# Parking on a ring, linear hashing

I’ve spent most of my doctorate trying to analyse how adding destructive dynamics affects the behaviour of a particular random growth process, the classical random graph. In this post I’m going to talk about another random growth process, which is slightly less natural, but for which one can show some similar qualitative properties.

The model, and the additive coalescent

Consider m places arranged in a circle, and for consistency of analogy we think of these as parking spaces. Some number n of cars will arrive one at a time. Each car will arrive at a space chosen uniformly at random. If it is empty they will park in it, otherwise they will look clockwise until they find an empty space, and park there. For now we are only interested in growth, so we assume cars never leave. We are interested in the sizes of blocks of consecutively parked cars.

The reason to consider this slightly unnatural statement is its equivalence to the problem of hashing with linear probing, apparently a key topic in computer science, which I won’t pretend that I know anything about. In any case, it’s a nice model, and it seems reasonable that it would have a basis in more realistic search algorithms.

So, how does the sequence of sizes of blocks of consecutively parked cars grow? Well, given the sequence of block sizes, it is reasonably easy to convince yourself that the order of the blocks around the circle is uniformly random, and the number of empty spaces between adjacent blocks is also uniformly random.

Assume for now that there are at least three blocks. A block of size x can merge with a block of size y with the arrival of the next car only if the blocks are adjacent, with exactly one empty space between them. The chance of this is uniform among all pairs of blocks. Now suppose this is the case, and that the block of size y lies clockwise from the block of size x. Then they will merge precisely if the next car arrives at any of the x occupied spaces in that block, or at the empty space between the pair of blocks. This has probability $\frac{x+1}{m}$. There’s also the opposite ordering to consider, where the block of size x lies clockwise from the other. The total probability of this merge $\{x,y\}\mapsto \{x+y+1\}$ is therefore proportional to (x+y+2).

So the process of block sizes looks a bit like the additive coalescent, at least for large blocks. This is in contrast to the random graph process, where the sequence of component sizes behaves exactly like a multiplicative coalescent, where blocks merge at a rate proportional to the product of their sizes.

Asymptotics

As in the random graph process, it’s interesting to ask roughly how large the largest block will be in such a configuration. Pittel [3] considers the case where the number of empty places $\ell = m-n \approx \beta m$, for some $\beta\in (0,1)$.

A less interesting model would be to choose the positions of the n cars uniformly at random. But then the size of a block is roughly geometric with parameter $\beta$, and there are $\Theta(m)$ blocks with high probability. Relatively straightforward calculations in extreme value theory suggest that the largest block is likely to have size on the order of $\log m$ in this setting.

Of course, the actual model is slightly more complicated, because the size of a block is self-reinforcing, since larger blocks are more likely to grow than smaller blocks. However, we can still get somewhere with naïve estimates. Let’s label the places clockwise. Then in order for there to be a block starting at 0 and stretching beyond $\alpha \log m$, a necessary condition is that at least $\alpha \log m$ cars arrive at those places. The number of cars which arrive at those places is binomial, since there are n cars, and each arrives at a place chosen uniformly, and independently of the other cars. So this event corresponds to

$\mathrm{Bin}(n,\frac{\alpha \log m}{m}) \ge \alpha \log m.$

Then, since $n\approx (1-\beta)n$, this event corresponds approximately to

$\mathrm{Po}((1-\beta)\alpha \log m) \ge \alpha \log m.$

The probability that a Poisson RV is at least a constant multiple larger than its mean decays exponentially with the mean, hence in this case the probability is asymptotically some negative power of m, depending on the value of $\alpha$. But there are $O(m)$ possible places for such a block to start, so whether we can apply a union bound usefully or not depends on whether the power of m is strictly less than -1.

Since all of this depends on $\alpha$, it is reasonable that everything is fine, and the largest block does have size at least $\alpha \log m$ when $\alpha$ is small, and very unlikely when $\alpha$ is large. This heuristic argument fits with Pittel’s theorem. Indeed, his result shows much stronger concentration: that the fluctuations of the size of the largest block are O(1).

Critical regime and empirical processes

The following is a paraphrase of the introduction and some methods from [2].

Obviously, once m=m cars have arrived, there’s no room for manoeuvre and definitely all the places are taken in one giant block. But it’s not obvious in general what scaling for the number of gaps will give rise to giant blocks of $\Theta(m)$ cars.

As for the random graph, we can find a process similar to the exploration process of a (random) graph which encodes much of the information we care about. Let $Y_k$ be the number of cars which arrive at place k. So the sum of the $Y_k$s will be n, the total number of cars. Now consider the process

$C_0=0, \ldots, C_{k+1}=C_k + Y_{k+1}-1.$

A block has the property that the number of arrivals within that set of places is equal to the number of places. So every time this *empirical process* C drops below its previous running minimum, this indicates the end of a block. To make this equivalence precise, we need to be a bit careful about where we start counting. It works exactly if we start at the beginning of a block. If not, it might introduce some unwanted divisions within the first block.

What we have is a process that looks roughly like a random walk that is constrained to pass through the point (m,n-m), which is equal to (m,-l). Even if we aren’t totally precise about how this is like a random walk, we would expect to see Brownian fluctuations after rescaling. Indeed, we might expect to see a Brownian bridge added to a deterministic linear function with negative gradient. But this is only meaningful if the random part is at least as large as the deterministic part, and since the fluctuations have order $\sqrt{m}$, if l is much larger than this, the rescaled empirical process is essentially deterministic, so we won’t see any macroscopic excursions above the minimum.

If l is substantially smaller than $\sqrt{m}$, then there is no real difference between (m,-l) and (m,0), and what we see is just a Brownian bridge. At this point, where we choose to start the process is actually important. If we were to start it at the minimum of the Brownian bridge instead, we would have seen a Brownian excursion, which corresponds to one block occupying (almost) all of the places.

Unsurprisingly, the story is completed by considering $\ell=\Theta(\sqrt{m})$, where the rescaled empirical process looks like a slanted Brownian bridge, that is Brownian motion conditioned to pass through $(1,-\frac{\ell}{\sqrt{m})$. There isn’t an obvious fix to the question of where to start the process, but it turns out that the correct way is now adding a Brownian excursion onto the deterministic linear function with gradient $- \frac{\ell}{\sqrt{m}}$. It’s now reasonable that the excursions above the minimum should macroscopic.

This scaling limit works dynamically as well, where the same Brownian excursion is used for different gradients of the deterministic line, corresponding to $\ell$ moving through the critical window $m-\Theta(\sqrt{m})$. Finally, a direction to Bertoin’s recent paper [1] for the model with an additional destructive property. Analogous to the forest fire, blocks of cars are removed at a rate proportional to their size (as a result, naturally, of ‘Molotov cocktails’…). Similar effects of self-organised criticality are seen when the rate of bombs is scaled appropriately.

References

[1] – Bertoin – Burning cars in a parking lot (paper / slides)

[2] – Chassaing + Louchard – Phase transition for parking blocks, Brownian excursion and coalescence (arXiv)

[3] – Pittel – Linear probing: the probable largest search time grows logarithmically with the number of records

# Isolated Vertices and the Second-Moment Method

It’s back-to-school day or week for much of the UK. I’m sure for many, this brings resolutions of better work habits, and while I could always use some of those too as I try to finish my thesis, I also want to start blogging again when possible.

Right now, I’m in Haifa for the Mostly Markov Mixing summer school and workshop hosted at the Technion. The talks have been interesting so far, and the environment stimulating both for the discussion of new problems, and getting work done on existing research.

I wrote much of this post a while ago, after some of the UK olympiad students asked me to tell them about second-moment methods and I politely declined. I was reminded of this by an interesting problem introduced by Elchanan Mossel in the first lecture of his series. I’ll start with this.

Suppose we consider a symmetric inhomogeneous random graph with two types. That is, we divide the vertices into two equally-sized classes, and we connect vertices in the same class with probability p, and vertices in different classes with probability q, all independently. The question is: if we can see the resulting graph structure, with what accuracy can we recover the class division? [Note: this setup with symmetry between the types can be called the block model.]

This is a hard problem, and got us all thinking a great deal. In the most relevant regime, even the most sophisticated techniques do not allow us to identify the partition perfectly with high probability as the number of vertices goes to infinity. For now, though, I only want to use the most uninteresting regime as a starting point for the rest of this post. EM asked: what happens if we take sparse scaling, that is $p,q=\Theta(1/N)$, where N is the number of vertices? Do we have a chance to identify the classes correctly?

Well in this case the answer is easy, because it is ‘no’, and the reason is that in sparse random graphs, there is a positive proportion of isolated vertices. In particular, there is a positive proportion of isolated vertices of each type. And so, when we see an isolated vertex, for large N we can specify accurately the distribution of each type given this extra information, but in doing so, we admit that we certainly cannot partition the isolated vertices into their classes. So for the rest of the talk, EM focused on the regime where the random graph is connected with high probability.

There are two ideas with worthwhile and simple content here. Firstly, the fact that such a random graph has a positive proportion of isolated vertices. I am finishing off a result in a model where the base structure is the inhomogeneous random graph, and have just now proved a short lemma showing exactly this in a slightly more complicated context. It’s a good example of the second-moment method. Secondly, implicit in the final statement of the previous paragraph is that the absence of isolated vertices and connectivity are roughly equivalent in a random graph. This is something I’ve talked about briefly before, and in the interests of keeping the resolution and actually finishing this post, I won’t talk about it here.

Earlier this year, I gave a short talk by request to the UK olympiad students about first-moment methods. In particular, they were hoping that they might occasionally want to apply such approaches to the sort of combinatorics problems you encounter in olympiads. Typically, the best examples in this setting involve demonstrating the existence of a set without certain bad properties, by showing that the probability that all bad properties hold simultaneously is strictly less than one.

The students asked why I wasn’t talking about second-moment methods. Firstly, the models (such as this one) which are the best examples are less familiar to the students, and are really rooted in the randomness. They can’t easily be turned into a combinatorics problem. Secondly, looking for lower bounds in probability suggests we are aiming for convergence in probability, rather than convergence in expectation, and this is a distinction that is unlikely to be appreciated before some undergraduate probability courses have been taken.

Anyway, we want to show that the proportion of isolated vertices in $G(n,c/n)$ is bounded below in probability as $n\rightarrow\infty$. All the content of the argument is seen in this classical Erdos-Renyi setting. Any inhomogeneous example will merely demand extra notation.

So, first we deal with the expected number of isolated vertices. The event that a given vertex v is isolated demands that the n-1 edges potentially incident to v do not appear. The probability of this is $(1-\frac{c}{n})^{n-1} \rightarrow e^{-c}$. Thus the expected number of such vertices is $ne^{-c}$. We now want to show that the fluctuations (standard deviation if you will) of this quantity are small relative to its mean. To bound the variance, we look at pairs of vertices, v and w. Note that the events {v isolated} and {w isolated} are not independent, since knowing that v is isolated tells us that the edge from v to w is not present, which slightly increases the chance of w being isolated. (For a very concrete example, think of the case n=2.)

However, in a large graph, the events are almost independent. That’s a statement which we feel is true, but we need to quantify, so we ask for the probability that {v and w isolated}. This happens precisely if none of the edges incident to either v or w are present. There are 2n-3 such edges, and so the probability of this is $(1-\frac{c}{n})^{2n-3}\rightarrow e^{-2c}$. So now we can write

$\mathbb{E}\left[ \mathbf{1}(1\text{ isolated})+\ldots+\mathbf{1}(n\text{ isolated})\right]^2= \sum_{k=1}^n \mathbb{E}\mathbf{1}(k\text{ isolated})^2 + 2\sum_{j\ne k} \mathbb{E}\mathbf{1}(j,k\text{ isolated}).$

An indicator function takes the values 0 and 1, and so in the first sum we can remove the square sign to leave us with the expected number of isolated vertices. Secondly, the vertices are exchangeable and so we can replace each summand with the value we have already established for v and w. We can now calculate the variance of the number of isolated vertices, and we see that it is $o(n^2)$. With a little bit more care about the limits in n, we can check it is actually O(n). In particular, the variance of the proportion of isolated vertices is tends to zero.

For explicit lower bounds in probability on the proportion of isolated vertices we could appeal to Chebyshev’s inequality. However, since the variance vanishes, we have convergence in distribution to a constant, and thus convergence in probability.

Finally, a word on the end of EM’s talk. Having said that the sparse phase is not interesting because it is impossible, we might ask about the dense phase, where p and q are fixed. Just a for concrete example, suppose the probability of connection within the class is ½, and between classes is 1/3. Thus between any pair of vertices in the same class we expect to see roughly N/8+N/18= 13N/72 paths of length 2. The summands correspond to the middle vertex of the path in the same class, and in the opposite class respectively. However, between any pair of vertices in different classes, we expect to see roughly N/6 paths of length 2 for similar reasons. Both of these quantities will be highly concentrated on their means: consider the second-moment as the existence of each possible path is independent. Indeed, the chance that we see a proportion of paths closer to N/6 when it should be 13N/72 is a large deviations event, and so has exponential decay. As a result, the chance that we get the relative positions of any pair of vertices wrong with this method vanishes for large N.

In fact, the condition that (for p<q),

$N \mathbb{P}(\text{Bin}(N-1,p)\ge \text{Bin}(N-1,q)) \rightarrow 0$

should be enough to identify the partition with high probability, and indeed this is proved by several authors including EM. Note that the dense regime comfortable satisfies this condition, since it holds even without the factor of N. (The sparse regime, completely fails as the probability is roughly constant.) Even closer to the connectivity threshold remains interesting!

# Enumerating Forests

I’ve just got back from a visit to Budapest University of Technology, where it was very pleasant to be invited to give a talk, as well as continuing the discussion our research programme with Balazs. My talk concerned a limit for the exploration process of an Erdos-Renyi random graph conditioned to have no cycles. Watch this space (hopefully very soon) for a fully rigorous account of this. In any case, my timings were not as slick as I would like, and I had to miss out a chunk I’d planned to say about a result of Britikov concerning enumerating unrooted forests. It therefore feels like an excellent time to write something again, and explain this paper, which you might be able to find here, if you have appropriate journal rights.

We are interested to calculate $a_{n,m}$ the number of forests with vertex set [n] consisting of m unrooted trees. Recall that if we were interested in rooted trees, we could appeal to Prufer codes to show that there are $m n^{n-m-1}$ such forests, and indeed results of Pitman give a coalescent/fragmentation scheme as m varies between 1 and n-1. It seems that there is no neat combinatorial re-interpretation of the unrooted case though, so Britikov uses an analytic method.

We know that

$a_{n,m}= \frac{n!}{m!} \sum_{\substack{k_1+\ldots+k_m=n\\ k_i\ge 1}} \prod_{j=1}^m \frac{k_j^{k_j-2}}{k_j!}.$

To see this, observe that the $k_j$s correspond to the sizes of the m trees in the forest; $\frac{n!}{\prod k_j!}$ gives the multinomial number of ways to assign vertices to the trees; given the labels for a tree of size $k_j$, there are $k_j^{k_j-2}$ ways to make up the tree itself; and $\frac{1}{m!}$ accounts for the fact that the trees have no order.

What we would really like to do is to take the uniform distribution on the set of all labelled trees, then simulate m IID copies of this distribution, and condition the union to contain precisely n vertices. But obviously this is an infinite set, so we cannot choose uniformly from it. Instead, we can tilt so that large trees are unlikely. In particular, for each x we define

$\mathbb{P}(\xi=k) \propto \frac{k^{k-2} x^k}{k!}$,

and define the normalising constant

$B(x):= \sum_{k\ge 1} \frac{k^{k-2}x^k}{k!},$

whenever it exists. It turns out that $x\le e^{-1}$ is precisely the condition for $B(x)<\infty$. Note now that if $\xi_1,x_2,\ldots$ are IID copies of $\xi$, then

$\mathbb{P}(\xi_1+\ldots+\xi_m=n) = \frac{x^n}{B(x)^m} \sum_{k_1+\ldots + k_m=n} \prod_{j=1}^m \frac{k_j^{k_j-2}}{k_j!},$

and so we obtain

$a_{n,m}= \frac{n!}{m!} \frac{B(x)^m}{x^n} \mathbb{P}(\xi_1+\ldots + \xi_m=n).$

So asymptotics for $a_{n,m}$ might follows from laws of large numbers of this distribution $\xi$.

So far, we haven’t said anything about how to choose this value x. But observe that if you want to have lots of trees in the forest, then the individual trees should generally be small, so we take x small to tilt away from a preference for large trees. It turns out that there is a similar interpretation of criticality for forests as for general graphs, and taking x equal to 1/e, its radius of convergence works well for this setting. If you want even fewer trees, there is no option to take x larger than 1/e, but instead one can use large deviations machinery rather than laws of large number asymptotics.

We will be interested in asymptotics of the characteristic function of $\xi$ for x=1/e. In particular $\mathbb{E}[e^{it\xi}]=\frac{B(xe^{it})}{B(x)}$, and it will be enough to clarify the behaviour of this as $t\rightarrow 0$. It’s easier to work with a relation analytic function

$\theta(x)=\sum_{k\ge 1} \frac{k^{k-1}x^k}{k!},$

ie the integral of B. What now feels like a long time ago I wrote a masters’ thesis on the subject of multiplicative coalescence, and this shows up as the generating function of the solutions to Smoluchowski’s equations with monodisperse initial conditions, which are themselves closely related to the Borel distributions. In any case, several of the early papers on this topic made progress by establishing that the radius of convergence is 1/e, and that $\theta(x)e^{-\theta(x)}=x$ everywhere where $|x|\le 1/e$. We want to consider x=1/e, for which $\theta=1$.

Note that $\mathbb{E}\xi = \frac{\theta(x)}{B(x)}$, so we will make progress by relating $B(x),\theta(x)$ in two ways. One way involves playing around with contour integrals in a fashion that is clear in print, but involves quite a lot of notation. The second way is the Renyi relation which asserts that $\theta(x)=B(x)+\frac{\theta(x)^2}{2}$. We will briefly give a combinatorial proof. Observe that after multiplying through by factorials and interpreting the square of a generating function, this is equivalent to

$k^{k-1} = k^{k-2} + \frac12 \sum_{\substack{l+m=k\\l,m\ge 1}} l^{l-1}m^{m-1}\binom{k}{l},$

for all k. As we might expect from the appearance of this equality, we can prove it using a bijection on trees. Obviously on the LHS we have the size of the set of rooted trees on [k]. Now consider the set of pairs of disjoint rooted trees with vertex set [k]. This second term on the RHS is clearly the size of this set. Given an element of this set, join up the two roots, and choose whichever root was not initially in the same tree as 1 to be the new root. We claim this gives a bijection between this set, and the set of rooted trees on [k], for which 1 is not the root. Given the latter, the only pair of trees that leads to the right rooted tree on [k] under this mapping is given by cutting off the unique edge incident to the root that separates the root and vertex 1. In particular, since there is a canonical bijection between rooted trees for which 1 is the root, and unrooted trees (!), we can conclude the Renyi relation.

The Renyi relation now gives $\mathbb{E}\xi = \frac{\theta(x)}{B(x)}=2$ when x=1/e. If we wanted, we could show that the variance is infinite, which is not completely surprising, as the parameter x lies on the radius of convergence of the generating function.

Now, playing around with contour integrals, and being careful about which strands to take leads to the asymptotic as $t\rightarrow 0$

$\mathbb{E}[ e^{it\xi}] = 1+2it + \frac{2}{3}i |2t|^{3/2} (i\mathrm{sign}(t))^{3/2} + o(|t|^{3/2}).$

So from this, we can show that the characteristic function of the rescaled centred partial sum $\frac{\xi_1+\ldots+\xi_N-2N}{bN^{2/3}}$ converges to $\exp(-|t|^{3/2}\exp(\frac{i\pi}{4}\mathrm{sign} t))$, where $b= (32/9)^{1/3}$ is a constant arising out of the previous step.

We recognise this as the characteristic function of the stable distribution with parameters 3/2 and -1. In particular, we know now that $\xi$ is in the domain of attraction for a stable-3/2 distribution. If we wanted a version of the central limit theorem for such partial sums, we could have that, but since we care about the partial sums of the $\xi_i$s taking a specific value, rather than a range of values on the scale of the fluctuations, we actually need a local limit theorem.

To make this clear, let’s return to the simplest example of the CLT, with some random variables with mean $\mu$ and variance $\sigma^2<\infty$. Then the partial sums satisfy

$\mathbb{P}(\mu N + a\sigma\sqrt{N} \le S_N \le \mu_N+b\sigma\sqrt{N}) \rightarrow \int_a^b f_{\mathcal N}(x)dx,$

as $N\rightarrow\infty$. But what about the probability of $S_N$ taking a particular value m that lies between $\mu N+a\sigma \sqrt{N}$ and $\mu N + b\sigma \sqrt{N}$? If the underlying distribution was continuous, this would be uncontroversial – considering the probability of lying in a range that is smaller than the scale of the CLT can be shown in a similar way to the CLT itself. A local limit theorem asserts that when the underlying distribution is supported on some lattice, mostly naturally the integers, then these probabilities are in the limit roughly the same whenever m is close to $\mu N+a\sigma\sqrt{N}$.

In this setting, a result of Ibragimov and Linnik that I have struggled to find anywhere in print (especially in English) gives us local limit theory for integer-supported distributions in the domain of attraction of a stable distribution. Taking p( ) to be the density of this distribution, we obtain

$bm^{2/3}\mathbb{P}(\xi_1+\ldots+\xi_m=n) - p(\frac{n-2m}{b m^{2/3}}) \rightarrow 0$

as $n\rightarrow\infty$, uniformly on any set of m for which $z= \frac{n-2m}{bm^{2/3}}$ is bounded. Conveniently, the two occurrences of b clear, and Britikov obtains

$a_{n,m} = (1+o(1)) \frac{\sqrt{2\pi} n^{n-1/6}}{2^{n-m}(n-m)!} p(\frac{n-2m}{n^{2/3}},$

uniformly in the same sense as before.

# Multitype Branching Processes

One of the fundamental objects in classical probability theory is the Galton-Watson branching process. This is defined to be a model for the growth of a population, where each individual in a generation gives birth to some number (possibly zero) of offspring, who form the next generation. Crucially, the numbers of offspring of the individuals are IID, with the same distribution both within generations and between generations.

There are several ways one might generalise this, such as non-IID offspring distributions, or pairs of individuals producing some number of offspring, but here we consider the situation where each individual has some type, and different types have different offspring distributions. Note that if there are K types, say, then the offspring distributions should now be supported on $\mathbb{Z}_{\ge 0}^K$. Let’s say the offspring distribution from a parent of type i is $\mu^{(i)}$.

The first question to address is one of survival. Recall that if we want to know whether a standard Galton-Watson process has positive probability of having infinite size, that is never going extinct, we only need to know the expectation of the offspring distribution. If this is less than 1, then the process is subcritical and is almost surely finite. If it is greater than 1, then it is supercritical and survives with positive probability. If the expectation is exactly 1 (and the variance is finite) then the process is critical and although it is still almost surely finite, the overall population size has a power-law tail, and hence (or otherwise) the expected population size is infinite.

We would like a similar result for the multitype process, saying that we do not need to know everything about the distribution to decide what the survival probability should be.

The first thing to address is why we can’t just reduce the multitype change to the monotype setting. It’s easiest to assume that we know the type of the root in the multitype tree. The case where the type of the root is random can be reconstructed later. Anyway, suppose now that we want to know the offspring distribution of a vertex in the m-th generation. To decide this, we need to know the probability that this vertex has a given type, say type j. To calculate this, we need to work out all the type possibilities for the first m generations, and their probabilities, which may well include lots of complicated size-biasing. Certainly it is not easy, and there’s no reason why these offspring distributions should be IID. The best we can say is that they should probably be exchangeable within each generation.

Obviously if the offspring distribution does not depend on the parent’s type, then we have a standard Galton-Watson tree with types assigned in an IID manner to the realisation. If the types are symmetric (for example if M, to be defined, is invariant under permuting the indices) then life gets much easier. In general, however, it will be more complicated than this.

We can however think about how to decide on survival probability. We consider the expected number of offspring, allowing both the type of the parent and the type of the child to vary. So define $m_{ij}$ to be the expected number of type j children born to a type i parent. Then write these in a matrix $M=(m_{ij})$.

One generalisation is to consider a Galton-Watson forest started from some positive number of roots of various types. Suppose we have a vector $\nu=(\nu_i)$ listing the number of roots of each type. Then the expected number of descendents of each type at generation n is given by the vector $\nu M^n$.

Let $\lambda$ be the largest eigenvalue of M. As for the transition matrices of Markov chains, the Perron-Frobenius theorem applies here, which confirms that, because the entries of M are positive, the eigenvalue with largest modulus is simple and real, and the associated eigenvector has entirely positive entries. [In fact we need a couple of extra conditions on M, including that it is possible to get from any type to any other type – we say irreducible – but that isn’t worth going into now.]

So in fact the total number of descendents at generation n grows like $\lambda^n$ in expectation, and so we have the same description of subcriticality and supercriticality. We can also make a sensible comment about the left-$\lambda$-eigenvector of M. This is the limiting proportion of the different types of vertices.

It’s a result (eg. [3]) that the height profile of a depth-first search on a standard Galton-Watson tree converges to Brownian Motion. Another way to phrase this is that a GW tree conditioned to have some size N has the Brownian Continuum Random Tree as a scaling limit as N grows to infinity. Miermont [4] proves that this result holds for the multitype tree as well. In the remainder of this post I want to discuss one idea along the way to the proof, and one application.

I said initially that there wasn’t a trivial reduction of a multitype process to a monotype process. There is however a non-trivial embedding of a monotype process in a multitype process. Consider all the vertices of type 1, and all the paths between such vertices. Then draw a new tree consisting of just the type 1 vertices. Two of these are joined by an edge if there is no other type 1 vertex on the unique path between them in the original tree. If that definition is confusing, think of the most sensible way to construct a tree on the type 1 vertices from the original, and you’ve probably chosen this definition.

There are two important things about this new tree. 1) It is a Galton-Watson tree, and 2) if the original tree is critical, then this reduced tree is also critical. Proving 1) is heavily dependent on exactly what definitions one takes for both the multitype branching mechanism and the standard G-W mechanism. Essentially, at a type 1 vertex, the number of type 1 descendents is not dependent on anything that happened at previous generations, nor in other branches of the original tree. This gives IID offspring distributions once it is formalised. As for criticality, we note that by the matrix argument given before, under the irreducibility condition discussed, the expectation of the total population size is infinite iff the expected number of type 1 vertices is also infinite. Since the proportion of type 1 vertices is given by the first element of the left eigenvector, which is positive, we can make a further argument that the number of type 1 vertices has a power-law tail iff the total population size also has a power-law tail.

I want to end by explaining why I was thinking about this model at all. In many previous posts I’ve discussed the forest fire model, where occasionally all the edges in some large component are deleted, and the component becomes a set of singletons again. We are interested in the local limit. That is, what do the large components look like from the point of view of a single vertex in the component? If we were able to prove that the large components have BCRT as the scaling limit, this would answer this question.

This holds for the original random graph process. There are two sensible ways to motivate this. Firstly, given that a component is a tree (which it is with high probability if its size is O(1) ), its distribution is that of the uniform tree, and it is known that this has BCRT as a scaling limit [1]. Alternatively, we know that the components have a Poisson Galton-Watson process as a local limit by the same argument used to calculate the increments of the exploration process. So we have an alternative description of the BCRT appearing: the scaling limit of G-W trees conditioned on their size.

Regarding the forest fires, if we stop the process at some time T>1, we know that some vertices have been burned several times and some vertices have never received an edge. What is clear though is that if we specify the age of each vertex, that is, how long has elapsed since it was last burned; conditional on this, we have an inhomogeneous random graph. Note that if we have two vertices of ages s and t, then the probability that there is an edge between them is $1-e^{-\frac{s\wedge t}{n}}$, ie approximately $\frac{s\wedge t}{n}$. The function giving the probabilities of edges between different types of vertices is called the kernel, and here it is sufficiently well-behaved (in particular, it is bounded) that we are able to use the results of Bollobas et al in [2], where they discuss general sparse inhomogeneous random graphs. They show, among many other things, that in this setting as well the local limit is a multitype branching process.

So in conclusion, we have almost all the ingredients towards proving the result we want, that forest fire components have BCRT scaling limit. The only outstanding matter is that the Miermont result deals with a finite number of types, whereas obviously in the setting where we parameterise by age, the set of types is continuous. In other words, I’m working hard!

References

[1] Aldous – The Continuum Random Tree III

[2] Bollobas, Janson, Riordan – The phase transition in inhomogeneous random graphs

[3] Le Gall – Random Trees and Applications

[4] Miermont – Invariance principles for spatial multitype Galton-Watson trees

# Persistent Hubs

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

# Critical Components in Erdos-Renyi

In various previous posts, I’ve talked about the phase transition in the Erdos-Renyi random graph process. Recall the definition of the process. Here we will use the Gilbert model G(n,p), where we have n vertices, and between any pair of vertices we add an edge, independently of other pairs with probability p. We are interested in the sparse scaling, where the typical vertex has degree O(1) in n, and so p=c/n for constant c>0, and we assume throughout that n is large. We could alternatively have considered the alternative Erdos-Renyi model where we choose uniformly at random from the set of graphs with n vertices and some fixed number of edges. Almost all the results present work equally well in this setting.

As proved by Erdos and Renyi, the typical component structure of such a graph changes noticeably around the threshold c=1. Below this, in the subcritical regime, all the components are small, meaning of size at most order O(log n). Above this, in the supercritical regime, there is a single giant component on some non-zero proportion of the vertices. The rest of the graph looks subcritical. The case c=1 exhibits a phase transition between these qualitatively different behaviours. They proved that here, the largest component is with high probability O(n^2/3). It seems that they thought this result held whenever c=1-o(1), but it turns out that this is not the case. In this post, I will discuss some aspects of behaviour around criticality, and the tools needed to treat them.

The first question to address is this: how many components of size n^{2/3} are there? It might be plausible that there is a single such component, like for the subsequent giant component. It might also be plausible that there are n^1/3 such components, so O(n) vertices are on such critical components. As then it is clear how we transition out of criticality into supercriticality – all the vertices on critical components coalesce to form the new giant component.

In fact neither of these are correct. The answer is that for all integers k>0, with high probability the k-th largest component is on a size scale of n^2/3. This is potentially a confusing statement. It looks like there are infinitely many such components, but of course for any particular value of n, this cannot be the case. We should think of there being w(1) components, but o(n^b) for any b>0.

The easiest way to see this is by a duality argument, as we have discussed previously for the supercritical phase. If we remove a component of size O(n^2/3), then what remains is a random graph with n-O(n^2/3) vertices, and edge probability the same as originally. It might make sense to rewrite this probability 1/n as

$\frac{1}{n-O(n^{2/3})}\cdot \frac{n-O(n^{2/3})}{n}=\frac{1-O(n^{-1/3})}{n-O(n^{2/3})}.$

The approximation in the final numerator is basically the same as

$1-o\left(n-O(n^{2/3})\right).$

Although we have no concrete reasoning, it seems at least plausible that this should look similar in structure to G(n,1/n). In particular, there should be another component of size

$O\left([n-O(n^{2/3})]^{2/3}\right)=O(n^{2/3})$.

In fact, the formal proof of this proceeds by an identical argument, only using the exploration process. Because I’ve described this several times before, I’ll be brief. We track how far we have gone through each component in a depth-first walk. In both the supercritical and subcritical cases, when we scale correctly we get a random path which is basically deterministic in the limit (in n). For exactly the same reasons as visible CLT fluctuations for partial sums of RVs with expectation zero, we start seeing interesting effects at criticality.

The important question is the order of rescaling to choose. At each stage of the exploration process, the number of vertices added to the stack is binomial. We want to distinguish between components of size $O(n^{2/3})$ so we should look at the exploration process at time $sn^{2/3}$. The drift of the exploration process is given by the expectation of a binomial random variable minus one (since we remove the current vertex from the stack as we finish exploring it). This is given by

$\mathbb{E}=\left[n-sn^{2/3}\right]\cdot \frac{1}{n}-1=-sn^{-1/3}.$

Note that this is the drift in one time-step. The drift in $n^{2/3}$ time-steps will accordingly by $sn^{1/3}$. So, if we rescale time by $n^{2/3}$ and space by $n^{1/3}$, we should get a nice stochastic process. Specifically, if Z is the exploration process, then we obtain:

$\frac{1}{n^{1/3}}Z^{(n)}_{sn^{2/3}} \rightarrow_d W_s,$

where W is a Brownian motion with inhomogeneous drift -s at time s. The net effect of such a drift at a fixed positive time is given by integrating up to that time, and hence we might say the process has quadratic drift, or is parabolic.

We should remark that our binomial expectation is not entirely correct. We have discounted those $sn^{2/3}$ vertices that have already been explored, but we have not accounted for the vertices currently in the stack. We should also be avoiding considering these. However, we now have a heuristic for the approximate number of these. The number of vertices in the stack should be $O(n^{1/3})$ at all times, and so in particular will always be an order of magnitude smaller than the number of vertices already considered. Therefore, they won’t affect this drift term, though this must be accounted for in any formal proof of convergence. On the subject of which, the mode of convergence is, unsurprisingly, weak convergence uniformly on compact sets. That is, for any fixed S, the convergence holds weakly on the random functions up to time $sn^{2/3}$.

Note that this process will tend to minus infinity almost surely. Component sizes are given by excursions above the running minimum. The process given by the height of the original process above the running minimum is called reflected. Essentially, we construct the reflected process by having the same generator when the current value is positive, and forcing the process up when it is at zero. There are various ways to construct this more formally, including as the scaling limit of some simple random walks conditioned never to stay non-negative.

The cute part of the result is that it holds equally well in a so-called critical window either side of the critical probability 1/n. When the probability is $\frac{1+tn^{-1/3}}{n}$, for any $t\in \mathbb{R}$, the same argument holds. Now the drift at time s is t-s, though everything else still holds.

This result was established by Aldous in [1], and gives a mechanism for calculating distributions of component sizes and so on through this critical window.

In particular, we are now in a position to answer the original question regarding how many such components there were. The key idea is that because whenever we exhaust a component in the exploration process, we choose a new vertex uniformly at random, we are effectively choosing a component according to the size-biased distribution. Roughly speaking, the largest components will show up near the beginning. Note that a critical $O(n^{2/3})$ component will not necessarily be exactly the first component in the exploration process, but the components that are explored before this will take up sufficiently few vertices that they won’t show up in the scaling of the limit.

In any case, the reflected Brownian motion ‘goes on forever’, and the drift is eventually very negative, so there cannot be infinitely wide excursions, hence there are infinitely many such critical components.

If we care about the number of cycles, we can treat this also via the exploration process. Note that in any depth-first search we are necessarily only interested in a spanning tree of the host graph. Anyway, when we are exploring a vertex, there could be extra edges to other vertices in the stack, but not to vertices we’ve already finished exploring (otherwise the edge would have been exposed then). So the expected number of excess edges into a vertex is proportional to the height of the exploration process at that vertex. So the overall expected number of excess edges, conditional on the exploration process is the area under the curve. This carries over perfectly well into the stochastic process limit. It is then a calculation to verify that the area under the curve is almost surely infinite, and thus that we expect there to be infinitely many cycles in a critical random graph.

REFERENCES

[1] Aldous D. – Brownian excursions, critical random graphs and the multiplicative coalescent

# Recent Research Activity

I’ve spent this week in Luminy, near Marseille, attending a summer school run by ALEA, the organisation of French probabilists. We’ve been staying in CIRM, a dedicated maths research conference centre at the edges of the calanques, the area of mountains and jagged coastal inlets between Marseille and Cassis. The walking possibilities have been excellent, as have the courses and lectures, on a range of topics in probability theory.

Anyway, the time here has been an excellent moment to reflect on my research progress, and try to come up with the sort of fresh ideas that are perhaps slightly inhibited by sitting at a desk with an endless supply of paper on which to try calculations. When I get back, I have to submit a first-year report, so at least for a little while I will have to suppress the desire to make further progress and instead diligently assemble the progress I have made.

The Model

I’ve defined some of these processes in past posts, but I see no harm in doing so again. We take the standard Erdos-Renyi random graph process, where edges are added one-at-a-time uniformly at random between n vertices, and amend it by adding a deletion mechanism. The aim is to arrive at a process which looks in equilibrium more like the critical random graph than either the subcritical or supercritical regimes, where the components are very small, and dominated by one giant component respectively. Rath, Toth and others have studied the process where each vertex is hit by lightning at uniform rate. When this happens, we delete all the edges in the component containing that vertex. Naturally, big components will be hit by lightning more often than small components, and so this acts as a mechanism to prevent the formation of giant components, if scaled correctly.

We take a different approach. We observe that criticality in the original random graph process is denoted by the first appearance of a giant component, but also by the first appearance of a) lots of cycles, and b) large cycles. In particular, it is very unlikely that a giant component could form without containing any cycles. We will therefore use the appearance of a cycle to trigger some form of deletion mechanism.

Our final goal is to treat the so-called ‘Cycle Deletion’ model. Here, whenever a cycle appears, we delete all the edges in that cycle immediately. There are several challenges in treating this model, because the rate at which cycles emerge in a tree is a function of the tree structure. The trees in this model will not be Uniform Spanning Trees (though it is very possible that they will be ‘almost USTs’ in some sense – we need to investigate this further) so it will be hard to make nice statements about the rates. For the standard random graph process, if we are only interested in the sizes of the components, we are actually allowed to ignore the graph structure entirely. The component sizes evolve as a discrete, stochastic version of the multiplicative coalescent (sometimes called a Marcus-Lushnikov process). We would like a deletion mechanism that has a nice interpretation as a fragmentation operation in the same sense. The rate at which a component fragments will be quadratic in the size of the component, since there are $O(k^2)$ possible edges between k vertices forming a component, and adding any of precisely these will create a cycle.

I’ve talked previously about how to overcome the problems with the tree structure in Cycle Deletion with the so-called Uniform Cycle Deleting model. In any case, as a starting point we might consider the Cycle-Induced Forest Fire model. Here, whenever a cycle appears, we delete all the edges, including the new one, in the whole component which contains the cycle.

We suspect this model may resemble the critical random graph at all times. The main characteristic of G(n,1/n) is that the largest component is of size O(n^2/3), and indeed there are arbitrarily many components of this size, with high probability in the limit. Since CIFF is recurrent for any fixed n, meaning that it will visit any state infinitely often (rather than tending to infinity or similar), we should ask what the largest component is typically in the equilibrium distribution. Our aim is to prove that it is O(n^2/3). We might suspect that the typical size of the largest component will be greater in the Cycle Deletion model, since each fragmentation event is less severe there, removing fewer edges.

An Upper Bound

The nice thing about Markov chains is that they have an ergodic property, which means that if you run them for long enough, the proportion of time spent in any state is given by the stationary probability of being in that state. It doesn’t matter whether or not you start in equilibrium, since it will converge anyway. Thus it is meaningful to talk about properties like the average number of isolated vertices as a time-average as well as an average with respect to some distribution.

This quantity is the key to an upper bound. We can equally talk about the average change in the number of isolated vertices in a time-step. This will increase when a component fragments, and will decrease when an isolated vertex coalesces with another component. In particular, the largest possible decrease in the number of isolated vertices in a single time-step is 2, corresponding to an edge appearing between two isolated vertices.

Suppose that with probability $\Theta(1)$ there is a component of size $n^\alpha$ for some $\alpha>2/3$. Then such a component makes a contribution to the expected change in the number of isolated vertices of

$\Theta(1) n^\alpha \left(\frac{n^\alpha}{n}\right)^2.$ (*)

Where does this come from? Well, we are tracking the contributions from the event that the largest component is of this size and that it fragments, giving $n^\alpha$ new isolated vertices. So the $\Theta(1)$ accounts for the probability that there is such a component to begin with. Then, conditional on that, the probability that it gets fragmented in the next time-step is the probability that both ends of the next edge added lie in that component. Since the edge is chosen uniformly at random, the probability of this is $n^\alpha/n$. Note that this is under a slightly odd definition of an edge, that allows loops. Basically, I don’t want to have lots of correction terms involving $\binom{n}{2}$ floating around. However, it would make no difference to the orders of magnitude if we to do it with these.

So, this is only one contribution to the typical rate of gain of isolated vertices. Now note that if $\alpha>2/3$, then this expression is >> 1. This is bad since the negative contributions to this expected flux in the number of isolated vertices is O(1). So this suggests that over time, the number of isolated vertices will keep growing. This is obviously ridiculous since a) we are in equilibrium, so the expected flux should be 0 and b) the number of isolated vertices cannot exceed n, for clear reasons.

This gives us an upper bound of n^2/3 as the typical scale of the largest component. We can come up with a similar argument for the cycle deleting model. The most helpful thing to track there is the number of edges in the graph. Note that since the graph is at all times a forest on n vertices, the number of edges is equal to n minus the number of (tree) components. We use the fact that the typical fragmentation of a component of size k creates $O(\sqrt{k})$ new components. It is possible to argue via isolated vertices here too, but the estimates are harder, or at least less present in the literature.

Lower Bounds?

The problem with lower bounds is that it is entirely possible that the flux in the number of isolated vertices is not driven by typical behaviour. Suppose for example we had a different rule. We begin a random graph process, and the first time we see a cycle in a component with size larger than n^2/3, we delete all the edges in the whole graph. Then we will see a sequence of random graph processes starting with the empty graph and stopped at some point close to criticality (in fact, with high probability in the *critical window*), and these will all be glued together. So then, most of the time the process will look subcritical, but the gains in isolated vertices will occur only during the critical periods, which are only an asymptotically small proportion of the time.

At the moment, my approach to the lower bound is instead to prove that the upper bound is tight. I mean this in the following sense. Suppose we wanted to be sure that (*) was in fact equal to the average rate of gain of isolated vertices. We would have to check the following:

• That the total contributions from all other components were similar or smaller than from the component(s) of size roughly $n^{\alpha}$.
• That there were only a few components of size $n^{\alpha}$. In particular, the estimate would be wrong if there were $n^\epsilon$ such components for any $\epsilon>0$.
• That it cannot be the case that for example, some small proportion of the time there is a component of size roughly $n^{\alpha+\epsilon}$, and over a large enough time these make a greater contribution to the average gain in isolated vertices.

A nice way to re-interpret this is to consider some special vertex and track the size of its component in time. It will be involved in repeated fragmentations over the course of time, so it is meaningful to talk about the distribution of the size of the component containing the vertex when it is fragmented. Our aim is to show that this distribution is concentrated on the scaling $O(n^\alpha)$.

So this has turned out to be fairly hard. Rather than try to explain some of the ideas I’ve employed in attempting to overcome this, I will finish by giving one reason why it is hard.

We have seen that the component sizes in random graphs evolve as the multiplicative coalescent, but at a fixed moment in time, we can derive good estimates from an analogy with branching processes. We might like to do that here. If we know what the system looks like most of the time, we might try to ‘grow’ a multiplicative coalescent, viewing it like a branching process, with distribution given by the typical distribution. The problem is that when I do this, I find that the expectation of the offspring distribution is $\Theta(1)$. This looks fine, since 1 is the threshold for extinction with probability 1. However, throughout the analysis, I have only been paying attention to the exponent of n in all the time and size estimates. For example, I view $n^\alpha$ and $n^\alpha \log n$ as the same. This is a problem, as when I say the expectation is $\Theta(1)$, I am really saying it is $\sim n^0$. This means it could be $\frac{1}{\log n}$ or $\log n$. Of course, there is a massive difference between these, since a branching process grows expectationally!

So, this approach appears doomed in its current form. I have some other ideas, but a bit more background may be required before going into those. I’m going to be rather busy with teaching on my return to the office, so unfortunately it is possible that there may be many posts about second year probability and third year applied probability before anything more about CIFF.

# The Configuration Model

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

REFERENCES

Janson – A New Approach to the Giant Component Problem

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

Janson – The Probability that  Random Multigraph is Simple

# Characterisations of Geometric Random Graphs

Continuing the LMS-EPSRC summer school on Random Graphs, Geometry and Asymptotic Structure, we’ve now had three of the five lectures by Mathew Penrose on Geometric Random Graphs.

The basic idea is that instead of viewing a graph entirely abstractly, we now place the vertices in the plane, or some other real space. In many network situations, we would expect connectivity to depend somehow on distance. Agents or sites which are close together might be considered more likely to have the sort of relationship indicated by being connected with an edge. In the model discussed in this course, this dependence is deterministic. We have some parameter r, and once we have chosen the location of all the vertices, we connect a pair of vertices if the distance between them is less than r.

For the purposes of this, we work in a compact space [0,1]^d, and we are interested in the limit as the number of vertices n grows to infinity. To avoid the graph getting too connected, as in the standard random graph model, we take r to be a decreasing function of n. Anyway, we place the n points into the unit hypercube uniformly at random, and then the edges are specified by the adjacency rule above. In general, because r_n will be o(1), we won’t have to worry too much above boundary effects. The number of vertices within r_n of the boundary of the cube will be o(1). For some results, this is a genuine problem, when it may be easier to work on the torus.

In G(n,p), the order of np in the limit determines the qualitative structure of the graph. This is the expected degree of a given fixed vertex. In this geometric model, the relevant parameter is $nr_n^d$, where d is the dimension of the hypercube. If this parameter tends to 0, we say the graph is sparse, and dense if it tends to infinity. The intermediate case is called a thermodynamic limit. Note that the definition of sparse here is slightly different from G(n,p).

Much of the content of the first three lectures has been verifying that the distributions of various quantities in the graph, for example the total number of edges, are asymptotically Poisson. Although sometimes arguments are applicable over a broad spectrum, we also sometimes have to use different calculations for different scaling windows. For example, it is possible to show convergence to a Poisson distribution for the number of edges in the sparse case, from which we get an asymptotic normal approximation almost for free. In the denser regimes, the argument is somewhat more technical, with some substantial moment calculations.

A useful tool in these calculations are some bounds derived via Stein’s method for sums of ‘almost independent’ random variables. For example, the presence or non-presence of an edge between two pairs of vertices are independent in this setting if the pairs are disjoint, and the dependence is still only mild if they share a vertex. An effective description is via a so-called dependency graph, where we view the random variables as the vertices of a graph, with an edge between them if there is some dependence. This description doesn’t have any power in itself, but it does provide a concise notation for what would otherwise be very complicated, and we are able to show versions of (Binomials converge to Poisson) and CLT via these that are exactly as required for this purpose.

In particular, we are able to show that if $E_n$ is the total number of edges, under a broad set of scaling regimes, if $\lambda_n$ is the expected total number of edges, then $d_{TV}(E_n,\mathrm{Po}(\lambda_n))\rightarrow 0$, as n grows. This convergence in total variation distance is as strong a result as one could hope for, and when the sequence of $\lambda_n$ is O(1), we can derive a normal approximation as well.

At this point it is worth discussing an alternative specification of the model. Recall that for a standard homogenous random graph, we have the choice of G(n,m) and G(n,p) as definitions. G(n,m) is the finer measure, and G(n,p) can be viewed as a weighted mix of G(n,m). We can’t replicate this directly in the geometric setting because the edges and non-edges are a deterministic function of the vertex locations. What we can randomise is the number of vertices. Since we are placing the vertices uniformly at random, it makes sense to consider as an alternative a Poisson Point Process with intensity n. The number of vertices we get overall will be distributed as Po(n), which is concentrated near n, in the same manner as G(n,c/n).

As in G(n,p), this is a less basic model because it is a mixture of the fixed-vertex models. Let’s see if how we would go about extending the total variation convergence result to this slightly different setting without requiring a more general version of the Poisson Approximation Lemma. To avoid having to define everything again, we add a ‘ to indicate that we are talking about the Poisson Point Process case. Writing d(.,.) for total variation distance, the result we have is:

$\lim_{n\rightarrow\infty} d(E_n,\mathrm{Po}(\lambda_n))=0.$

We want to show that

$\lim_{n\rightarrow\infty}d(E_n',\mathrm{Po}(\lambda_n'))=0,$

which we can decompose in terms of expectations in the original model by conditioning on $N_n$

$\leq \lim_{n\rightarrow\infty}\mathbb{E}\Big[\mathbb{E}[d(E_{N_n},\mathrm{Po}(\lambda_n')) | N_n]\Big],$

where the outer expectation is over N. The observation here, is that the number of points given by the Poisson process induces a measure on distributions, the overwhelming majority of which look quite like Poisson distributions with parameter n. The reason we have a less than sign is that we are applying the triangle inequality in the sum giving total variation distance:

$d(X,Y)=\sum_{k\geq 0}|\mathbb{P}(X=k)-\mathbb{P}(Y=k)|.$

From this, we use the triangle inequality again:

$\lim_{n\rightarrow\infty} \mathbb{E}\Big[\mathbb{E}[d(E_{N_n},\mathrm{Po}(\lambda_{N_n})) | N_n]\Big]$

$+\lim_{n\rightarrow\infty}\mathbb{E}\Big[\mathbb{E}[d(\mathrm{Po}(\lambda_{N_n}),\mathrm{Po}(\lambda_n')) | N_n]\Big].$

Then, by a large deviations argument, we have that for any $\epsilon>0$, $\mathbb{P}(|N_n-n|\geq \epsilon n)\rightarrow 0$ exponentially in n. Also, total variation distance is, by definition, bounded above by 1. In the first term, the inner conditioning on N_n is irrelevant, and we have that $E_{N_n}$ converges to the Poisson distribution for any fixed $N_n\in (n(1-\epsilon),n(1+\epsilon))$. Furthermore, we showed in the proof of the non-PPP result that this convergence is uniform in this interval. (This is not surprising – the upper bound is some well-behaved polynomial in 1/n.) So with probability $1- e^{-\Theta(n)}$ N_n is in the region where this convergence happens, and elsewhere, the expected TV distance is bounded below 1, so the overall expectation tends to 0. With a similar LD argument, for the second term it suffices to prove that when $\lambda\rightarrow\mu$, we must have $d(\mathrm{Po}(\lambda),\mathrm{Po}(\mu))\rightarrow 0$. This is ‘obviously’ true. Formally, it is probably easiest to couple the distributions $\mathrm{Bin}(n,\lambda/n),\mathrm{Bin}(n,\mu/n)$ in the obvious way, and carry the convergence of TV distance as the parameter varies through the convergence in n.

That all sounded a little bit painful, but is really just the obvious thing to do with each term – it’s only the language that’s long-winded!

Anyway, I’m looking forward to seeing how the course develops. In particular, when you split the space into small blocks, the connectivity properties resemble those of (site) percolation, so I wonder whether there will be concrete parallels. Also, after reading about some recent results concerning the metric structure of the critical components in the standard random graph process, it will be interesting to see how these compare to the limit of a random graph process which comes equipped with metric structure for free!

# Long Paths and Expanders

I’m in Birmingham this week for the LMS-EPSRC summer school on Random Graphs, Geometry and Asymptotic Structure. The event consists of three five-hour mini-courses, a plenary lecture, leaving plenty of time for problem sheet and discussion. I thought it would be worth trying to say a couple of interesting things each day – I do not know whether this will succeed, but I might as well try.

Today, a few thoughts on the first two lectures of Michael Krivelevich’s course on Long Paths and Hamiltonicity in Random Graphs. The aim is to develop tools to investigate the threshold for the presence of a Hamiltonian cycle in G(n,p). In this first part of the course, we were mainly thinking about long paths.

One tool we used a lot was the Depth-First Search algorithm. This is very similar to the exploration process I’ve talked about before. Essentially, here we consider trying to explore the graph in a depth-first way, but instead of viewing all the edges incident to a vertex we have just arrived at, we only look to see whether there is an edge out of the new vertex. If there is, we explore it, then come back eventually to look for more. It really comes down to a difference in the information we are storing. In this DFS, we store the vertices which we haven’t finished exploring, which is the set of vertices on the explored path between the root and the current vertex. So the size of this set evolves like the contour process. In particular, we can read off the sizes of paths from this description. These dynamics are useful in particular because we know there are no edges between the set of vertices we have finished exploring, and the ones we have yet to explore. The stack of ‘processing’ vertices must glue everything else together.

We can translate one of the arguments back into the language for the old exploration process. Recall the increments of the exploration process are $\mathrm{Bin}(\alpha n,\frac{c}{n}) -1$ once we have explored $\alpha n$ vertices. We don’t need to worry about the -1 bit for now. Observe that because we are exploring in a depth-first way, if a subsequence of the Binomial variables of length k are all positive, this corresponds to a path of length (k-1).

So to prove, for example, that the longest path in a subcritical random graph is O(log n), it suffices to prove that there are O(log n) consecutive positive entries in the sequence of n binomial entries. Since the distribution changes continuously, it is convenient to prove that there are O(log n) consecutive positive entries in the first $\epsilon n$ binomial entries. The probability that any of these entries is positive is bounded below by some p, so it suffices to consider instead a sequence of Bernoulli RVs with parameter p. So if we never have clog n consecutive, this gives control of the sequence of geometric random variables corresponding to the gaps between 0s in the sequence. Precisely, these are Geom(q), and we must have $\frac{\epsilon n}{c\log n}$ of them independently being less than clog n. We have to chase a few constants, and use the fact that if $f(n)\rightarrow\infty, \frac{g(n)}{f(n)}\rightarrow\infty$, then

$(1-\frac{1}{f(n)})^{g(n)}\rightarrow 0,$

by comparison with the standard asymptotic result for $e^{-x}$. In any case, we get that this probability tends to 0 if we choose c small enough, and so with high probability there is a path of length clog n.

This is interesting, because we knew already that the largest component in a subcritical random graph had size O(log n). But we also knew that all the components were trees, or ‘almost trees’, and were uniformly chosen from the set of trees (or trees + an edge or two) with appropriate size. And the largest path in a UST on n vertices is $O(n^{1/2})$ with high probability. So we learn that there are enough components of size $\geq c\log n$ that it is actually very probable that one of them will have the unlikely property of being much more path-like than a typical tree.

Krivelevich also showed a pleasant elementary proof of the result that a supercritical random graph has a path of length O(n), using a similar idea.

The other definition of major interest was an expander graph. Often when doing calculations about neighbourhoods of sets of vertices, we run into the problem that the neighbourhoods may overlap, and so we cannot get the total outer neighbourhood (or outer boundary) just by summing over the individual neighbourhood sizes. In an expander graph, we demand that all small sets of vertices have neighbourhood at least as large as some constant multiple of the set size, essentially giving us a bound on the above problem. Concretely, G is a $(k,\alpha)$-expander is for any set of vertices $|U|\leq k, |N(U)|\geq \alpha |U|$.

There’s a very nice argument using Posa’s lemma, where we consider all the possible ways to rearrange the vertices in some longest path into a different longest path, and then focus on the endpoints of all these paths. With this so-called rotation-extension technique, we can show that a (k,2)-expander has a path of length at least 3k-1.

There are structural similarities between expander graphs and regular graphs, so it seems natural that there will be some interesting spectral properties. I don’t know much about this, but perhaps it will come up later in the week. But, returning to the random graph long path problem, it now suffices to show subcritical G(n,p) is a (clog n,2)-expander for some c. Expander properties are in some sense the opposite of clustering properties, and independence of a RG inhibit most clustering properties (as discussed in much greater detail in some of the posts about network models). Unfortunately, this doesn’t actually work, as in a subcritical graph, the typical expansion coefficient, even of a small set will be c, for G(n,c/n), which is not large enough. However, if you chose the constants carefully, such an argument should work for c>2, so long as you chose k=an, with a small enough that the probability of a vertex elsewhere in the graph being joined to (at least) two of the k vertices in the set, was small compared with (c-2).

REFERENCES

The course notes are not available, though chapter 3 from these 2010 notes by the same lecturer are related and interesting.