Barabasi–Albert trees are hypoenergetic

Abstract

We prove that graphs following the model of Barabasi–Albert tree with n vertices are hypoenergetic in the large n limit.

Appendix 1: Number of neighbors with degree 2

Let us denote by $n_{kl}$, the proportion of edges that connect a pair of nodes of degrees k and l with $k\le l$ in our graph G. In [13] it was proven that as $n\rightarrow \infty$ we have

$$\begin{aligned} E(n_{kl})&= \frac{4(l-1)}{k(k+1)(k+l)(k+l+1)(k+l+2)}\\&\quad + \frac{12(l-1)}{k(k+l-1)(k+l)(k+l+1)(k+l+2)}+o(1). \end{aligned}$$

In particular, $E(n_{2,2})=\frac{1}{45}+o(1)$.

In this appendix we prove that with probability tending to 1, as $n\rightarrow \infty$ we have

$$\begin{aligned} n_{2,2}=\frac{1}{45}+o(1), \end{aligned}$$

which is required in the proof or our main theorem. More precisely, we will prove the following theorem

Theorem A.1

Let $T_n$ be a Barabasi–Albert tree on n vertices and let $n_{2,2}(n)$ the number of edges connecting two vertices of degree 2. Then

$$\begin{aligned}\mathbb {P}(|n_{2,2}(n)-1/45|>\sqrt{ \log n/n})\rightarrow 0\end{aligned}$$

as n tends to infinity.

We follow the strategy of [4] by using Azuma–Hoeffding inequality.

Proposition A.2

(Azuma–Hoeffding inequality) Let $(X_t)_{t=0}^n$ be a martingale with respect to the filtration $\mathcal {F}_t$. Assume that there are predictable processes $(A_t)$ and $(B_t)$ (i.e., $A_t, B_t \in \mathcal {F}_{t-1}$) and a constant $0< c < + \infty$ such that for all $t \ge 1$, almost surely

$$\begin{aligned}A_t \le X_t -X_{t-1} \le B_t\end{aligned}$$

and

$$\begin{aligned}B_t-A_t \le c.\end{aligned}$$

Then for all $\beta > 0$

$$\begin{aligned}\mathbb {P}(|X_t - X_0| \ge \beta ) \le 2 \exp \left( - \frac{ \beta ^2}{2c^2n}\right) .\end{aligned}$$

To prove the claim we will build a martingale $(X_n)^n_{i=0}$, such that $X_n=n_{2,2}(n)$, $X_0=E(n_{2,2})\approx 1/45$ and $|X_i-X_{i-1}|\le 4/n$. If this is the case then by applying the Azuma–Hoeffding inequality to $X_n$ with $c=4/n$ and $\beta =\sqrt{\log n}$ we obtain

$$\begin{aligned}\mathbb {P}(|X_n - E(X_n)| \ge \sqrt{\log n/n})\le 2 \exp \big (- \frac{ \log n }{32}\big ) =2n^{-32},\end{aligned}$$

which together with the fact the $E(X_n)\approx 1/45$, as $n\rightarrow \infty$, give the desired result.

To construct such martingale we consider the Barabasi–Albert tree $T_t$ as embedded in the process $\{T_t\}_{i\ge 0}$, where $T_{t}$ is built from $T_{t-1}$ by adding joining the vertex $v_t$ to the vertex $v_i$ with probability proportional to $\mathrm{{deg}}(v_i)$ as described above.

Now let $\mathcal {F}_t$ be the $\sigma$-algebra generated by $\{T_1,...T_t\}$. We claim that $X_t=E(n_{2,2}(n)|F_t)$ satisfies the above properties. The fact that it is a martingale with $X_n=n_{2,2}(n)$ and $X_0=E(n_{2,2}(n))$ is clear. So the main thing to check is the $|X_{t}-X_{t+1}|\le 4$. To see this note as in [4] that whether at stage t we join $v_t$ to $v_i$ or $v_j$ does not affect the degrees at later times of vertices $v_k$, $k\notin \{i,j\}$.

More precisely, the joint distribution of all other degrees is the same in either case.

So we are only interested in the number of vertices v attached to $v_i$ or $v_j$ with degree 2, provided that $v_i$ or $v_j$ have degree 2. There are at most 2 such vertices for in either case and thus, choosing $v_i$, instead of $v_j$ changes the number of edges joining vertices of degree 2 in at most 4 and then $n_{22}(n)$ changes in at most 4/n. Thus $|X_t-X_{t+1}|\le 4/n$ as desired.