# 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

# Persistent Hubs

This post is based on the paper “Existence of a persistent hub in the convex preferential attachment model” which appeared on arXiv last week. It can be found here. My aim is to explain (again) the model; the application-based motivation for the result; and a heuristic for the result in a special case. In particular, I want to stress the relationship between PA models and urns.

The preferential attachment model attempts to describe the formation of large complex networks. It is constructed dynamically: vertices are introduced one at a time. For each new vertex, we have to choose which existing vertices to join to. This choice is random and reinforced. That is, the new vertex is more likely to join to an existing vertex with high degree than to an existing vertex with degree 1. It’s clear why this might correspond well to the evolution of, say, the world wide web. New webpages are much more likely to link to an established site, eg Wikipedia or Google, than to a uniformly randomly chosen page.

The model is motivated also by a desire to fit a common property of real-world networks that is not exhibited by, among others, the Erdos-Renyi random graph model. In such a network, we expect a few nodes to have much greater connectivity than average. In a sense these so-called hubs drive connectivity of the system. This makes sense in practice. If you are travelling by train around the South-East of England, it is very likely you will pass through at least one of Reading, East Croydon, or about five major terminus in London. It would be absurd for every station to be of equal significance to the network. By contrast, the typical vertex degree in the sparse Erdos-Renyi model is O(1), and has a limiting Poisson distribution, with a super-exponential tail.

So, this paper addresses the following question. We know that the PA model, when set up right, has power-law tails for the degree distribution, and so has a largest degree that is an order of magnitude larger than the average degree. Let’s call this the ‘hub’ for now. But the model is dynamic, so we should ask how this hub changes in time as we add extra vertices. In particular, is it the case that one vertex should grow so large that it remains as the dominant hub forever? This paper answers this question in the affirmative, for a certain class of preferential attachment schemes.

We assign a weighting system to possible degrees, that is a function from N to R+. In the case of proportional PA, this function could be f(n)=n. In general, we assume it is convex. Note that the more convex this weight function is, the stronger the preference a new vertex feels towards existing dominant vertices. Part of the author’s proof is a formalisation of this heuristic, which provides some machinery allowing us to treat only really the case f(n)=n. I will discuss only this case from now on.

I want to focus on the fact that we have another model which describes aspects of the degree evolution rather well. We consider some finite fixed collection of vertices at some time, and consider the evolution of their degrees. We will be interested in limiting properties, so the exact time doesn’t matter too much. We look instead at the jump chain, ie those times when one of the degrees changes. This happens when a new vertex joins to one of the chosen vertices. Given that the new vertex has joined one of the chosen vertices, the choice of which of the chosen vertices is still size-biased proportional to the current degrees. In other words, the jump chain of this degree sequence is precisely Polya’s Urn.

This is a powerful observation, as it allows us to make comments about the limiting behaviour of finite quantities almost instantly. In particular, we know that from any starting arrangement, Polya’s Urn converges almost surely. This is useful to the question of persistence for the following reason.

Recall that in the case of two colours, starting with one of each, we converge to the uniform distribution. We should view this as a special case of the Dirichlet distribution, which is supported on partitions into k intervals of [0,1]. In particular, for any fixed k, the probability that two of the intervals have the same size is 0, as the distribution is continuous. So, since the convergence of the proportions in Polya’s Urn is almost sure, with probability one all of the proportions are with $\epsilon>0$ of their limit, and so taking epsilon small enough, given the limit, which we are allowed to do, we can show that the colour which is largest in the limit is eventually the largest at finite times.

Unfortunately, we can’t mesh these together these finite-dimensional observations particularly nicely. What we require instead is a result showing that if a vertex has large enough degree, then it can never be overtaken by any new vertex. This proved via a direct calculation of the probability that a new vertex ‘catches up’ with a pre-existing vertex of some specified size.

That calculation is nice and not too complicated, but has slightly too many stages and factorial approximations to consider reproducing or summarising here. Instead, I offer the following heuristic for a bound on the probability that a new vertex will catch up with a pre-existing vertex of degree k. Let’s root ourselves in the urn interpretation for convenience.

If the initial configuration is (k,1), corresponding to k red balls and 1 blue, we should consider instead the proportion of red balls, which is k/k+1 obviously. Crucially (for proving convergence results if nothing else), this is a martingale, which is clearly bounded within [0,1]. So the expectation of the limiting proportion is also k/k+1. Let us consider the stopping time T at which the number of red balls is equal to the number of blue balls. We decompose the expectation by conditioning on whether T is finite.

$\mathbb{E}X_\infty=\mathbb{E}[X_\infty|T<\infty]\mathbb{P}(T<\infty)+\mathbb{E}[X_\infty|T=\infty]\mathbb{P}(T=\infty)$

$\leq \mathbb{E}[X_\infty | X_T,T<\infty]\mathbb{P}(T<\infty)+(1-\mathbb{P}(T=\infty))$

using that $X_\infty\leq 1$, regardless of the conditioning,

$= \frac12 \mathbb{P}(T<\infty) + (1-\mathbb{P}(T<\infty))$

$\mathbb{P}(T<\infty) \leq \frac{2}{k+1}.$

We really want this to be finite when we sum over k so we can use some kind of Borel-Cantelli argument. Indeed, Galashin gets a bound of $O(k^{-3/2})$ for this quantity. We should stress where we have lost information. We have made the estimate $\mathbb{E}[X_\infty|T=\infty]=1$ which is very weak. This is unsurprising. After all, the probability of this event is large, and shouldn’t really affect the limit that much when it does not happen. The conditioned process is repelled from 1/2, but that is of little relevance when starting from k/k+1. It seems likely this expectation is in fact $\frac{k}{k+1}+O(k^{-3/2})$, from which the result will follow.

# The Chinese Restaurant Process

A couple of months ago I wrote a post about Polya’s Urn, the simplest example of self-reinforcing process. Recall that we have a bag containing black and white balls, and sequentially we draw a ball then replace it together with an additional ball of the same colour. The process is self-reinforcing in the sense that if there is a surplus of black balls, the dynamics will reinforce this by adding more black balls than white balls. Alternatively, you can think of a natural limit process when the number of balls is large, for which any distribution is an invariant distribution. We have seen models such as the Preferential Attachment dynamics for network creation, where the degrees of vertices clearly have this self-reinforcing property. New vertices are more likely to join to existing vertices with large degrees.

One difference between the Polya Urn and some of the models we might be interested in for applications is that for the urn model, the number of classes (in this context colours of balls) is fixed. In many applications, we will want to allow new classes to appear. In the process which follows, we will allow this, and the new classes will have initial size equal to 1, so will be at a disadvantage for the self-reinforcing dynamics. Nonetheless, some will show up in a meaningful way in the limit. It is worth emphasising that Polya’s Urn gave us the Dirichlet distribution in the limit, and this can be thought of as a partition of [0,1]. These more general processes will give us a more interesting family of partitions, called the Poisson-Dirichlet distributions. These will turn up in a wide variety of contexts, and this is perhaps the friendliest way to introduce them.

The model is this. We start with a single diner who sits at the first table. Then whenever the (n+1)th diner arrives, they join a table with k diners already with probability k/n+1, and they start a new table with probability 1/n+1.

(Aside: I’m not exactly sure how this relates to a Chinese restaurant? It seems more reminiscent of a university dining hall during freshers’ week, but I guess that would be a less catchy name for a model.)

Anyway, the interest in this description lies not in organising seating arrangements. Consider choosing uniformly at random from the set of permutations on [n+1]. Suppose x maps to n+1 and n+1 maps to y. Consider taking the permutation of [n] formed by instead mapping x to y and ignoring n+1. This has the uniform distribution on the set of permutations of [n]. By reversing this procedure, we can construct a uniform permutation of [n+1] from a uniform permutation of [n]. When you do this as a process for n growing, observe that the orbits correspond exactly to tables in the Chinese Restaurant Process. If we wanted the CRP to give all the information about the permutation, we could specify the ordering round each table, by saying that with probability 1/n+1 the new diner sits to the left of any given existing diner.

As a starting point for why this is a useful description of the uniform permutation distribution, observe that the size of the component containing the element 1 evolves as a Polya Urn with initial vector (1,1). The second 1 in the initial vector corresponds to the possibility of starting a new table, which is maintained at every stage. This tells us immediately that as n grows to infinity, the proportion of elements in the same cycle as 1 in the uniform permutation converges in distribution to U[0,1]. The construction also allows for an easy proof that the expected number of cycles is roughly log n for large n, since on each pass of the process, the probability that there is a new cycle formed is 1/k.

In this case, the partition induced on [n] by the process is clearly exchangeable given our permutation description. However, this will turn out to hold in greater generality. Note also,, that conditional on the size of the cycle containing 1, the sizes of the remaining cycles are given by a uniform permutation on a smaller number of elements. So the limiting result holds jointly in the first k cycle sizes for all k. More precisely, if $(N_1,N_2,\ldots)$ are the cycle sizes ordered by least element, then the frequencies converge to:

$(U_1,(1-U_1)U_2,(1-U_1)(1-U_2)U_3,\ldots),$

where the Us are independent U[0,1] RVs. This is known as a stick-breaking procedure, where at each step we break off some proportion of the stick according to a fixed distribution, and assemble the pieces into a partition.

We generalise this process to get a two-parameter version. The standard notation for the parameters is $(\alpha,\theta)$. Then we amend the dynamics. We now have to take into account how many tables are occupied when the (n+1)th diner arrives. If k tables are occupied, and the ith table has $n_i$ diners, then the new one will join this table with probability $\frac{n_i-\alpha}{n+\theta}$, and will start a new table otherwise, so with probability $\frac{\theta+k\alpha}{n+\theta}$. The original process therefore corresponds to parameters (0,1).

First we examine which parameters are possible. If $\alpha<0$, and $m|\alpha|<\theta<(m+1)|\alpha|$, then with high probability the (m+1)th table will eventually be occupied, whereafter the probability of forming a further table will be negative. So we have to demand instead that $\theta$ is an integer multiple of $-\alpha$. Then the number of tables is bounded by this multiple, so for large n, the probability of joining one of the k (fixed) tables is roughly $\frac{n_i}{n}$, so this should behave roughly like the standard Polya Urn. And indeed, the induced frequencies do converge to the Dirichlet distribution with k equal parameters.

Obviously $\alpha$ cannot be greater than 1, otherwise the probability of the second diner joining the first table is negative. If it is equal to 1, then every diner starts a new table, which isn’t very interesting. So we care about $\alpha\in[0,1)$, and for the probability of the second diner starting a new table to be non-negative we require $\theta>-\alpha$.

It turns out that the partitions induced by this process are exchangeable also. We also have a stick-breaking construction, although now the broken proportions are not IID, but distributed as

$U_i\sim \mathrm{Beta}(1-\alpha,\theta+i\alpha),$

with the same notation otherwise. It turns out that under mild assumptions, these are all the infinite exchangeable random partitions with this stick-breaking property.

My initial struggle with this process was to understand what roles $(\alpha,\theta)$ played in a more precise way. It turns out this is best explained through the limit of the partition, but Pitman’s Exercise 3.2.2 does at least give an idea of how such a process with parameters (1/2,0) might naturally arise as a version of an urn model.

3.2.2. Let an urn initially contain two balls of different colours. Draw 1 is a simple draw from the urn with replacement. Thereafter, balls are drawn from the urn, with replacement of the ball drawn, and addition of two more balls as follows. If the ball drawn is of a colour never drawn before, it is replaced together with two additional balls of two distinct new colours, different to the colours of balls already in the urn. Whereas if the ball drawn is of a colour that has been drawn before, it is replaced together with two balls of its own colour.

Let $n_1$ be the number of times a ball of the colour of the first ball drawn (and replaced) is drawn. Let $n_2,n_3,\ldots$ be the number of times balls of each other colour are drawn. Suppose after n draws, we have drawn k colours. (There will be other colours in the bag not yet drawn.) Then, for each drawn colour i, there are $2n_i-1$ balls of that colour in the bag, giving 2n-k in total. But there should be 2n balls in total, so there are k other balls. Then the probability that we see a new colour is k/2n, and the probability that we see colour i again is $\latex \frac{2n_i-1}{2n}=\frac{n_i-1/2}{n}$, which exactly corresponds to the dynamics for PD(1/2,0).

The other question I was puzzled by initially is where does the dust come from in the limit? Recall that in an infinite exchangeable partition, the sum of the frequencies does not need to be 1. The difference between this sum and 1 gives the probability that an element is in a block by itself. Obviously, when the number of tables is bounded (as when $\alpha<0$) this is not an issue, but for positive $\alpha$, this won’t hold. So we need to account for these singletons. The temptation is to imagine that these correspond to tables which are started but never joined. But this use of ‘never’ is not ideal. For each k, the k-th table will eventually include arbitrarily large numbers of diners. But for any finite n, there will likely be some proportion of people dining alone, some in pairs, and so on. So the sum of all of these proportions in the limit gives this dust.

Generalising Polya’s Urn in another direction, if I have time, I might write something about a model which I recently read about on arXiv where the classes are vertices of a graph, and there is dependence between them based on the presence of edges. This might also be a good moment to explain some other generalisations and stochastic approximation methods used to treat them.

REFERENCES

This post is almost entirely a paraphrase of Sections 3.1 and 3.2 from Pitman’s Combinatorial Stochastic Processes, available online here.

# Urns and the Dirichlet Distribution

As I’ve explained in some posts from a while ago, I’ve been thinking about some models related to random graph processes, where we ensure the configuration stays critical by deleting any cycles as they appear. Under various assumptions, this behaves in the limit as the number of vertices grows to infinity, like a coagulation-fragmentation process, with multiplicative coalescence and quadratic fragmentation rate, where the fragmentation kernel is the Poisson-Dirichlet distribution, PD(1/2,1/2). I found it quite hard to find accessible notes on these, partly because the theory is still relatively recently, and also because it seems to be one of those topics where you can’t understand anything properly until you kind of understand everything.

This post was motivated and is based on chapter 3 of Pitman’s Combinatorial Stochastic Processes, and the opening pair of lectures from Pierre Tarres’s TCC course on Self-Interaction and Learning.

It makes sense to begin by discussing the Dirichlet distribution, and there to start with the most simple case, the Beta distribution. As we learned in the Part A Statistics course while trying some canonical examples of posterior distributions, it is convenient to ignore the normalising constants of various distributions until right at the end. This is particularly true of the Beta distribution, which is indeed often used as a prior in such situations. The density function of $\text{Beta}(\alpha,\beta)$ is $x^{\alpha-1}(1-x)^{\beta-1}$. If these are natural numbers, we have a quick proof by induction using integration by parts, otherwise a slightly longer but still elementary argument gives the normalising constant as

$\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}.$

(note that the ‘base case’ is the definition of the Gamma function.) For the generalisation we are about to make, it is helpful to think of this Beta density as a distribution not on [0,1], but on partitions of [0,1] into two parts. That is pairs (x,y) such that x+y=1. Why? Because then the density has the form $x^{\alpha-1}y^{\beta-1}$ and is clear how we might generalise this.

Indeed the Dirichlet distribution with parameters $(\alpha_1,\ldots,\alpha_m)$ is a random variable supported on the subset of $\mathbb{R}^m$ with $\sum p_k=1$ with density $\propto \prod p_k^{\alpha_k-1}$. For similar reasons, the correct normalising constant in the general case is

$\frac{\Gamma(\sum \alpha_k)}{\prod \Gamma(\alpha_k)}.$

You can prove this by inducting on the number of variables, using the Beta distribution as a base case.

In many situations, it is useful to be able to express some distribution as a function of IID random variables with a simpler distribution. We can’t quite do that for the Dirichlet distribution, but we can express it very simply as function of independent RVs from the same family. It turns out that the family of Gamma distributions is a wise choice. Recall that the gamma distribution with parameters $(\alpha,\beta)$ has density:

$\frac{1}{\beta^\alpha \Gamma(\alpha)}y^{\alpha-1}e^{-y/\beta},\quad y>0.$

Anyway, define independent RVs $Y_k$ as gamma distributions with parameters $(\alpha_k,1)$, then we can specify $(X_1,\ldots,X_m)$ as the Dirichlet distribution with these parameters by:

$(X_1,\ldots,X_m)\stackrel{d}{=}\frac{(Y_1,\ldots,Y_m)}{\sum Y_k}.$

In other words, the Dirichlet distribution gives the ratio between independent gamma RVs. Note the following:

– the sum of the gamma distributions, ie the factor we have to scale by to get back to a ratio, is a gamma distribution itself.

– If we wanted, we could define it in an identical way using Gamma with parameters $(\alpha_k,\beta)$ for some fixed $\beta$.

– More helpfully, because the gamma distribution is additive in the first argument, we can take a limit to construct a gamma process, where the increments have the form required. This will be a useful interpretation when we take a limit, as largest increments will correspond to largest jumps.

Polya’s Urn

This is one of the best examples of a self-reinforcing process, where an event which has happened in the past is more likely to happen again in the future.

The basic model is as follows. We start with one white ball and one black ball in a bag. We draw a ball from the bag uniformly at random then replace it along with an additional ball of the same colour. Repeat this procedure.

The first step is to look at the distribution at some time n, ie after n balls have been added, so there are n+2 in total. Note that there are exactly n+1 possibilities for the state of the bag at this time. We must have between 1 and n+1 black balls, and indeed all of these are possible. In general, part of the reason why this process is self-reinforcing is that any distribution is in some sense an equilibrium distribution.

What follows is a classic example of a situation which is a notational nightmare in general, but relatively straightforward for a fixed finite example.

Let’s example n=5, and consider the probability that the sequence of balls drawn is BBWBW. This probability is:

$\frac12\times \frac 23\times \frac14\times \frac 35\times \frac26.$

So far this isn’t especially illuminating, especially if we start trying to cancel these fractions. But note that the denominator of the product will clearly be 6!. What about the numerator? Well, the contribution to the numerator of the product from black balls is 1x2x3=3! while the contribution from white balls is 1×2=2!. In particular, the contribution to the numerator from each colour is independent of the order of whites and blacks. It depends only on the number of whites and blacks. So we can conclude that the probability that we end up a particular ordering of k+1 whites and (n-k)+1 blacks is

$\frac{k! (n-k)!}{(n+1)!},$

and so the probability that we end up with k+1 whites where we no longer care about ordering is

$\binom{n}{k}\frac{k!(n-k)!}{(n+1)!}=\frac{1}{n+1}.$

In other words, the distribution of the number of white balls in the bag after n balls have been added is uniform on [1, n+1].

That looks like it might be something of a neat trick, so the natural question to ask is what happens if we adjust the initial conditions. Suppose that instead we start with $a_1,\ldots,a_m$ balls of each of m colours. Obviously, this is going to turn into a proof by suggestive notation. In fact, the model doesn’t really rely on the $(a_i)$ being positive integers. Everything carries through with $a_i\in\mathbb{R}$ if we view the vector as the initial distribution.

As before, the order in which balls of various colours are drawn doesn’t matter hugely. Suppose that the first n balls drawn feature $n_i$ balls of colour i. The probability of this is:

$\binom{n}{n_1,\ldots,n_k}\frac{\prod_i \alpha_i(\alpha_i+1)\ldots (\alpha_i+n_i-1)}{\alpha(\alpha+1)\ldots(\alpha+(n-1))}$

where $\alpha=\sum_i \alpha_i$. Then for large n, assuming for now that the $\alpha_i\in\mathbb{N}$ we have

$\frac{\alpha(\alpha+1)\ldots(\alpha+n-1)}{n!}=\frac{[\alpha_i+(n_i-1)]!}{n_i! (\alpha_i-1)!}\approx \frac{n_i^{\alpha_i-1}}{(\alpha_i-1)!}.$

The denominator will just be a fixed constant, so we get that overall, the probability above is approximately

$\frac{\prod_i n_i^{\alpha_i-1}}{n^{\alpha-1}}=\prod (\frac{n_i}{n})^{\alpha_i-1},$

which we recall is the pdf of the distribution distribution with parameters $(\alpha_i)$ as telegraphed by our choice of notation. With some suitable martingale machinery, you can also prove that this convergence happens almost surely, for a suitable limit RV defined on the tail sigma algebra.

Next time I’ll introduce a more complicated family of self-reinforcing processes, and discuss some interesting limits of the Dirichlet distribution that relate to such processes.