Persistence of hubs in growing random networks

Abstract

We consider models of evolving networks $\left\{ {\mathcal {G}}_n:n\ge 0\right\}$ modulated by two parameters: an attachment function $f:{\mathbb {N}}_0 \rightarrow {\mathbb {R}}_+$ and a (possibly random) attachment sequence $\left\{ m_i:i\ge 1\right\}$. Starting with a single vertex, at each discrete step $i\ge 1$ a new vertex $v_i$ enters the system with $m_i\ge 1$ edges which it sequentially connects to a pre-existing vertex $v\in {\mathcal {G}}_{i-1}$ with probability proportional to $f(\text{ degree }(v))$. We consider the problem of emergence of persistent hubs: existence of a finite (a.s.) time $n^*$ such that for all $n\ge n^*$ the identity of the maximal degree vertex (or in general the K largest degree vertices for $K\ge 1$) does not change. We obtain general conditions on f and $\left\{ m_i:i\ge 1\right\}$ under which a persistent hub emerges, and also those under which a persistent hub fails to emerge. In the case of lack of persistence, for the specific case of trees ($m_i\equiv 1$ for all i), we derive asymptotics for the maximal degree and the index of the maximal deg ree vertex (time at which the vertex with current maximal degree entered the system) to understand the movement of the maximal degree vertex as the network evolves. A key role in the analysis is played by an inverse rate weighted martingale constructed from a continuous time embedding of the discrete time model. Asymptotics for this martingale, including concentration inequalities and moderate deviations form the technical foundations for the main results.

Appendix A. Verifying Assumptions C1, C2 and C3 for classes I and II in Remark 4.1

Class I As $f_r$ is positive and regularly varying with index $\alpha \in [0, 1/2)$, taking $\epsilon \in (0, \alpha )$ such that $2\alpha + \epsilon < 1$ and using [10, Theorem 1.5.6], we obtain

$$\begin{aligned} \lim _{k \rightarrow \infty }\frac{f_r(k)}{k^{2\alpha + \epsilon }} = 0. \end{aligned}$$

Recalling $f = f_rf_b \le b_2f_r$, we conclude $\Phi _2(\infty ) = \infty$, and hence, Assumption C1 holds.

Define $\Phi _{i,r} (t) := \int _0^t \frac{1}{f^i_r(z)}dz, \ t \ge 0, i=1,2$, and ${\mathcal {K}}_r(\cdot ) := \Phi _{2,r} \circ \Phi _{1,r}^{-1}(\cdot )$. Note that, using a simple u-substitution,

$$\begin{aligned} {\mathcal {K}}(t) := \Phi _2 \circ \Phi _1^{-1}(t) = \int _0^t \frac{1}{{\bar{f}}(z)}dz \end{aligned}$$

where ${\bar{f}}(\cdot ) := f \circ \Phi _1^{-1}(\cdot )$. Thus, for any $\delta >0, t \ge 0$, recalling $f \ge b_1 f_r$,

$$\begin{aligned} {\mathcal {K}}((1+\delta )t) = {\mathcal {K}}(t) + \int _t^{(1+\delta )t} \frac{1}{{\bar{f}}(z)}dz \le {\mathcal {K}}(t) + \frac{1}{b_1}\int _t^{(1+\delta )t} \frac{1}{f_r\circ \Phi _1^{-1}(z)}dz.\nonumber \\ \end{aligned}$$

(A.1)

Further, note that $\frac{1}{b_2} \Phi _{1,r}(s) \le \Phi _1(s) \le \frac{1}{b_1} \Phi _{1,r}(s)$ for $s \ge 0$, and hence, $\Phi ^{-1}_{1,r}(b_1s) \le \Phi ^{-1}_1(s) \le \Phi ^{-1}_{1,r}(b_2s), s \ge 0$. Thus,

$$\begin{aligned} 1 \le \frac{\Phi ^{-1}_1(s)}{\Phi ^{-1}_{1,r}(b_1s)} \le \frac{\Phi ^{-1}_{1,r}(b_2s)}{\Phi ^{-1}_{1,r}(b_1s)}, \ s \ge 0. \end{aligned}$$

(A.2)

By Theorem 1.5.12 of Bingham et al. [10], $\Phi ^{-1}_{1,r}(\cdot )$ is regularly varying with index $(1-\alpha )^{-1}$ and hence,

$$\begin{aligned} \lim _{s \rightarrow \infty } \frac{\Phi ^{-1}_{1,r}(b_2s)}{\Phi ^{-1}_{1,r}(b_1s)} = \left( \frac{b_2}{b_1}\right) ^{1/(1-\alpha )}. \end{aligned}$$

Using this in (A.2), we obtain $s_0>0$ such that for all $s \ge s_0$,

$$\begin{aligned} 1 \le \frac{\Phi ^{-1}_1(s)}{\Phi ^{-1}_{1,r}(b_1s)} \le 2 \left( \frac{b_2}{b_1}\right) ^{1/(1-\alpha )}. \end{aligned}$$

(A.3)

By Theorem 1.5.2 of [10], there exists $s_1 \ge s_0$ such that for all $s \ge s_1$,

$$\begin{aligned} \left| \frac{f_r(x\Phi ^{-1}_{1,r}(b_1s))}{f_r(\Phi ^{-1}_{1,r}(b_1s))} - x^{\alpha }\right| < \frac{x^{\alpha }}{2} \ \text { for all } \ x \in \left[ 1, 2 \left( \frac{b_2}{b_1}\right) ^{1/(1-\alpha )}\right] . \end{aligned}$$

In particular, by (A.3), for all $s \ge s_1$,

$$\begin{aligned} \frac{1}{2} \le \frac{f_r(\Phi ^{-1}_{1}(s))}{f_r(\Phi ^{-1}_{1,r}(b_1s))} \le \frac{3}{2}\left[ 2\left( \frac{b_2}{b_1}\right) ^{1/(1-\alpha )}\right] ^{\alpha } =: b(\alpha ). \end{aligned}$$

(A.4)

From (A.1) and (A.4), for $t \ge s_1$,

$$\begin{aligned} {\mathcal {K}}((1+\delta )t)&\le {\mathcal {K}}(t) + \frac{2}{b_1}\int _t^{(1+\delta )t} \frac{1}{f_r\circ \Phi _{1,r}^{-1}(b_1z)}dz\nonumber \\&= {\mathcal {K}}(t) + \frac{2}{b_1^2}\int _{b_1t}^{b_1(1+\delta )t} \frac{1}{f_r\circ \Phi _{1,r}^{-1}(z)}dz = {\mathcal {K}}(t) \nonumber \\&+ \frac{2\left( {\mathcal {K}}_r(b_1(1+\delta )t) - {\mathcal {K}}_r(b_1 t)\right) }{b_1^2}. \end{aligned}$$

(A.5)

Moreover, recalling $f \le b_2 f_r$ and again using (A.4), for $t \ge s_1$,

$$\begin{aligned} {\mathcal {K}}(t) \ge \frac{1}{b_2}\int _{s_1}^{t} \frac{1}{f_r\circ \Phi _1^{-1}(z)}dz\ge & {} \frac{1}{b_2 b(\alpha )}\int _{s_1}^{t}\frac{1}{f_r\circ \Phi _{1,r}^{-1}(b_1z)}dz\nonumber \\= & {} \frac{{\mathcal {K}}_r(b_1 t) - {\mathcal {K}}_r(b_1s_1)}{b_1 b_2 b(\alpha )}. \end{aligned}$$

(A.6)

Hence, by (A.5) and (A.6),

$$\begin{aligned} 1 \le \limsup _{t \rightarrow \infty } \frac{{\mathcal {K}}((1+\delta )t)}{{\mathcal {K}}(t)} \le 1 + \frac{2b_2b(\alpha )}{b_1}\limsup _{t \rightarrow \infty }\left( \frac{{\mathcal {K}}_r(b_1(1+\delta )t)}{{\mathcal {K}}_r(b_1t)} - 1\right) . \end{aligned}$$

(A.7)

By Proposition 1.5.8, Theorem 1.5.7 and Theorem 1.5.12 of [10], ${\mathcal {K}}_r(\cdot )$ is regularly varying with index $\theta _{\alpha } := \frac{1-2\alpha }{1-\alpha }$. Using this observation in (A.7), we conclude that for any $\delta >0$,

$$\begin{aligned} 1 \le \limsup _{t \rightarrow \infty } \frac{{\mathcal {K}}((1+\delta )t)}{{\mathcal {K}}(t)} \le 1 + \frac{2b_2b(\alpha )}{b_1}\left( (1+\delta )^{\theta _{\alpha }} - 1\right) . \end{aligned}$$

(A.8)

Taking a limit as $\delta \rightarrow 0$ in (A.8) shows that Assumption C2 is satisfied. Taking $\delta = 2$ in (A.8) shows that Assumption C3 is satisfied with $D = 2\left[ 1 + \frac{2b_2b(\alpha )}{b_1}\left( 3^{\theta _{\alpha }} - 1\right) \right]$ and some $t'>0$.

Class II From the assumptions $\sum _{k=0}^{\infty }\frac{1}{g^2(k)} = \infty$ and $\lim _{k \rightarrow \infty } h(k)/g(k) = 0$, it readily follows that $\Phi _2(\infty ) = \infty$, and hence, Assumption C1 holds. To verify Assumptions C2 and C3, extend g, h as f to all of $[0, \infty )$. Note that for any $\delta \ge 0$,

$$\begin{aligned} {\mathcal {K}}((1+\delta )t) = \int _0^{\Phi ^{-1}_1((1+\delta )t)}\frac{1}{f^2(z)}dz = (1+\delta )\int _0^t\frac{1}{f \circ \Phi ^{-1}_1((1+\delta )u)}du \nonumber \\ \end{aligned}$$

(A.9)

where the last step follows by a u-substitution with $u= \Phi _1(z)/(1+\delta )$.

Take any $\delta >0$. As $h(\cdot )$ is non-negative, $g(t) \le f(t)$ for all $t \ge 0$. Moreover, as $\lim _{t \rightarrow \infty } h(t)/g(t) = 0$ we obtain $t_{\delta }>0$ such that $f(t) \le (1+\delta )g(t)$ for all $t \ge t_{\delta }$. Hence, by (A.9), for all $t \ge \Phi _1(t_{\delta })$,

$$\begin{aligned} {\mathcal {K}}((1+\delta )t)&\le (1+\delta )\int _0^t\frac{1}{g \circ \Phi ^{-1}_1((1+\delta )u)}du \le (1+\delta )\int _0^t\frac{1}{g \circ \Phi ^{-1}_1(u)}du\\&\le (1+\delta )\int _0^{\Phi _1(t_{\delta })}\frac{1}{g \circ \Phi ^{-1}_1(u)}du + (1+\delta )^2\int _{\Phi _1(t_{\delta })}^t\frac{1}{f \circ \Phi ^{-1}_1(u)}du\\&\le (1+\delta )\int _0^{\Phi _1(t_{\delta })}\frac{1}{g \circ \Phi ^{-1}_1(u)}du + (1+\delta )^2\int _{0}^t\frac{1}{f \circ \Phi ^{-1}_1(u)}du\\&= (1+\delta )\int _0^{\Phi _1(t_{\delta })}\frac{1}{g \circ \Phi ^{-1}_1(u)}du + (1+\delta )^2{\mathcal {K}}(t). \end{aligned}$$

Hence, for any $\delta >0$,

$$\begin{aligned} \limsup _{t \rightarrow \infty } \frac{{\mathcal {K}}((1+\delta )t}{{\mathcal {K}}(t)} \le (1+\delta )^2. \end{aligned}$$

Assumptions C2 and C3 follow from this by respectively taking $\delta \rightarrow 0$ and $\delta = 2$.