Kernels of critical graph components

This post is motivated by G(N,p), the classical Erdos-Renyi random graph, specifically its critical window, when p=p(N)=\frac{1}{N}(1+\lambda N^{-1/3}).

We start with the following observation, which makes no restriction on p. Suppose a component of G(N,p) is a tree. Then, the graph geometry of this component is that of a uniform random tree on the appropriate number of vertices. This is deliberately informal. To be formal, we’d have to say “condition on a particular subset of vertices forming a tree-component” and so on. But the formality is broadly irrelevant, because at the level of metric scaling limits, if we want to describe the structure of a tree component, it doesn’t matter whether it has \log N or \frac{1}{7}N vertices, because in both cases the tree structure is uniform. The only thing that changes is the scaling factor.

In general, when V vertices form a connected component of a graph with E edges, we define the excess to be E-V+1. So the excess is non-negative, and is zero precisely when the component is a tree. I’m reluctant to say that the excess counts the number of cycles in the component, but certainly it quantifies the amount of cyclic structure present. We will sometimes, in a mild abuse of notation, talk about excess edges. But note that for a connected component with positive excess, there is a priori no way to select which edges would be the excess edges. In a graph process, or when there is some underlying exploration of the component, there sometimes might be a canonical way to classify the excess edges, though it’s worth remarking that the risk of size-biasing errors is always extremely high in this sort of situation.

Returning to the random graph process, as so often there are big changes around criticality. In the subcritical regime, the components are small, and most of them, even the largest with high probability, are trees. In the supercritical regime, the giant component has excess \Theta(N), which is qualitatively very different.

It feels like every talk I’ve ever given has begun with an exposition of Aldous’s seminal paper [Al97] giving a distributional scaling limit of the sizes of critical components in the critical window, and a relation between the process on this time-scale and the multiplicative coalescent. And it remains relevant here, because the breadth-first exploration process can also be used to track the number of excess edges.

In a breadth-first exploration, we have a stack of vertices we are waiting to explore. We pick one and look its neighbours restricted to the rest of the graph, that is without the vertices we have already fully explored, and also without the other vertices in the stack. That’s the easiest way to handle the total component size. But we can simultaneously track how many times we would have joined to a neighbour within the stack, which leads to an excess edge, and Aldous derives a joint distributional scaling limit for the sizes of the critical components and their excesses. (Note that in this case, there is a canonical notion of excess edge, but it depends not just on the graph structure, but also on the extra randomness of the ordering within the breadth-first search.)

Roughly speaking, we consider the reflected exploration process, and its scaling limit, which is a reflected parabolically-drifting Brownian motion (though the details of this are not important at this level of exposition, except that it’s a well-behaved non-negative process that hits zero often). The component sizes are given by the widths of the excursions above zero, scaled up in a factor N^{1/3}. Then conditional on the shape of the excursion, the excess is Poisson with parameter the area under the excursion, with no rescaling. That is, a critical component has \Theta(1) excess.

So, with Aldous’s result in the background, when we ask about the metric structure of these critical components, we are really asking: “what does a uniformly-chosen connected component with fixed excess look like when the number of vertices grows?”

I’ll try to keep notation light, but let’s say T(n,k) is a uniform choice from connected graphs on n vertices with excess k.

[Note, the separation of N and n is deliberate, because in the critical window, the connected components have size n = \Theta(N^{2/3}), so I want to distinguish the two problems.]

In this post, we will mainly address the question: “what does the cycle structure of T(n,k) look like for large n?” When k=0, we have a uniform tree, and the convergence of this to the Brownian CRT is now well-known [CRT2, LeGall]. We hope for results with a similar flavour for positive excess k.

2-cores and kernels

First, we have to give a precise statement of what it means to study just the cycle structure of a connected component. From now on I will assume we are always working with a connected graph.

There are several equivalent definitions of the 2-core C(G) of a graph G:

  • When the excess is positive, there are some cycles. The 2-core is the union of all edges which form part of some cycle, and any edges which lie on a path between two edges which both form part of some cycle.
  • C(G) is the maximal induced subgraph where all degrees are at least two.
  • If you remove all the leaves from the graph, then all the leaves from the remaining graph, and continue, the 2-core is the state you arrive at where there are no leaves.

It’s very helpful to think of the overall structure of the graph as consisting of the 2-core, with pendant trees ‘hanging off’ the 2-core. That is, we can view every vertex of the 2-core as the root of a (possibly size 1) tree. This is particular clear if we remove all the edges of the 2-core from the graph. What remains is a forest, with one tree for each vertex of the 2-core.

In general, the k-core is the maximal induced subgraph where all degrees are at least k. The core is generally taken to be something rather different. For this post (and any immediate sequels) I will never refer to the k-core for k>2, and certainly not to the traditional core. So I write ‘core’ for ‘2-core’.

As you can see in the diagram, the core consists of lots of paths, and topologically, the lengths of these paths are redundant. So we will often consider instead the kernel, K(G), which is constructed by taking the core and contracting all the paths between vertices of degree greater than 2. The resulting graph has minimal degree at least three. So far we’ve made no comment about the simplicity of the original graphs, but certainly the kernel need not be simple. It will regularly have loops and multiple edges. The kernel of the graph and core in the previous diagram is therefore this:

Kernels of critical components

To recap, we can deconstruct a connected graph as follows. It has a kernel, and each edge of the kernel is a path length of some length in the core. The rest of the graph consists of trees hanging off from the core vertices.

For now, we ask about the distribution of the kernel of a T(n,K). You might notice that the case k=1 is slightly awkward, as when the core consists of a single cycle, it’s somewhat ambiguous how to define the kernel. Everything we do is easily fixable for k=1, but rather than carry separate cases, we handle the case k\ge 2.

We first observe that fixing k doesn’t confirm the number of vertices or edges in the kernel. For example, both of the following pictures could correspond to k=3:

However, with high probability the kernel is 3-regular, which suddenly makes the previous post relevant. As I said earlier, it can introduce size-biasing errors to add the excess edges one-at-a-time, but these should be constant factor errors, not scaling errors. So imagine the core of a large graph with excess k=2. For the sake of argument, assume the kernel has the dumbbell / handcuffs shape. Now add an extra edge somewhere. It’s asymptotically very unlikely that this is incident to one of the two vertices with degree three in the core. Note it would need to be incident to both to generate the right-hand picture above. Instead, the core will gain two new vertices of degree three.

Roughly equivalently, once the size of the core is fixed (and large) we have to make a uniform choice from connected graphs of this size where almost every vertex has degree 2, and \Theta(1) of the rest have degree 3 or higher. But the sum of the degrees is fixed, because the excess is fixed. If there are n vertices in the core, then there are \Theta(n) more graphs where all the vertices have degree 2 or 3, than graphs where a vertex has degree at least 4. Let’s state this formally.

Proposition: The kernel of a uniform graph with n vertices and excess k\ge 2 is, with high probability as n\rightarrow\infty, 3-regular.

This proved rather more formally as part of Theorem 7 of [JKLP], essentially as a corollary after some very comprehensive generating function setup; and in [LPW] with a more direct computation.

In the previous post, we introduced the configuration model as a method for constructing regular graphs (or any graphs with fixed degree sequence). We observe that, conditional on the event that the resulting graph is simple, it is in fact uniformly-distributed among simple graphs. When the graph is allowed to be a multigraph, this is no longer true. However, in many circumstances, as remarked in (1.1) of [JKLP], for most applications the configuration model measure on multigraphs is the most natural.

Given a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges, and K a uniform choice from the configuration model with these parameters, we have

\mathbb{P}\left( K \equiv H \right) \propto \left(2^{t(H)} \prod_{e\in E(H)} \mathrm{mult}(e)! \right)^{-1},

where t(H) is the number of loops in H, and mult(e) the multiplicity of an edge e. This might seem initially counter-intuitive, because it looks we are biasing against graphs with multiple edges, when perhaps our intuition is that because there are more ways to form a set of multiple edges we should bias in favour of it.

I think it’s most helpful to look at a diagram of a multigraph as shown, and ask how to assign stubs to edges. At a vertex with degree three, all stub assignments are different, that is 3!=6 possibilities. At the multiple edge, however, we care which stubs match with which stubs, but we don’t care about the order within the multi-edge. Alternatively, there are three choices of how to divide each vertex’s stubs into (2 for the multi-edge, 1 for the rest), and then two choices for how to match up the multi-edge stubs, ie 18 in total = 36/2, and a discount factor of 2.

We mention this because in fact K(T(n,k)) converges in distribution to this uniform configuration model. Once you know that K(T(n,k)) is with high probability 3-regular, then again it’s probably easiest to think about the core, indeed you might as well condition on its total size and number of degree 3 vertices. It’s then not hard to convince yourself that a uniform choice induces a uniform choice of kernel. Again, let’s state that as a proposition.

Proposition: For any H a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges as before,

\lim_{n\rightarrow\infty}\mathbb{P}\left( K(T(n,k)) \equiv H \right) \propto \left(2^{t(H)} \prod_{e\in E(H)} \mathrm{mult}(e)! \right)^{-1}.

As we said before, the kernel describes the topology of the core. To reconstruct the graph, we need to know the lengths in the core, and then how to glue pendant trees onto the core. But this final stage depends on k only through the total length of paths in the core. Given that information, it’s a combinatorial problem, and while I’m not claiming it’s easy, it’s essentially the same as for the case with k=1, and is worth treating separately.

It is worth clarifying a couple of things first though. Even the outline of methods above relies on the fact that the size of the core diverges as n grows. Again, the heuristic is that up to size-biasing errors, T(n,k) looks like a uniform tree with some uniformly-chosen extra edges. But distances in T(n,k) scale like n^{1/2} (and thus in critical components of G(N,p) scale like N^{1/3}). And the core will be roughly the set of edges on paths between the uniformly-chosen pairs of vertices, and so will also have length \Theta(n^{1/2}).

Once you have conditioned on the kernel structure, and the (large) number of internal vertices on paths in the core (ie the length of the core), it is natural that the assignment of the degree-2 vertices to core paths / kernel edges is uniform. A consequence of this is that if you record (Y_1,\ldots,Y_m) the lengths of paths in the core, where m=3(k-1), then

\frac{(Y_1,\ldots,Y_m)}{\sum Y_i} \stackrel{d}\rightarrow \mathrm{Dirichlet}(1,1,\ldots,1).

This is stated formally as Corollary 7 b) of [ABG09]. It’s worth noting that this confirms that the lengths of core paths are bounded in probability away from zero after the appropriate rescaling. In seeking a metric scaling limit, this is convenient as it means there’s so danger that two of the degree-3 vertices end up in ‘the same place’ in the scaling limit object.

To recap, the only missing ingredients now to give a complete limiting metric description of T(n,k) are 1) a distributional limit of the total core length; 2) some appropriate description of set of pendant trees conditional on the size of the pendant forest. [ABG09] show the first of these. As remarked before, all the content of the second of these is encoded in the unicyclic k=1 case, which I have written about before, albeit slightly sketchily, here. (Note that in that post we get around size-biasing by counting a slightly different object, namely unicyclic graphs with an identified cyclic edge.)

However, [ABG09] also propose an alternative construction, which you can think of as glueing CRTs directly onto the stubs of the kernel (with the same distribution as before). The proof that this construction works isn’t as painful as one might fear, and allows a lot of the other metric distributional results to be read off as corollaries.

References

[ABG09] – Addario-Berry, Broutin, Goldschmidt – Critical random graphs: limiting constructions and distributional properties

[CRT2] – Aldous – The continuum random tree: II

[Al97] – Aldous – Brownian excursions, critical random graphs and the multiplicative coalescent

[JKLP] – Janson, Knuth, Luczak, Pittel – The birth of the giant component

[LeGall] – Le Gall – Random trees and applications

[LPW] – Luczak, Pittel, Wierman – The structure of a random graph at the point of the phase transition

 

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Random 3-regular graphs

A graph is d-regular if every vertex has degree d. Probably the easiest examples of d-regular graphs are the complete graph on (d+1) vertices, and the infinite d-ary tree. A less trivial example is the Petersen graph, which is 3-regular. 3-regular graphs will be the main focus for some of this post, but initially we lose nothing by considering general d.

Throughout, a necessary condition for the existence of a d-regular graph with N vertices is that at least one of d and N is even, as the sum of the degrees of a graph must be even. We will always assume that this holds, so that when d=3, we are always taking N to be even.

A natural pair of questions for a probabilist is ‘can we sample a d-regular graph with N vertices uniformly at random?’ and ‘what does a typical large d-regular graph look like?’

In a rather old post, I addressed some aspects of the first question, but revisit it briefly here. A good idea, due to Bollobas [B80] is to assign to all the vertices d stubs (or half-edges), and choose a matching of the Nd stubs uniformly at random. This works as a method to generate a random graph with any fixed degree sequence.

If you want your graphs to be simple, this can go wrong, because there’s a chance you get loops (that is, an edge from a vertex v to itself) and multiple edges between the same pair of vertices. It would be nice the graph formed in this fashion was simple with high probability when N\rightarrow\infty. Unfortunately that’s not the case, however the probability that the graph is simple remains asymptotically bounded away from 0 and 1. Indeed, because the presence of a loop / multiple edge is asymptotically independent of the presence of a loop / multiple edge elsewhere, it’s unsurprising we have a Poisson limit for the number of such occurences. So from a sampling point of view, it’s reasonable to sample a graph in this way until you find a simple one. This takes O(1) steps, and it’s O(N) steps to check whether a given multigraph is simple.

It’s clear that conditional on the graph generated in this fashion being simple, its distribution is uniform on the set of simple graphs with the correct degree distribution. If you are happy for your graphs to have loops, then it’s a little bit more complicated, because if an edge has multiplicity k, these can appear in k! ways in the configuration construction.

Other asymptotic properties

Loops and multiple edges can be thought of as cycles of length 1 and 2 respectively if you want. We might ask about other small cycles. A calculation in expectation is relatively straightforward. Given three vertices, the probability they form a triangle (in at least one way) is \Theta(N^{-3}), and there are \Theta(N^3) ways to choose three vertices. Thus the expected number of triangles is \Theta(1). Finally, the edge structure induced on disjoint triples is asymptotically independent, and hence a Poisson limit. (See [J06] for details, including more detail on the general configuration construction.) The same result holds for the same reasons for cycles of any fixed finite length.

We might also ask about connectivity. At a heuristic level, there are two ways for the graph to be disconnected: it could have some small components; or it could have two components of size \Theta(N). The smallest possible component is K_4, and an argument like for the cycles above shows that the number of copies of K_4 vanishes in expectation. Now, consider having two components of size roughly N/2. There are \binom{N}{N/2} \sim 2^{2N} ways to make this choice. However, given such a choice, we can handle the probability that all the stubs from one class match within that class by going through the class one stub at a time:

\frac{\frac{3N}{2}-1}{3N-1} \times \frac{\frac{3N}{2}-3}{3N-3} \times \cdots \times \frac{1}{\frac{3N}{2}+1}.

We approximate this as

\frac{\sqrt{(3N/2)!}}{\sqrt{ (3N)!}} \sim  e^{3N/2} 2^{-3N/2} \left(3N\right)^{-3N/2},

and this dominates the number of choices powerfully enough that we might believe it remains valid for a broader range of class sizes. In fact we have a much stronger statement, namely that G(N,3) is 3-connected with high probability. This means that the graph cannot be disconnected by removing two vertices, or equivalently that there are three vertex-disjoint paths between any pair of vertices in the graph, essentially one emerging from each stub. See this note by David Ellis for a quick proof. We might return to this later.

You might ask about planarity. It’s clear from degree consideration that there are no induced copies of K_5 in any random 3-regular graph, and since K_{3,3} contains a cycle of length 4, and with high probability G(N,3) doesn’t, that takes care of that possibility too. However, there might be minors of this form. This seemed a good example of the Kuratowski criterion not actually being that useful, since I certainly don’t find the minors of the 3-regular graph an obvious structure to handle.

However, we can use Euler’s formula V – E + F = 2 for planar graphs. Here V = N, E = 3N/2. Faces are described by (a subset of the) cycles, and we there are asymptotically O(1) small cycles, so most faces include a large number of edges. But each edge corresponds to at most two faces. So we have F \ll E, and so with high probability Euler’s formula can’t hold in G(N,3) for large N.

We can also ask about the local limit of G(N,3). Since the vertices are exchangeable, we don’t need to worry about whether we choose the root uniformly at random (often referred to as the Benjamini-Schramm sense) or by some other method.

The root has up to three neighbours, and with high probability it has exactly three neighbours. These neighbours have at most two other neighbours themselves. However, we’ve already seen that there are asymptotically O(1) cycles, and so with high probability there are no small cycles near a fixed root vertex. So the six neighbours-of-neighbours are with high probability different to the root and the root’s neighbours and to each other. We can make this argument at arbitrary finite radius from the root, to conclude that the local limit of G(N,3) is the infinite 3-ary tree.

Spectral expansion

[Caveat – this is something I read about and wanted to mention, but I really don’t know much at all about any of this theory, and it’s definitely not certain that what follows wouldn’t be better replaced by a set of links.]

This straightforward local limit offers good heuristics on some of the more global properties. Almost by definition, the d-ary tree expands as rapidly as is possible away from the root among infinite d-regular graphs. There are a number of ways to measure the expansion of a graph, and some methods transfer better to the infinite setting than others. The adjacency matrix of an infinite graph can be defined similarly to that of a finite graph, and it remains possible to talk about eigenfunctions and spectrum. As for the finite setting, d is an eigenvalue because the tree is d-regular, and -d is an eigenvalue because it is also bipartite.

The next largest eigenvalue \lambda_2 governs the spectral gap d-\lambda_2 which is a measure of the expansion of a graph. A graph is a good (spectral) expander if all the non-trivial eigenvalues are close to zero. A priori, all we know is that |\lambda_2|\le d. For the infinite d-ary tree, we have \lambda_2 = 2\sqrt{d-1}. This blog post by Luca Trevisan gives a very readable proof.

A key result is that finite graphs can have \lambda_2 \le 2\sqrt{d-1}, but not asymptotically. That is, taking N to be the number of vertices:

\lambda_2 \ge 2\sqrt{d-1} - o_N(1).

This is the content of the Alon-Boppana theorem [Al86]. In fact the error can be quantified as O(\frac{1}{\log N}) – the diamater of the graph is relevant here. A finite d-regular graph for which \lambda_2\le 2\sqrt{d-1} is called a Ramanujan graph. The existence of Ramanujan graphs has been much studied, and various constructions often rely on number theoretic properties of N, and lie at the interface of disparate branches of mathematics where my understanding is zero rather than epsilon.

Now return to our view of the d-ary tree as the local limit of a d-regular graph on N vertices for large N. We might expect from everything above that the uniform d-regular graph is a good expander. Bollobas shows that in the sense of edge-expansion, asymptotically almost all d-regular graphs have edge-expansion bounded away from zero. (See Section 2 of [Ell], including history of the d=3 case.) Friedman [Fri08] proves the conjecture of Alon that for every \epsilon>0, a.a.s. \lambda_2 for G(N,d) is at most 2\sqrt{d-1}+\epsilon. In this sense, G(N,d) is asymptotically ‘almost Ramanujan’. (See also [Bor17] for another proof and an introduction including history, context and references.)

Some other links: The Wikipedia page on expanders, which includes a discussion of the different descriptions of expansion, and the Cheeger inequalities and other relations between them; slides for a talk by Spielman on spectra and Ramanujan graphs; a survey by Murty on Ramanujan graphs;.

What next?

This post took a slightly different direction from what I had intended, and rather than make a halting U-turn back to my planned finale, I’ll postpone this. However, a short overture is that I’m interested in the structure of critical components of random graphs during the critical window. This is the window during which the largest components first have cycles with probability \Theta(1). Indeed, the critical components have size \Theta(N^{2/3}) and \Theta(1) surplus edges. Conditional on their size, and number of surplus edges, the choice of the graph structure on the component is uniform among such (connected) graphs.

Addario-Berry, Broutin and Goldschmidt [ABG09] study scaling limits of such components. Central to this analysis is the 2-core of such components, which can be described in terms of 3-regular (multi)graphs. Various processes we are now interested in running on the critical components of critical RGs can then be studied in terms of related processes on random 3-regular graphs.

References

[ABG09] – Addario-Berry, Broutin, Goldschmidt – Critical random graphs: limiting constructions and distributional properties

[Al86] – Alon – Eigenvalues and expanders

[B80] – Bollobas – A probabilistic proof of an asymptotic formula for the number of labelled regular graphs

[B88] – Bollobas – The isoperimetric number of random regular graphs

[Bor17] – Bordenave – A new proof of Friedman’s second eigenvalue theorem and its extension to random lifts. Arxiv.

[Ell] – Ellis – The expansion of random regular graphs

[Fri08] – Friedman – A proof of Alon’s second eigenvalue conjecture and related problems

[J06] – Janson – The probability that a random multigraph is simple by

Random transpositions

We study a procedure for generating a random sequence of permutations of [N]. We start with the identity permutation, and then in each step, we choose two elements uniformly at random, and swap them. We obtain a sequence of permutations, where each term is obtained from the previous one by multiplying by a uniformly-chosen transposition.

Some more formality and some technical remarks:

  • This is a Markov chain, and as often with Markov chains, it would be better it was aperiodic. As described, the cycle will alternate between odd and even permutations. So we allow the two elements chosen to be the same. This laziness slows down the chain by a factor N-1/N, but removes periodicity. We will work over timescales where this adjustment makes no practical difference.
  • Let \tau_1,\tau_2,\ldots be the sequence of transpositions. We could define the sequence of permutations by \pi_m= \tau_m\cdot\tau_{m-1}\cdot \ldots\cdot \tau_1. I find it slightly more helpful to think of swapping the elements in places i and j, rather the elements i and j themselves, and so I’ll use this language, for which \pi_m = \tau_1\cdot \tau_2\cdot\ldots \cdot \tau_m is the appropriate description. Of course, transpositions and the identity are self-inverse permutations, so it makes no difference to anything we might discuss.
  • You can view this as lazy random walk on the Cayley graph of S_N generated by the set of transpositions. That is, the vertices of the graph are elements of S_N, and two are connected by an edge if one can be obtained from the other by multiplying by a transposition. Note this relation is symmetric. Hence random transposition random walk.
  • Almost everything under discussion would work in continuous time too.

At a very general level, this sort of model is interesting because sometimes the only practical way to introduce ‘global randomness’ is repeatedly to apply ‘local randomness’. This is not the case for permutations – it is not hard to sample uniformly from S_N. But it is a tractable model in which to study relevant questions about the generating randomness on a complicated set through iterated local operations.

Since it is a Markov chain with a straightforward invariant distribution, we can ask about the mixing time. That is, the correct scaling for the number of moves before the random permutation is close in distribution (say in the sense of total variation distance) to the equilibrium distribution. See this series of posts for an odd collection of background material on the topic. Diaconis and Shahshahani [DS81] give an analytic argument for mixing around \frac{N\log N}{2} transpositions. Indeed they include a constant because it is a sharp cutoff, where the total variation distance drops from approximately 1 to approximately 0 in O(N) steps.

Comparison with Erdos-Renyi random graph process

In the previous result, one might observe that m=\frac{N\log N}{2} is also the threshold number of edges to guarantee connectivity of the Erdos-Renyi random graph G(N,m) with high probability. [ER59] Indeed, there is also a sharp transition around this threshold in this setting too.

We explore this link further. We can construct a sequence of random graphs simultaneously with the random transposition random walk. When we multiply by transposition (i j), we add edge ij in the graph. Laziness of RTRW and the possibility of multiple edges mean this definition isn’t literally the same as the conventional definition of a discrete-time Erdos-Renyi random graph process, but again this is not a problem for any of the effects we seek to study.

The similarity between the constructions is clear. But what about the differences? For the RTRW, we need to track more information than the random graph. That is, we need to know what order the transpositions were added, rather than merely which edges were added. However, the trade-off is that a permutation is a simpler object than a graph in the following sense. A permutation can be a described as a union of disjoint cycles. In an exchangeable setting, all the information about a random permutation is encoded in the lengths of the these cycles. Whereas in a graph, geometry is important. It’s an elegant property of the Erdos-Renyi process that we can forget about the geometry and treat it as a process on component sizes (indeed, a multiplicative coalescent process), but there are other questions we might need to ask for which we do have to study the graph structure itself.

Within this analogy, unfortunately the word cycle means different things in the two different settings. In a permutation, a cycle is a directed orbit, while in a graph it has the usual definition. I’m going to write graph-cycle whenever relevant to avoid confusion.

A first observation is that, under this equivalence, the cycles of the permutation form a finer partition than the components of the graph. This is obvious. If we split the vertices into sets A and B, and there are no edges between them, then nothing in set A will ever get moved out of set A by a transposition. (Note that the slickness of this analogy is the advantage of viewing a transposition as swapping the elements in places i and j.)

However, we might then ask under what circumstances is a cycle of the permutation the same as a component of the graph (rather than a strict subset of it). A first answer is the following:

Lemma: [Den59] The permutation formed by a product of transpositions corresponding in any order to a tree in the graph has a single cycle.

We can treat this as a standalone problem and argue in the following predictable fashion. (Indeed, I was tempted to set this as a problem during selection for the UK team for IMO 2017 – it’s perfectly suitable in this context I think.) The first transposition corresponds to some edge say ab, and removing this edge divides the vertices into components A \ni a, B\ni b. Since no further transposition swaps between places in A and places in B, the final permutation maps a into B and b into A, and otherwise preserves A and B.

This argument extends to later transpositions too. Now, suppose there are multiple cycles. Colour one of them. So during the process, the coloured labels move around. At some point, we must swap a coloured label with an uncoloured label. Consider this edge, between places a and b as before, and indeed the same conclusion holds. WLOG we move the coloured label from a to b. But then at the end of the process (ie in the permutation) there are more coloured labels in B than initially. But the number of coloured labels should be the same, because they just cycle around in the final permutation.

We can learn a bit more by trying thinking about the action on cycles (in the permutation) of adding a transposition. In the following pair of diagrams, the black arrows represent the original permutation (note it’s not helpful to think of the directed edges as having anything to do with transpositions now), the dashed line represents a new transposition, and the new arrows describe the new permutation which results from this product.

It’s clear from this that adding a transposition between places corresponding to different cycles causes the cycles to merge, while adding a transposition between places already in the same cycle causes the cycle to split into two cycles. Furthermore the sizes of the two cycles formed is related to the distance in the cycle between the places defining the transposition.

This allows us to prove the lemma by adding the edges of the tree one-at-a-time and using induction. The inductive claim is that cycles of the permutation exactly correspond to components of the partially-built tree. Assuming this claim guarantees that the next step is definitely a merge, not a split (otherwise the edge corresponding to the next step would have to form a cycle). If all N-1 steps are merges, then the number of cycles is reduced by one on each step, and so the final permutation must be a single cycle.

Uniform split-merge

This gives another framework for thinking about the RTRW itself, entirely in terms of cycle lengths as a partition of [N]. That is, given a partition, we choose a pair of parts in a size-biased way. If they are different, we merge them; and if it is the same part, with size k, we split them into two parts, with sizes chosen uniformly from { (1,k-1), (2,k-2), …  (k-1,1) }.

What’s nice about this is that it’s easy to generalise to real-valued partitions, eg of [0,1]. Given a partition of [0,1], we sample two IID U[0,1] random variables U_1,U_2. If these correspond to different parts, we replace these parts by a single part with size given by the sum. If these correspond to the same part, with size \alpha, we split this part into two parts with sizes |U_1-U_2| and \alpha - |U_1-U_2|. This is equivalent in a distributional sense to sampling another U[0,1] variable U and replacing \alpha with (\alpha U, \alpha(1-U)). We probably want our partition to live in \ell^1_\searrow, so we might have to reorder the parts afterwards too.

These uniform split-merge dynamics have a (unique) stationary distribution, the canonical Poisson-Dirichlet random partition, hereafter PD(0,1). This was first shown in [DMZZ04], and then in a framework more relevant to this post by Schramm [Sch08].

Conveniently, PD(0,1) is also the scaling limit of the cycle lengths in a uniform random permutation (scaled by N). The best way to see this is to start with the observation that the length of the cycle containing 1 in a permutation chosen uniformly from S_N has the uniform distribution on {1,…,N}. This matches up well with the uniform stick-breaking construction of PD(0,1), though other arguments are available too. Excellent background on Poisson-Dirichlet distributions and this construction and equivalence can be found in Chapter 3 of Pitman’s comprehensive St. Flour notes [CSP]. Also see this post, and the links within, with the caveat that my understanding of the topic was somewhat shaky then (as presently, for now).

However, Schramm says slightly more than this. As the Erdos-Renyi graph passes criticality, there is a well-defined (and whp unique) giant component including \Theta(N) vertices. It’s not clear that the corresponding permutation should have giant cycles. Indeed, whp the giant component has \Theta(N) surplus edges, so the process of cycle lengths will have undergone O(N) splits. Schramm shows that most of the labels within the giant component are contained in giant cycles in the permutation. Furthermore, the distribution of cycle lengths within the giant component, rescaled by the size of the giant component, converges in distribution to PD(0,1) at any supercritical time \frac{(1+\epsilon)N}{2}

This is definitely surprising, since we already know that the whole permutation doesn’t look close to uniform until time \frac{N\log N}{2}. Essentially, even though the size of the giant component is non-constant (ie it’s gaining vertices), the uniform split-merge process is happening to the cycles within it at rate N. So heuristically, at the level of the largest cycles, at any supercritical time we have a non-trivial partition, so at any slightly later time (eg \frac{(1+\epsilon/2)N}{2} and \frac{(1+\epsilon)N}{2} ), mixing will have comfortably occurred, and so the distribution is close to PD(0,1).

This is explained very clearly in the introduction of [Ber10], in which the approach is extended to a random walk on S_N driven by a uniform choice from any conjugacy class.

So this really does tell us how the global uniform randomness emerges. As the random graph process passes criticality, we have a positive mass of labels in a collection of giant cycles which are effectively a continuous-space uniform split-merge model near equilibrium (and thus with PD(0,1) marginals). The remaining cycles are small, corresponding to small trees which make up the remaining (subcritical by duality) components of the ER graph. These cycles slowly get absorbed into the giant cycles, but on a sufficiently slow timescale relevant to the split-merge dynamics that we do not need to think of a separate split-merge-with-immigration model. Total variation distance on permutations does feel the final few fixed points (corresponding to isolated vertices in the graph), hence the sharp cutoff corresponding to sharp transition in the number of isolated vertices.

References

[Ber10] – N. Berestycki – Emergence of giant cycles and slowdown transition in random transpositions and k-cycles. [arXiv version]

[CSP] – Pitman – Combinatorial stochastic processes. [pdf available]

[Den59] – Denes – the representation of a permutation as a product of a minimal number of transpositions, and its connection with the theory of graphs

[DS81] – Diaconis, Shahshahani – Generating a random permutation with random transpositions

[DMZZ04] – Diaconis, Mayer-Wolf, Zeitouni, Zerner – The Poisson-Dirichlet distribution is the unique invariant distribution for uniform split-merge transformations [link]

[ER59] – Erdos, Renyi – On random graphs I.

[Sch08] – Schramm – Compositions of random transpositions [book link]

DGFF 4 – Properties of the Green’s function

I’m at UBC this month for the PIMS probability summer school. One of the long courses is being given by Marek Biskup about the Discrete Gaussian Free Field (notes and outline here) so this seems like a good moment to revive the sequence of posts about the DGFF. Here’s DGFF1, DGFF2, DGFF3 from November.

The first draft of this post was about the maximum of the DGFF in a large box V_N, and also about the Green’s function G^{V_N}(x,y), which specifies the covariance structure of the DGFF. This first draft also became too long, so I’m splitting it into two somewhat shorter ones. As we’ll see, some understanding and standard estimates of the Green’s function is enough to say quite a bit about the maximum. In this first post, we’ll explore some ‘low-hanging fruit’ concerning the Green’s function, as defined through a simple random walk, which are useful, but rarely explained in the DGFF literature.

Symmetry of Green’s function

We start with one of these low-hanging fruit. If G^{V_N} is to be a covariance matrix, it has to be symmetric. In the first post, showing that the definition of the DGFF as a random field with given Hamiltonian is equivalent to \mathcal{N}(0,G^{V_N}) certainly can be viewed as a proof of symmetry. However, it would be satisfying if there was a direct argument in the language of the definition of the Green’s function.

To make this self-contained, recall the random walk definition of G^{V_N}(x,y). Let (S_m)_{m\ge 0} be simple random walk on V_N, and \mathbb{P}_x,\,\mathbb{E}_x denote starting the random walk at x\in V_N. As usual, let \tau_y,\,\tau_A denote the hitting time of a vertex y or a set A respectively. Then

G^{V_N}(x,y):= \mathbb{E}_x \left[ \sum_{m=0}^{\tau_{\partial V_N}}1_{(S_m=y) }\right].

That is, G^{V_N}(x,y) is the expected number of visits to y by a random walk from x, before it exits V_N.

Let’s drop the superscript for now, as everything should hold for a more general subset of the lattice. I don’t think it’s immediately obvious at the level of Markov chains why G(x,y)=G(y,x). In particular, it’s not the case that

\mathbb{P}_x(\tau_y < \tau_{D^c}) = \mathbb{P}_y(\tau_x <\tau_{D^c}),

and it feels that we can’t map between paths x \to \partial D and y\to \partial D in a way that preserves the number of visits to y and x, respectively. However, we can argue that for any m

\mathbb{P}_x(S_m=y, \tau_{D^c}>m) = \mathbb{P}_y(S_m=x, \tau_{D^c}>m),

by looking at the suitable paths of (S_m). That is, if we have a path x=S_0,S_1,\ldots,S_m=y that stays within D, then the probability of seeing this path starting from x and its reverse direction starting from y are equal. Why? Because

\mathbb{P}_x(S_0=x,S_1=v_1,\ldots,S_{m-1}=v_{m-1},S_m=y) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_\ell)},

and

\mathbb{P}_y(S_0=y,S_1=v_{m-1},\ldots,S_{m-1}=v_1, S_m=x) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_{m-\ell})} = \prod_{\ell=1}^m \frac{1}{\mathrm{deg}(v_\ell)}.

Since D\subset \mathbb{Z}^d and x,y are in the interior of D, we must have \mathrm{deg}(x)=\mathrm{deg}(y), and so these two expressions are equal. Summing over all such two-way paths, and then all m gives the result.

Fixing one argument

We now focus on G^D(\cdot,y), where the second argument is fixed. This is the solution to the Poisson equation

\Delta G^D(\cdot,y) = -\delta_y(\cdot),\quad G^D(x,y)=0,\; \forall x\in \partial D.

To see this, can use a standard hitting probability argument (as here) with the Markov property. This is harmonic in D\backslash \{y\}, and since we know

G^D(y,y)= \frac{1}{\mathbb{P}_y(\text{RW hits }\partial D\text{ before returning to }y)},

this uniquely specifies G^D(\cdot,y). Anyway, since harmonic functions achieve their maxima at the boundary, we have G(y,y)\ge G(x,y) for all x\in D. We can also see this from the SRW definition as

G(x,y)=G(y,x) = \mathbb{P}_y (\tau_x < \tau_{\partial D} ) G(x,x) \le G(x,x).

Changing the domain

Now we want to consider nested domains D\subset E, and compare G^D(\cdot,\cdot) and G^E(\cdot,\cdot) on DxD. The idea is that for SRW started from x\in D, we have \tau_{\partial D}\le \tau_{\partial E}, since one boundary is contained within the other. From this, we get

G^D(x,y)\le G^E(x,y),\quad \forall x,y\in D,

and we will use the particular case y=x.

For example, if x\in V_N, the box with width N, then the box with width 2N centred on x contains the whole of V_N. So, if we set \bar {V}_{2N}:= [-N,N]^d, then with reference to the diagram, we have

G^{V_N}(x,x)\le G^{\bar{V}_{2N}}(0,0),\quad x\in V_N.

As we’ll see when we study the maximum of the DGFF on V_N, uniform control over the pointwise variance will be a useful tool.

Maximising the Green’s function

The idea of bounding G^{V_N}(x,x) by G^{\bar V_{2N}}(0,0) for any x\in V_N is clever and useful. But a more direct approach would be to find the value of x that maximises G^{V_N}(x,x). We would conjecture that when V_N has a central vertex, then this is the maximiser.

We can prove this directly from the definition of the Green’s function in terms of random walk occupation times. Let’s assume we are working with \bar{V}_N for even N, so that 0 is the central vertex. Again, since

G^D(x,x)=\frac{1}{\mathbb{P}_x(\text{RW hits }\partial D\text{ before returning to }x)}, (*)

it would suffice to show that this probability is minimised when x=0. This feels right, since 0 is furthest from the boundary. Other points are closer to the boundary in some directions but further in others, so we can’t condition on the maximum distance from its start point achieved by an excursion of SRW (we’re vertex-transitive, so these look the same from all starting points), as even allowing for the four possible rotations, for an excursion of diameter slightly larger than N, starting at the centre is maximally bad.

However, intuitively it does feel as if being closer to the boundary makes you more likely to escape earlier. In fact, with a bit more care, we can couple the SRW started from 0 and the SRW started from r=(r^x,r^y)\ne 0 such that the latter always exits first. For convenience we’ll assume also that r^x,r^y are both even.

I couldn’t find any reference to this, so I don’t know whether it’s well-known or not. The following argument involves projecting into each axis, and doing separate couplings for transitions in the x-direction and transitions in the y-direction. We assume WLOG that x is in the upper-right quadrant as shown. Then, let 0=S_0,S_1,S_2,\ldots be SRW started from 0, and we will construct r=R_0,R_1,R_2,\ldots on the same probability space as (S_m)_{m\ge 0} as follows. For every m, we set the increment R_{m+1}-R_m to be \pm(S_{m+1}-S_m). It remains to specify the sign, which will be determined by the direction of the S-increment, and a pair of stopping times. The marginal is therefore again an SRW, started from r. Temporarily, we use the unusual notation S_m= (S^x_m,S^y_m) for the coordinates of S_m.

So, if S_{m+1}-S_m=(1,0), (-1,0), ie S moves left or right, then we set

R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m) &\quad \text{if }m<T^x\\ S_{m+1}-S_m&\quad \text{if }m>T^x.\end{cases} (*)

where T^x:= \min\{m\,:\, R^x_m=S^x_m\}. That is, R^x moves in the opposing direction to S^x until the first time when they are equal (hence the parity requirement), and then they move together. WLOG assume that r^x>0. Then suppose S^x_m=\pm N and such m is minimal. Then by construction, if m\ge T^x, then R^x_m=\pm N also. If m<T^x, then we must have S^x_m=-N, and so since R^x‘s trajectory is a mirror image of S^x‘s, in fact R^x_m = N+r^x>N, so R^x hit +N first. In both cases, we see that R^x hits \pm N at the same time or before S^x.

In other words, when S^x_m has non-negative x coordinate, the lazy random walk R^x follows the same trajectory as S^x, and when it has negative x coordinate, the R^x mirrors S^x. At some time, it may happen that S^x_m= R^x_m=0 (recall the parity condition on r). Call this time T^x. We then adjust the description of the coupling so that (*) is the mechanism for m<T^x, and then for m\ge T^x, we take S^x_m=R^x_m.

Similarly, if S_{m+1}-S_m =(0,1), (0,-1), ie S moves up or down, then we set

R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m)&\quad \text{ if }m<T^y\\  S_{m+1}-S_m&\quad \text{if }m\le T^y,\end{cases}

with corresponding definition of the stopping time T^y.

This completes the coupling, and by considering T^x\wedge T^y, we have shown what that the exit time for the walk started from zero dominates the exit time for walk started from r. Recall that so far we are in the case where the box has even width and r=(r^x,r^y) has even coordinates.

This exit time comparison isn’t exactly what we need to compare G^N(0,0) and G^N(x,x). It’s worth remarking at this stage that if all we cared about was the Green’s function on the integer line [-N,N], we would have an easier argument, as by the harmonic property of G(\cdot,y)

G^{[-N,N]}(0,r)=\frac{N-r}{N}G^{[-N,N]}(0,0),

G^{[-N,N]}(r,0) = \frac{N}{N+r}G^{[-N,N]}(r,r),

and so G(0,0)>G(r,r) follows by symmetry. To lift from 1D to 2D directly, we need a bit more than this. It’s possible that S returns in both x- and y- coordinates more often than R, but never at the same time. Fortunately, the coupling we defined slightly earlier does give us a bit more control.

Let \tau^x(S), \tau^x(R) be the first times that S^x, R^x hit \pm N. Under this coupling, for any m\ge 0

\mathbb{P}(S^x_m=0, m<T^x) = \mathbb{P}(R^x_m=r^x, m<T^x)

since these events are literally equal. Since we showed that \tau^x(R)\le \tau^x(S) almost surely, we can further deduce

\mathbb{P}(S^x_m=0,m<T^x\wedge \tau^x(S)) \ge \mathbb{P}(S^x_m=0,m<T^x\wedge \tau^x(R))

=\mathbb{P}(R^x_m=r^x, m <T^x \wedge \tau^x(R)).

To address the corresponding events for which m\ge T^x, we apply the strong Markov property at T^x, to obtain SRW Z_m started from r/2, and let \tau_{-N},\tau_{+N} be the hitting times of -N,+N respectively and \tau_{\pm N}=\tau_{-N}\wedge \tau_{+N}. It will now suffice to prove that

\mathbb{P}(Z_m=0, m< \tau_{\pm N}) \ge \mathbb{P}(Z_m=r,m<\tau_{\pm N}), (**)

as then we can apply the law of total probability and sum over values of T^x and m\ge 0.

To prove this result, we consider the following bijection between trajectories of length m from r/2 to {0,r}. We decompose the trajectories into excursions away from r/2, and then a final meander from r/2 to {0,r} that stays on the same side of r/2. We construct the new trajectory by preserving all the initial excursions, but reversing all the steps of the final meander. So if the original trajectory ended up at 0, the image ends up at r. Trivially, the initial excursions in the image only hit \pm N if the excursions in the original trajectory did this too. But it’s also easy to see, by a similar argument to the coupling at the start of this section, that if the original trajectory ends at r and does not hit \pm N, then so does the image. However, the converse is not true. So we conclude (**), and thus

\mathbb{P}(S_m^x=0) \ge \mathbb{P}(R_m^x=0)

for all m by combining everything we have seen so far. And so we can now lift to a statement about S_m itself, that is considering both coordinates separately.

 

The remaining cases for r require a little more care over the definition of T^x, though the same projection argument works, for fundamentally the same reason. (Note that in the above argument, if S^x_m=-N and m<T^x, then in fact R^x_m\ge N+2, and so it’s not hard to convince yourself that a sensible adjustment to the stopping time will allow a corresponding result with R^x_m\ge N+1 in the odd r^x case.) The case for N odd is harder, since in one dimension there are two median sites, and it’s clear by symmetry that we can’t couple them such that RW from one always exits at least as early as RW from the other. However, the distributions of exit times started from these two sites are the same (by symmetry), and so although we can’t find a coupling, we can use similar stopping times to obtain a result in probability.

In the next post, we’ll see how to apply this uniform bound on G^{V_N}(x,x) to control the maximum of the DGFF on V_N. In particular, we address how the positive correlations of DGFF influence the behaviour of the maximum by comparison with independent Gaussians at each site.

Generating uniform trees

A long time ago, I wrote quite a few a things about uniform trees. That is, a uniform choice from the n^{n-2} unrooted trees with vertex set [n]. This enumeration, normally called Cayley’s formula, has several elegant arguments, including the classical Prufer bijection. But making a uniform choice from a large set is awkward, and so we seek more probabilistic methods to sample such a tree, which might also give insight into the structure of a ‘typical’ uniform tree.

In another historic post, I talked about the Aldous-Broder algorithm. Here’s a quick summary. We run a random walk on the complete graph K_n started from a uniformly-chosen vertex. Every time we arrive at a vertex we haven’t visited before, we record the edge just traversed. Eventually we have visited all n vertices, so have recorded n-1 edges. It’s easy enough to convince yourself that these n-1 edges form a tree (how could there be a cycle?) and a bit more complicated to decide that the distribution of this tree is uniform.

It’s worth noting that this algorithm works to construct a uniform spanning tree on any connected base graph.

This post is about a few alternative constructions and interpretations of the uniform random tree. The first construction uses a Galton-Watson process. We take a Galton-Watson process where the offspring distribution is Poisson(1), and condition that the total population size is n. The resulting random tree has a root but no labels, however if we assign labels in [n] uniformly at random, the resulting rooted tree has the uniform distribution among rooted trees on [n].

Proof

This is all about moving from ordered trees to non-ordered trees. That is, when setting up a Galton-Watson tree, we distinguish between the following two trees, drawn extremely roughly in Paint:

That is, it matters which of the first-generation vertices have three children. Anyway, for such a (rooted) ordered tree T with n vertices, the probability that the Galton-Watson process ends up equal to T is

\mathbb{P}(GW = T) = \prod_{v\in T} \frac{e^{-1}}{C(v)!} = e^{-n} \prod_{v\in T}\frac{1}{C(v)!},

where C(v) is the number of children of a vertex v\in T. Then, since \mathbb{P}( |GW|=n ) is a function of n, we find

\mathbb{P}(GW=T \,\big|\, |GW|=n) = f(n)\prod_{v\in T} \frac{1}{C(v)!},

where f(n) is a function of n alone (ie depends on T only through its size n).

But given an unordered rooted tree t, labelled by [n], there are \prod_{v \in t} C(v)! ordered trees associated to t in the natural way. Furthermore, if we take the Poisson Galton-Watson tree conditioned to have total population size n, and label uniformly at random with [n], we obtain any one of these ordered trees with probability \frac{f(n)}{n!} \prod_{v\in t} \frac{1}{C(v)!}. So the probability that we have t after we forget about the ordering is \frac{f(n)}{n!}, which is a function of n alone, and so the distribution is uniform among the set of rooted unordered trees labelled by [n], exactly as required.

Heuristic for Poisson offspring distribution

In this proof, the fact that \mathbb{P}(C(v)=k)\propto \frac{1}{k!} exactly balances the number of orderings of the k children explains why Poisson(1) works out. Indeed, you can see in the proof that Poisson(c) works equally well, though when c\ne 1, the event we are conditioning on (namely that the total population size is n) has probability decaying exponentially in n, whereas for c=1, the branching process is critical, and the probability decays polynomially.

We can provide independent motivation though, from the Aldous-Broder construction. Both the conditioned Galton-Watson construction and the A-B algorithm supply the tree with a root, so we’ll keep that, and look at the distribution of the degree of the root as constructed by A-B. Let \rho=v_1,v_2,v_3,\ldots be the vertices [n], ordered by their discovery during the construction. Then \rho is definitely connected by an edge to v_2, but thereafter it follows by an elementary check that the probability \rho is connected to v_m is \frac{1}{n-1}, independently across all m. In other words, the distribution of the degree of \rho in the tree as constructed by A-B is

1+ \mathrm{Bin}\left(n-2,\frac{1}{n-1}\right) \approx 1+\mathrm{Poisson}(1).

Now, in the Galton-Watson process, conditioning the tree to have fixed, large size changes the offspring distribution of the root. Conveniently though, in a limiting sense it’s the same change as conditioning the tree to have size at least n. Since these events are monotone in n, it’s possible to take a limit of the conditioning events, and interpret the result as the Galton-Watson tree conditioned to survive. It’s a beautiful result that this interpretation can be formalised as a local limit. The limiting spine decomposition consists of an infinite spine, where the offspring distribution is a size-biased version of the original offspring distribution (and so in particular, always has at least one child) and where non-spine vertices have the original distribution.

In particular, the number of the offspring of the root is size-biased, and it is well-known and not hard to check that size-biasing Poisson(c) gives 1+Poisson(c) ! So in fact we have, in an appropriate limiting sense in both objects, a match between the degree distribution of the root in the uniform tree, and in the conditioned Galton-Watson tree.

This isn’t supposed to justify why a conditioned Galton-Watson tree is relevant a priori (especially the unconditional independence of degrees), but it does explain why Poisson offspring distributions are relevant.

Construction via G(N,p) and the random cluster model

The main reason uniform trees were important to my thesis was their appearance in the Erdos-Renyi random graph G(N,p). The probability that vertices {1, …, n} form a tree component in G(N,p) with some particular structure is

p^{n-1} (1-p)^{\binom{n}{2}-(n-1)} \times (1-p)^{n(N-m)}.

Here, the first two terms give the probability that the graph structure on {1, …, n} is correct, and the the final term gives the probability of the (independent) event that these vertices are not connected to anything else in the graph. In particular, this has no dependence on the tree structure chosen on [n] (for example, whether it should be a path or a star – both examples of trees). So the conditional distribution is uniform among all trees.

If we work in some limiting regime, where pn\rightarrow 0 (for example if n is fixed and p=\frac{1}{N}\rightarrow 0), then we can get away asymptotically with less strong conditioning. Suppose we condition instead just that [n] form a component. Now, there are more ways to form a connected graph with one cycle on [n] than there are trees on [n], but the former all require an extra edge, and so the probability that a given one such tree-with-extra-edge appears as the restriction to [n] in G(N,p) is asymptotically negligible compared to the probability that the restriction to [n] of G(N,p) is a tree. Naturally, the local limit of components in G(N,c/N) is a Poisson(c) Galton-Watson branching process, and so this is all consistent with the original construction.

One slightly unsatisfying aspect to this construction is that we have to embed the tree of size [n] within a much larger graph on [N] to see uniform trees. We can’t choose a scaling p=p(n) such that G(n,p) itself concentrates on trees. To guarantee connectivity with high probability, we need to take p> \frac{\log n}{n}, but by this threshold, the graph has (many) cycles with high probability.

At this PIMS summer school in Vancouver, one of the courses is focusing on lattice spin models, including the random cluster model, which we now briefly define. We start with some underlying graph G. From a physical motivation, we might take G to be \mathbb{Z}^d or some finite subset of it, or a d-ary tree, or the complete graph K_N. As in classical bond percolation (note G(N,p) is bond percolation on K_N), a random subset of the edges of G are included, or declared open. The probability of a given configuration w, with e open edges is proportional to

p^e (1-p)^{|E(G)| - e} q^{k(w)}, (*)

where the edge-weight p\in(0,1) as usual, and cluster weight q\in (0,\infty), and k(w) counts the number of connected components in configuration w. When q=1, we recover classical bond percolation (including G(N,p) ), while for q>1, this cluster-reweighting favours having more components, and q<1 favours fewer components. Note that in the case q\ne 1, the normalising constant (or partition function) of (*) is generally intractable to calculate explicitly.

As in the Erdos-Renyi graph, consider fixing the underlying graph G, and taking p\rightarrow 0, but also taking \frac{q}{p}\rightarrow 0. So the resulting graph asymptotically ‘wants to have as few edges as possible, but really wants to have as few components as possible’. In particular, 1) all spanning trees of G are equally likely; 2) any configuration with more than one component has asymptotically negligible probability relative to any tree; 3) any graph with a cycle has #components + #edges greater than that of a tree, and so is asymptotically negligible probability relative to any tree.

In other words, the limit of the distribution is the uniform spanning tree of G, and so this (like Aldous-Broder) is a substantial generalisation, which constructs the uniform random tree in the special case where G=K_n.

 

Random walks conditioned to stay positive

In this post, I’m going to discuss some of the literature concerning the question of conditioning a simple random walk to lie above a line with fixed gradient. A special case of this situation is conditioning to stay non-negative. Some notation first. Let (S_n)_{n\ge 0} be a random walk with IID increments, with distribution X. Take \mu to be the expectation of these increments, and we’ll assume that the variance \sigma^2 is finite, though at times we may need to enforce slightly stronger regularity conditions.

(Although simple symmetric random walk is a good example for asymptotic heuristics, in general we also assume that if the increments are discrete they don’t have parity-based support, or any other arithmetic property that prevents local limit theorems holding.)

We will investigate the probability that S_n\ge 0 for n=0,1,…,N, particularly for large N. For ease of notation we write T=\inf\{n\ge 0\,:\, S_n<0\} for the hitting time of the negative half-plane. Thus we are interested in S_n conditioned on T>N, or T=N, mindful that these might not be the same. We will also discuss briefly to what extent we can condition on T=\infty.

In the first paragraph, I said that this is a special case of conditioning SRW to lie above a line with fixed gradient. Fortunately, all the content of the general case is contained in the special case. We can repose the question of S_n conditioned to stay above n\alpha until step N by the question of S_n-n\alpha (which, naturally, has drift \mu-\alpha) conditioned to stay non-negative until step N, by a direct coupling.

Applications

Simple random walk is a perfectly interesting object to study in its own right, and this is a perfectly natural question to ask about it. But lots of probabilistic models can be studied via naturally embedded SRWs, and it’s worth pointing out a couple of applications to other probabilistic settings (one of which is the reason I was investigating this literature).

In many circumstances, we can desribe random trees and random graphs by an embedded random walk, such as an exploration process, as described in several posts during my PhD, such as here and here. The exploration process of a Galton-Watson branching tree is a particularly good example, since the exploration process really is simple random walk, unlike in, for example, the Erdos-Renyi random graph G(N,p), where the increments are only approximately IID. In this setting, the increments are given by the offspring distribution minus one, and the hitting time of -1 is the total population size of the branching process. So if the expectation of the offspring distribution is at most 1, then the event that the size of the tree is large is an atypical event, corresponding to delayed extinction. Whereas if the expectation is greater than one, then it is an event with limiting positive probability. Indeed, with positive probability the exploration process never hits -1, corresponding to survival of the branching tree. There are plenty of interesting questions about the structure of a branching process tree conditional on having atypically large size, including the spine decomposition of Kesten [KS], but the methods described in this post can be used to quantify the probability, or at least the scale of the probability of this atypical event.

In my current research, I’m studying a random walk embedded in a construction of the infinite-volume DGFF pinned at zero, as introduced by Biskup and Louidor [BL]. The random walk controls the gross behaviour of the field on annuli with dyadically-growing radii. Anyway, in this setting the random walk has Gaussian increments. (In fact, there is a complication because the increments aren’t exactly IID, but that’s definitely not a problem at this level of exposition.) The overall field is decomposed as a sum of the random walk, plus independent DGFFs with Dirichlet boundary conditions on each of the annuli, plus asymptotically negligible corrections from a ‘binding field’. Conditioning that this pinned field be non-negative up to the Kth annulus corresponds to conditioning the random walk to stay above the magnitude of the minimum of each successive annular DGFF. (These minima are random, but tightly concentrated around their expectations.)

Conditioning on \{T > N\}

When we condition on \{T>N\}, obviously the resulting distribution (of the process) is a mixture of the distributions we obtain by conditioning on each of \{T=N+1\}, \{T=N+2\},\ldots. Shortly, we’ll condition on \{T=N\} itself, but first it’s worth establishing how to relate the two options. That is, conditional on \{T>N\}, what is the distribution of T?

Firstly, when \mu>0, this event always has positive probability, since \mathbb{P}(T=\infty)>0. So as N\rightarrow\infty, the distribution of the process conditional on \{T>N\} converges to the distribution of the process conditional on survival. So we’ll ignore this for now.

In the case \mu\le 0, everything is encapsulated in the tail of the probabilities \mathbb{P}(T=N), and these tails are qualitatively different in the cases \mu=0 and \mu<0.

When \mu=0, then \mathbb{P}(T=N) decays polynomially in N. In the special case where S_n is simple symmetric random walk (and N has the correct parity), we can check this just by an application of Stirling’s formula to count paths with this property. By contrast, when \mu<0, even demanding S_N=-1 is a large deviations event in the sense of Cramer’s theorem, and so the probability decays exponentially with N. Mogulskii’s theorem gives a large deviation principle for random walks to lie above a line defined on the scale N. The crucial fact here is that the probabilistic cost of staying positive until N has the same exponent as the probabilistic cost of being positive at N. Heuristically, we think of spreading the non-expected behaviour of the increments uniformly through the process, at only polynomial cost once we’ve specified the multiset of values taken by the increments. So, when \mu<0, we have

\mathbb{P}(T\ge(1+\epsilon)N) \ll \mathbb{P}(T= N).

Therefore, conditioning on \{T\ge N\} in fact concentrates T on N+o(N). Whereas by contrast, when \mu=0, conditioning on \{T\ge N\} gives a nontrivial limit in distribution for T/N, supported on [1,\infty).

A related problem is the value taken by S_N, conditional on {T>N}. It’s a related problem because the event {T>N} depends only on the process up to time N, and so given the value of S_N, even with the conditioning, after time N, the process is just an unconditioned RW. This is a classic application of the Markov property, beloved in several guises by undergraduate probability exam designers.

Anyway, Iglehart [Ig2] shows an invariance principle for S_N | T>N when \mu<0, without scaling. That is S_N=\Theta(1), though the limiting distribution depends on the increment distribution in a sense that is best described through Laplace transforms. If we start a RW with negative drift from height O(1), then it hits zero in time O(1), so in fact this shows that conditonal on \{T\ge N\}, we have T= N +O(1) with high probability. When \mu=0, we have fluctuations on a scale \sqrt{N}, as shown earlier by Iglehart [Ig1]. Again, thinking about the central limit theorem, this fits the asymptotic description of T conditioned on T>N.

Conditioning on T=N

In the case \mu=0, conditioning on T=N gives

\left[\frac{1}{\sqrt{N}}S(\lfloor Nt\rfloor ) ,t\in[0,1] \right] \Rightarrow W^+(t), (*)

where W^+ is a standard Brownian excursion on [0,1]. This is shown roughly simultaneously in [Ka] and [DIM]. This is similar to Donsker’s theorem for the unconditioned random walk, which converges after rescaling to Brownian motion in this sense, or Brownian bridge if you condition on S_N=0. Skorohod’s proof for Brownian bridge [Sk] approximates the event \{S_N=0\} by \{S_N\in[-\epsilon \sqrt{N},+\epsilon \sqrt{N}]\}, since the probability of this event is bounded away from zero. Similarly, but with more technicalities, a proof of convergence conditional on T=N can approximate by \{S_m\ge 0, m\in[\delta N,(1-\delta)N], S_N\in [-\epsilon \sqrt{N},+\epsilon\sqrt{N}]\}. The technicalities here emerge since T, the first return time to zero, is not continuous as a function of continuous functions. (Imagine a sequence of processes f^N for which f^N(x)\ge 0 on [0,1] and f^N(\frac12)=\frac{1}{N}.)

Once you condition on T=N, the mean \mu doesn’t really matter for this scaling limit. That is, so long as variance is finite, for any \mu\in\mathbb{R}, the same result (*) holds, although a different proof is in general necessary. See [BD] and references for details. However, this is particularly clear in the case where the increments are Gaussian. In this setting, we don’t actually need to take a scaling limit. The distribution of Gaussian *random walk bridge* doesn’t depend on the mean of the increments. This is related to the fact that a linear transformation of a Gaussian is Gaussian, and can be seen by examining the joint density function directly.

Conditioning on T=\infty

When \mu>0, the event \{T=\infty\} occurs with positive probability, so it is well-defined to condition on it. When \mu\le 0, this is not the case, and so we have to be more careful.

First, an observation. Just for clarity, let’s take \mu<0, and condition on \{T>N\}, and look at the distribution of S_{\epsilon N}, where \epsilon>0 is small. This is approximately given by

\frac{S_{\epsilon N}}{\sqrt{N}}\stackrel{d}{\approx}W^+(\epsilon).

Now take \epsilon\rightarrow\infty and consider the RHS. If instead of the Brownian excursion W^+, we instead had Brownian motion, we could specify the distribution exactly. But in fact, we can construct Brownian excursion as the solution to an SDE:

\mathrm{d}W^+(t) = \left[\frac{1}{W^+(t)} - \frac{W^+(t)}{1-t}\right] \mathrm{d}t + \mathrm{d}B(t),\quad t\in(0,1) (**)

for B a standard Brownian motion. I might return in the next post to why this is valid. For now, note that the first drift term pushes the excursion away from zero, while the second term brings it back to zero as t\rightarrow 1.

From this, the second drift term is essentially negligible if we care about scaling W^+(\epsilon) as \epsilon\rightarrow 0, and we can say that W^+(\epsilon)=\Theta(\sqrt{\epsilon}).

So, returning to the random walk, we have

\frac{S_{\epsilon N}}{\sqrt{\epsilon N}}\stackrel{d}{\approx} \frac{W^+(\epsilon)}{\sqrt{\epsilon}} = \Theta(1).

At a heuristic level, it’s tempting to try ‘taking N\rightarrow\infty while fixing \epsilon N‘, to conclude that there is a well-defined scaling limit for the RW conditioned to stay positive forever. But we came up with this estimate by taking N\rightarrow\infty and then \epsilon\rightarrow 0 in that order. So while the heuristic might be convincing, this is not the outline of a valid argument in any way. However, the SDE representation of W^+ in the \epsilon\rightarrow 0 regime is useful. If we drop the second drift term in (**), we define the three-dimensional Bessel process, which (again, possibly the subject of a new post) is the correct scaling limit we should be aiming for.

Finally, it’s worth observing that the limit \{T=\infty\}=\lim_{N\rightarrow\infty} \{T>N\} is a monotone limit, and so further tools are available. In particular, if we know that the trajectories of the random walk satisfy the FKG property, then we can define this limit directly. It feels intuitively clear that random walks should satisfy the FKG inequality (in the sense that if a RW is large somewhere, it’s more likely to be large somewhere else). You can do a covariance calculation easily, but a standard way to show the FKG inequality applies is by verifying the FKG lattice condition, and unless I’m missing something, this is clear (though a bit annoying to check) when the increments are Gaussian, but not in general. Even so, defining this monotone limit does not tell you that it is non-degenerate (ie almost-surely finite), for which some separate estimates would be required.

A final remark: in a recent post, I talked about the Skorohod embedding, as a way to construct any centered random walk where the increments have finite variance as a stopped Brownian motion. One approach to conditioning a random walk to lie above some discrete function is to condition the corresponding Brownian motion to lie above some continuous extension of that function. This is a slightly stronger conditioning, and so any approach of this kind must quantify how much stronger. In Section 4 of [BL], the authors do this for the random walk associated with the DGFF conditioned to lie above a polylogarithmic curve.

References

[BD] – Bertoin, Doney – 1994 – On conditioning a random walk to stay nonnegative

[BL] – Biskup, Louidor – 2016 – Full extremal process, cluster law and freezing for two-dimensional discrete Gaussian free field

[DIM] – Durrett, Iglehart, Miller – 1977 – Weak convergence to Brownian meander and Brownian excursion

[Ig1] – Iglehart – 1974 – Functional central limit theorems for random walks conditioned to stay positive

[Ig2] – Iglehart – 1974 – Random walks with negative drift conditioned to stay positive

[Ka] – Kaigh – 1976 – An invariance principle for random walk conditioned by a late return to zero

[KS] – Kesten, Stigum – 1966 – A limit theorem for multidimensional Galton-Watson processes

[Sk] – Skorohod – 1955 – Limit theorems for stochastic processes with independent increments

Doob inequalities and Doob-Meyer decomposition

The first post I wrote on this blog was about martingales, way back in 2012 at a time when I had known what a martingale was for about a month. I now don’t have this excuse. So I’m going to write about a couple of properties of (discrete-time) martingales that came up while adjusting a proof which my thesis examiners suggested could be made much shorter as part of their corrections.

Doob’s submartingale inequality

When we prove that some sequence of processes converges to some other process, we typically want to show that this holds in some sense uniformly over a time-interval, rather than just at some fixed time. We don’t lose much at this level of vagueness by taking the limit process to be identically zero. Then, if the convergent processes are martingales or closely similar, we want to be able to bound \sup_{k\le n} |Z_k| in some sense.

Doob’s submartingale inequality allows us to do this. Recall that a submartingale has almost-surely non-negative conditional increments. You might think of it heuristically as ‘more increasing than a martingale’. If Z_n is a martingale, then |Z_n| is a submartingale. This will be useful almost immediately.

The statement is that for (Z_n) a non-negative submartingale,

\mathbb{P}\left( \sup_{k\le n} Z_k \ge \lambda\right) \le \frac{\mathbb{E}\left[Z_n\right]}{\lambda}.

The similarity of the statement to the statement of Markov’s inequality is no accident. Indeed the proof is very similar. We consider whether the event in question happens, and find lower bounds on the expectation of Z_n under both possibilities.

Formally, for ease of notation, let Z_n^* be the running maximum \sup_{k\le n}Z_k. Then, we let T:= n\wedge \inf\{k\le n, M_j\ge \lambda\} and apply the optional stopping theorem for submartingales at T, which is by construction at most n. That is

\mathbb{E}[Z_n]\ge \mathbb{E}[Z_T]=\mathbb{E}\left[Z_T\mathbf{1}_{Z_n^*<\lambda}\right] + \mathbb{E}\left[Z_T \mathbf{1}_{Z_n^*\ge \lambda}\right].

The first of these summands is positive, and the second is at least \lambda \mathbb{P}\left( Z_N^* \ge \lambda \right), from which the result follows.

We’ve already said that for any martingale Z_n, |Z_n| is a submartingale, but in fact f(Z_n) is a submartingale whenever f is convex, and \mathbb{E}|f(Z_n)|<\infty for each n. Naturally, this continues to hold when Z_n is itself a submartingale.

[Note that Z_n^* is also a submartingale, but this probably isn’t as interesting.]

A particularly relevant such function f is f(x)=x^p, for p>1. If we take Z_n a non-negative submartingale which is uniformly bounded in L^p, then by applying Holder’s inequality and this submartingale inequality, we obtain

\mathbb{E}\left( \sup_{k\le n}Z_n^p \right) \le \left(\frac{p}{p-1}\right)^p \mathbb{E}\left[ Z_n^p \right].

Since Z_n^p is a submartingale, then a limit in n on the RHS is monotone, and certainly a limit in n on the LHS is monotone, so we can extend to

\mathbb{E}\left( \sup_{k\le n}Z_\infty^p \right) \le \left(\frac{p}{1-p}\right)^p \mathbb{E}\left[ Z_\infty^p \right].

Initially, we have to define \mathbb{E}\left[ Z_\infty^p \right] through this limit, but in fact this result, Doob’s L^p inequality, shows that Z_\infty:= \lim Z_n exists almost surely as well.

Naturally, we will often apply this in the case p=2, and in the third of these three sections, we will see why it might be particularly straightforward to calculate \mathbb{E}\left[Z_\infty^2\right].

Remark: as in the case of Markov’s inequality, it’s hard to say much if the submartingale is not taken to be non-negative. Indeed, this effect can be seen even if the process is only defined for a single time step, for which the statement really is then Markov’s inequality.

Doob-Meyer decomposition

Unfortunately, most processes are not martingales. Given an discrete-time process X_n adapted to \mathcal{F}=(\mathcal{F}_n), it is a martingale if the conditional expectations of the increments are all almost surely zero. But given a general adapted process X_n which is integrable (so the increments have well-defined finite expectation), we can iteratively construct a new process M_n, where the increments are centred versions of X_n‘s increments. That is,

M_{n+1}-M_n:= X_{n+1}-X_n - \mathbb{E}\left[ X_{n+1}-X_n \,\big|\, \mathcal{F}_n\right] = X_{n+1}-\mathbb{E}\left[X_{n+1} \,\big|\, \mathcal{F}_n\right]. (*)

Then it’s immediately clear from the definition that M_n is a martingale.

There’s a temptation to tie oneself up in knots with the dependence. We might have that increments of the original process X_n depend on the current value of the process. And is it necessarily clear that we can recover the current value of the original process from the current value of M_n? Well, this is why we demand that everything be adapted, rather than just Markov. It’s not the case that M_n should be Markov, but it clearly is adapted.

Now we look at the middle expression in (*), and in particular the term we are subtracting, namely the conditional expectation. If we define, in the standard terminology, A_0=0 and

A_{n+1}-A_n:= \mathbb{E}\left[ X_{n+1}-X_n \,\big|\, \mathcal{F}_n\right],

then we have decomposed the original process X_n as the sum of a martingale M_n, and this new process A_n. In particular, note that the increment A_{n+1}-A_n given above is adapted to \mathcal{F}_n, which is a stronger condition than being adapted to \mathcal{F}_{n+1} as we would expect a priori. This property of the process (A_n) is called predictability (or possibly previsibility).

This decomposition X_n=X_0+M_n+A_n as just defined is called the Doob-Meyer decomposition, and there is a unique such decomposition where M_n is a martingale, and A_n is predictable. The proof of uniqueness is very straightforward. We look at the equalities given above as definitions of M_n,A_n, but then work in the opposite direction to show that they must hold if the decomposition holds.

I feel a final heuristic is worthwhile, using the term drift, more normally encountered in the continuous-time setting to describe infinitissimal expected increments. The increments of A_n represent the drift of X_n, and the increments of M_n are what remains from X_n after subtracting the drift. In general, the process to be subtracted to turn a non-martingale into a martingale is called a compensator, and the existence or otherwise of such processes is important but challenging for some classes of continuous-time processes.

In particular, note that when X_n is itself a martingale, then A_n\equiv 0. However, probably the most useful case is when X_n is a submartingale, as then the drift is always non-negative, and so A_n is almost surely increasing. The converse holds too.

This is relevant because this Doob-Meyer decomposition is obviously only a useful tool for treating X_n if we can handle the two processes M_n,A_n easily. We have tools to bound the martingale term, but this previsible term might in general be tricky, and so the case where X_n is a submartingale is good, as increasing processes are much easier than general processes, since bounding the whole process might involve only bounding the final term in many contexts.

Predictable quadratic variation

A particularly relevant example is the square of a martingale, that is X_n=M_n^2, where M_n is a martingale. By the convexity condition discussed earlier, X_n is a submartingale (provided it is integrable, ie M_n is square-integrable), and so the process A_n in its Doob-Meyer decomposition is increasing. This is often called the (predictable) quadratic variation of (X_n).

This predictable quadratic variation is sometimes denoted \langle X_n\rangle. This differs from the (regular) quadratic variation which is defined as the sum of the squares of the increments, that is [X_n]:= \sum_{k=0}^{n-1} (X_{k+1}-X_k)^2. Note that this is adapted, but obviously not previsible. The distinction between these two processes is more important in continuous time. There, they are almost surely equal for a continuous local martingale, but not for eg a Poisson process. (For a Poisson process, the PQV is deterministic, indeed linear, while the (R)QV is almost surely equal to the Poisson process itself.) In the discrete time setting, the regular quadratic variation is not relevant very often, while the predictable quadratic variation is useful, precisely because of this decomposition.

Whenever we have random variables which we then centre, there is a standard trick to apply when treating their variance. That is

A_{n+1}-A_n= \mathbb{E}\left[ M^2_{n+1}-M^2_n \,\big|\, \mathcal{F}_n\right]
= \mathbb{E}\left[ M^2_{n+1}\,\big|\, \mathcal{F}_n\right] - 2M_n^2 +M_n^2
= \mathbb{E}\left[ M^2_{n+1}\,\big|\, \mathcal{F}_n\right] - 2M_n \mathbb{E}\left[ M_{n+1}\,\big|\, \mathcal{F}_n\right] + M_n^2
= \mathbb{E}\left[ \left(M_{n+1}-M_n\right)^2\,\big|\, \mathcal{F}_n\right].

One consequence is seen by taking an ‘overall’ expectation. Because M_n^2-A_n is a martingale,

\mathbb{E}\left[M_n^2\right] = \mathbb{E}\left[A_n\right] = \mathbb{E}\left[M_0^2\right] + \sum_{k=0}^{n-1} \mathbb{E}\left[A_{k+1}-A_k\right]
= \mathbb{E}\left[ M_0^2\right] + \sum_{k=0}^{n-1}\mathbb{E}\left[ \left(M_{k+1}-M_k\right)^2 \right]. (**)

This additive (Pythagorean) property of the square of a martingale is useful in applications where there is reasonably good control on each increment separately.

We can also see this final property without the Doob-Meyer decomposition. For a martingale it is not the case that the increments on disjoint intervals are independent. However, following Williams 12.1 [1], disjoint intervals are orthogonal, in the sense that

\mathbb{E}\left[(M_t-M_s)(M_v-M_u)\right]=0,

whenever s\le t\le u\le v. Then, when we square the expression M_n=M_0+\sum M_{k+1}-M_k, and take expectations, all the cross terms vanish, leaving precisely (*).

References

[1] Williams – Probability with Martingales

I also followed the notes I made in 2011/12 while attending Perla Sousi’s course on Advanced Probability, and Arnab Sen’s subsequent course on Stochastic Calculus, though I can’t find any evidence online for the latter now.

Skorohod embedding

Background

Suppose we are given a standard Brownian motion (B_t), and a stopping time T. Then, so long as T satisfies one of the regularity conditions under which the Optional Stopping Theorem applies, we know that \mathbb{E}[B_T]=0. (See here for a less formal introduction to OST.) Furthermore, since B_t^2-t is a martingale, \mathbb{E}[B_T^2]=\mathbb{E}[T], so if the latter is finite, so is the former.

Now, using the strong Markov property of Brownian motion, we can come up with a sequence of stopping times 0=T_0, T_1, T_2,\ldots such that the increments T_k-T_{k-1} are IID with the same distribution as T. Then 0,B_{T_1},B_{T_2},\ldots is a centered random walk. By taking T to be the hitting time of \{-1,+1\}, it is easy to see that we can embed simple random walk in a Brownian motion using this approach.

p1020956_compressedEmbedding simple random walk in Brownian motion.

The Skorohod embedding question asks: can all centered random walks be constructed in this fashion, by stopping Brownian motion at a sequence of stopping time? With the strong Markov property, it immediately reduces the question of whether all centered finite-variance distributions X can be expressed as B_T for some integrable stopping time T.

The answer to this question is yes, and much of what follows is drawn from, or at least prompted by Obloj’s survey paper which details the problem and rich history of the many approaches to its solution over the past seventy years.

Applications and related things

The relationship between random walks and Brownian motion is a rich one. Donsker’s invariance principle asserts that Brownian motion appears as the scaling limit of a random walk. Indeed, one can construct Brownian motion itself as the limit of a sequence of consistent random walks with normal increments on an increasingly dense set of times. Furthermore, random walks are martingales, and we know that continuous, local martingales can be expressed as a (stochastically) time-changed Brownian motion, from the Dubins-Schwarz theorem.

The Skorohod embedding theorem can be used to prove results about random walks with general distribution by proving the corresponding result for Brownian motion, and checking that the construction of the sequence of stopping times has the right properties to allow the result to be carried back to the original setting. It obviously also gives a coupling between a individual random walk and a Brownian motion which may be useful in some contexts, as well as a coupling between any pair of random walks. This is useful in proving results for random walks which are much easier for special cases of the distribution. For example, when the increments are Gaussian, or when there are combinatorial approaches to a problem about simple random walk. At the moment no aspect of this blog schedule is guaranteed, but I plan to talk about the law of the iterated logarithm shortly, whose proof is approachable in both of these settings, as well as for Brownian motion, and Skorohod embedding provides the route to the general proof.

At the end, we will briefly compare some other ways to couple a random walk and a Brownian motion.

Adding extra randomness

One thing we could do is sample a copy of X independently from the Brownian motion, then declare T= \tau_{X}:= \inf\{t\ge 0: B_t=X\}, the hitting time of (random value) X. But recall that unfortunately \tau_x has infinite expectation for all non-zero x, so this doesn’t fit the conditions required to use OST.

Skorohod’s original method is described in Section 3.1 of Obloj’s notes linked above. The method is roughly to pair up positive values taken by X appropriately with negative values taken by X in a clever way. If we have a positive value b and a negative value a, then \tau_{a,b}, the first hitting time of \mathbb{R}\backslash (a,b) is integrable. Then we choose one of these positive-negative pairs according to the projection of the distribution of X onto the pairings, and let T be the hitting time of this pair of values. The probability of hitting b conditional on hitting {a,b} is easy to compute (it’s \frac{-a}{b-a}) so we need to have chosen our pairs so that the ‘probability’ of hitting b (ie the density) comes out right. In particular, this method has to start from continuous distributions X, and treat atoms in the distribution of X separately.

The case where the distribution X is symmetric (that is X\stackrel{d}=-X) is particularly clear, as then the pairs should be (-x,x).

However, it feels like there is enough randomness in Brownian motion already, and subsequent authors showed that indeed it wasn’t necessary to introduce extra randomness to provide a solution.

One might ask whether it’s possible to generate the distribution on the set of pairs (as above) out of the Brownian motion itself, but independently from all the hitting times. It feels like it might be possible to make the distribution on the pairs measurable with respect to

\mathcal{F}_{0+} = \bigcap\limits_{t>0} \mathcal{F}_t,

the sigma-algebra of events determined by limiting behaviour as t\rightarrow 0 (which is independent of hitting times). But of course, unfortunately \mathcal{F}_{0+} has a zero-one law, so it’s not possible to embed non-trivial distributions there.

Dubins solution

The exemplar for solutions without extra randomness is due to Dubins, shortly after Skorohod’s original argument. The idea is to express the distribution X as the almost sure limit of a martingale. We first use the hitting time of a pair of points to ‘decide’ whether we will end up positive or negative, and then given this information look at the hitting time (after this first time) of two subsequent points to ‘decide’ which of four regions of the real interval we end up in.

I’m going to use different notation to Obloj, corresponding more closely with how I ended up thinking about this method. We let

a_+:= \mathbb{E}[X \,|\, X>0], \quad a_- := \mathbb{E}[X\,|\, X<0], (*)

and take T_1 = \tau_{\{a_-,a_+\}}. We need to check that

\mathbb{P}\left( B_{T_1}=a_+\right) = \mathbb{P}\left(X>0\right),

for this to have a chance of working. But we know that

\mathbb{P}\left( B_{T_1}=a_+\right) = \frac{a_+}{a_+-a_-},

and we can also attack the other side using (*) and the fact that \mathbb{E}[X]=0, using the law of total expectation:

0=\mathbb{E}[X]=\mathbb{E}[X\,|\, X>0] \mathbb{P}(X>0) + \mathbb{E}[X\,|\,X<0]\mathbb{P}(X<0) = a_+ \mathbb{P}(X>0) + a_- \left(1-\mathbb{P}(X>0) \right),

\Rightarrow\quad \mathbb{P}(X>0)=\frac{a_+}{a_+-a_-}.

Now we define

a_{++}=\mathbb{E}[X \,|\, X>a_+],\quad a_{+-}=\mathbb{E}[X\,|\, 0<X<a_+],

and similarly a_{-+},a_{--}. So then, conditional on B_{T_1}=a_+, we take

T_2:= \inf_{t\ge T_1}\left\{ B_t\not\in (a_{+-},a_{++})  \right\},

and similarly conditional on B_{T_1}=a_-. By an identical argument to the one we have just deployed, we have \mathbb{E}\left[B_{T_2} \,|\,\mathcal{F}_{T_1} \right] = B_{T_1} almost surely. So, although the a_{+-+} notation now starts to get very unwieldy, it’s clear we can keep going in this way to get a sequence of stopping times 0=T_0,T_1,T_2,\ldots where B_{T_n} determines which of the 2^n regions of the real line any limit \lim_{m\rightarrow\infty} B_{T_m} should lie in.

A bit of work is required to check that the almost sure limit T_n\rightarrow T is almost surely finite, but once we have this, it is clear that B_{T_n}\rightarrow B_T almost surely, and B_T has the distribution required.

Komlos, Major, Tusnady coupling

We want to know how close we can make this coupling between a centered random walk with variance 1, and a standard Brownian motion. Here, ‘close’ means uniformly close in probability. For large times, the typical difference between one of the stopping times 0,T_1,T_2,\ldots in the Skorohod embedding and its expectation (recall \mathbb{E}[T_k]=k) is \sqrt{n}. So, constructing the random walk S_0,S_1,S_2,\ldots from the Brownian motion via Skorohod embedding leads to

\left |S_k - B_k \right| = \omega(n^{1/4}),

for most values of k\le n. Strassen (1966) shows that the true scale of the maximum

\max_{k\le n} \left| S_k - B_k \right|

is slightly larger than this, with some extra powers of \log n and \log\log n as one would expect.

The Komlos-Major-Tusnady coupling is a way to do a lot better than this, in the setting where the distribution of the increments has a finite MGF near 0. Then, there exists a coupling of the random walk and the Brownian motion such that

\max_{k\le n}\left|S_k- B_k\right| = O(\log n).

That is, there exists C such that

\left[\max_{k\le n} \left |S_k-B_k\right| - C\log n\right] \vee 0

is a tight family of distributions, indeed with uniform exponential tail. To avoid digressing infinitely far from my original plan to discuss the proof of the law of iterated logarithm for general distributions, I’ll stop here. I found it hard to find much coverage of the KMT result apart from the challenging original paper, and many versions expressed in the language of empirical processes, which are similar to random walks in many ways relevant to convergence and this coupling, but not for Skorohod embedding. So, here is a link to some slides from a talk by Chatterjee which I found helpful in getting a sense of the history, and some of the modern approaches to this type of normal approximation problem.

DGFF 3 – Gibbs-Markov property for entropic repulsion

In the previous post, we saw that it isn’t much extra effort to define the DGFF with non-zero boundary conditions, by adding onto the zero-BC DGFF the unique (deterministic) harmonic function which extends the boundary values into the domain. We also saw how a Gibbs-Markov property applies, whereby the values taken by the field on some sub-region A\subset D depend on the values taken on D\backslash A only through values taken on \partial A.

In this post, we look at how this property and some other methods are applied by Deuschel [1] to study the probability that the DGFF on a large box in \mathbb{Z}^d is positive ‘everywhere’. This event can be interpreted in a couple of ways, all of which are referred to there as entropic repulsion. Everything which follows is either taken directly or paraphrased directly from [1]. I have tried to phrase this in a way which avoids repeating most of the calculations, instead focusing on the methods and the motivation for using them.

Fix dimension d\ge 2 throughout. We let P^0_N be the law of the DGFF on V_N:=[-N,N]^d\subset \mathbb{Z}^d with zero boundary conditions. Then for any subset A\subset \mathbb{Z}^d, in an intuitively-clear abuse of notation, we let

\Omega^+(A):= \{ h_x\ge 0, x\in A\},

be the event that some random field h takes only non-negative values on A. The goal is to determine P^0_N ( \Omega^+(V_N)). But for the purposes of this post, we will focus on showing bounds on the probability that the field is non-negative on a thin annulus near the boundary of V_N, since this is a self-contained step in the argument which contains a blog-friendly number of ideas.

We set (L_N) to be a sequence of integers greater than one (to avoid dividing by zero in the statement), for which \frac{L_N}{N}\rightarrow 0. We now define for each N, the annulus

W_N = \{v\in V_N: L_N\le d_{\mathbb{Z}^d}(v, V_N^c)\le 2L_N \}

with radius L_N set a distance L_N inside the box V_N. We aim to control P^N_0 (\Omega^+(W_N)). This forms middle steps of Deuschel’s Propositions 2.5 and 2.9, which discuss P^N_0(\Omega^+(V_{N-L_N})). Clearly there is the upper bound

P^N_0(\Omega^+(V_{N-L_N})) \le P^N_0(\Omega^+(W_N)) (1)

and a lower bound on P^N_0(\Omega^+(V_{N-L_N})) is obtained in the second proposition by considering the box as a union of annuli then combining the bounds on each annulus using the FKG inequality.

Upper bound via odds and evens

After removing step (1), this is Proposition 2.5:

\limsup_{N\rightarrow \infty} \frac{L_N}{N^{d-1} \log L_N} \log P^N_0(\Omega^+(W_N)) < 0. (2)

This is giving a limiting upper bound on the probability of the form L_N^{-CN^{d-1}/L_N}, though as with all LDP estimates, the form given at (2) is more instructive.

Morally, the reason why it is unlikely that the field should be non-negative everywhere within the annulus is that the distribution at each location is centred, and even though any pair of values are positively correlated, this correlation is not strong enough to avoid this event being unlikely. But this is hard to corral into an upper bound argument directly. In many circumstances, we want to prove upper bounds for complicated multivariate systems by projecting to get an unlikely event for a one-dimensional random variable, or a family of independent variables, even if we have to throw away some probability. We have plenty of tools for tail probabilities in both of these settings. Since the DGFF is normal, a one-dimensional RV that is a linear combination (eg the sum) of all the field heights is a natural candidate. But in this case we would have thrown away too much probability, since the only way we could dominate is to demand that the sum \sum_{x\in W_N}h^N_x\ge 0, which obviously has probability 1/2 by symmetry. (3)

So Deuschel splits W_N into W_N^o,W_N^e, where the former includes all vertices with odd total parity in W_N and the latter includes all the vertices with even total parity in the interior of W_N. (Recall that \mathbb{Z}^d is bipartite in exactly this fashion). The idea is to condition on h^N\big|_{W^o_N}. But obviously each even vertex is exactly surrounded by odd vertices. So by the Gibbs-Markov property, conditional on the odd vertices, the values of the field at the even vertices are independent. Indeed, if for each v\in W_N^e we define \bar h_v to be the average of its neighbours (which is measurable w.r.t to the sigma-algebra generated by the odd vertices), then

\{h_v: v\in W_N^e \,\big|\, \sigma(h_w: w\in W_N^o)\},

is a collection of independent normals with variance one, and where the mean of h_v is \bar h_v.

To start finding bounds, we fix some threshold m=m_N\gg 1 to be determined later, and consider the odd-measurable event A_N that at most half of the even vertices v have \bar h_v\ge m. So A_N^c\cap \Omega^+(W_N) says that all the odd vertices are non-negative and many are quite large. This certainly feels like a low-probability event, and unlike at (3), we might be able to obtain good tail bounds by projection into one dimension.

In the other case, conditional on A_N, there are a large number of even vertices with conditional mean at most m, and so we can control the probability that at least one is negative as a product

(1-\varphi(m))^{\frac12 |W_N^e|}. (4)

Note that for this upper bound, we can completely ignore the other even vertices (those with conditional mean greater than m).

So we’ll go back to A_N^c \cap \Omega^+(W_N). For computations, the easiest one-dimensional variable to work with is probably the mean of the \bar h_vs across v\in W_N^e, since on A_N^c\cap \Omega^+(W_N) this is at least \frac{m}{2}. Rather than focus on the calculations themselves involving

\bar S^e_N:= \frac{1}{|W_N^e|} \sum\limits_{v\in W_N^e} \bar h_v,

let us remark that it is certainly normal and centered, and so there are many methods to bound its tail, for example

P^0_N \left( \bar S^e_N \ge \frac{m}{2} \right) \le \exp\left( \frac{-m^2}{8\mathrm{Var}(\bar S^e_N)} \right), (5)

as used by Deuschel just follows from an easy comparison argument within the integral of the pdf. We can tackle the variance using the Green’s function for the random walk (recall the first post in this set). But before that, it’s worth making an observation which is general and useful, namely that \bar S^e_N is the expectation of

S^e_N:= \sum{1}{|W_N^e|}\sum\limits_{v\in W_N^e} h_v

conditional on the odds. Directly from the law of total variance, the variance of any random variable X is always larger than the variance of \mathbb{E}[X|Y].

So in this case, we can replace \mathrm{Var}(\bar S^e_N) in (5) with \mathrm{Var}(S^e_N), which can be controlled via the Green’s function calculation.

Finally, we choose m_N so that the probability at (4) matches the probability at (5) in scale, and this choice leads directly to (2).

In summary, we decomposed the event that everything is non-negative into two parts: either there are lots of unlikely local events in the field between an even vertex and its odd neighbours, or the field has to be atypically large at the odd sites. Tuning the parameter m_N allows us to control both of these probabilities in the sense required.

Lower bound via a sparse sub-lattice

To get a lower bound on the probability that the field is non-negative on the annulus, we need to exploit the positive correlations in the field. We use a similar idea to the upper bound. If we know the field is positive and fairly large in many places, then it is increasingly likely that it is positive everywhere. The question is how many places to choose?

We are going to consider a sub-lattice that lives in a slightly larger region than W_N itself, and condition the field to be larger than m=m_N everywhere on this lattice. We want the lattice to be sparse enough that even if we ignore positive correlations, the chance of this happening is not too small. But we also want the lattice to be dense enough that, conditional on this event, the chance that the field is actually non-negative everywhere in W_N is not too small either.

To achieve this, Deuschel chooses a sub-lattice of width \lfloor\epsilon L_N^{2/d}\rfloor, and sets \Lambda_N(\epsilon) to be the intersection of this with the annulus with radii [N-\frac{5}{2}L_N, N-\frac{1}{2}L_N], to ensure it lives in a slightly larger region than W_N itself. The scaling of this sub-lattice density is such that when a random walk is started at any v\in W_N, the probability that the RW hits \Lambda_N(\epsilon) before \partial V_N is asymptotically in (0,1). (Ie, not asymptotically zero or one – this requires some definitely non-trivial calculations.) In particular, for appropriate (ie large enough) choice of \epsilon, this probability is at least 1/2 for all v\in W_N. This means that after conditioning on event B_N:=\{h_v\ge m : v\in \Lambda_N(\epsilon)\}, the conditional expectation of h_w is at least \frac{m}{2} for all w\in W_N\backslash \Lambda_N(\epsilon). Again this uses the Gibbs-Markov property and the Gaussian nature of the field. In particular, this conditioning means we are left with the DGFF on V_N\backslash \Lambda_N(\epsilon), ie with boundary \partial V_N\cup \Lambda_N(\epsilon), and then by linearity, the mean at non-boundary points is given by the harmonic extension, which is linear (and so increasing) in the boundary values.

At this point, the route through the calculations is fairly clear. Since we are aiming for a lower bound on the probability of the event \Omega^+(W_N), it’s enough to find a lower bound on P^0_N(\Omega^+(W_N)\cap B).

Now, by positive correlation (or, formally, the FKG inequality) we can control P^0_N(B) just as a product of the probabilities that the field exceeds the threshold at each individual site in \Lambda_N(\epsilon). Since the value of the field at each site is normal with variance at least 1 (by definition), this is straightforward.

Finally, we treat P^0_N(\Omega^+(W_N) \,\big|\, B). We’ve established that, conditional on B, the mean at each point of W_N\backslash \Lambda_N(\epsilon) is at least \frac{m}{2}, and we can bound the variance above too. Again, this is a conditional variance, and so is at most the corresponding original variance, which is bounded above by \sigma_N^2:=\mathrm{Var}(h^N_0). (This fact that the variance is maximised at the centre is intuitively clear when phrased in terms of occupation times, but the proof is non-obvious, or at least non-obvious to me.)

Since each of the event h_v^N\ge 0 for v\in W_N\backslash \Lambda_N(\epsilon) is positively correlated with B, we can bound the probability it holds for all v by the product of the probabilities that it holds for each v. But having established that the conditional mean is at least \frac{m_N}{2} for each v, and the variance is uniformly bounded above (including in N), this gives an easy tail bound of the form we require.

Again it just remains to choose the sequence of thresholds m_N to maximise the lower bound on the probability that we’ve found in this way. In both cases, it turns out that taking m_N= \sqrt{C\log N} is sensible, and this turns out to be linked to the scaling of the maximum of the DGFF, which we will explore in the future.

References

[1] – J-D Deuschel, Entropic Repulsion of the Lattice Free Field, II. The 0-Boundary Case. Available at ProjectEuclid.