# Discontinuous Phase Transitions

Yesterday, Demeter Kiss from Cambridge gave a seminar in Oxford about a model for self-destructive percolation on $\mathbb{Z}^2$ that had implications for the (non-)existence of an infinite-parameter forest fire model on the same lattice. I enjoyed talking about this and his recent work on the related model of frozen percolation on $\mathbb{Z}^2$. Considering these models in the lattice setting present a whole range of interesting geometric challenges that are not present in the mean-field case that has mainly occupied my research direction so far.

The afternoon’s discussion included lots of open problems about percolation. Several of these are based around continuity of the phase transition, so I thought I would write a quite post about some simple examples of this, and one example where it does not hold.

A helpful base example is bond percolation on the lattice $\mathbb{Z}^2$. Here, we specify some probability p in [0,1], and we declare edges of the lattice open with probability p, independently of each other. We then consider the graph induced by the open edges. We say that percolation occurs if the origin is contained in an infinite open component. The terminology arises from the interpretation as fluid being added at the origin and flowing down open edges. We define $\theta(p)$ to be the probability that the origin is in an infinite component when the parameter is p. By translation-invariance, we can get some sort of 0-1 law, to conclude that there is an infinite component somewhere in the system with probability either 0 or 1, depending on whether $\theta(p)$ is positive or zero. Indeed, we can further show that if it is positive, then with probability 1 there is a unique infinite component.

We define the critical probability $p_c:= \inf\{\theta(p)>0\}$. A question worth asking is then, what is $\theta(p_c)$? In some examples, we can find $p_c$, but we cannot prove that $\theta(p)$ is continuous around $p_c$. In the case of $\mathbb{Z}^2$ this is known, and it is known from work of Kesten that $p_c=1/2$. See below for a plot of $\theta(p)$ in this setting (obtained from this blog, though possibly originating elsewhere).

The aim is to find an example where we do not have such a continuous phase transition. The original work on frozen percolation took place on trees, and one of Kiss’s results is confirms that these show qualitatively different phenomena to the same process on the lattice. In some sense, trees lie halfway between a lattice and a mean-field model, since there is often some independence when we look down the tree from a given generation, if it is well-defined to use such language.

Anyway, first we consider percolation on an infinite regular rooted k-ary tree. This means we have a root, which has k children, each of which in turn has k children, and so on. As before we consider bond percolation with parameter p. In this setting, we have a language to describe the resulting open component of the root. The offspring distribution of any vertex in the open component is given by Bin(k,p) independently of everything else, so we can view this component as the realisation of a Galton-Watson tree with this offspring distribution. This distribution has finite mean kp, and so we can state explicitly when the survival probability is positive. This happens when the mean is greater than 1, ie p>1/k.

For our actual example, we will consider the survival probability, but the technicalities are easier to explain if we look at the extinction probability, now using the language of branching processes. Suppose the offspring distribution has pgf given by

$f(x)=p_0+p_1x+p_2x^2+\ldots.$

Then the extinction probability q satisfies f(q)=q. I want to pause to consider what happens if this equation has multiple solutions. Indeed, in most interesting cases it will have multiple solutions, since f(1) will always be 1 if it is a non-defective offspring distribution. It is typically cited that: the extinction probability q is the smallest solution to this equation. I want to discuss why that is the case.

To approach this, we have to consider what extinction means. It is the limit in the event sense of the events {we are extinct after n generations}. Let the probabilities of these events be $q_n$, so $q_0=0$. Then by a straightforward coupling argument, we must have

$0=q_0\le q_1\le q_2 \le\ldots\le q:= \lim q_n \le 1.$

But, by the same generating function argument as before, $q_{n+1}=f(q_n)\ge q_n$. So if we split [0,1] into regions A where $f(x)\ge x$ and B where $f(x), all the $(q_n)$s must occur in the former, and so since it is closed, their limit must be in A also. Note that if f(x) intersects x lots of times, then region A is not necessarily connected. In the diagram below, in moving from $q_n$ to $q_{n+1}$ we might jump across part of B.

This is bad, as we are trying to prove that q is the right boundary of the connected component of A containing 0. But this cannot happen, as f is monotonic. So if one of the roots of f(x)=x in between the hypothesised $q_n is called z, then $f(q_n)< f(z)=z < q_{n+1}$, a contradiction.

Ok, so now we are ready to consider our counterexample to continuity over the percolation threshold. See references for a link to the original source of this example. We have to choose a slightly more complicated event than mere survival or extinction. We consider bond percolation as before on the infinite ternary tree, where every vertex has precisely 3 offspring. Our percolation event is now that the root is the root of an infinite binary tree. That is, the root has at least two children, each of which have at least two children, each of which, and so on.

If we set this probability equal to q, and the probability of an edge being open equal to p, then we have the recurrence:

$q=3p^2(1-p)q^2+p^3[3q^2(1-q)+q^3].$

The first term corresponds to the root having two open edges to offspring, and the second to the root having all three open edges to offspring. After manipulating, we end up with

$q\left[2p^3q^2-3p^2q+1\right]=0.$

We are therefore interested in roots of the quadratic lying between 0 and 1. The discriminant can be evaluated as

$\Delta=p^3(9p-8),$

and so there are no real roots where p<8/9. But when p=8/9, we have a repeated root at q=27/32, which is obviously not zero!

This equation is qualitatively different to the previous one for the extinction probability of a Galton-Watson tree. There, we had a quadratic, with one root at 1. As we varied p, the other root moved continuously from greater than one to less than one, so it passed through 1, giving continuity at the critical probability. Here, we have a cubic, again with one root at 1. But now the other roots are complex for small p, meaning that the local minimum of the cubic lies above the x-axis. As p gets to the critical value, it the local minimum passes below the x-axis, and suddenly we have a repeated root, not at zero.

I would like to have a neat probabilistic heuristic for this result, without having to make reference to generating functions. At the moment, the best I can come up with is to say that the original problem is simple, in the sense that the critical probability is as small as it could be while still making sense in expectation. To be concrete, when the mean of the offspring generation is less than 1, the expected size of the nth generation tends to zero, so there certainly could not be positive probability of having an infinite component.

Whereas in the binary tree example, we only require p=2/3 to have, in expectation, the right number of open edges to theoretically allow an infinite binary tree. If we think of percolation as a dynamic process by coupling in p, essentially as we move from p=2/3 to p=8/9 we need to add enough edges near the origin to be able to take advantage of the high density of edges available far from the origin. The probability of this working given you start from n vertices grows much faster (as n grows) than in the original problem, so you might expect a faster transition.

This is so content-free I’m reluctant even to call it a heuristic. I would be very interested to hear of any more convincing argument for this phenomenon!

REFERENCES

Dekking, Pakes – On family trees and subtrees of simple branching processes (link)

# The Configuration Model

In the past, I’ve talked about limitations of the Erdos-Renyi model of homogeneous random graphs for applications in real-world networks. In a previous post, I’ve discussed a dynamic model, the Preferential Attachment mechanism, that ‘grows’ a graph dynamically by adding edges from new vertices preferentially to existing vertices with high degree. The purpose of this adjustment is to ensure that the distribution of the degrees is not concentrated around some fixed value (which would be c in G(n,c/n) ) but rather exhibits a power-law tail such as observed in many genuine examples.

In this post, we introduce some aspects of the configuration model, which achieves this property more directly. This idea probably first arose in the guise of regular graphs. Recall a regular graph has all degrees equal. How would we construct a random d-regular graph on a large number of vertices?

What we probably want to do is to choose uniformly at random from the set of such graphs, but it is not clear even how large this set is, let alone how one would order its elements to make it possible to make this uniform choice. Instead, we try the following. Assign to each vertex d so-called stubs, which will end up being ‘half-edges’. We then choose two stubs uniformly at random, and glue them together. More formally, we construct an edge between the host vertices, and then delete the chosen stubs. We then continue.

The construction makes no reference to the distribution of stubs, so we are free to choose this as we please. We could for example specify some sequence of degrees which approximates a power-law, so we could sample a random sequence of degrees in some way. So long as we have a sequence of stub set sizes before we start building the edges of the graph we will be able to use the above algorithm.

So what might go wrong? There seem to me to be three potential problems that might arise with this construction.

Firstly, there might be a stub left over, if the sum of the stub set sizes is odd. Recall that in a graph the sum of the degrees is twice the sum of the number of edges, and so in particular the sum of the degrees should be even. But this is a small problem. When the degree sequence is deterministic we can demand that it have even sum, and if it is random, we will typically be working in a large N regime, and so deleting the solitary stub, if such a thing exists, will not affect the sort of properties of the graph we are likely to be interested in.

The second and third objections are perhaps more serious. If we glue together stubs naively, we might end up with loops, that is, edges that ‘begin’ and ‘end’ at the same vertex. These are not allowed in the standard definition of a graph. Alternatively, we might end up with more than one edge between the same pair of vertices.

Our overall aim is that this mechanism gives a convenient way of simulating the uniform distribution on simple graphs with a given degree sequence. At present we have the uniform distribution on potential multigraphs, with a weighting of 1/k! for every multi-edge with multiplicity k, and a weighting of 1/2 for every loop. The latter can be seen because there is an initial probability proportional to $d(v_i)d(v_j)$ that vertices v_i and v_j will be joined, whereas a probability proportional (with the same constant) to $d(v_i)^2$ that v_i will receive a loop. The multi-edge weighting justification is similar.

However, conditional on getting a simple graph, the distribution is uniform on the set of simple graphs with that degree sequence. So it remains to investigate the probability that a graph generated in this way is simple. So long as this probability does not tend to 0 as n grows, we will probably be happy.

The strongest results on this topic are due to Janson. First observe that if the sum of the degrees grows faster than the number of vertices n, we fail to get a graph without loops with high probability. Heuristically, note that on the first pass, we are taking two picks from the set of vertices, biased by the number of stubs. By Cauchy-Schwarz, Rearrangement Inequality or just intuition, the probability of getting the same vertex is greater than if we picked uniformly from the set of vertices without biasing. So the probability of getting no loop on the first pass is $\le (1-\frac{1}{n})$. Take some function a(n) that grows faster than n, but slower than the sum of the degrees. Then after a(n) passes, the degree distribution is still roughly the same. In particular, the sum of the degrees is still an order of magnitude greater than n. So we obtain:

$\mathbb{P}(\text{no loops})\leq (1-\frac{1}{n})^{a(n)}\approx e^{-\frac{a(n)}{n}}\rightarrow 0.$

So, since isolated vertices have no effect on the simplicity or otherwise, we assume the sum of the degrees is $\Theta(n)$. Then, Janson shows that the further condition

$\sum_{i=1}^n d_i^2=O(n),$

is essentially necessary and sufficient for simplicity. We can see why this might be true by looking at the probability that the first edge added is a loop, which is roughly

$\frac{d_1^2+d_2^2+\ldots+d_n^2}{2(\sum d_i)^2}.$

We have to consider $O(\sum d_i)$ edges, so if the above expression is much larger than this, we can perform a similar exponential estimate to show that the probability there are no loops is o(1). The technical part is showing that this probability doesn’t change dramatically as the first few stubs disappear.

Note that in both cases, considering only loops is sufficient for simplicity. Although it looks like loop appearance is weaker than multiplicity of edges, in fact they have the same threshold. It should also be pointed out that, like the uniform random forests, an alternative approach is simply to count the number of simple graphs and multigraphs with a given degree sequence. Good asymptotics can then be found for the probability of simplicity.

In the case of G(n,c/n), we were particularly interested in the emergence of the giant component at time c=1. While first-moment methods can be very effective in demonstrating such results, a branching process local limit representation is probably easiest heuristic for this phase transition.

So long as the degree sequences converge in a natural way, we can apply a similar approach to this configuration model. Concretely, we assume that the proportion of vertices with degree i is $\lambda_i$ in the limit. Although the algebra might push through, we should be aware that this means we are not explicitly specifying how many vertices have degree, eg $\Theta(n^{1/2})$. For now assume the $\lambda_i$s sum to 1, so specify a probability distribution for degree induced by choosing a vertex uniformly at random.

So we start at a vertex, and look at its neighbours. The expected number of neighbours of this root vertex is $\sum i\lambda i$. Thereafter, when we consider a child vertex, based on how the stubs are paired up (and in particular the fact that the order of the operations does not matter – the choice of partner of a given stub is chosen uniformly at random), we are really choosing a stub uniformly at random. This corresponds to choosing a vertex at random, biased by the number of stubs available. The quantity of interest is how many additional stubs (other than the one that led to the vertex) are attached to this vertex. We assume we don’t need to worry too much about repeating vertices, in a similar way to G(n,c/n). So the expected number of additional stubs is

$\frac{1}{\sum i\lambda_i}\sum i\lambda_i(i-1).$

For an infinite component, we required the expectation to be > 1, which is equivalent to

$\sum \lambda_i i(i-2)>0.$

This was proven by Molloy and Reed (95), then with fewer conditions by Janson (07). The latter also shows how to use this construction to derive the giant component for G(n,c/n) result.

REFERENCES

Janson – A New Approach to the Giant Component Problem

Molloy, Reed – A Critical Point for Random Graphs with a Given Degree Sequence

Janson – The Probability that  Random Multigraph is Simple

# Large Deviations 6 – Random Graphs

As a final instalment in this sequence of posts on Large Deviations, I’m going to try and explain how one might be able to apply some of the theory to a problem about random graphs. I should explain in advance that much of what follows will be a heuristic argument only. In a way, I’m more interested in explaining what the technical challenges are than trying to solve them. Not least because at the moment I don’t know exactly how to solve most of them. At the very end I will present a rate function, and reference properly the authors who have proved this. Their methods are related but not identical to what I will present.

Problem

Recall the two standard definitions of random graphs. As in many previous posts, we are interested in the sparse case where the average degree of a vertex is o(1). Anyway, we start with n vertices, and in one description we add an edge between any pair of vertices independently and with fixed probability $\frac{\lambda}{n}$. In the second model, we choose uniformly at random from the set of graphs with n vertices and $\frac{\lambda n}{2}$ edges. Note that if we take the first model and condition on the number of edges, we get the second model, since the probability of a given configuration appearing in G(n,p) is a function only of the number of edges present. Furthermore, the number of edges in G(n,p) is binomial with parameters $\binom{n}{2}$ and p. For all purposes here it will make no difference to approximate the former by $\frac{n^2}{2}$.

Of particular interest in the study of sparse random graphs is the phase transition in the size of the largest component observed as $\lambda$ passes 1. Below 1, the largest component has size on a scale of log n, and with high probability all components are trees. Above 1, there is a unique giant component containing $\alpha_\lambda n$ vertices, and all other components are small. For $\lambda\approx 1$, where I don’t want to discuss what ‘approximately’ means right now, we have a critical window, for which there are infinitely many components with sizes on a scale of $n^{2/3}$.

A key observation is that this holds irrespective of which model we are using. In particular, this is consistent. By the central limit theorem, we have that:

$|E(G(n,\frac{\lambda}{n}))|\sim \text{Bin}\left(\binom{n}{2},\frac{\lambda}{n}\right)\approx \frac{n\lambda}{2}\pm\alpha,$

where $\alpha$ is the error due to CLT-scale fluctuations. In particular, these fluctuations are on a scale smaller than n, so in the limit have no effect on which value of $\lambda$ in the edge-specified model is appropriate.

However, it is still a random model, so we can condition on any event which happens with positive probability, so we might ask: what does a supercritical random graph look like if we condition it to have no giant component? Assume for now that we are considering $G(n,\frac{\lambda}{n}),\lambda>1$.

This deviation from standard behaviour might be achieved in at least two ways. Firstly, we might just have insufficient edges. If we have a large deviation towards too few edges, then this would correspond to a subcritical $G(n,\frac{\mu n}{2})$, so would have no giant components. However, it is also possible that the lack of a giant component is due to ‘clustering’. We might in fact have the correct number of edges, but they might have arranged themselves into a configuration that keeps the number of components small. For example, we might have a complete graph on $Kn^{1/2}$ vertices plus a whole load of isolated vertices. This has the correct number of edges, but certainly no giant component (that is an O(n) component).

We might suspect that having too few edges would be the primary cause of having no giant component, but it would be interesting if clustering played a role. In a previous post, I talked about more realistic models of complex networks, for which clustering beyond the levels of Erdos-Renyi is one of the properties we seek. There I described a few models which might produce some of these properties. Obviously another model is to take Erdos-Renyi and condition it to have lots of clustering but that isn’t hugely helpful as it is not obvious what the resulting graphs will in general look like. It would certainly be interesting if conditioning on having no giant component were enough to get lots of clustering.

To do this, we need to find a rate function for the size of the giant component in a supercritical random graph. Then we will assume that evaluating this near 0 gives the LD probability of having ‘no giant component’. We will then compare this to the straightforward rate function for the number of edges; in particular, evaluated at criticality, so the probability that we have a subcritical number of edges in our supercritical random graph. If they are the same, then this says that the surfeit of edges dominates clustering effects. If the former is smaller, then clustering may play a non-trivial role. If the former is larger, then we will probably have made a mistake, as we expect on a LD scale that having too few edges will almost surely lead to a subcritical component.

Methods

The starting point is the exploration process for components of the random graph. Recall we start at some vertex v and explore the component containing v depth-first, tracking the number of vertices which have been seen but not yet explored. We can extend this to all components by defining:

$S(0)=0, \quad S(t)=S(t-1)+(X(t)-1),$

where X(t) is the number of children of the t’th vertex. For a single component, S(t) is precisely the number of seen but unexplored vertices. It is more complicated in general. Note that when we exhaust the first component S(t)=-1, and then when we exhaust the second component S(t)=-2 and so on. So in fact

$S_t-\min_{0\leq s\leq t}S_s$

is the number of seen but unexplored vertices, with $\min_{0\leq s\leq t}S_s$ equal to (-1) times the number of components already explored up to time t.

Once we know the structure of the first t vertices, we expect the distribution of X(t) – 1 to be

$\text{Bin}\Big(n-t-[S_t-\min_{0\leq s\leq t}S_s],\tfrac{\lambda}{n}\Big)-1.$

We aren’t interested in all the edges of the random graph, only in some tree skeleton of each component. So we don’t need to consider the possibility of edges connecting our current location to anywhere we’ve previously visited (as such an edge would have been consider then – it’s a depth-first exploration), hence the -t. But we also don’t want to consider edges connecting our current location to anywhere we’ve seen, since that would be a surplus edge creating a cycle, hence the -S_s. It is binomial because by independence even after all this conditioning, the probability that there’s an edge from my current location to any other vertex apart from those discounted is equal to $\frac{\lambda}{n}$ and independent.

For Mogulskii’s theorem in the previous post, we had an LDP for the rescaled paths of a random walk with independent stationary increments. In this situation we have a random walk where the increments do not have this property. They are not stationary because the pre-limit distribution depends on time. They are also not independent, because the distribution depends on behaviour up to time t, but only through the value of the walk at the present time.

Nonetheless, at least by following through the heuristic of having an instantaneous exponential cost for a LD event, then products of sums becoming integrals within the exponent, we would expect to have a similar result for this case. We can find the rate function $\Lambda_\lambda^*(x)of$latex \text{Po}(\lambda)-1\$ and thus get a rate function for paths of the exploration process

$I_\lambda(f)=\int_0^1 \Lambda_{(1-t-\bar{f}(t))\lambda}^*(f')dt,$

where $\bar{f}(t)$ is the height of f above its previous minimum.

Technicalities and Challenges

1) First we need to prove that it is actually possible to extend Mogulskii to this more general setting. Even though we are varying the distribution continuously, so we have some sort of ‘local almost convexity’, the proof is going to be fairly fiddly.

2) Having to consider excursions above the local minima is a massive hassle. We would ideally like to replace $\bar{f}$ with f. This doesn’t seem unreasonable. After all, if we pick a giant component within o(n) steps, then everything considered before the giant component won’t show up in the O(n) rescaling, so we will have a series of macroscopic excursions above 0 with widths giving the actual sizes of the giant components. The problem is that even though with high probability we will pick a giant component after O(1) components, then probability that we do not do this decays only exponentially fast, so will show up as a term in the LD analysis. We would hope that this would not be important – after all later we are going to take an infimum, and since the order we choose the vertices to explore is random and in particular independent of the actual structure, it ought not to make a huge difference to any result.

3) A key lemma in the proof of Mogulskii in Dembo and Zeitouni was the result that it doesn’t matter from an LDP point of view whether we consider the linear (continuous) interpolation or the step-wise interpolation to get a process that actually lives in $L_\infty([0,1])$. In this generalised case, we will also need to check that approximating the Binomial distribution by its Poisson limit is valid on an exponential scale. Note that because errors in the approximation for small values of t affect the parameter of the distribution at larger times, this will be more complicated to check than for the IID case.

4) Once we have a rate function, if we actually want to know about the structure of the ‘typical’ graph displaying some LD property, we will need to find the infimum of the integrated rate function with some constraints. This is likely to be quite nasty unless we can directly use Euler-Lagrange or some other variational tool.

Papers by O’Connell and Puhalskii have found the rate function. Among other interesting things, we learn that:

$I_{(1+\epsilon)}(0)\approx \frac{\epsilon^3}{6},$

while the rate function for the number of edges:

$-\lim\tfrac{1}{n}\log\mathbb{P}\Big(\text{Bin}(\tfrac{n^2}{2},\tfrac{1+\epsilon}{n})\leq\tfrac{n}{2}\Big)\approx \frac{\epsilon^2}{4}.$

So in fact it looks as if there might be a significant contribution from clustering after all.

# Coalescence 1: What is it, and why do we care?

As part of Part III, instead of sitting an extra exam paper I am writing an essay. I have chosen the topic of ‘Multiplicative Coalescence’. I want to avoid contravening plagiarism rules, which don’t allow you to quote your own words without a proper citation, which I figure is tricky on a blog, nor open publishing of anything you intend to submit. So just to be absolutely sure, I’m going to suppress this series of posts until after May 4th, when everything has to be handed in.

———–

Informal Description

Coalescence refers to a process in which particles join together over time. An example might be islands of foam on the surface of a cup of coffee. When two clumps meet, they join, and will never split. In this example, a model would need to take into account the shape of all the islands, their positions, their velocities, and boundary properties. To make things tractable, we need to distance ourselves from the idea that particles merge through collisions, which are highly physical and complicated, and instead just consider that they merge.

Description of the Model

When two particles coalesce, it is natural to assume that mass is conserved, as this will be necessary in any physical application. With this in mind, it makes sense to set up the entire model using only the masses of particles. Define the kernel K(x,y) which describes the relative rate or likelihood of the coalescence {x,y} -> x+y. This has a different precise meaning in different contexts. Effectively, we are making a mean-field assumption that all the complications of a physical model as described above can be absorbed into this coalescent kernel, either because the number of particles is large, or because the other effects are small.

When there is, initially, a finite number of particles, the process is stochastic. Coalescence almost surely happen one at a time, and so we can view the process as a continuous time Markov Chain with state space the set of relevant partitions of the total mass present. The transition rate p(A,B) is given by K(x,y) when the coalescence {x,y} -> x+y transforms partition into B, and 0 otherwise. An observation is that the process recording the number of {x,y} -> x+y coalescences is an inhomogeneous Poisson process with local intensity n(x,t)n(y,t)K(x,y) where n(x,t) is the number of particles with mass at time t.

This motivates the move to an infinite-volume setting. Suppose that there are infinitely many particles, so that coalescences are occurring continuously. The rate of {x,y} -> x+y coalescences is still n(x,t)n(y,t)K(x,y) but now n(x,t) specifies the density of particles with mass at time t. Furthermore, because of the continuum framework, this rate is now deterministic rather than stochastic. This is extremely important, as by removing the probability from a probabilistic model, it can be treated as a large but simple ODE.

Two Remarks

1) Once this introduction is finished, we shall be bringing our focus onto multiplicative coalescence, where K(x,y) = xy. In particular, this is a homogeneous function, as are the other canonical kernels. This means that considering K(x,y) = cxy is the same as performing a constant factor time-change when K(x,y) = xy. Similarly, it is not important how the density n(x,t) is scaled as this can also be absorbed with a time-change. In some contexts, it will be natural and useful to demand that the total density be 1, but this will not always be possible. In general it is convenient to absorb as much as possible into the time parameter, particularly initial conditions, as will be discussed.

2) Working with an infinite volume of particles means that mass is no longer constrained to finitely many values. Generally, it is assumed that the masses are discrete, taking values in the positive integers, or continuous, taking values in the positive reals. In this case, the rate of coalescences between particles with masses in (x, x+dx) and (y,y+dy) is n(x,t)n(y,t)K(x,y)dxdy. The main difference between these will arise when we try to view the process as limits of finite processes. Continue reading