# Lecture 10 – the configuration model

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we enter the final stages of the semester, I want to discuss some extensions to the standard Erdos-Renyi random graph which has been the focus of most of the course so far. Although we will not get far into the details during this course, the overall goal is to develop models which are close to Erdos-Renyi in terms of ease of analysis, while also allowing more of the features characteristic of networks observed in the real world.

One of the more obvious deficiencies of the sparse regime of Erdos-Renyi random graphs for modelling ‘real-world phenomena’ concerns the degree sequence. Indeed, the empirical degree distribution of G(n,c/n) converges to Poisson(c). By contrast, in real-world networks, a much wider range of degrees is typically observed, and in many cases it is felt that these should follow a power law, with a small number of a very highly connected agents.

One way around this problem to construct random graphs where we insist that the graph has a given sequence of degrees. The configuration model, which is the subject of this lecture and this post (and about which I’ve written before), offers one way to achieve this.

Definition and notes

Let $n\ge 1$ and let $d=(d_1,d_2,\ldots,d_n)$ be a sequence of non-negative integers such that $\sum_{i=1}^n d_i$ is even. Then the configuration model with degree sequence d is a random multigraph with vertex set [n], constructed as follows:

• To each vertex $i\in[n]$, assign $d_i$ half-edges;
• Then, take a uniform matching of these half-edges;
• Finally, for each pair of half-edges in the matching, replace the two half-edges with a genuine edge, to obtain the multigraph $CM_n(d)$, in which, by construction, vertex i has degree $d_i$.

One should note immediately that although the matching is uniform, the multigraph is not uniform amongst multigraphs with that degree sequence. Note also that the condition on the sums of the degrees is necessary for any graph, and in this context means that the number of half-edges is even, without which it would not be possible to construct a matching.

This effect is manifest in the simplest possible example, when n=2 and d=(3,3). There are two possible graphs, up to isomorphism, which are shown below:

For obvious reasons, we might refer to these as the handcuffs and the theta , respectively. It’s helpful if we, temporarily, assume the half-edges are distinguishable at the moment we join them up in the configuration model construction. Because then there are 3×3=9 ways to join them up to form the handcuffs (think of which half-edge ends up forming the edge between the two vertices) while there are 3!=6 ways to pair up the half-edges in the theta.

In general, for multigraphs H with the correct degree sequence, we have

$\mathbb{P}( CM_n(d)\simeq H) \propto \left( 2^{\# \text{loops}(H)} \prod_{e\in E(H)} \text{mult}(e)! \right),$

where $\text{mult}(e)$ is the multiplicity with which a given edge e appears in H.

Note: it might seem counterintuitive that this procedure is biased against multiple edges and self-loops, but it is really just saying that there are more ways to form two distinct edges than to form two equal edges (ie a multiedge pair) when we view the half-edges as distinguishable. (See this post for further discussion of this aspect in the 3-regular setting.)

However, a consequence of this result is that if we condition on the event that $CM_n(d)$ is simple, then the resulting random graph is uniform on the set of simple graphs satisfying the degree property. Note that the same example as above shows that there’s no guarantee that there exists a simple graph whose degrees are some given sequence.

d-regular configuration model

In general, from a modelling point of view, we are particularly interested in simple, connected graphs, and so it is valuable to study whether the large examples of the configuration model are likely to have these properties. In this lecture, I will mainly focus on the case where the multigraphs are d-regular, meaning that all the vertices have degree equal to d. For the purposes of this lecture, we denote by $G^d(n)$, the d-regular configuration model $CM_n(d,\ldots,d)$.

• d=1: to satisfy the parity condition on the sums of degrees, we must have n even. But then $G^1(n)$ will consist of n/2 disjoint edges.
• d=2: $G^2(n)$ will consist of some number of disjoint cycles, and it is a straightforward calculation to check that when n is large, with high probability the graph will be disconnected.

In particular, I will focus on the case when d=3, which is the first interesting case. Most of the results we prove here can be generalised (under various conditions) to more general examples of the configuration model. The main goal of the lecture is revision of some techniques of the course, plus one new one, in a fresh setting, and the strongest possible versions of many of these results can be found amongst the references listed at the end.

Connectedness

In the lecture, we showed that $G^3(2n)$ is connected with high probability. This is, in fact, a very weak result, since in fact $G^d(n)$ is d-connected with high probability for $d\ge 3$ [Bol81, Wor81]. Here, d-connected means that one must remove at least d vertices in order to disconnect the graph, or, equivalently, that there are d disjoint paths between any pair of vertices. Furthermore, Bollobas shows that for $d\ge 3$, $G^d(n)$ is a (random) expander family [Bol88].

Anyway, for the purposes of this course, the main tool is direct enumeration. The matching number $M_{2k}$ satisfies

$M_{2k}=(2k-1)\times (2k-3)\times\ldots\times 3\times 1 = \frac{(2k)!}{2^k \cdot k!},$

and so Stirling’s approximation gives the asymptotics

$M_{2k} = (\sqrt{2}+o(1)) \left(\frac{2}{e}\right)^k k^k,$

although it will be useful to use the true bounds

$c \left(\frac{2}{e}\right)^k k^k \le M_{2k}\le C\left(\frac{2}{e}\right)^k k^k,\quad \forall k,$

instead in some places. Anyway, in $G^3(2n)$, there are 6n half-edges in total, and so the probability that the graph may be split into two parts consisting of $2\ell,2m$ vertices, with $2\ell+2m=2n$, and with no edges between the classes is $\frac{\binom{2n}{2\ell} M_{6\ell}M_{6m}}{M_{6n}}.$ Continue reading

# Lecture 9 – Inhomogeneous random graphs

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we enter the final stages of the semester, I want to discuss some extensions to the standard Erdos-Renyi random graph which has been the focus of most of the course so far. In doing so, we can revisit material that we have already covered, and discover how easily one can extend this directly to more exotic settings.

The focus of this lecture was the model of inhomogeneous random graphs (IRGs) introduced by Soderberg [Sod02] and first studied rigorously by Bollobas, Janson and Riordan [BJR07]. Soderberg and this blog post address the case where vertices have a type drawn from a finite set. BJR address the setting with more general typespaces, in particular a continuum of types. This generalisation is essential if one wants to use IRGs to model effects more sophisticated than those of the classical Erdos-Renyi model G(n,c/n), but most of the methodology is present in the finite-type setting, and avoids the operator theory language which is perhaps intimidating for a first-time reader.

Inhomogeneous random graphs

Throughout, $k\ge 2$ is fixed. A graph with k types is a graph G=(V,E) together with a type function $V\to \{1,\ldots,k\}$. We will refer to a $k\times k$ symmetric matrix with non-negative entries as a kernel.

Given $n\in\mathbb{N}$ and a vector $p=(p_1,\ldots,p_k)\in\mathbb{N}_0^k$ satisfying $\sum p_i=n$, and $\kappa$ a kernel, we define the inhomogeneous random graph $G^n(p,\kappa)$ with k types as:

• the vertex set is [n],
• types are assigned uniformly at random to the vertices such that exactly $p_i$ vertices have type i.
• Conditional on these types, each edge $v\leftrightarrow w$ (for $v\ne w\in [n]$) is present, independently, with probability

$1 - \exp\left(-\frac{\kappa_{\mathrm{type}(v),\mathrm{type}(w)} }{n} \right).$

Notes on the definition:

• Alternatively, we could assign the types so that vertices $\{1,\ldots,p_1\}$ have type 1, $\{p_1+1,\ldots,p_1+p_2\}$ have type 2, etc etc. This makes no difference except in terms of the notation we have to use if we want to use exchangeability arguments later.
• An alternative model considers some distribution $\pi$ on [k], and assigns the types of the vertices of [n] in an IID fashion according to $\pi$. Essentially all the same results hold for these two models. (For example, this model with ‘random types’ can be studied by quenching the number of each type!) Often one works with whichever model seems easier for a given proof.
• Note that the edge probability given is $\approx \frac{\kappa_{\mathrm{type}(v),\mathrm{type}(w)}}{n}$. The exponential form has a more natural interpretation if we ever need to turn the IRGs into a process. Additionally, it avoids the requirement to treat small values of n (for which, a priori, $k/n$ might be greater than 1) separately.

In the above example, one can see that, roughly speaking, red vertices are more likely to be connected to each other than blue vertices. However, for both colours, they are more likely to be connected to a given vertex of the same colour than a vertex of the opposite colour. This might, for example, correspond to the kernel $\begin{pmatrix}3&1\\1&2\end{pmatrix}$.

The definition given above corresponds to a sparse setting, where the typical vertex degrees are $\Theta(1)$. Obviously, one can set up an inhomogeneous random graph in a dense regime by an identical argument.

From an applications point of view, it’s not hard to imagine that an IRG of some flavour might be a good model for many phenomena observed in reality, especially when a mean-field assumption is somewhat appropriate. The friendships of boys and girls in primary school seems a particularly resonant example, though doubtless there are many others.

One particular application is to recover the types of the vertices from the topology of the graph. That is, if you see the above picture without the colours, can you work out which vertices are red, and which are blue? (Assuming you know the kernel.) This is clearly impossible to do with anything like certainty in the sparse setting – how does one decide about isolated vertices, for example? The probabilities that a red vertex is isolated and that a blue vertex is isolated differ by a constant factor in the $n\rightarrow\infty$ limit. But in the dense setting, one can achieve this with high confidence. When studying such statistical questions, these IRGs are often referred to as stochastic block models, and the recent survey of Abbe [Abbe] gives a very rich history of this type of problem in this setting.

Poisson multitype branching processes

As in the case of the classical random graph G(n,c/n), we learn a lot about the IRG by studying its local structure. Let’s assume from now on that we are given a sequence of IRGs $G^n(p^n,\kappa)$ for which $\frac{p^n}{n}\rightarrow \pi$, where $\pi=(\pi_1,\ldots,\pi_k)\in[0,1]^k$ satisfies $||\pi||_1=1$.

Now, let $v^n$ be a uniformly-chosen vertex in [n]. Clearly $\mathrm{type}(v^n)\stackrel{d}\rightarrow \pi$, with the immediate mild notation abuse of viewing $\pi$ as a probability distribution on [k].

Then, conditional on $\mathrm{type}(v^n)=i$:

• when $j\ne i$, the number of type j neighbours of $v^n$ is distributed as $\mathrm{Bin}\left(p_j,1-\exp\left(-\frac{\kappa_{i,j}}{n}\right)\right)$.
• the number of type i neighbours of $v^n$ is distributed as $\mathrm{Bin}\left( p_i-1,1-\exp\left(-\frac{\kappa_{i,i}}{n}\right)\right)$.

Note that $p_j\left[1-\exp\left(-\frac{\kappa_{i,j}}{n}\right)\right]\approx \frac{p_j\cdot \kappa_{i,j}}{n} \approx \kappa_{i,j}\pi_j$, and similarly in the case j=i, so in both cases, the number of neighbours of type j is distributed approximately as $\mathrm{Poisson}(\kappa_{i,j}\pi_j)$.

This motivates the following definition of a branching process tree, whose vertices have k types. Continue reading

# Lecture 7 – The giant component

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we edge into the second half of the course, we are now in a position to return to the question of the phase transition between the subcritical regime $\lambda<1$ and the supercritical regime $\lambda>1$ concerning the size of the largest component $L_1(G(n,\lambda/n))$.

In Lecture 3, we used the exploration process to give upper bounds on the size of this largest component in the subcritical regime. In particular, we showed that

$\frac{1}{n}\big| L_1(G(n,\lambda/n)) \big| \stackrel{\mathbb{P}}\rightarrow 0.$

If we used slightly stronger random walk concentration estimates (Chernoff bounds rather than 2nd-moment bounds from Chebyshev’s inequality), we could in fact have shown that with high probability the size of this largest component was at most some logarithmic function of n.

In this lecture, we turn to the supercritical regime. In the previous lecture, we defined various forms of weak local limit, and asserted (without attempting the notationally-involved combinatorial calculation) that the random graph $G(n,\lambda/n)$ converges locally weakly in probability to the Galton-Watson tree with $\text{Poisson}(\lambda)$ offspring distribution, as we’ve used informally earlier in the course.

Of course, when $\lambda>1$, this branching process has strictly positive survival probability $\zeta_\lambda>0$. At a heuristic level, we imagine that all vertices whose local neighbourhood is ‘infinite’ are in fact part of the same giant component, which should occupy $(\zeta_\lambda+o_{\mathbb{P}}(1))n$ vertices. In its most basic form, the result is

$\frac{1}{n}\big|L_1(G(n,\lambda/n))\big|\;\stackrel{\mathbb{P}}\longrightarrow\; \zeta_\lambda,\quad \frac{1}{n}\big|L_2(G(n,\lambda/n))\big| \;\stackrel{\mathbb{P}}\longrightarrow\; 0,$ (*)

where the second part is a uniqueness result for the giant component.

The usual heuristic for proving this result is that all ‘large’ components must in fact be joined. For example, if there are two giant components, with sizes $\approx \alpha n,\approx \beta n$, then each time we add a new edge (such an argument is often called ‘sprinkling‘), the probability that these two components are joined is $\approx 2ab$, and so if we add lots of edges (which happens as we move from edge probability $\lambda-\epsilon\mapsto \lambda$ ) then with high probability these two components get joined.

It is hard to make this argument rigorous, and the normal approach is to show that with high probability there are no components with sizes within a certain intermediate range (say between $\Theta(\log n)$ and $n^\alpha$) and then show that all larger components are the same by a joint exploration process or a technical sprinkling argument. Cf the books of Bollobas and of Janson, Luczak, Rucinski. See also this blog post (and the next page) for a readable online version of this argument.

I can’t find any version of the following argument, which takes the weak local convergence as an assumption, in the literature, but seems appropriate to this course. It is worth noting that, as we shall see, the method is not hugely robust to adjustments in case one is, for example, seeking stronger estimates on the giant component (eg a CLT).

Anyway, we proceed in three steps:

Step 1: First we show, using the local limit, that for any $\epsilon>0$,

$\frac{1}{n}\big|L_1(G(n,\lambda/n))\big| \le \zeta_\lambda+\epsilon,$ with high probability as $n\rightarrow\infty$.

Step 2: Using a lower bound on the exploration process, for $\epsilon>0$ small enough

$\frac{1}{n}\big|L_1(G(n,\lambda/n))\big| \ge \epsilon,$ with high probability.

Step 3: Motivated by duality, we count isolated vertices to show

$\mathbb{P}(\epsilon n\le |L_1| \le (\zeta_\lambda-\epsilon)n) \rightarrow 0.$

Step 1

This step is unsurprising. The local limit gives control on how many vertices are in small components of various sizes, and so gives control on how many vertices are in small components of all finite sizes (taking limits in the right order). This gives a bound on how many vertices can be in the giant component. Continue reading

# Lecture 6 – Local limits

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

By this point of the course, we’ve studied several aspects of the Erdos-Renyi random graph, especially in the sparse setting $G(n,\frac{\lambda}{n})$. We’ve also taken a lengthy detour to revise Galton-Watson trees, with a particular focus on the case of Poisson offspring distribution.

This is deliberate. Note that a given vertex v of $G(n,\frac{\lambda}{n})$ has some number of neighbours distributed as $\mathrm{Bin}(n-1,\frac{\lambda}{n})\stackrel{d}\approx\mathrm{Po}(\lambda)$, and the same approximation remains valid as we explore the graph (for example in a breadth-first fashion) either until we have seen a large number of vertices, or unless some ultra-pathological event happens, such as a vertex having degree n/3.

In any case, we are motivated by the notion that the local structure of $G(n,\frac{\lambda}{n})$ is well-approximated by the Galton-Watson tree with $\mathrm{Po}(\lambda)$ offspring, and in this lecture and the next we try to make this notion precise, and discuss some consequences when we can show that this form of convergence occurs.

Deterministic graphs

Throughout, we will be interested in rooted graphs, since by definition we have to choose a root vertex whose local neighbourhood is to be studied. Usually, we will study a sequence of rooted graphs $(G_n,\rho_n)$, where the vertex set of $G_n$ is [n], or certainly increasing in n (as in the first example).

For some rooted graph $(G,\rho)$, we say such a sequence $(G_n,\rho_n)$ converges to $(G,\rho)$ locally if for all radii $r\ge 1$, we have $B_r^{G_n}(\rho_n)\simeq B_r^G(\rho)$. In words, the neighbourhood around $\rho_n$ in $G_n$ is the same up to radius r as the neighbourhood around $\rho$ in $G$, so long as n is large enough (for given r).

This is best illustrated by an example, such as $T_n$, the binary tree to depth n.

If we take $\rho_n$ to be the usual root, then the trees are nested, and converge locally to the infinite binary tree $T_\infty$. Slightly less obviously, if we take $\rho_n$ to be one of the leaves, then the trees are still nested (up to labelling – ie in the sense of isomorphisms of rooted trees), and converge locally to the canopy tree, defined by a copy of $\mathbb{Z}_{\ge 0}$ with nearest-neighbour edges, and where each vertex $n\ge 1$ is connected to the root of a disjoint copy of $T_{n-1}$, as shown below:

Things get more interesting when the root is chosen randomly, for example, uniformly at random, as this encodes more global information about the graphs $G_n$. In the case where the $G_n$ are vertex-transitive, then if we only care about rooted graphs up to isomorphism, then it doesn’t matter how we choose the root.

Otherwise, we say that $G_n$ converges in the local weak sense to $(G,\rho)$ if, for all $r\ge 1$ and for all rooted graphs $(H,\rho_H)$,

$\mathbb{P}\left( B^{G_n}_r(\rho_n)\simeq (H,\rho_H) \right) \longrightarrow \mathbb{P}\left( B_r^G(\rho)\simeq H\right),$

as $n\rightarrow\infty$.

Alternatively, one can phrase this as a result about convergence of rooted-graph-valued distributions.

A simple non-transitive example is $G_n\simeq P_n$, the path of length n. Then, the r-neighbourhood of a vertex is isomorphic to $P_{2r}$unless that vertex is within graph-distance (r-1) of one of the leaves of $G_n$. As $n\rightarrow\infty$, the proportion of such vertices vanishes, and so, $\mathbb{P}\left( B^{P_n}_r(\rho_n)\simeq P_{2r}\right)\rightarrow 1$, from which we conclude the unsurprising result that $P_{n}$ converges in the local weak sense to $\mathbb{Z}$. (Which is vertex-transitive, so it doesn’t matter where we select the root.)

The binary trees offer a slightly richer perspective. Let $\mathcal{L}_n$ be the set of leaves of $T_n$, and we claim that when $\rho_n$ is chosen uniformly from the vertices of $T_n$, then $d_{T_n}(\rho_n,\mathcal{L}_n)$ converges in distribution. Indeed, $\mathbb{P}\left( d_{T_n}(\rho_n,\mathcal{L}_n)=k\right) = \frac{2^{n-k}}{2^{n+1}-1}$, whenever $n\ge k$, and so the given distance converges in distribution to the Geometric distribution with parameter 1/2 supported on {0,1,2,…}.

This induces a random local weak limit, namely the canopy tree, rooted at one of the vertices we denoted by $\mathbb{Z}_{\ge 0}$, with the choice of this vertex given by Geometric(1/2). Continue reading

# Lecture 2 – Connectivity threshold

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

The goal of the second lecture was to establish the sharp phase transition for the connectivity of the random graph G(n,p(n)) around the critical regime $p(n)\sim \frac{\log n}{n}$. In the end, we showed that when $\omega(n)$ is any diverging sequence, and $p(n)=\frac{\log n-\omega(n)}{n}$, then we have that G(n,p(n)) is with high probability not connected.

In the next lecture, we will finish the classification by studying $p(n)=\frac{\log n+\omega(n)}{n}$, and show that for this range of p, the graph G(n,p(n)) is with high probability connected.

The details of the lecture, especially the calculation, are not presented fully here. There, I followed van der Hofstad’s recent book fairly closely, sometimes taking different approximations and routes through the algebra, though all versions remain fairly close to the original enumerations by Renyi.

Immediate remarks

• One is allowed to be surprised that for almost all scalings of p(n), G(n,p) is either whp connected or whp not connected. The speed of the transition is definitely interesting.
• As defined in lectures, the property that a graph is connected is an increasing property, meaning that it is preserved when you add additional edges to the graph.
• Because of the natural coupling between G(n,p) and G(n,q), the fact that connectedness is an increasing property makes life easier. For example, we can insist temporarily that $\omega(n)\ll \log n$, or whatever scaling turns out to be convenient for the proof, but conclude the result for all diverging $\omega(n)$. This avoids the necessity for an annoying case distinction.

Heuristics – Isolated vertices

It turns out that the ‘easiest’ way for such a graph to be disconnected is for it to have an isolated vertex. In determining that the graph has a cut into classes of sizes a and b with no edges between them, there is a trade-off between the number of ways to choose the partition (which increases with min(a,b) ) and the probabilistic penalty from banning the ab edges between the classes (which decreases with min(a,b) ). It turns out that the latter effect is slightly stronger, and so (1,n-1) dominates.

Method 1: second-moment method

In the case $p(n)=\frac{\log n - \omega(n)}{n}$, we use a second-moment method argument to establish that G(n,p) contains an isolated vertex with high probability. Note that a given vertex v is isolated precisely if n-1 edges are not present. Furthermore, two given vertices v,w are both isolated, precisely if 2n-3 edges are not present. So in fact, both the first moment and the second moment of the number of isolated vertices are straightforward to evaluate.

It turns out that the number of isolated vertices, $Y_n$, satisfies

$\mathbb{E}[Y_n]= \exp(\omega(n)+o(1))\rightarrow\infty.$ (*)

As always, we have to eliminate the possibility that this divergent expectation is achieved by the graph typically having no isolated vertices, but occasionally having very many. So we turn to the second moment, and can show

$\mathrm{Var}(Y_n)= (1+o(1))\mathbb{E}[Y_n],$

and so by Chebyshev’s inequality, we have $\mathbb{P}(Y_n=0)\rightarrow 0$.

Method 2: first-moment method

Counter-intuitively, although the case $p(n)=\frac{\log n + \omega(n)}{n}$ requires only a first-moment method, it is more technical because it involves the non-clear direction of the informal equivalence:

$\text{Connected}\; \iff ''\; \text{no isolated vertices}.$

At the time we showed (*), we also showed that for this regime of p(n), G(n,p) whp has no isolated vertices. It remains to show that it has no splits into (unions of) connected components of sizes k and n-k. Continue reading

# Random Graphs – Lecture 1

My plan is to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics about the course can be found here.

In the first lecture, we revised some basic definitions about graphs, focusing on those which are most relevant to a first study of the Erdos-Renyi random graph G(n,p) which will be the focus of the lecture course. We discussed in abstract why the independence of the (potential) edges makes the model easier to analyse, but reduces its suitability as a direct model for lots of networks one might see in the real world, where knowledge that A is directly connected to both B and C affects the probability that B is directly connected to C, in either direction. Thinking about the Facebook friendship graph is one of the best examples, where in this case, we expect this extra information to increase the probability that B and C are connected. Even as the world moves away from heteronormativity, it realistically remains the case that in a graph of the dating history amongst a well-defined community we would likely observe the opposite effect.

All of these more complicated phenomena can be captured by various random graphs, but G(n,p) remains the corner stone, evinced by the $>10^5$ citations towards one of Erdos and Renyi’s original papers on the topic.

Somewhat paraphrasing, one of their (well, mostly Renyi’s) original questions was: when n is large, what should p be so that there’s a good chance that G(n,p) is connected?

The answer to this question lies in Lecture 2, but to cement understanding of the model, and explore some key methods for proofs in discrete probability (as well as play around with the big-O and little-o notation), we investigated the following two situations, which are very far from interesting as far as connectivity of G(n,p) is concerned.

Dense regime

When p is fixed, there are many interesting questions one could ask about the asymptotic properties of G(n,p), but connectivity is not one of them. In particular, for $p\in(0,1)$ we claim:

Proposition: $\mathrm{diam}(G(n,p)) \stackrel{\mathbb{P}}\rightarrow 2$ as $n\rightarrow\infty$.

Note that if $\mathrm{diam}(G(n,p))=1$, then $G(n,p)\simeq K_n$, the complete graph on n vertices. In other words, every possible edge is actually present. But the probability of this event is $p^{\binom{n}{2}}\rightarrow 0$, so long as p<1.

It then suffices to prove that $\mathbb{P}(G(n,p)>2) \rightarrow 0$. We use a union bound, where we study the probability that the graph distance $d_{G(n,p}(v,w)>2$ for two fixed vertices $v\ne w$ first, and then sum over all such pairs. Of course, there is a probability p that the two vertices are directly connected by an edge. Then, there are (n-2) other vertices with the potential to be a common neighbour of v and w, which would ensure that the graph distance between them is at most two. So

$\mathbb{P}(d_{G(n,p)}(v,w)>2)=(1-p)[1-p^2]^{n-2} .$

Note that we are using independence throughout this calculation. Then comes the union bound:

$\mathbb{P}(\mathrm{diam}(G(n,p)) >2) \le \sum_{v\ne w \in[n]} \mathbb{P}(d_{G(n,P)}(v,w)>2)$

$\le \binom{n}{2} (1-p)[1-p^2]^{n-2} \rightarrow 0,$

since exponential decay ‘kills’ polynomial growth.

Ultra-sparse regime

In general, we work in the setting where p=p(n) depends on n. If p(n) decays fast enough (see Exercise 2), then with high probability G(n,p) has no edges at all. However, when $p(n)=o(n^{-3/2})$ we have

Proposition: $\mathbb{P}(\text{edges of }G(n,p)\text{ form a matching}) \rightarrow 1$ as $n\rightarrow\infty$.

A matching is a collection of edges with no vertices in common. So if the edge set of the graph is a matching, we have essentially no interesting connectivity structure at all. The longest path has length one, for example.

To prove this, note that the edge set of the graph fails to be a matching precisely if one of the vertices has degree at least two. But since a vertex v is connected to each of the (n-1) other vertices in the graph independently with probability p, we have

$\mathrm{deg}_{G(n,p)}(v) \sim \mathrm{Bin}(n-1,p),$

and so we can directly make the crude approximation

$\mathbb{P}(\mathrm{deg}_{G(n,p)}(v) =k) = \binom{n-1}{k}p^k(1-p)^{n-1-k}\le n^k p^k.$

We’ve made this very weak bound to make life easier when we sum:

$\mathbb{P}(\mathrm{deg}_{G(n,p)}(v) \ge 2) \le \sum_{k\ge 2}(np)^k = \frac{(np)^2}{1-np}.$

Since $p=o(n^{-3/2})$, we have $\frac{1}{1-np}=\frac{1}{1-o(1)}=1+o(1)$, and overall we obtain

$\mathbb{P}(\mathrm{deg}_{G(n,p)}(v) \ge 2) = o(\frac{1}{n}).$

Again, we finish with a union bound, considering this event across all vertices $v\in[n]$.

$\mathbb{P}(E(G(n,p))\text{ not a matching}) \le \sum_{v\in[n]} \mathbb{P}(\mathrm{deg}_{G(n,p)}(v)\ge 2)$

$=n\mathbb{P}(\mathrm{deg}_{G(n,p)}(1)\ge 2) = o(1),$

as required.

Next time

In the next lecture, we’ll study the regime $p(n)\sim \frac{\log n}{n}$, where G(n,p) experiences a phase transition from probably not connected to probably connected. Part of this involves making the notion probably connected precise, which will be useful throughout the rest of the course, as well as establishing the language for comparing G(n,p) and G(n,q).

The proof itself requires some more sophisticated versions of calculations from Lecture 1, and more sophisticated probabilistic tools (first- and second-moment methods) to convert them into statements about convergence in probability. This will be an advertisement for the more classical enumerative methods that underpinned much of the early work on random graphs.

The rest of the course will exploit much more some comparisons and embeddings involving branching processes and exploration processes, so don’t worry – it won’t be 26 hours of counting trees!

# 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

# 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]

# 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 $G=K_n$.

# Azuma-Hoeffding Inequality

It’s (probably) my last Michaelmas term in Oxford, at least for the time being, and so also the last time giving tutorials on either of the probability courses that students take in their first two years. This time, I’m teaching the second years, and as usual the aim of the majority of the first half of the course is to acquire as sophisticated an understanding as possible of the Central Limit Theorem. I feel a key step is appreciating that CLT tells you about the correct scaling for the deviations from the mean of these partial sums of IID random variables. The fact that these deviations on this correct scaling converge in law to a normal distribution, irrespective (apart from mild conditions) on the underlying distribution, is interesting, but should be viewed as a secondary, bonus, property.

Emphasising the scaling of deviations in CLT motivates the next sections of this (or any) course. We develop tools like Markov’s inequality to control the probability that a random variable is much larger than its expectation, and experiment with applying this to various functions of the random variable to get stronger bounds. When the moment generating function exists, this is an excellent choice for this analysis. We end up with a so-called Chernoff bound. For example, we might consider the probability that when we toss N coins, at least a proportion ¾ are Heads. A Chernoff bound says that this probability decays exponentially in N.

One direction to take is to ask how to control precisely the parameter of this exponential decay, which leads to Cramer’s theorem and the basis of the theory of Large Deviations. An alternative direction is to observe that the signed difference between the partial sums of independent random variables and their means is an example of a martingale, albeit not a very interesting one, since in general the increments of a martingale are not independent. So we might ask: under what circumstances can we show exponential tail bounds on the deviation of a martingale from its mean (that is, its initial value) at a fixed (perhaps large) time?

Azuma-Hoeffding inequality

The following result was derived and used by various authors in the 60s, including Azuma and Hoeffding (separately), but also others.

Let $X_0,X_1,X_2,\ldots$ be a martingale with respect to some filtration, and we assume that the absolute value of each increment $|X_i-X_{i-1}|$ is bounded almost surely by some $c_i<\infty$. Then, recalling that $\mathbb{E}[X_n|\mathcal{F}_0]=X_0$, we have

$\mathbb{P}(X_n \ge X_0+t) \le \exp\left( -\frac{t^2}{2\sum_{i=1}^n c_i^2}\right).$

Proof

We apply a Chernoff argument to each increment. First, observe that for Y a distribution supported on [-1,1] with mean zero, by convexity $\mathbb{E}[e^{tY}]$ is maximised by taking Y equal to +1 and -1 each with probability ½. Thus

$\mathbb{E}[e^{tY}]\le \frac12 e^t + \frac 12 e^{-t}=\cosh(t) \le e^{-t^2/2},$

where the final inequality follows by directly comparing the Taylor series.

We’ll use this shortly. Before that, we start the usual argument for a Chernoff bound on $X_n-X_0$.

$\mathbb{P}(X_n-X_0\ge t) = \mathbb{P}(e^{\theta(X_n-X_0)}\ge e^{\theta t})\le e^{-\theta t} \mathbb{E}[e^{\theta(X_n-X_0)}]$

$= e^{-\theta t} \mathbb{E}[\mathbb{E}[e^{\theta((X_n-X_{n-1}) +X_{n-1}-X_0)} | \mathcal{F}_{n-1}]]$

$= e^{-\theta t} \mathbb{E}[e^{\theta(X_{n-1}-X_0)} \mathbb{E}[e^{\theta(X_n-X_{n-1})}|\mathcal{F}_{n-1}] ],$

and our preliminary result allows us to control this inner expectation

$\le e^{-\theta t} e^{\theta^2c_n^2/2} \mathbb{E}[e^{\theta(X_{n-1}-X_0)}].$

So now we can apply this inductively to obtain

$\mathbb{P}(X_n-X_0\ge t) \le e^{-\theta t+ \theta^2 \sum_{i=1}^n c_i^2}.$

Finally, as usual in such an argument, we need to choose a sensible value of the free parameter $\theta$, and naturally we want to choose it to make this RHS as small as possible, which is achieved when $\theta = \frac{t}{\sum_{i=1}^n c_i^2}$, and leads exactly to the statement of the inequality.

Applications

Unsurprisingly, we can easily apply this to the process of partial sums of IID random variables with mean zero and bounded support, to recover a Chernoff bound.

A more interesting example involves revealing the state (ie open or closed) of the edges of an Erdos-Renyi graph one at a time. We need to examine some quantitative property of the graph which can’t ever be heavily influenced by the presence or non-presence of a single given edge. The size of the largest clique, or the largest cut, are good examples. Adding or removing an edge can change these quantities by at most one.

So if we order the edges, and let the filtration $\mathcal{F}_k$ be generated by the state of the first k edges in this ordering, then $X_k=\mathbb{E}[\text{max cut}| \mathcal{F}_k]$ is a martingale. (A martingale constructed backwards in this fashion by conditioning a final state on a filtration is sometimes called a Doob martingale.) Using A-H on this shows that the deviations from the mean are of order $\sqrt{N}$, where N is the size of the graph. In the sparse case, it can be justified fairly easily that the maximum cut has size $\Theta(N)$, since for example there will always be some positive proportion of isolated vertices. However, accurate asymptotics for the mean of this quantity seem (at least after a brief search of the literature – please do correct me if this is wrong!) to be unknown. So this might be an example of the curious situation where we can control the deviations around the mean better than the mean itself!

Beyond bounded increments

One observation we might make about the proof is that it is tight only if all the increments $X_i-X_{i-1}$ are supported on $\{-c_i,+c_i\}$, which is stronger than demanding that the absolute value is bounded. If in fact we have $X_i-X_{i-1}\in[-d_i,c_i]$ almost surely, then, with a more detailed preliminary lemma, we can have instead a bound of $\exp\left( -\frac{2t^2}{\sum_{i=1}^n (c_i+d_i)^2} \right)$.

While it isn’t a problem in these examples, in many settings the restriction to bounded increments is likely to be the obstacle to applying A-H. Indeed, in the technical corner of my current research problem, this is exactly the challenge I faced. Fortunately, at least in principle, all is not necessarily lost. We might, for example, be able to establish bounds $(c_i)$ as described, such that the probability that any $|X_i-X_{i-1}|$ exceeds its $c_i$ is very small. You could then construct a coupled process $(Y_i)$, that is equal to $X_i$ whenever the increments are within the given range, and something else otherwise. For Y to fit the conditions of A-H, the challenge is to ensure we can do this such that the increments remain bounded (ie the ‘something else’ also has to be within $[-c_i,c_i]$ ) and also that Y remains a martingale. This total probability of a deviation is bounded above by the probability of Y experiencing that deviation, plus the probability of Y and X decoupling. To comment on the latter probability is hard in general without saying a bit more about the dependence structure in X itself.