# Convex ordering on Galton-Watson trees

This blog was recently revived, via a post about convex ordering and its relevance to the problem of sampling with and without replacement that forms part of the potpourri of related results all sometimes referred to as Hoeffding’s inequality.

The previous post had been lying almost-complete but dormant for over two years. I revisited it because of a short but open-ended question about trees posed in our research group meeting by Serte Donderwinkel, one of our high-flying doctoral students.

Simplified Question:

For a Galton-Watson tree, can one obtain upper bounds in probability on the height of the tree, uniformly across all offspring distributions with mean $\mu$?

Note that in this setting, it is helpful to have in mind the classical notation $(Z_0,Z_1,Z_2,\ldots)$ for a Galton-Watson process, where typically $Z_0=1$, and $Z_{n+1}$ is the sum of $Z_n$ IID copies of the offspring distribution. Then we have

$\mathbb{P}(\mathrm{height}(\mathcal{T}) < k) = \mathbb{P}(Z_k=0).$

1) Subcritical case. When $\mu<1$, we certainly have $\mathbb{P}(Z_k>0)\le \mathbb{E}[Z_k]=\mu^k$.

Furthermore, if we’re studying all such offspring distributions, this is the best possible upper bound, by considering the offspring distribution given by $Z_1=1$ with probability $\mu$ and zero otherwise.

2) In the critical or supercritical case, $\mu\ge 1$ it is possible that the height is infinite with probability one.

So neither case is especially interesting for now.

Refined question:

What if instead we aren’t trying to obtain a bound uniformly across all offspring distributions with given mean $\mu$, but instead across a subset $\mathcal{X}$ of these distributions? How do we determine which distribution in $\mathcal{X}$ maximises the probability of reaching height k?

This is the question Serte was asking in our group meeting, in the setting where $\mu=1+o(1)$ and the height k has a particular scaling. Also, as far as I understand, the approach outlined in this post didn’t provide strong enough bounds in this particular context. Happily, Serte has recently tied up all the corners of this project concerning the supercritical Galton-Watson forest, and interested readers can find her preprint here on the Arxiv.

Nonetheless the interpretation via convex ordering feels perfect for a blog post, rather than being lost forever.

Convex ordering for offspring distributions

The main observation is that given two offspring distributions X and Y, such that $X\le_{cx} Y$ (which recall means that the means are the same but X is more concentrated) then a number of distributions associated to the Galton-Watson trees for X and Y also satisfy convex ordering relations.

As a warm-up, and because it was the original genesis, we first study heights. We will use the notation

$(Z_0^X,Z_1^X,Z_2^X,\ldots), (Z_0^Y,Z_1^Y,Z_2^Y,\ldots),$

to denote the two Galton-Watson processes. We shall compare $\mathbb{P}(Z^X_k=0)$ and $\mathbb{P}(Z^Y_k=0)$. If we write $\delta_0(\cdot)$ for the function defined on the non-negative integers such that

$\delta_0(0)=1,\quad \delta_0(n)=0,\,n\ge 1$,

it holds that $\delta_0(\cdot)$ is convex. In particular, if $X\le_{cx}Y$, then $\mathbb{E}[\delta_0(X)]\le \mathbb{E}[\delta_0(Y)]$, which exactly says that

$\mathbb{P}(Z^X_1 = 0)\le \mathbb{P}(Z^Y_1 = 0).$

We can then prove that $\mathbb{P}(Z^X_k=0)\le \mathbb{P}(Z^Y_k=0)$ by induction on $k\ge 1$. Note that $\mathbb{P}(Z^X_k=0)^n$ is a convex function of n, regardless of the value of this probability, and so we have

$\mathbb{P}(Z^X_{k+1}=0) = \mathbb{E}\left[ (\mathbb{P}(Z^X_k=0))^X\right] \le \mathbb{E}\left[(\mathbb{P}(Z^X_k=0))^Y\right].$

By the induction hypothesis, this final quantity is at most

$\mathbb{E}\left[(\mathbb{P}(Z^Y_k=0))^Y\right] = \mathbb{P}(Z^Y_{k+1}=0)$.

In conclusion, we have shown that $\mathbb{P}(Z^X_k=0)\le \mathbb{P}(Z^Y_k=0)$ holds for all k, and thus

$\mathrm{height}(\mathcal{T}^X) \ge_{st} \mathrm{height}(\mathcal{T}^Y).$

To return to the original context, suppose we have a large class of offspring distributions $\mathcal{Y}$ and a subclass $\mathcal{X}\subseteq \mathcal{Y}$ such that for all $Y\in\mathcal{Y}$, there exists $X\in \mathcal{X}$ such that $X\le_{cx} Y$. Then one can obtain uniform bounds on the heights of Galton-Watson trees with offspring distributions drawn from $\mathcal{Y}$ by checking those generated from distributions in $\mathcal{X}$ (which is particularly convenient if, for example, $\mathcal{X}$ is finite).

Convex ordering of generation sizes

The above argument solves the original problem, but brushes over the natural question: is it true that $Z^X_k \le_{cx} Z^Y_k$?

The answer is yes. Here’s a proof.

This follows from the following general statement:

Lemma 1: Suppose $X\le_{cx} Y$ are non-negative valued RVs and the non-negative integer valued RVs $M,N$ also satisfy $M \le_{cx} N$. Then

$X_1+\ldots+X_M \le_{cx} Y_1+\ldots Y_N,$

where $X_1,X_2,\ldots$ are IID copies of X and, independently, $Y_1,Y_2,\ldots$ are IID copies of Y.

Lemma 2: Suppose $W_1\le_{cx}Z_1$ and $W_2\le_{cx} Z_2$, and the four random variables are independent. Then $W_1+W_2\le_{cx}Z_1+Z_2$.

Proof of Lemma 2: First, note that for any random variable X, and convex function f

$\mathbb{E}\left[f(Z+x)\right]$ is a convex function of x.

(Indeed, this holds since “f(z+x) is convex” holds for every z, and any definition of convex will pass to the expectation.)

Now we can attack the lemma directly, we may write

$\mathbb{E}\left[ f(W_1+W_2)\right]=\mathbb{E}\left[\, \mathbb{E}[f(W_1+W_2) \mid W_2 ] \,\right] \le \mathbb{E}\left[\, \mathbb{E}[f(W_1+Z_2)\mid Z_2 ] \, \right].$

But then for any $z_2$, we know $f(\cdot+z_2)$ is convex, so $\mathbb{E}[f(W_1+z_2)]\le \mathbb{E}[f(Z_1+z_2)]$, and it follows that

$\mathbb{E}\left[ f(W_1+W_2)\right]\le \mathbb{E} \left[ f(Z_1+Z_2)\right],$

which proves the lemma.

Corollary 3: When $W_1,\ldots,W_m, Z_1,\ldots,Z_m$ are independent, and satisfy $W_i \le_{cx} Z_i$, then we have $W_1+\ldots+W_m\le_{cx} Z_1+\ldots+Z_m$.

Proof of Lemma 1: Note that

$\mathbb{E}\left[ f(X_1+\ldots+X_M)\mid M=n\right] \le \mathbb{E}\left[ f(Y_1+\ldots+Y_N)\mid N=n\right],$

follows from Corollary 3. So a useful question to consider is whether $\mathbb{E}\left[f(Y_1+\ldots+Y_n)\right]$ (*) is a convex function of n?

Denote this quantity by $F(n)$. To check convexity of a function defined on the integers, it suffices to verify that $F(n+1)-F(n)\ge F(n)-F(n-1)$.

There is a canonical coupling between the RVs used to define all of $F(n-1),F(n),F(n+1)$, but it will be convenient to adjust the coupling, and write:

$F(n+1)-F(n)= \mathbb{E}\left[ f(Y_1+\ldots+Y_n + Y^*) - f(Y_1+\ldots+Y_n)\right],$

$F(n)-F(n-1)=\mathbb{E}\left[f(Y_1+\ldots+Y_{n-1}+Y^*) - f(Y_1+\ldots+Y_{n-1})\right],$

where $Y^*$ is a further independent copy of $Y$. But note that for any choice $C\ge c$ and $y\in \mathbb{R}$,

$f(C+y) - f(C) - f(c+y) + f(c)\ge 0.$ (*)

(Essentially, this says that the ‘chord’ of f on the interval $[c,C+y]$ lies above the chord on interval $[C,c+y]$ or $[c+y,C]$, which some people choose to call Karamata’s inequality, but I think is more helpful to think of as part of the visual definition of convexity.)

In any case, setting $y=Y^*, c=Y_1+\ldots+Y_{n-1}, C=Y_1+\ldots+Y_n$ and taking expectations, we obtain

$\mathbb{E}\left[ f(Y_1+\ldots+Y_n+Y^*) - f(Y_1+\ldots+Y_n)\right.$

$\left.- f(Y_1+\ldots+Y_{n-1}+Y^*) + f(Y_1+\ldots+Y_{n-1})\right]\ge 0,$

as required. So $F(n)$ is convex. We may now finish off as

$\mathbb{E}\left[ X_1+\ldots+X_M\right] = \mathbb{E}\left[ \,\mathbb{E}[X_1+\ldots+X_M\mid M]\,\right] \le \mathbb{E}\left[\, \mathbb{E}[Y_1+\ldots+Y_M\mid M]\,\right] = \mathbb{E}[f(M)]\le \mathbb{E}[f(N)] = \mathbb{E}[Y_1+\ldots+Y_N],$

completing the proof of Lemma 1.

• The analysis in this post is not sufficient to study the total population sizes of two Galton-Watson trees generated by X and Y. Note that in Lemma 2, it is important that the random variables are independent. Otherwise, we could, for example, consider $\mathbb{E}[X]=\mathbb{E}[Y]=0$ with $X\le_{cx}Y$ but clearly it should not hold that $X_1+X_2 \le_{cx} Y + (-Y) = 0$. So for total population size, since $(Z^X_k,\,k\ge 1)$ are not independent, an alternative approach would be required.
• A further characterisation of convex ordering is given by Strassen’s theorem [Str65], which is touched on in the previous post, and to which I may return to in a future post on this topic. This may be a more promising avenue for established a convex ordering result on total population size.
• Lemma 1 requires that X,Y are non-negative. Note that during the argument we set $y=Y^*, c=Y_1+\ldots+Y_{n-1}, C=Y_1+\ldots+Y_n$, and when we relax the non-negative support condition, it is no longer guaranteed that $C\ge c$, which is crucial for the step which follows.
• In a recent article in ECP addressing Lemma 1 by a different method, Berard and Juillet [BJ20] provide a simple example showing that the non-negative assumption is genuinely necessary. Consider the random variable $\tau\in \{0,2\}$ with equal probability so $1\le_{cx} \tau$. But then, taking both X and Y to be simple random walk on $\mathbb{Z}$, we do not have $S_1\le_{cx}S_{\tau}$.

References

[BJ20] – Berard, Juillet – A coupling proof of convex ordering for compound distributions, 2020

[Str65] – Strassen – The existence of probability measures with given marginals, 1965

# Lecture 9 – Inhomogeneous random graphs

I am aiming to write a short post about each lecture in my ongoing course onÂ 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 aÂ type function $V\to \{1,\ldots,k\}$. We will refer to a $k\times k$ symmetric matrix with non-negative entries as aÂ 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 aÂ 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 asÂ 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 onÂ 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 theÂ 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 sameÂ 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 aÂ 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:Â 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:Â 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:Â 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 onÂ 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 inÂ rootedÂ 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 $(G_n,\rho_n)$Â 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 theÂ 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 theÂ 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 $P_{2r}$,Â 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 4 – Hitting time theorem

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

This lecture consisted of revision of the most relevant theory of Galton-Watson trees, with a focus on the case where the offspring distribution is Poisson, since, as we have seen in previous lectures, this is a strong candidate to approximate the structure of $G(n,\lambda/n)$. It makes sense to cover the theory of the trees before attempting to make rigorous the sense of approximation.

Given a Galton-Watson tree T, it is natural to label the vertices in a breadth-first order as $\varnothing=v_1,v_2,\ldots,v_{|T|}$. This is easiest if we have constructed the Galton-Watson tree as a subset of the infinite Ulam-Harris tree, where vertices have labels like (3,5,17,4), whose parent is (3,5,17). If this child vertex is part of the tree, then so are (3,5,17,1), (3,5,17,2), and (3,5,17,3). This means our breadth-first order is canonically well-defined, as we have a natural ordering of the children of each parent vertex.

Note:Â one advantage of using breadth-first order rather thanÂ depth-first order (which corresponds to the usual dictionary, orÂ lexicographic ordering of the labels) is that if the tree is infinite, we don’t explore all of it during a depth-first search. (In the sense that there exist vertices which are never given a finite label.) For breadth-first search, a similar problem arises precisely when some vertex has infinitely many children. For a conventional Galton-Watson tree, the latter situation is much less of a problem than the infinite total population problem, which happens with positive probability whenever $\mu=\mathbb{E}[X]>1$.

Anyway, given the depth-first order, one can consider anÂ exploration process $S_0,S_1,S_2,\ldots,S_{|T|}$ given by

$S_0=1,\quad S_i=S_{i-1}+(X_i-1),$

where $X_i$ is the number of children of $v_i$. In this way, we see that

$S_i=\big| \Gamma(v_1)\cup\ldots\cup\Gamma(v_i)\backslash \{v_1,\ldots,v_i\}\big|,\quad i\ge 1,$

records the number of vertices in someÂ stack containing those which we have ‘seen but not explored’. Some authors prefer to start from 0, in which case one ends up with a similar but slightly different interpretation of the ‘stack’, but that’s fine since we aren’t going to define formally what ‘seen’ and ‘explored’ means in this post.

Essentially, we exhaust the vertices of the tree whenever $S_t=0$, and so the condition that $|T|=n$ requires

$S_n=0,\quad S_m\ge 1,\; m=0,1,\ldots,n-1.$

Conveniently, so long as we have avoiding ordering ambiguity, for example by insisting that trees live within the Ulam-Harris tree, we can reconstruct T uniquely from $(S_0,S_1,\ldots,S_{|T|})$.

Furthermore, if T is a Galton-Watson process, then the numbers of children $X_i$ are IID, and so in fact this exploration process is a random walk, and the size of the tree can be recovered as the hitting time of zero.

Note:Â making fully rigorous the argument that children in the GW tree are independent of the breadth-first walk fully rigorous is somewhat technical, and not to be dismissed lightly, though not of principle interest at the level of this topics course. See Proposition 1.5 in Section 1.2 of Le Gall’s notes or Section 1.2.2 of my doctoral thesis for further discussion and argument.

TheÂ hitting time theoremÂ allows us to study the distribution of the hitting time of a random walk whose increments are bounded below by -1, in terms of the distribution of the value of the random walk.

Theorem:Â Let $(S_n,\, n\ge 0)$ be a random walk with $S_0=0$ and IID increments $(X_n,n\ge 1)$ satisfying $\mathbb{P}(X_n\ge -1)=1$. Let $H_{-k}=\inf \left\{n\,:\, S_n=-k\right\}$ be theÂ hitting timeÂ of -k.

Then $\mathbb{P}\big( H_{-k}=n\big) = \frac{k}{n}\mathbb{P}\big(S_n=-k)$.

Commentary:Â there are local central limit theorem estimates and large deviation estimates that allow good control of the probability on the RHS for a rich class of contexts. So at a meta-level, the hitting time theorem allows us to reduce a complicated (though still classical) problem, to a real classical problem, which is particularly helpful when the LHS is a device for capturing relevant information about our random tree model.

# The reflection principle and conditioned RWs

I haven’t published a post about probability for far too long. Several are queued, so perhaps this will be the start of a deluge.

Anyway, with my advisor at Technion, I’m still working on some problems concerning Gaussian random walk subject to some conditioning which is complicated, but in practice (we hope) only mildly different to conditioning the walk to stay positive. Our conditioning at step n depends on some external randomness, but also on the future trajectory of the walk (related to the embedding of the walk in a 2DÂ DGFF), thus ruining the possibility of applying the Markov property in any proof without significant preliminary work.

It seemed worth writing briefly about some of these results in a slightly simpler setting. The goal is to assemble many of the ingredients required to prove a local limit for Gaussian random walk conditioned to stay positive, in a sense which will be clarified towards the end. This is not the best way to go about getting scaling limits (as discussed briefly here, and for which see references [Ig74] and [Bo76]), and it’s probably not the best way to get local limits in the simplest setting, but it’s the method we are currently working to generalise, and follows the outline of [B-JD05], but in much less technical detail.

Probabilities via the reflection principle

We start with Brownian motion. The reflection principle, as described briefly in this post from the depths of history, is a classical technique for studying the maximum of Brownian motion. Roughly speaking, we exploit the fact that $(W_t,t\ge 0)\stackrel{d}=(-W_t,t\ge 0)$, but we then apply this at the hitting time of a particular positive value, using the Strong Markov Property.

Let $S_t=\max_{0\le s\le t}W_s$ be the running maximum of the Brownian motion $W_t$, and $\tau_b$ the hitting time ofÂ b. Then

$\mathbb{P}(S_t\ge b, B_t\le a)=\mathbb{P}(\tau_b

which, by SMP at $\tau_b$ and the reflection invariance of a standard BM, is equal to

$\mathbb{P}(\tau_b

This obviously assumed $b\ge a$, but if we set $b=a$, we find

$\mathbb{P}(S_t\ge b)=\mathbb{P}(B_t>b)+\mathbb{P}(S_t\ge b,B_t\le b)=2\mathbb{P}(B_t\ge b).$

Or, in other words, $S_t\stackrel{d}=|B_t|$.

While we can’t derive such nice equalities in distribution, the reflection principle is robust enough to treat more complicated settings, such as Brownian bridge.

We might want to ask about the maximum of a standard Brownian bridge, but we can be more general, and ask about the maximum of a Brownian bridge with drift (let’s say general bridge here). It’s important to remember that a general Brownian bridge has the same distribution as a linear transformation of a standard Brownian bridge. Everything is Gaussian, after all. So asking whether the maximum of a general Brownian bridge is less than a particular value is equivalent to asking whether a standard Brownian bridge lies below a fixed line. Wherever possible, we make such a transformation at the start and perform the simplest version of the required calculation.

So, suppose we have a bridge B from (0,0) to (t,a), and we want to study $\max_{s\in[0,t]} B_s$. Fix some $b>a$, and work with a standard Brownian motion $W_s$. By a similar argument to before,

$\mathbb{P}(\tau_b\le t, W_t\in[a,a+\mathrm{d}x]) = \mathbb{P}(W_t\in [2b-a-\mathrm{d}x,2b-a]) = \frac{\mathrm{d}x}{\sqrt{2\pi t}}e^{-(2b-a)^2/2t},$

and

$\mathbb{P}(W_t\in[a,a+\mathrm{d}x])=\frac{\mathrm{d}x}{\sqrt{2\pi t}}e^{-a^2/2t}.$

So

$\mathbb{P}(\max_{s\in[0,t]}B_t\ge b) = \exp\left(\frac{-(2b-a)^2 + a^2}{2t}\right).$

Random walk conditioned to stay positive

Our main concern is conditioning to stay above zero. Let $\mathbb{P}_{0,x}^{t,y}$ be some complete if cumbersome notation for a Brownian bridge B from (0,x) to (t,y). Then another simple transformation of the previous result gives

$\mathbb{P}_{0,x}^{t,y}(B_s\ge 0,\,s\in[0,t])=1-\exp\left( \frac{-(x+y)^2 + (x-y)^2}{2t} \right)= 1-\exp\left(-\frac{2xy}{t}\right).$

Then, if $xy\ll t$, we can approximate this by $\frac{2xy}{t}$. (*)

Extend the notation so $\mathbb{P}_{0,x}$ describes Brownian motion started from (0,x). Then integrating over y, gives

$\mathbb{P}_{0,x}(B_s\ge 0,\, s\in[0,t] ) = \frac{x}{t}\mathbb{E}[B_t\vee 0] = \sqrt{\frac{2}{\pi}} \frac{x}{\sqrt{t}}.$

(It might appear that we have integrated the approximation (*) over parts of the range where it is not a valid approximation, but the density of $B_t=\Theta(t)$ vanishes exponentially fast, and so actually it’s not a problem.)

We now want to extend this to random walks. Some remarks:

• We used the Gaussian property of Brownian motion fairly heavily throughout this derivation. In general random walks are not Gaussian, though we can make life easier by focusing on this case.
• We also used the continuity property of Brownian motion when we applied the reflection principle. For a general random walk, it’s hopeless to consider the hitting times of individual values. We have to consider instead the hitting times of regions $\tau_{(-\infty,b]}$, and so on. One can still apply SMP and a reflection principle, but this gives bounds rather than equalities. (The exception is simple random walk, for which other more combinatorial methods may be available anyway.)
• On the flip side, if we are interested in Brownian motion/bridge staying positive, we can’t start it from zero, as then the probability of this event is zero, by Blumenthal’s 0-1 law. By contrast, we can certainly ask about random walk staying positive when started from zero without taking a limit.

A useful technique will be the viewpoint of random walk as the values taken by Brownian motion at a sequence of stopping times. This Skorohod embedding is slightly less obvious when considering a general random walk bridge inside a general Brownian bridge, but is achievable. We want to study quantities like

$\mathbb{P}(S_k\ge 0,\, k=1,\ldots,n \big| S_0=x,S_n=y),$

where for simplicity let’s just take $(S_k,k\ge 0)$ to be a random walk with standard Gaussian increments. It’s possible we might want to take a scaling limit inÂ x andÂ y as functions ofÂ n. But first if we takeÂ x,y fixed, and embed the random walk bridge with these endpoints into the corresponding Brownian bridge with $t\approx n$, we are then faced with the question:

What’s the probability that the Brownian bridge goes below zero, but the embedded RW with n steps does not?

If the Brownian bridge conditioned to go below zero spends time $\Theta_p(n)$ below zero, then for large n it’s asymptotically very unlikely that the n places at which we embed the random walk avoids this set of intervals.

Several technical estimates are required to make this analysis rigorous. The conclusion is that there exists a function $f(x)$ for which $f(x)=x(1+o(1))$ as $x\rightarrow\infty$, such that

$q_n(x,y):=\mathbb{P}(S_k\ge 0,\, k=0,1,\ldots,n \,\big|\, S_0=x,S_n=y) \sim \frac{2f(x)f(y)}{n},$

$\text{and}\quad q_n(x):=\mathbb{P}(S_k\ge 0,\,k=0,1,\ldots,n\,\big|\, S_0=x)\sim \sqrt{\frac{2}{\pi}}\frac{f(x)}{\sqrt{n}}.$

As earlier, the second is obtained from the first by integrating over suitableÂ y. This function $f$ has to account for the extra complications when either end-point is near zero, for which the event where the Brownian motion goes negative without the random walk going negative requires additional care.

Limits for the conditioned random walk

In the previous post on this topic, we addressed scaling limits in space and time for conditioned random walks. But we don’t have to look at the classical Donsker scaling to see the effects of conditioning to stay positive. In our setting, we are interested in studying the distribution of $S_m$ conditional on the event $(S_1\ge 0,S_2\ge 0,\ldots, S_n\ge 0)$, with limits taken in the order $n\rightarrow\infty$Â and then $m\rightarrow\infty$.

(At a more general level, it is meaningful to describe the random walkÂ conditioned on staying positive forever. Although this would a priori require conditioning on an event of probability zero, it can be handled formally as an example of an h-transform.)

As explained in that previous post, the scaling invariance of the Bessel process $W^+$ (which it’s not unreasonable to think of as ‘Brownian motion conditioned to stay non-negative’) suggests that this limit should exist, and be given by the entrance law of $W^+$. But this is hard to extract from a scaling limit.

However, we can use the previous estimate to start a direct calculation.

$\mathbb{P}(S_m\in \mathrm{d}y \,\big|\, S_k\ge 0,\, k=1,\ldots,n) = \frac{q_m(0,y) q_{n-m}(y) \mathbb{P}(S_m\in\mathrm{d}y)}{q_n(0)}.$

Here, we used the Markov property at time m to split the event that $S_m=y$ and the walk stays positive into two time-intervals. We will later take m large, so we may approximate as

$\frac{2f(0)f(y)/m \times \sqrt{\frac{2}{\pi}}f(y)/\sqrt{n-m}\times \mathbb{P}(S_m\in\mathrm{d}y) } { \sqrt{\frac{2}{\pi}}f(0)/\sqrt{n}}\stackrel{n\rightarrow\infty}=\frac{2f(y)^2}{m}\mathbb{P}(S_m\in\mathrm{d}y).$

This final probability emphasises that as $m\rightarrow\infty$ we only really have to consider $y=\Theta(\sqrt{m})$, so set $y=z\sqrt{m}$, and we obtain

$\lim_{n\rightarrow\infty}\mathbb{P}(\frac{S_m}{\sqrt{m}}\in\mathrm{d}z\,\big|\, S_k\ge 0,\,k=1,\ldots,n)$

$\sim \sqrt{m}\cdot\frac{2z^2m}{m}\cdot \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{m}}e^{-z^2/2} = \sqrt{\frac{2}{\pi}}z^2 e^{-z^2/2}.$

This is precisely the entrance law of the 3-dimensional Bessel process, usually denoted $R$. This process is invariant under time-rescaling in the same fashion as Brownian motion. Indeed, one representation of R is as the radial part of a three-dimensional Brownian motion, given by independent BMs in each coordinate. (See [Pi75] for explanation of the relation to ‘BM conditioned to stay non-negative’.) We could complete the analogy by showing that $q_n(x,y)$ converges to the transition density of R as well. (Cf the prelude to Theorem 2.2 of [B-JD05].)

Final remarks

The order of taking limits is truly crucial. We can also obtain a distributional scaling limit at time n under conditioning to stay non-negative up to time n. But then this is the size-biased normal distribution $\sim ze^{-z^2/2}$ (the Rayleigh distribution), rather than the square-size-biased normal distribution we say in this setting. And we can exactly see why. Relative to the normal distribution which applies in the absence of conditioning, we require size-biasing to account for the walk staying positive up to time m,Â and then alsoÂ size-biasing to account for the walk staying positive for the rest of time (or up to n in the $n\rightarrow\infty$ limit if you prefer).

The asymptotics for $q_n(x,y)$ were the crucial step, for which only heuristics are present in this post. It remains the case that estimates of this kind form the crucial step in other more exotic conditioning scenarios. This is immediately visible (even if the random walk notation is rather exotic) in, for example, Proposition 2.2 of [CHL17], of which we currently require a further level of generalisation.

References

[Bo76] – Bolthausen – On a functional central limit theorem for random walks conditioned to stay positive

[B-JD05] – Bryn-Jones, Doney – A functional limit theorem for random walk conditioned to stay non-negative

[CHL17] – Cortines, Hartung, Louidor – The structure of extreme level sets in branching Brownian motion

[Ig74] – Iglehart – Functional central limit theorems for random walks conditioned to stay positive

[Pi75] – Pitman – One-dimensional Brownian motion and the three-dimensional Bessel process

# Kernels of critical graph components

This post is motivated by G(N,p), the classical Erdos-RenyiÂ 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 theÂ 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 aboutÂ 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 componentsÂ andÂ 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,Â 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 theÂ 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 theÂ 2-core C(G)Â 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 treesÂ ‘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, theÂ k-core is the maximal induced subgraph where all degrees are at least k. TheÂ 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 theÂ 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 postÂ 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:Â 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 biasingÂ 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 theÂ 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:Â For any H a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges as before,

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

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 graphsÂ 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 –Â 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 inÂ 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. HenceÂ 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 aÂ 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 merelyÂ 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 wordÂ 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]Â 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), …Â  (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)Â 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 andÂ 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] –Â 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 anÂ 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 decayingÂ exponentially in n, whereas for c=1, the branching process is critical, and the probability decaysÂ 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 treeÂ conditioned to survive. It’s a beautiful result that this interpretation can be formalised as a local limit. The limitingÂ 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 declaredÂ 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 weightÂ $q\in (0,\infty)$, 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 moreÂ components, and q<1 favoursÂ fewerÂ 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 $p\rightarrow 0$, butÂ also taking $\frac{q}{p}\rightarrow 0$. So the resulting graph asymptotically ‘wants to have as few edges as possible, butÂ 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 $G=K_n$.

# Real Trees – Root Growth and Regrafting

Two weeks ago in our reading group meeting, Raphael told us about Chapter Five which introduces root growth and regrafting. One of the points of establishing the Gromov-Hausdorff topology in this book was to provide a more natural setting for a discussion of tree-valued processes. Indeed in what follows, one can imagine how to start the construction of a similar process for the excursions which can be used to encode real trees, involving cutting off sub-excursions above one-sided local minima, then glueing them back in elsewhere. But taking account of the equivalence structure will be challenging, and it is much nicer to be able to describe cutting a tree in two by removing a single point without having to worry about quotient maps.

We have seen in Chapter Two an example of a process defined on the family of rooted trees with n labelled vertices which has the uniform rooted tree as an invariant distribution. Given a rooted tree with root p, we choose uniformly at random a vertex p’ in [n] to be the new root. Then if p’=p we do nothing, otherwise we remove the unique first edge in the path from p’ to p, giving two trees. Adding an edge from p to p’ completes the step and gives a new tree with p’ as root. We might want to take a metric limit of these processes as n grows and see whether we end up with a stationary real tree-valued process whose marginals are the BCRT.

To see non-trivial limiting behaviour, it is most interesting to consider the evolution of a particular subtree (which includes the root) through this process. If the vertex chosen for cutting lies in our observed subtree, then the subtree undergoes a prune and regraft operation. On the other hand, if the vertex chosen for cutting does not lie in the subtree, then we do not see any effect of the pruning, except the addition of a new vertex below the original root, which becomes the new root. So essentially, from the point of view of our observed subtree, the root is growing.

Now we can think about interpreting the dynamics of a natural limit process acting on real trees. The key idea is that we don’t change the set on which the tree is supported much, but instead just change the metric. In particular, we will keep the original tree, and add on length at unit rate. Of course, where this length gets added on entirely determines the metric structure of the tree, but that doesn’t stop us giving a simple ‘name’ for the extra length. If we consider a process $X^T$ starting from a particular finite subtree T, then at time t, the tree $X^T_t$ is has vertex set $T \coprod (0,t]$. (Finite subtree here means that it has finite total length.)

Root regrafting should happen at a rate proportional to the total length of the current observed tree. This is reasonable since after all it is supported within a larger tree, so in the discrete case the probability of a prune-regrafting event happening within a given observed subtree is proportional to the number of vertices in that subtree, which scales naturally as length in the real tree limit. It turns out that to get unit rate root growth with $\Theta(1)$ rate prune-regrafting, we should consider subtrees of size $\sqrt{n}$ within a host tree of size n as $n\rightarrow\infty$. We also rescale the lengths by $\frac{1}{\sqrt{n}}$, and time by $\sqrt{n}$ so we actually see prune-regraft events.

Furthermore, if the subtree is pruned, the location of the pruning is chosen uniformly by length of the current observed subtree. So we can view the pruning process as being driven by a Poisson point process with intensity given by the instantaneous length measure of the tree, which at time t has vertex set $T\coprod (0,t]$. It will turn out to be consistent that there is a ‘piecewise isometry’ for want of a better phrase between the metric (and thus length measure) on $X^T_t$ and the canonical induced measure on $T\coprod (0,t]$, so we can describe the instances and locations of the pruning events via a pair of PPPs. The first is supported on $T \times [0,\infty)$, and the second on $\{(t,x): 0 \le x \le t\}$, since we only ‘notice’ pruning at the point labelled x if the pruning happens at some time t after x was created.

If we start from a compact tree T, then the total intensity of this pair is finite up to some time t, and so we have a countable sequence $\tau_0=0<\tau_1<\tau_2<\ldots$ of times for pruning events. It is easy to describe (but a bit messy to notate) the evolution of the metric between these pruning times. Essentially the distance between any pair of points in the observed tree at time $\tau_m$ with root $\rho_{\tau_m}$ is constant between times $\tau_m,\tau_{m+1}$, and new points are added so that the distance between $\rho_{\tau_m}$ and any new point $a\in(\tau_m,\tau_{m+1}]$ is $a-\tau_m$, and everything thing else follows from straightforward consideration of geodesics.

When a pruning event happens at point $x_m$ at time $\tau_m$, distances are preserved within the subtree above $x_m$ in $X^T_{\tau_m -}$, and within the rest of the tree. Again, an expression for the cross distances is straightforward but requires a volume of notation not ideally suited to this medium.

The natural thing to consider is the coupled processes started from different subtrees (again both must contain the original root) of the same host tree. Say $T^1,T^2\le T$, then it is relatively easy to check that $X^{T^1}_t,X^{T^2}_t \le X^T_t \,\forall t$, when we drive the processes by consistent coupled Poisson processes. Furthermore, it is genuinely obvious that the Hausdorff distance between $X^{T^1}_t,X^{T^2}_t$, here viewed as compact subsets of $(X^T_t, d^T_t)$ remains constant during root growth phase.

Less obvious but more important is that the Hausdorff distance decreases during regrafting events. Suppose that just before a regrafting event, the two subtrees are T’ and T”, and the Hausdorff distance between them is $\epsilon$. This Hausdorff distance is with respect to the metric on the whole tree T. [Actually this is a mild abuse of notation – I’m now taking T to be the whole tree just before the regraft, rather than the tree at time 0.]

So for any $a\in T'$, we can choose $b\in T''$ such that $d_T(a,b)\le \epsilon$. This is preserved under the regraft unless the pruning point lies on the geodesic segment (in T) between a and b. But in that case, the distance between a and the pruning point is again at most $\epsilon$, and so after the regrafting, a is at most $\epsilon$ away from the new root, which is in both subtrees, and in particular the regrafted version of T”.

This is obviously a useful first step on the path to proving any kind of convergence result. There are some technicalities which we have skipped over. It is fairly natural that this leads to a Markov process when the original tree is finite, but it is less clear how to define these dynamics when the total tree length is infinite, as we don’t want regrafting events to be happening continuously unless we can bound their net effect in some sense.

Last week, Franz showed us how to introduce the BCRT into matters. Specifically, that BCRT is the unique stationary distribution for this process. After a bit more work, the previous result says that for convergence properties it doesn’t matter too much what tree we start from, so it is fine to start from a single point. Then, the cut points and growth mechanism corresponds very well to the Poisson line-breaking construction of the BCRT. With another ‘grand coupling’ we can indeed construct them simultaneously. Furthermore, we can show weak convergence of the discrete-world Markov chain tree algorithm to the process with these RG with RG dynamics.

It does seem slightly counter-intuitive that a process defined on the whole of the discrete tree converges to a process defined through subtrees. Evans remarks in the introduction to the chapter that this is a consequence of having limits described as compact real trees. Then limitingly almost all vertices are close to leaves, so in a Hausdorff sense, considering only $\sqrt{n}$ of the vertices (ie a subtree) doesn’t really make any difference after rescaling edge lengths. I feel I don’t understand exactly why it’s ok to take the limits in this order, but I can see why this might work after more checking.

Tomorrow, we will have our last session, probably discussing subtree prune-and-regraft, where the regrafting does not necessarily happen at the root.