\eta _i^{t+1})=\frac{s_i(t)}{2t}, $$

we get the difference equation

$$\begin{aligned}&\mathbb {E}\left( \varDelta _i(t+1)|G_t\right) \nonumber \\&\qquad =\,\left( \frac{(s_i(t)+m)^2}{(d_i(t)+1)^3}-\frac{s_i^2(t)}{d_i^3(t)}\right) \frac{d_i(t)}{2t}+\left( \frac{(s_i(t)+1)^2}{(d_i(t)+1)^3}-\frac{s_i^2(t)}{d_i^3(t)}\right) \frac{s_i(t)}{2t} \nonumber \\&\quad =\,\frac{ms_i(t)d_i(t)}{t(d_i(t)+1)^3}+\frac{m^2d_i(t)}{2t(d_i(t)+1)^3}-\frac{3s_i^2(t)}{2t(d_i(t)+1)^3}-\frac{3s_i^2(t)}{2td_i(t)(d_i(t)+1)^3} \nonumber \\&\qquad \qquad \qquad \quad -\,\frac{s_i^2(t)}{2td_i^2(t)(d_i(t)+1)^3}+\frac{s_i^2(t)}{td_i^3(t)}+\frac{s_i(t)}{2td_i^3(t)} \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\,-\frac{s_i^2(t)}{2td_i^3(t)}+\frac{ms_i(t)}{td_i^2(t)}+O\left( \frac{s_i^2(t)}{td_i^4(t)}\right) . \end{aligned}$$

(10)

Using Lemma 1, we get the following approximate first order differential equation:

$$ \frac{df(t)}{dt}=-\frac{f(t)}{2t} + \frac{m i^{\frac{1}{2}}}{2t^{\frac{3}{2}}}\log t + \frac{bmi^{\frac{1}{2}}}{2t^{\frac{3}{2}}}, $$

the solution of which is $f(t)=m \left( \frac{i}{t}\right) ^{\frac{1}{2}}\left( \frac{1}{4}\log ^2 t+\frac{b}{2}\log t+c\right)$, where c is a constant.

   $\square$

Proof

(of Lemma 2) We have

$$\begin{aligned}&\varDelta \alpha _i^2(t+1):=\alpha _i^2(t+1)-\alpha _i^2(t) \nonumber \\&\qquad \qquad \,\, =\,\left( \frac{(s_i(t)+m)^2}{(d_i(t)+1)^2}-\frac{s_i^2(t)}{d_i^2(t)}\right) \xi _i^{t+1}+\left( \frac{(s_i(t)+1)^2}{(d_i(t)+1)^2}-\frac{s_i^2(t)}{d_i^2(t)}\right) \eta _i^{t+1}. \end{aligned}$$

(11)

Since

$$ \mathbb {E}(\xi _i^{t+1})=\frac{d_i(t)}{2t},\ \mathbb {E}(\eta _i^{t+1})=\frac{s_i(t)}{2t}, $$

we get the difference equation

$$\begin{aligned}&\quad \mathbb {E}\left( \varDelta \alpha _i^2(t+1)|G_t\right) \nonumber \\&\qquad \quad =\,\left( \frac{(s_i(t)+m)^2}{(d_i(t)+1)^2}-\frac{s_i^2(t)}{d_i^2(t)}\right) \frac{d_i(t)}{2t}+\left( \frac{(s_i(t)+1)^2}{(d_i(t)+1)^2}-\frac{s_i^2(t)}{d_i^2(t)}\right) \frac{s_i(t)}{2t} \nonumber \\&=\,\frac{ms_i(t)d_i(t)}{t(d_i(t)+1)^2}+\frac{m^2}{2t(d_i(t)+1)^2}-\frac{s_i^2(t)}{t(d_i(t)+1)^2}-\frac{s_i^2(t)}{2td_i(t)(d_i(t)+1)^2}+\frac{s_i^2(t)}{td_i^2(t)}+\frac{s_i(t)}{2td_i(t)} \nonumber \\&\qquad \qquad \,\, =\,\frac{ms_i(t)}{td_i(t)} +\frac{3s_i^2(t)}{2td_i^3(t)}-\left( 2m-\frac{1}{2}\right) \frac{s_i(t)}{td_i^2(t)}+\frac{m^2}{2td_i(t)}+O\left( \frac{s_i^2(t)}{td_i^4(t)}\right) . \end{aligned}$$

(12)

Using Lemmas 1, 3 and 4, we get the following approximate first order differential equation:

$$\begin{aligned}&\frac{df(t)}{dt}= \frac{m}{t} \left( \frac{m}{2}\log t+a-\frac{1}{2} \left( \frac{i}{t}\right) ^{\frac{1}{2}}\log t+\frac{1}{2} b\left( \frac{i}{t}\right) ^{\frac{1}{2}}\right) \nonumber \\&\qquad \qquad \qquad \qquad +\,\frac{3m}{2t} \left( \frac{i}{t}\right) ^{\frac{1}{2}}\left( \frac{1}{4}\log ^2 t+\frac{b}{2}\log t+c\right) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad -\,\left( 2m-\frac{1}{2}\right) \frac{1}{2} \left( \frac{i}{t}\right) ^{\frac{1}{2}}(\log t+b) + \frac{mi^{\frac{1}{2}}}{2t^{\frac{3}{2}}}, \end{aligned}$$

(13)

the solution of which follows

$$ f(t)=\frac{m^2}{4}\log ^2 t+am\log t +d-\frac{3}{4}\,m \left( \frac{i}{t}\right) ^{\frac{1}{2}} \log ^2 t+O\left( \frac{\log t}{t^{\frac{1}{2}}}\right) , $$

where d is a constant.

   $\square$