Lecture 4. Giant component (1)

Consider the Erdös-Rényi graph model G(n,p). In previous lectures, we focused on the “high complexity regime”, i.e., as n goes to infinity, p is fixed. We discussed topics such as clique numbers and chromatic numbers. From now on, we shall consider the “low complexity regime”, where as n goes to infinity, p=\frac{c}{n} for a fixed constant c>0. As before, let \eta=(\eta_{i,j})_{i,j\in[n]} be the adjacency matrix of G(n, \frac{c}{n}). Then \{\eta_{ij}: i,j\in[n], i<j\} are i.i.d. Bernoulli random variables with success probability p=\frac{c}{n}.

Theorem 1 Let C_v be the connected component of G(n, \frac{c}{n}) that contains v\in [n].

  1. If c<1, then \max_v|C_v|= O(\log n) in probability.
  2. If c>1, then \max_v|C_v|\sim (1-\rho) n in probability, for some 0<\rho<1.
  3. If c=1, then \max_v|C_v|\sim n^{2/3} in distribution.

In the following lectures, we will aim to prove at least parts 1 and 2.

The exploration process

How to study |C_v|? We will explore C_v by starting an “exploration process'' at v that moves around C_v until all its sites have been visited. This walk will be constructed so that it hits each site once. So, the time it takes to explore all of C_v is exactly |C_v|. As a consequence, studying |C_v| reduces to studying a hitting time of a certain random walk, and to the latter we can apply martingale theory.

At each time t=0,1,2,\ldots, we maintain three sets of vertices:

    \begin{eqnarray*} 	R_t &=& \{\text{removed sites}\},\\ 	A_t &=& \{\text{active sites}\},\\  	U_t &=& \{\text{unexplored sites}\}. \end{eqnarray*}

Below is an illustration of how these sets are updated on a simple example.

  • At t=0, initialize A_0=\{v\}, U_0=[n]\backslash \{v\} and R_0=\varnothing. Namely, only v is active, all the vertices other than v are unexplored, and no vertices have been removed.

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  • At t=1, update A_1=A_0\backslash \{v\}\cup \{\text{neighbors of } v\}, U_1=U_0\backslash A_1 and R_1=\{v\}. Namely, all neighbors of v are moved from the unexplored set to the active set, and v itself is removed.

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  • At t=2, pick some x\in A_1 and update A_2=A_1\backslash \{x\}\cup \{\text{unexplored neighbors of } x\}, U_2=U_1\backslash A_2 and R_2=R_1\cup \{x\}. Namely, all unexplored neighbors of x are moved into the active set, and x itself is removed.

    Rendered by QuickLaTeX.com

  • At time t, we pick some vertex x from the current active set A_{t-1}, activate all unexplored neighbors of x and remove x itself.

This is a sort of local search along the connected component C_v: much like playing a game of Minesweeper! At each t, the choice of x\in A_{t-1} can be made arbitrarily (e.g., selecting the vertex with the smallest index or randomly selecting a vertex in A_{t-1}). The only requirement is that it is nonanticipating (only depending on the edges visited up to time t). For example, we cannot pick the vertex in A_{t-1} which has the largest number of unexplored neighbors, as this choice relies on unexplored edges.

A formal description of the “exploration process'':

  • Initialize A_0=\{v\}, U_0=[n]\backslash \{v\} and R_0=\varnothing.
  • For t\geq 0, we set

        \begin{eqnarray*} 	R_{t+1} &=& R_t\cup \{i_t\},\\ 	A_{t+1} &=& A_t\backslash \{i_t\}\cup \{w\in U_t: \eta_{wi_t}=1\},\\ 	U_{t+1} &=& U_t\backslash \{w\in U_t: \eta_{wi_t}=1\}, \end{eqnarray*}

    where i_t\in A_t is a nonanticipating but otherwise arbitrary choice.

This process stops when there are no more active vertices. It hits each vertex in C_v once and only once. At each time t, we remove one vertex in C_v. So the stopping time is exactly equal to the size of C_v:

    \[ \tau\equiv \inf\{t: |A_t|=0 \} = |C_v|. \]

So, we only need to study the stopping time \tau.

Recall that \eta_{ij} indicates whether there is an edge between vertices i and j, and \eta_{ij}\overset{\mathrm{i.i.d.}}{\sim} \text{Bernoulli}(\frac{c}{n}). By construction,

    \[ |A_{t+1}| = |A_t| - 1 + \sum_{w\in U_t} \eta_{wi_t}. \]

Now, let's do a thought experiment (wrong, but intuitive). Let's forget for the moment that some sites were previously visited, and assume that in each step all neighbors of i_t are unvisited still (note that when n is really large and t is relatively small, this assumption makes sense). Then |A_{t+1}|-|A_t|+1 is the sum of n independent \text{Bernoulli}(\frac{c}{n}) variables, which has a \text{Binomial}(n, \frac{c}{n}) distribution. This binomial variable is independent of the past because it only depends on unexplored edges; in addition, its distribution does not depend on |A_t|. Therefore, |A_t| would be a random walk with increment distribution \text{Binomial}(n, \frac{c}{n})-1\approx\text{Poisson}(c)-1. Then, studying |C_v| boils down to studying first time a Poisson random walk hits zero! Of course, we cannot really ignore previously visited sites, but this rough intuition nonetheless captures the right idea as n\to\infty and will serve as a guiding principle for the proof.

A comparison random walk

The reason that |A_t| is not a random walk is that there are only |U_t| edges (not n) to explore at time t. We can artificially create a random walk by adding n-|U_t| “fictitious'' points at each time t as follows.

Let \tilde{\eta}_{v}^t be i.i.d. \text{Bernoulli}(\frac{c}{n}) for t\geq 0, v\in[n], which are independent of (\eta_{ij}). Define

    \[ 	S_0 = 1, \qquad S_{t+1} = S_t - 1 +  	\sum_{w\in U_t}  	\eta_{wi_t} + \sum_{w\in [n]\backslash  	U_t}\tilde{\eta}^t_{w}. \]

(When t\ge\tau, then A_t=\varnothing and thus i_t is undefined; in this case, we simply add all n variables \tilde\eta^t_w.)
Note that \sum_{w\in U_t} \eta_{wi_t} is the sum of edges from A_t to U_t. Since we have not explored U_t yet, those edges are independent of all edges explored up to time t (here we use that i_t is nonanticipating). We therefore see that (S_t) is indeed a random walk with increment

    \[ S_{t}- S_{t-1} \sim \text{Binomial}(n, \tfrac{c}{n})-1. \]

Moreover, since all \tilde{\eta}_{w}^t are nonnegative,

    \[ S_{t+1} - S_t \geq |A_{t+1}| - |A_t| \]

as long as t<\tau. It follows that |A_t| is dominated by the genuine random walk S_t, that is,

    \[|A_t| \le S_t \quad\mbox{for all }t\le \tau.\]

We can now obtain bounds on |C_v| by analyzing hitting times of the random walk S_t.

The subcritical case c<1

Define the first time the comparison walk hits zero as

    \[T\equiv\inf\{t: S_t=0\}.\]

Since |A_t|\leq S_t for t\le\tau, it is obvious that

    \[|C_v|=\tau\leq T.\]

Now we study T. The intuition is that as \mathbb{E}[S_{t+1}-S_t]=c-1, (S_t) is a random walk with negative drift in the subcritical case c<1. Thus \mathbb{P}[T<\infty]=1, and in fact the hitting time T has nice tails!

Lemma 2 Let c<1 and \alpha=c-1-\log c>0. Then for any positive integer k,

    \[ \mathbb{P}[T\geq k]\leq \frac{1}{c} e^{-\alpha k}. \]

We will prove this lemma below. Using the lemma, we immediately obtain:

Corollary 3 If c<1, then for any a>\frac{1}{\alpha}=\frac{1}{c-1-\log c}

    \[ \mathbb{P}\big[\max_v|C_v|\geq a\log n\big] \xrightarrow{n\to\infty} 0.  \]

Proof. Applying the Lemma 2 and the union bound,

    \begin{eqnarray*} \mathbb{P}\big[\max_v|C_v|\geq a\log n\big] &\leq & \sum_{v\in[n]} \mathbb{P}[|C_v|\geq a\log n]\\ &\leq & \sum_{v\in[n]} \mathbb{P}[T\geq a\log n] \\ &\leq & \frac{n}{c} n^{-a\alpha} \to 0. \qquad\square \end{eqnarray*}

This corollary proves part 1 of Theorem 1. In fact, it turns out that the constant \frac{1}{\alpha} is tight: by using the second moment method, one can prove a matching lower bound on \max_v|C_v| (see, for example, the lecture notes by van der Hofstad), which implies that in fact \max_v|C_v|\sim \frac{1}{\alpha}\log n in probability. The proof is not much more difficult, but we prefer to move on to the supercritical case.

Remark. It might seem somewhat surprising that the result we obtain is so sharp, considering that we have blindly replaced |A_t| by the larger quantity S_t. However, in going from |A_t| to S_t we do not lose as much as one might think at first sight. When n is large and t is relatively small, the excess term \sum_{w\in[n]\backslash U_t}\tilde{\eta}^t_w in the definition of S_t is zero with high probability, as most vertices are unexplored and the Bernoulli success probability \frac{c}{n} of the \tilde{\eta}^t_w is very small. With a bit of work, one can show that S_t and |A_t| will actually stick together for times t\lesssim\log n with probability going to one as n\to\infty. Thus, in the subcritical case where the random walk only lives for \sim\log n time steps, nothing is lost in going from |A_t| to S_t, and our rough intuition that |A_t| should behave like a random walk as n\to\infty is vindicated.

To wrap up the subcritical case, it remains to prove the lemma.

Proof of Lemma 2. By the Markov inequality,

    \[ \mathbb{P}[T\geq k] = \mathbb{P}[e^{\alpha T}\geq e^{\alpha k}] \leq e^{-\alpha k} \mathbb{E}[e^{\alpha T}].  \]

It remains to bound \mathbb{E}[e^{\alpha T}]\le\frac{1}{c}, which is a standard exercise in martingale theory.

Recall that

    \[S_t = 1+\sum_{k=1}^t X_k,\]

where X_k are i.i.d. Define the moment generating function \phi(\beta)= \log\mathbb{E}[e^{\beta X_k}], and let

    \[ M_t \equiv e^{\beta S_t-\phi(\beta)t}, \qquad \text{for }t\geq 0.  \]

Since e^{\beta S_t}= e^{\beta}\prod_{k=1}^t e^{\beta X_k} and X_t is independent of M_0,M_1,\cdots,M_{t-1},

    \[ 	\mathbb{E}[M_t|M_0,\cdots, M_{t-1}] = M_{t-1}\, 	\mathbb{E}[\tfrac{M_t}{M_{t-1}}|M_0,\cdots, M_{t-1}]  	= M_{t-1}\, \mathbb{E}[e^{\beta X_t-\phi(\beta)}] = M_{t-1}, \]

where we have used \mathbb{E}[e^{\beta X_k - \phi(\beta)}]=1. So (M_t) is a martingale.

In the case \beta>0 and \phi(\beta)<0,

    \[\mathbb{E}[e^{-\phi(\beta)T}] = \mathbb{E}\big[\lim_{n\to\infty} M_{T\wedge n}\big] \leq\liminf_{n\to\infty} \mathbb{E}[M_{T\wedge n}]= M_0 = e^{\beta}.\]

The inequality is by Fatou's lemma and the second equality is by the optional stopping theorem. To see the first equality, note that if T<\infty, then S_{T\wedge n}\to S_T=0 and T\wedge n\to T as n\to\infty, while if T=\infty, then S_{T\wedge n}>0 for all n and T\wedge n\to\infty. Thus e^{-\phi(\beta)T}=\lim_{n\to\infty}e^{\beta S_{T\wedge n}-\phi(\beta)(T\wedge n)}=\lim_{n\to\infty}M_{T\wedge n}.

Next, we compute \phi(\beta). Recall that (X_k+1)\sim \text{Binomial}(n, \frac{c}{n}). It has the same distribution as the sum of n i.i.d. \text{Bernoulli}(\frac{c}{n}) variables. For Y\sim\text{Bernoulli}(p), we have \mathbb{E}[e^{\beta Y}]= 1+(e^{\beta}-1)p. Therefore,

    \begin{eqnarray*} -\phi(\beta) &=& - \log \mathbb{E}[e^{\beta X_t}]\\ &=& -\log\big( e^{-\beta} (1+(e^{\beta}-1)\tfrac{c}{n})^n \big)\\ &=& \beta - n\log\big(1+(e^{\beta}-1)\tfrac{c}{n}\big)\\ &\geq& \beta - c(e^{\beta}-1), \end{eqnarray*}

where the last inequality is because -\log(1+x)\geq -x for any x. Now, by setting \beta=-\log c, we obtain that -\phi(\beta)\geq c-1-\log c=\alpha. Thus we have shown \mathbb{E}[e^{\alpha T}]\le\mathbb{E}[e^{-\phi(\beta)T}]\le e^\beta=\frac{1}{c}. \quad\square

The supercritical case c>1

The goal of the following lectures is to prove part 2 of Theorem 1. More precisely, we will prove:

Theorem 4 Let c>1. Then

    \[ \frac{\max_v|C_v|}{n}\xrightarrow{n\to\infty} 1 - \rho \]

in probability, where \rho is the smallest positive solution of the equation \rho = e^{c(\rho-1)}. Moreover, there is \beta>0 such that all but one of the components have size \leq \beta\log n, with probability tending to 1 as n\to\infty.

This theorem says that with probability tending to 1, there is a unique giant component whose size is (1-\rho)n, and all other components are small with size \leq \beta\log n.

Here we provide some vague intuition for this theorem. When c>1, the random walk (S_t) satisfies \mathbb{E}[S_t-S_{t-1}]=c-1>0, i.e., (S_t) has positive drift. Then \mathbb{P}[T<\infty]<1! In fact, the further away it starts from 0, the smaller the probability it will ever hit 0. Consider the two situations:

  • S_t dies quickly: this implies that the component is small.
  • S_t lives long: then it must live very long, as once it gets far away from 0, the probability of returning is very small. This implies that the component must be very large (if we pretend that S_t=|A_t|).

Of course, S_t is not |A_t| (obviously |A_t| eventually hits 0). But the intuition explains that there cannot be components of intermediate size: given any vertex v, either |C_v| is small (\lesssim \log n), or it must get very large (\gtrsim n^{2/3}, say). In fact, we will find that all components of size \geq \beta\log n must grow all the way to \gtrsim n^{2/3}. However, any pair of large components must intersect with high probability, as there are many potential edges between them! Therefore, all vertices v with |C_v|\geq \beta\log n should be in the same giant component. We then show that the number of such vertices is (1-\rho)n with high probability.

Many thanks to Tracy Ke for scribing this lecture!

06. April 2013 by Ramon van Handel
Categories: Random graphs | Comments Off on Lecture 4. Giant component (1)