## Fall 2013: Information theoretic methods

While information theory has traditionally been based on probabilistic methods, ideas and methods from information theory have recently played an increasingly important role in various areas of probability theory itself (as well as in statistics, combinatorics, and other areas of mathematics). The goal of these informal lectures is to introduce some topics and tools at the intersection of probability theory and information theory. No prior knowledge of information theory will be assumed. Potential topics include: entropic central limit theorems, entropic Poisson approximation, and related information-theoretic and probabilistic inequalities; connections to logarithmic Sobolev inequalities and Stein’s method; entropic inequalities for additive, matroidal, and tree structures, and their applications; transportation cost-information inequalities and their relation to concentration of measure; basic large deviations theory.

**Prerequisites:** Probability at the level of ORF 526 is assumed.

**Time and location:** Thursdays, 4:30-6:00, Bendheim Center classroom 103.

The first lecture will be on September 19.

**References:**

- N. Gozlan and C. Leonard, “Transport Inequalities. A Survey”
- O. Johnson, “Information Theory and the Central Limit Theorem”
- M. Ledoux, “The Concentration of Measure Phenomenon”
- C. Villani, “Topics in Optimal Transportation”

## End of Spring 2013 seminar

That was it for the Spring 2013 stochastic analysis seminar! We will be back in the Fall. The topic and location/time slot will be announced in September.

## Lectures 7/8. Games on random graphs

*The following two lectures by Rene Carmona are on games on random graphs.
Many thanks to Patrick Rebeschini for scribing these lectures!*

In the following we discuss some results from the paper “Connectivity and equilibrium in random games” by Daskalakis, Dimakis, and Mossel. We define a random game on a random graph and we characterize the graphs that are likely to exhibit Nash equilibria for this game. We show that if the random graph is drawn from the Erdös-Rényi distribution, then in the high connectivity regime the law of the number of pure Nash equilibria converges toward a Poisson distribution, asymptotically, as the size of the graph is increased.

Let be a simple (that is, undirected and with no self-edges) graph, and for each denote by the set of neighbors of , that is, . We think of each vertex in as a *player* in the game that we are about to introduce. At the same time, we think of each edge as a *strategic interaction* between players and .

Definition(Game on a graph).For each let represent theset of strategiesfor player , assumed to be a finite set. We naturally extend this definition to include families of players: for each , let be the set of strategies for each player in . For each , denote by thereward functionfor player . Agameis a collection .

The above definition describes a game that is *static*, in the sense that the game is played only once, and *local*, in the sense that the reward function of each player depends only on its own strategy and on the strategy of the players in its neighbors. We now introduce the notion of pure Nash equilibrium.

Definition(Pure Nash equilibrium).We say that is apure Nash equilibrium(PNE) if for each we have

A pure Nash equilibrium represents a state where no player can be better off by changing his own strategy if he is the only one who is allowed to do so. In order to investigate the existence of a pure Nash equilibrium it suffices to study the *best response function* defined below.

Definition(Best response function).Given a reward function for player , we define thebest response functionfor as

Clearly, is a pure Nash equilibrium if and only if for each . We now define the type of random games that we will be interested in; in order to do so, we need to specify the set of strategies and the reward function for each player.

Definition(Random game on a fixed graph).For a graph and an atomless probability measure on , let be the associatedrandom gamedefined as follows:

- for each ;
- is a collection of independent identically distributed random variables with distribution .

**Remark.** For each game the family is a collection of independent random variables that are uniformly distributed in , and for each , we have almost surely. In fact, note that if and only if and this event has probability since the two random variables appearing on both sides of the inequality sign are independent with the same law and is atomless. As far as the analysis of the existence of pure Nash equilibria is concerned, we could take the present notion of best response functions as the definition of our random game on a fix graph. In fact, note that the choice of in does not play a role in our analysis, and we would obtain the same results by choosing different (atomless) distributions for sampling (independently) the reward function of each player.

Denote by the distribution of a Erdös-Rényi random graph with vertices where each edge is present independently with probability . We now introduce the notion of a random game on a random graph.

Definition(Random game on a ramon graph).For each , and each probability measure on , do the following:

- choose a graph from ;
- choose a random game from for the graph .

Henceforth, given a random variable let represent its distribution. Given two measures and on a measurable space, define the *total variation distance* between and as

where the supremum is taken over measurable functions such that the supremum norm is less than or equal to .

We are now ready to state the main theorem that we will prove in the following (Theorem 1.9 in Daskalakis, Dimakis, and Mossel).

Theorem 1(High connectivity regime).Let be the number of pure Nash equilibria in the random game on random graph defined above. Let define the high-connectivity regime as

where satisfies the following two properties:

Then, we have

where denotes the conditional probability given the graph and is a Poisson random variable with mean . In particular,

which shows that in this regime a pure Nash equilibrium exists with probability converging to as the size of the network increases.

**Remark.** Using the terminology of statistical mechanics, the first result in Theorem 1 represents a *quenched*-type result since it involves the conditional distribution of a system (i.e., the game) given its environment (i.e., the graph). On the other hand, the second result represents an *annealed*-type result, where the unconditional probability is considered.

In order to prove Theorem 1 we need the following lemma on Poisson approximations. The lemma is adapted from the results of R. Arratia, L. Goldstein, and L. Gordon (“Two moments suffice for Poisson approximations: the Chen-Stein method“, Ann. Probab. 17, 9-25, 1989) and it shows how the total variation distance between the law of a sum of Bernoulli random variables and a Poisson distribution can be bounded by the first and second moments of the Bernoulli random variables. This result is a particular instance of the Stein’s method in probability theory.

Lemma 2(Arratia, Goldstein, and Gordon, 1989).Let be a collection of Bernoulli random variables with . For each let be such that is independent of . Define

where . Define and . If is a Poisson random variable with mean , then

**Proof.** We define the following operators that act on each function :

We point out that characterizes in the sense that if and only if is a Poisson random variable with mean ; is an example of *Stein’s operator*. First of all, we show that for each we have . In fact, if we have

and if we have

For each define and . The following properties hold:

- ,
- ,
- .

In what follows, consider any given function such that . Define the function as for each , and let . From what was seen above we have and we get

The first term is bounded above by while the third term is equal to since is independent of . In order to bound the second term we want to rewrite each term as a telescoping sum. In what follow fix , label the elements of as and define

Noticing that , we have

and we get

Therefore, combining all together we get

Since the total variation distance can be characterized in terms of sets as

from this point on we restrict our analysis to indicator functions, which are easier to deal with than generic functions. For each define , and . The previous result yields

and the proof of the Lemma is concluded if we show that for each . In fact, in what follows we will show that

where the right hand side is clearly upper bounded by . The proof that we are going to present is contained in the Appendix of “Poisson Approximation for Some Statistics Based on Exchangeable Trials” by Barbour and Eagleson.

First of all, note that for each we have and

for each . From this expression it is clear that for each , which suggests that we can restrict our analysis to singletons. For each we have

and from the series expansion of the Poisson probabilities it is easy seen that the function is negative and decreasing if and is positive and decreasing if . Hence, the only positive value taken by the difference function corresponds to the case that can be bounded as follows:

Therefore, for each , we have

Noticing that , then for each , we also get

and we proved that .

We now introduce the notation that will naturally allow us to use Lemma 2 to prove Theorem 1. We label the *pure strategy profiles* in as , where and as always . Often times, it will be convenient to use the labels to enumerate the vertices of the graph . Accordingly, one can think of the strategy profiles defined in a specific way, for example by positing that is the binary decomposition of . In particular becomes the strategy where each player plays zero, that is, for each , and the strategy where each player plays one, that is, for all . For each define

Clearly the quantity identifies the number of pure Nash equilibria and corresponds to the existence of a pure Nash equilibrium. We recall that both the randomness in the choice of the graph and the randomness in the choice of the game are embedded in the random variables . Note that conditionally on a given graph sampled form we have, for each ,

from which it follows that

That is, the current definition of a game on a fixed graph implies that the expected number of pure Nash equilibria is for any given graph. It follows that also . Notice that Theorem 1 adds more information on the table since it describes the asymptotic distribution of on a particular regime of the Erdös-Rényi random graph.

In the way we set up the stage it seems tempting to apply Lemma 2 to the random variables just defined. However, this approach is not fruitful since, apart from trivial cases, any given random variable has a neighborhood of dependence that coincides with the entire set . To see this, consider any two strategy profiles and . As long as there exists such that , then we can always find a realization of the graph such that does not have any edges attached to it, that is, ; this implies that and, consequently, it implies that and are not independent. Therefore, only and are independent, where and for each . However, Lemma 2 can be fruitfully applied to the random variables when we look at them conditionally on a given graph realization, as the following Lemma demonstrates (Lemma 2.2 of Daskalakis, Dimakis, Mossel).

Lemma 3.Let be a graph. Define

and for each define

where

and is the exclusive-or operation. Then, for each we have that is independent of .

**Proof.**We first show that is independent of . In order to make the independency structure manifest, we characterize . Recall that is the index set of the pure strategy profiles for which there exists a player having all his neighbors playing . Therefore, each strategy profile corresponding to an index set in is characterized by the fact that each player has at least one neighbor who is playing . Hence, for each we have for all and, consequently, the events

are independent of the events

which proves our claim.

We now generalize this result to show that is independent of , for each . For any , note that the exclusive-or map with respect to preserves the differences in the strategy profiles (and, of course, it also preserves the equalities). That is, if and are such that for some , then also and are such that . Therefore,

holds true if and only if

holds true. Equivalently stated, if and only if , where is the index set such that . Hence, the proof is concluded once we notice that and that is defined as the index set of the pure strategy profiles that are obtained by an exclusive-or map with respect to of a strategy profile in .

For a given graph with , define , where and represents the conditional probability conditionally on the graph . Define as in Lemma 3; then, conditionally on we have that is independent of . Define

where . Define and recall that . If is a Poisson random variable with mean , then Lemma 2 yields

At this point, let us introduce the following two lemmas (Lemmas 2.4 and 2.5 of Daskalakis, Dimakis, Mossel).

Lemma 4.If is sampled from the Erdös-Rényi distribution , we have

Lemma 5.Under the assumptions of Theorem 1 there exists and such that

We now show how the proof of Theorem 1 follows easily from the two lemmas above.

**Proof of Theorem 1.** Let and as in Lemma 5. Define , and . Clearly, by Lemma 5 we have

Define the event

By the Markov inequality and the previous asymptotic bounds, for each we have

where we used the result of Lemma 4 and the above estimates for and . Since for some , then there clearly exists such that

Hence, we have that for . Let us now define and such that and

Then

which proves the first statement in Theorem 1. In fact, we have

since by definition of , on the event we have

By the properties of conditional expectations we can now prove the convergence in total variation of the unconditional law of to the law of . In fact, for we have

from which it follows that

Since convergence in total variation implies convergence in distribution, the previous result implies that converges in distribution to , which concludes the proof of Theorem 1.

We now provide the proof of Lemma 4, while we refer the reader to Daskalakis, Dimakis, Mossel for the proof of Lemma 5.

**Proof of Lemma 4.** We begin with the study of . By the symmetry of the model we have

Since , we have . By the symmetry of the Erdös-Rényi distribution we also have that if and have the same number of players playing (equivalently, the same number of players playing ).

Therefore, if we label the vertices of the graph as , we have

where for each the index set is such that the strategy satisfies if and if . Hence, the bound for in the statement of the Lemma is proved if we show that . In fact, by definition of we have

We now study the term . Proceeding as above, by symmetry we have

We now analyze the term . As noticed above, if and only if the graph is such that there exists a player such that . In the case in which such also satisfies the property , then and for each , and it follows that . In fact, the event corresponds to the realizations where both strategy and strategy are pure Nash equilibria, that is, for each . But it can not be that both and are best responses for the player when all player in the neighbor play . Hence, is different from on the event

and on the event

Define and . Note that we have . On the events and we have and we get

Because of the independency structure in the Erdös-Rényi random graph we have

Furthermore, we have

This follows immediately from the definition of pure Nash equilibrium in terms of the best response functions once noticed that on the event there are exactly players (each such that ) such that , while for the remaining players we have . Putting everything together we get

Using the fact that , it clearly follows that .

## Giant component: final remarks

The past three lectures were devoted to the giant component theorem:

TheoremLet be the connected component of that contains .

- If , then in probability.
- If , then in probability, for some .
- If , then in distribution.

We proved only the first (subcritical) and second (supercritical) cases: our presentation was largely inspired by the treatment of Janson, Luczak, and Rucinski and Durrett. We have omitted the critical case, however, as the last two lectures of the semester will be on another topic. The goal of this post is to provide some final remarks and references on the giant component theorem.

**Retrospective**

At first sight, the “double jump” in the giant component theorem looks quite shocking. In hindsight, however, this does not seem quite so miraculous, as it mirrors an elementary phenomenon that is covered in many introductory probability courses: given a (nice) random walk with initial condition , define the hitting time for some . Then there are three cases:

- If has
*negative drift*, then . In fact, the random variable has a light (exponential) tail. - If has
*positive drift*, then . - If has
*zero drift*, then a.s. but . That is, the random variable has a heavy tail.

This “double jump” in the behavior of the hitting probabilities of a random walk is directly analogous to the behavior of the connected components of an Erdös-Rényi graph, and this was indeed the basic idea behind the proofs given in the previous lectures. Of course, it remains a bit of a miracle that the random walk approximation of the exploration process, which only holds for small times, is sufficiently powerful that it describes so completely the behavior of the random graph.

**The critical case**

In the *subcritical* case, the size of the largest component is of order because the hitting time of a random walk with negative drift has an *exponential tail*: that is, we proved

which goes to zero as for .

Similarly, we would expect that we can obtain the size of the largest component in the *critical* case if we understand the *heavy tail* behavior of the hitting time of a random walk with zero drift. This is in fact the case. Indeed, when there is zero drift, one can show that

The crude union bound argument used above now does not give the correct answer, but an only slightly better argument is needed. Indeed, note that

Therefore, . With some further work, a corresponding lower bound can also be proved. See the paper by Nachmias and Peres or the notes by van der Hofstad for the details.

It turns out that in the critical case is not only bounded in probability, but in fact converges weakly to some limiting distribution. This distribution, and much more, is beautifully described by Aldous in terms of Brownian excursions. This is an interesting example of the application of stochastic analysis to discrete probability; unfortunately, we do not have the time to cover it.

In a different direction, it turns out that various additional phase transitions appear when we consider a finer scaling, for example, in the “critical window” . For an overview of the various transitions, see, for example, section 11.1 in Alon and Spencer.

**Connectivity threshold**

Rather than considering the size of the largest component, one could ask when the *entire* Erdös-Rényi graph is connected. Note that when , the constant in the size of the giant component is always strictly positive, so the graph is not connected. Therefore, in order for the entire graph to be connected, we must let (that is, the edge probability must be superlinear). It turns out that the appropriate scaling for this question is , and another phase transition arises here.

Theorem.Let . If , then the Erdös-Rényi graph is connected with probability tending to as . If , the graph is connected with probability tending to as .

To get some intuition, consider the probability that a vertex is isolated (that is, disconnected from every other vertex):

Thus for , we have

In particular, if , then there must exist an isolated vertex with positive probability as , in which case the graph is not connected (in fact, it is not hard to show that the variance of the number of isolated components is of the same order as its mean, so that the probability that the graph is connected tends to zero). Somewhat miraculously, it turns out that when there are no isolated vertices, then the graph must already be connected, so that we do indeed obtain the sharp transition described in the Theorem above. For a proof of this fact (by a clever combinatorial argument) see, for example, the lecture notes by van der Hofstad. Alternatively, one can use a random walk argument entirely in the spirit of the proofs in the previous lectures to prove that the random graph is connected for : by running simultaneous exploration processes from different vertices as we did in the proof of the supercritical case, one can show that *all* connected components must intersect when and thus the entire graph must be connected. See section 2.8 in Durrett for such an argument.

## Lecture 6. Giant component (3)

Let us begin with a brief recap from the previous lecture. We consider the Erdös-Rényi random graph in the supercritical case . Recall that denotes the connected component of the graph that contains the vertex . Our goal is to prove the existence of the giant component with size , while the remaining components have size .

Fix sufficiently large (to be chosen in the proof), and define the set

of vertices contained in “large” components. The proof consists of two parts:

**Part 1:****Part 2:**in probability.

Part 1 states that all the sufficiently large components must intersect, forming the giant component. Part 2 counts the number of vertices in the giant component. Part 2 was proved in the previous lecture. The goal of this lecture is to prove Part 1, which completes the proof of the giant component.

**Overview**

As in the previous lectures, the central idea in the study of the giant component is the *exploration process* , where

We have seen that , where is a random walk with increments

When , we have . Thus is approximately a random walk with *positive drift*. The intuitive idea behind the proof of Part 1 is as follows. Initially, the random walk can hit rapidly, in which case the component is small. However, if the random walk drifts away from zero, then with high probability it will never hit zero, in which case the component must keep growing until the random walk approximation is no longer accurate. Thus there do not exist any components of intermediate size: each component is either very small () or very large (we will show , but the precise exponent is not important).

We now want to argue that any pair of large components must necessarily intersect. Consider two disjoint sets and of vertices of size . As each edge is present in the graph with probability , the probability that there is no edge between and is

We therefore expect that any pair of large components must intersect with high probability. The problem with this argument is that we assumed that the sets and are nonrandom, while the random sets themselves depend on the edge structure of the random graph (so the events and are highly correlated). To actually implement this idea, we therefore need a little bit more sophisticated approach.

To make the proof work, we revisit more carefully our earlier random walk argument. The process has positive drift as . Thus the process is still approximately a random walk with positive drift! Applying the above intuition, either dies rapidly (the component is small), or grows linearly in as is illustrated in the following figure:

This means that the exploration process for a component of size will not only grow large () with high probability, but that the exploration process will also possess a large number of active vertices (. To prove that all large components intersect, we will run different exploration processes simultaneously starting from different vertices. We will show that if two of these processes reach a large number of active vertices then there must be an edge between them with high probability, and thus the corresponding components must coincide. This resolves the dependence problem in our naive argument, as the edges between the sets of active vertices have not yet been explored and are therefore independent of the history of the exploration process.

**The component size dichotomy**

We now begin the proof in earnest. We will first show the dichotomy between large and small components: either the component size is , or the number of active vertices grows linearly up to time . To be precise, we consider the following event:

Our goal is to show that is large.

Define the stopping time

We can write

Now suppose and . Then , as exploration process is alive at time and stays alive until time . We can therefore write

To bound the probabilities inside the sum, we compare to a suitable random walk.

**The random walk argument**

To bound the probability that , we must introduce a comparison random walk that lies *beneath* . We use the same construction as was used in the previous lecture. Let

where , are i.i.d. random variables independent of , (the same used in the exploration process), and is the set of the first components of (if , then and thus is undefined; then we simply add variables ).

As in the previous lecture, we have:

- is a random walk with increments.
- whenever and .

Now suppose that and . Then

We therefore obtain for

Thus computing reduces to compute the tail probability of a random walk (or, in less fancy terms, a sum of i.i.d. random variables). That is something we know how to do.

Lemma(Chernoff bound).Let . Then

**Proof.** Let . Then

The result follows by optimizing over .

Note that . We therefore have by the Chernoff bound

for all (here depends only on and ). In particular, we have

provided is sufficiently large. Thus we can estimate

which goes to zero as provided that is chosen sufficiently large. In particular, the component size dichotomy follows: choosing any , we obtain

**Remark:** Unlike in the proof of Part 2 in the previous lecture, here we *do* need to choose sufficiently large for the proof to work. If is too small, then the random walk cannot move sufficiently far away from zero to ensure that it will never return. In particular, even in the supercritical case, the second largest component has size of order .

**Large components must intersect**

To complete the proof, it remains to show that all large components must intersect. To do this, we will run several exploration processes at once starting from different vertices. If the sets of active vertices of two of these processes grow large, then there must be an edge between them with high probability, and thus the corresponding components intersect. Let us make this argument precise.

In the following, we denote by the exploration process started at . For each such process, we define the corresponding set that we have investigated above:

We have shown above that, provided , we have

We can therefore estimate

Now note that by time , the exploration process has only explored edges where (or ), and similarly for . It follows that

In particular, if are disjoint subsets of vertices, then

On the other hand, implies that must be disjoint at every time . Thus if , there can be no edges between vertices in and at any time (if such an edge exists, then the vertices connected by this edge will eventually be explored by both exploration processes, and then the sets of removed vertices will no longer be disjoint). Therefore,

Thus we finally obtain

and the proof of the giant component theorem is complete.

*Many thanks to Quentin Berthet for scribing this lecture!*

## Lecture 5. Giant component (2)

Consider the Erdös-Rényi graph model , and denote as usual by the connected component of the graph that contains the vertex . In the last lecture, we focused mostly on the subcritical case , where we showed that . Today we will begin developing the supercritical case , where for a suitable constant . In particular, our aim for this and next lecture is to prove the following theorem.

Theorem.Let . Then

where is the smallest positive solution of the equation . Moreover, there is a such that all but one of the components have size with probability tending to as .

**Beginning of the proof.** Define the set

The proof of the Theorem consists of two main parts:

**Part 1:**.**Part 2:**in probability.

Part 1 states that all “large” components of the graph must intersect, forming one *giant component*. Some intuition for why this is the case was given at the end of the previous lecture. Part 2 computes the size of this giant component. In this lecture, we will concentrate on proving Part 2, and we will find out where the mysterious constant comes from. In the next lecture, we will prove Part 1, and we will develop a detailed understanding of why all large components must intersect.

Before we proceed, let us complete the proof of the Theorem assuming Parts 1 and 2 have been proved. First, note that with probability tending to one, the set is itself a connected component. Indeed, if and then must lie in disjoint connected components by the definition of . On the other hand, with probability tending to one, all must lie in the same connected component by Part 1. Therefore, with probability tending to one, the set forms a single connected component of the graph. By Part 2, the size of this component is , while by the definition of , all other components have size . This completes the proof.

The remainder of this lecture is devoted to proving Part 2 above. We will first prove that the claim holds on average, and then prove concentration around the average. More precisely, we will show:

- .
- .

Together, these two claims evidently prove Part 2.

**Mean size of the giant component**

We begin by writing out the mean size of the giant component:

where we note that does not depend on the vertex by the symmetry of the Erdös-Rényi model. Therefore, to prove convergence of the mean size of the giant component, it suffices to prove that

This is what we will now set out to accomplish.

In the previous lecture we defined exploration process . We showed that

and that for

where is an arbitrary nonanticipating choice, say, (recall that denotes the adjacency matrix of the random graph). As are i.i.d. and as edges emanating from the set of unexplored vertices have not yet appeared in previous steps, the process is “almost'' a random walk: it fails to be a random walk as we only add Bernoulli variables in each iteration, rather than a constant number. In the last lecture, we noted that we can estimate from above by a genuine random walk by adding some fictitious vertices. To be precise, we define

where are i.i.d. independent of the (if , then and thus is undefined; in this case, we simply add all variables ). In the present lecture, we also need to bound from below. To this end, we introduce another process as follows:

where is the set consisting of the first elements of in increasing order of the vertices (if , we add variables ). The idea behind these processes is that is engineered, by including “fictitious'' vertices, to always add i.i.d. Bernoulli variables in every iteration, while is engineered, by including “fictitious'' vertices when is small and omitting vertices when is large, to always add i.i.d. Bernoulli variables in every iteration. The following facts are immediate:

- is a random walk with i.i.d. increments.
- is a random walk with i.i.d. increments.
- for all .
- for all on the event .

To see the last property, note that the exploration process can only explore as many vertices as are present in the connected component , so that for all ; therefore, in this situation only the second possibility in the definition of occurs, and it is obvious by construction that then (nonetheless, the first possibility in the definition must be included to ensure that is a random walk).

We now define the hitting times

Then we evidently have

(Note how we cleverly chose the random walk precisely so that whenever ). We have therefore reduced the problem of computing to computing the hitting probabilities of random walks. Now we are in business, as this is something we know how to do using martingales!

**The hitting time computation**

Let us take a moment to gather some intuition. The random walks and have increments distributed as and , respectively. As , both increment distributions converge to a distribution, so we expect that where is the first hitting time of the Poisson random walk. On the other hand, as , we expect that . The problem then reduces to computing the probability that a Poisson random walk *ever* hits the origin. This computation can be done explicitly, and this is precisely where the mysterious constant comes from!

We now proceed to make this intuition precise. First, we show that the probability can indeed be replaced by , as one might expect.

Lemma.as .

**Proof.** We need to show that

Note that as when ,

We can evidently write

where and

Choosing , we obtain for . Therefore,

This completes the proof.

By the above Lemma, and a trivial upper bound, we obtain

To complete computation of the mean size of the giant component, it therefore remains to show that and converge to . In fact, we can compute these quantities exactly.

Lemma.Let . Then

where is the smallest positive solution of .

**Proof.** Recall the martingale used in last lecture:

Suppose that and . Then

The first equality holds since if then and , while if then and . The second equality holds by dominated convergence since , and the third equality is by the optional stopping theorem.

Now suppose we can find such that as . Then we have

by dominated convergence. Thus, evidently, it suffices to find with the requisite properties. Now note that as , , and for , we evidently must have

We can find such by inspecting the following illustration:

Evidently the requisite assumptions are satisfied when is the smallest root of the equation (but not for the larger root at !)

**Remark.** Note that the supercritical case is essential here. If then the equation for has no solutions , and the argument in the proof does not work. In fact, when , we have .

By an immediate adaptation of the proof of the previous lemma, we obtain

where is the smallest positive solution of . Letting , we see that

where is the smallest solution of the equation (which is precisely the probability that the Poisson random walk hits zero, by the identical proof to the lemma above). We have therefore proved

**Variance of the giant component size**

To complete the proof of Part 2 of the giant component theorem, it remains to show that

To this end, let us consider

To estimate the terms in this sum, we condition on one of the components:

To proceed, note that the event can be written as

In particular, the event is independent of the edges for . Therefore, for , the conditional law of given coincides with the (unconditional) law of , the conncted component containing in the induced subgraph on the vertices :

As this quantity only depends on by the symmetry of the Erdös-Rényi model, we can evidently write

for , . In particular, we obtain

Now note that, by its definition, is distributed precisely as the component containing vertex in the random graph model. We can therefore show, repeating exactly the proof of the mean size of the giant component above, that

We have therefore shown that

which evidently implies

This is what we set out to prove.

**Remark.** It should be noted that the proof of Part 2 did not depend on the value of , or even on the rate, in the definition of the set : any sequence that grows sublinearly to infinity would have given the same result. This suggests that all but a vanishing fraction of vertices are contained in connected components of order or . We find out only in the next lecture why the rate (for sufficiently large!) is important: only sufficiently large connected components are guaranteed to intersect, while there might (and do) exist components of order that are disjoint from the giant component. If we do not exclude the latter, we will not be able to prove Part 1.

*Many thanks to Weichen Wang for scribing this lecture!*

## Lecture 4. Giant component (1)

Consider the Erdös-Rényi graph model . In previous lectures, we focused on the “high complexity regime”, i.e., as goes to infinity, is fixed. We discussed topics such as clique numbers and chromatic numbers. From now on, we shall consider the “low complexity regime”, where as goes to infinity, for a fixed constant . As before, let be the adjacency matrix of . Then are i.i.d. Bernoulli random variables with success probability .

Theorem 1Let be the connected component of that contains .

- If , then in probability.
- If , then in probability, for some .
- If , then in distribution.

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

**The exploration process**

How to study ? We will explore by starting an “exploration process'' at that moves around 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 is exactly . As a consequence, studying reduces to studying a hitting time of a certain random walk, and to the latter we can apply martingale theory.

At each time , we maintain three sets of vertices:

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

- At , initialize , and . Namely, only is active, all the vertices other than are unexplored, and no vertices have been removed.
- At , update , and . Namely, all neighbors of are moved from the unexplored set to the active set, and itself is removed.
- At , pick some and update , and . Namely, all unexplored neighbors of are moved into the active set, and itself is removed.
- At time , we pick some vertex from the current active set , activate all unexplored neighbors of and remove itself.

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

A formal description of the “exploration process'':

- Initialize , and .
- For , we set
where is a nonanticipating but otherwise arbitrary choice.

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

So, we only need to study the stopping time .

Recall that indicates whether there is an edge between vertices and , and . By construction,

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 are unvisited still (note that when is really large and is relatively small, this assumption makes sense). Then is the sum of independent ) variables, which has a distribution. This binomial variable is independent of the past because it only depends on unexplored edges; in addition, its distribution does not depend on . Therefore, would be a random walk with increment distribution . Then, studying 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 and will serve as a guiding principle for the proof.

**A comparison random walk**

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

Let be i.i.d. for , , which are independent of . Define

(When , then and thus is undefined; in this case, we simply add all variables .)

Note that is the sum of edges from to . Since we have not explored yet, those edges are independent of all edges explored up to time (here we use that is nonanticipating). We therefore see that is indeed a random walk with increment

Moreover, since all are nonnegative,

as long as . It follows that is dominated by the genuine random walk , that is,

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

**The subcritical case **

Define the first time the comparison walk hits zero as

Since for , it is obvious that

Now we study . The intuition is that as , is a random walk with *negative* drift in the subcritical case . Thus , and in fact the hitting time has nice tails!

Lemma 2Let and . Then for any positive integer ,

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

Corollary 3If , then for any

**Proof.** Applying the Lemma 2 and the union bound,

This corollary proves part 1 of Theorem 1. In fact, it turns out that the constant is tight: by using the second moment method, one can prove a matching lower bound on (see, for example, the lecture notes by van der Hofstad), which implies that in fact 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 by the larger quantity . However, in going from to we do not lose as much as one might think at first sight. When is large and is relatively small, the excess term in the definition of is zero with high probability, as most vertices are unexplored and the Bernoulli success probability of the is very small. With a bit of work, one can show that and will actually stick together for times with probability going to one as . Thus, in the subcritical case where the random walk only lives for time steps, nothing is lost in going from to , and our rough intuition that should behave like a random walk as is vindicated.

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

**Proof of Lemma 2.** By the Markov inequality,

It remains to bound , which is a standard exercise in martingale theory.

Recall that

where are i.i.d. Define the moment generating function , and let

Since and is independent of ,

where we have used . So is a martingale.

In the case and ,

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 , then and as , while if , then for all and . Thus .

Next, we compute . Recall that . It has the same distribution as the sum of i.i.d. variables. For , we have . Therefore,

where the last inequality is because for any . Now, by setting , we obtain that . Thus we have shown .

**The supercritical case **

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

Theorem 4Let . Then

in probability, where is the smallest positive solution of the equation . Moreover, there is such that all but one of the components have size , with probability tending to as .

This theorem says that with probability tending to , there is a *unique* giant component whose size is , and all other components are small with size .

Here we provide some vague intuition for this theorem. When , the random walk satisfies , i.e., has *positive* drift. Then ! In fact, the further away it starts from , the smaller the probability it will ever hit . Consider the two situations:

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

Of course, is not (obviously eventually hits ). But the intuition explains that there cannot be components of *intermediate* size: given any vertex , either is small (), or it must get very large (, say). In fact, we will find that all components of size must grow all the way to . However, any pair of large components must intersect with high probability, as there are many potential edges between them! Therefore, all vertices with should be in the *same* giant component. We then show that the number of such vertices is with high probability.

*Many thanks to Tracy Ke for scribing this lecture!*

## Next lecture: April 4

Just a reminder that next week (March 21) is Spring Break, while the week after (March 28) there will be no lecture due to the ORFE/PACM colloquium of Elchanan Mossel.

The next Stochastic Analysis Seminar lecture will be on April 4. We will start fresh with a new topic: the study of the giant component of an Erdös-Rényi graph.

## Lecture 3. Chromatic number

Let be set of integers that represent colors. A *vertex coloring* of a graph is an assignment of a color to each vertex . Furthermore a vertex coloring is said to be *proper* if for all . The *chromatic number* of a graph is the smallest number of colors needed for a proper vertex coloring to exist. In this lecture, any coloring is a proper vertex coloring. All logs are in base .

Theorem[Bollobas, 1988]

The chromatic number of a random graph satisfies

**1. Basic properties**

In this section, we begin by reviewing some basic properties of the chromatic number, and in particular how it relates to independent sets. Recall that an *independent set* of a graph is a set of vertices such that . Independent sets are sometimes called *stable sets*. Denote by the *complement graph* of where iff . If is an independent set of , then it is a clique in . The largest independent set in is called *independence number* of and denoted by .

PropositionFor any graph with independence number , clique number and chromatic number the following holds

*Proof:* Trivial.

**2. Chromatic, clique and independence numbers**

Our goal is to study the chromatic number of a random graph . It actually builds upon the clique number that we studied during last lecture. Indeed, we will use the following observation: if then so does the complement graph of . It implies that the independence number has the same distribution as the clique number . We know from last lecture that . Therefore, it follows from the above proposition that

Our main theorem suggests that this simple bound is, in fact, tight. Proving this requires a stronger lower bound on clique of a random graph.

Define and let be a subset of vertices and observe that induced subgraph restricted to vertices in has distribution .

Define

That is: every vertex subgraph contains an independent set of size .

Consider now the following coloring procedure to produce a proper coloring. Pull out independent sets of size one by one and give each a distinct color. According to the definition of , as long as there are at least vertices, this is possible. If there are less than vertices left, give each a distinct color.

The above algorithm gives one possible way of coloring graph and thus the output number is an upper bound on the chromatic number . Let us now compute this upper bound. By definition of , we have

Therefore, to find an upper bound on , we need to find a lower bound on for an appropriately chosen . To that end, observe that

Therefore, we need to prove a bound on , uniformly in . This can be done using concentration properties of a suitable martingale.

**3. Azuma-Hoeffding inequality**

Martingales have useful concentration properties. Specifically the following lemma holds.

Lemma[Azuma-Hoeffding inequality.]

Let be a filtered space and let be a random variable on . Assume that the martingale is such that for all ,Then for any positive integer and any , it holds,

and

*Proof:* We assume the following inequality due to Hoeffding. For any random variable such that a.s. and any , it holds

Now, using a Chernoff bound, we get for any ,

where . The above two displays yield that

The second part of the proof follows by applying the same argument to the martingale .

Consider the lexicographic ordering of and let be iid Bernoulli random variables with parameter indicating the presence of an edge in a graph . Define the -algebra . We are going to apply the Azuma-Hoeffding inequality to the (Doob) martingale where is maximal number of edge-disjoint -cliques in .

Note first that a.s., since adding/removing one edge can add/remove at most one edge disjoint -clique. Moreover, this inequality would not hold if we had chosen to be the number cliques of size (not necessarily disjoint). Indeed, adding or removing an edge can create or destroy many overlapping cliques.

It follows that,

Therefore, applying the Azuma-Hoeffding inequality, we get

It remains to show that is sufficiently large. Note first that , where is is the number of -cliques that do not share an edge with any other -clique of the graph.

**4. A strong lower bound on the clique number**

Fix positive integers and

and let denote the number of cliques in that do not share an edge with any other -clique of the graph. Then

for all . Here the asymptotic notation notations are as .

*Proof:* Define (for large enough) for some to be chosen later.

Moreover, note that for large enough,

so that where is the number of cliques in that do not share an edge with any other -clique of the graph . Let .

For any define the indicator:

Moreover for any set , let denote the number of -cliques in , other than itself, that share at least two vertices with . Observe that

We bound as follows

The following bound holds for all ,

It yields

Moreover, for , , it holds

Hence,

Therefore,

Thus,

Moreover,

since .

To conclude the proof, observe that by Stirling, since ,

Let and observe that

We can therefore apply the previous proposition together with~(2), to get

* *

**5. Proof of the main theorem**

Since

we get for ,

Thus . Together with~(1), it yields

Which completes the proof of our theorem.

*Lecture and scribing by Philippe Rigollet*

## Lecture 2. Clique number

In this lecture all logs are in base . We will prove the following.

TheoremFor the centered and normalized clique number

That is, as , the clique number of the Erdös-Rényi graph is .

*Proof:* The proof is divided into two parts. First, we show that the clique number cannot be *too big* using a union bound. Then we show that the clique number cannot be *too small* using the second moment method. In the following, denotes an Erdös-Rényi graph .

For the first part one has

Thus, choosing , we obtain

for every , where we used that for large .

For the second part it is useful to introduce some notation. Define

.

In particular, iff . Thus we want to show that for one has . Using the trivial observation that implies , we get

where we have used Markov’s inequality. Note that by linearity of the expectation, . Furthermore, we can write

As are boolean random variables we have and thus

which tends to for (see the first part of the proof and use the inequality ). Thus it remains to show that the following quantity tends to :

First note that, by the independence of the edges, for with we have that and are independent, so that in the numerator of the above quantity one can restrict to with . Now by an elementary reasoning we have (with being an arbitrary subset of vertices)

Thus we are now left with proving that the following quantity goes to :

Clearly one has

which shows that (1) can be rewritten as

Now note that since

one has

Using one obtains that (2) is bounded from above by

As is a convex function and as , one has . Thus for large enough the exponent in the above display is bounded from above by , and for this latter is bounded by . Thus we proved that (2) is bounded from above by

which tends to as tends to infinity, concluding the proof.