# Random transpositions

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

Some more formality and some technical remarks:

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

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

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

Comparison with Erdos-Renyi random graph process

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

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

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

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

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

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

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

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

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

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

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

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

Uniform split-merge

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

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

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

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

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

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

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

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

References

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

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

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

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

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

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

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

# Random Mappings for Cycle Deletion

In previous posts here and here, I’ve talked about attempts to describe a cycle deleting process. We amend the dynamics of the standard random graph process by demanding that whenever a cycle is formed in the graph we delete all the edges that lie on the cycle. The aim of this is to prevent the system growing giant components, and perhaps give a system that displays the characteristics of self-organised criticality. In the posts linked to, we discuss the difficulties caused by the fact that the tree structure of components in such a process is not necessarily uniform.

Today we look in the opposite direction. It gives a perfectly reasonable model to take a multiplicative coalescent with quadratic fragmentation (this corresponds to cycle deletion, since there are $O(n^2)$ edges which would give a cycle if added to a tree on n vertices) and a fragmentation kernel corresponding to adding an extra edge to a uniform spanning tree on n vertices then deleting the edges of the unique cycle. The focus of the rest of this post, we consider this fragmentation mechanism, in particular thinking about how we would sample from it most practically. Not least, without going through Prufer codes or some other clever machinery, it is not trivial to sample a uniform spanning tree.

First, we count the number of unicyclic graphs on n labelled vertices. If we know that the vertices on the cycle are $v_1,\ldots,v_k$, then the number of cycles with an identified edge is

$u_1=1,\quad u_k=\frac{k!}{2},\, k\ge 2.$

If we know that the tree coming off the cycle from vertex v_i has size m, say, then each of the possible rooted labelled trees with size m is equally likely. So taking $w_j=j^{j-1}$, the number of rooted trees on j labelled vertices, we get $B_n(u_\bullet,w_\bullet)$ for the number of such unicyclic graphs on [n]. Recall $B_n$ is the nth Bell polynomial, which gives the size of a compound combinatorial structure, where we have some structure on blocks and some other structure within blocks. Then the random partition of [n] given by the tree sizes has the distribution $\text{Gibbs}_n(u_\bullet,w_\bullet)$.

Consider now a related object, the so-called random mapping digraph. What follows is taken from Chapter 9 of Combinatorial Stochastic Processes. We can view any mapping $M_n:[n]\rightarrow[n]$ as a digraph where every vertex has out-degree 1. Each such digraph contains a collection of directed cycles, supported on those elements x for which $M_n^k(x)=x$ for some k. Such an element x is called a cyclic point. Each cyclic point can be viewed as the root of a labelled tree.

In an identical manner to the unicyclic graph, the sizes of these directed trees in the digraph decomposition of a uniform random mapping is distributed as $\text{Gibbs}_n(\bullet !,w_\bullet)$. So this is exactly the same as the cycle deletion kernel, apart from in the probability that the partition has precisely one block. In practice, for large n, the probability of this event is very small in both cases. And if we wanted to sample the cycle deletion kernel exactly, we could choose the trivial partition with some probability p, and otherwise sample from the random mapping kernel, where p is chosen such that

$p+\frac{1-p}{B_n(\bullet !, w_\bullet)}=\frac{1}{B_n(u_\bullet,w_\bullet)}.$

At least we know from the initial definition of a random mapping, that $B_n(\bullet !,w_\bullet)=n^n$. The number of unicyclic graphs with an identified edge is less clear. It turns out that the partition induced by the random mapping has a nice limit, after rescaling, as the lengths of excursions away from 0 in the standard Brownian bridge on [0,1].

The time for a fuller discussion of this sort of phenomenon is in the context of Poisson-Dirichlet distributions, as the above exchangeable partition turns out to be PD(1/2,1/2). However, for now we remark that the jumps of a subordinator give a partition after rescaling. The case of a stable subordinator is particularly convenient, as calculations are made easier by the Levy-Khintchine formula.

A notable example is the stable-1/2 subordinator, which can be realised as the inverse of the local time process at zero of a Brownian motion. The jumps of this process are then the excursion lengths of the original Brownian motion. A calculation involving the tail of the w_j’s indicates that 1/2 is the correct parameter for a subordinator to describe the random mappings. Note that the number of blocks in the partition corresponds to the local time at zero of the Brownian motion. (This is certainly not obvious, but it should at least be intuitively clear why a larger local time roughly indicates more excursions which indicates more blocks.)

So it turns out, after checking some of the technicalities, that it will suffice to show that the rescaled number of blocks in the random mapping partition $\frac{|\Pi_n|}{\sqrt{n}}$ converges to the Raleigh density, which is a size-biased Normal random variable (hence effectively first conditioned to be positive), and which also is the distribution of the local time of the standard Brownian bridge.

After that very approximate description, we conclude by showing that the distribution of the number of blocks does indeed converge as we require. Recall Cayley’s formula $kn^{n-k-1}$ for the number of labelled forests on [n] with a specified set of k roots. We also need to know how many labelled forests there are with any set of roots. Suppose we introduce an extra vertex, labelled 0, and connect it only to the roots of a rooted labelled forest on [n]. This gives a bijection between unlabelled trees on {0,1,…,n} and labelled forests with a specified set of roots on [n]. So we can use Cayley’s original formula to conclude there are $(n+1)^{n-1}$ such forests. We can do a quick sanity check that these are the same, which is equivalent to showing

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

This odd way of writing it is well-motivated. The form of the LHS is reminiscent of a generating function, and the additional k suggests taking a derivative. Indeed, the LHS is the derivative

$\frac{d}{dx}(1+x)^n,$

evaluated at $\frac{1}{n}$. This is clearly the same as the RHS.

That said, having established that the random mapping partition is essentially the same, it is computationally more convenient to consider that instead. By the digraph analogy, we again need to count forests with k roots on n vertices, and multiply by the number of permutations of the roots. This gives:

$\mathbb{P}(|\Pi_n|=k)=\frac{kn^{n-k-1}\cdot k! \binom{n}{k}}{n^n}=\frac{k}{n}\prod_{i=1}^{k-1}\left(1-\frac{i}{n}\right).$

Now we can consider the limit. Being a bit casual with notation, we get:

$\lim \mathbb{P}(\frac{|\Pi_n|}{\sqrt{n}}\in dl)\approx \sqrt{n}dl \mathbb{P}(|\Pi_n|=l\sqrt{n}).$

Since the Raleigh distribution has density $l\exp(-\frac12 l^2)dl$, it suffices for this informal verification to check that

$\prod_{i=1}^{l\sqrt{n}}(1-\frac{i}{n})\approx \exp(-\frac12 l^2).$ (*)

We take logs, so the LHS becomes:

$\log(1-\frac{1}{n})+\log(1-\frac{2}{n})+\ldots+\log(1-\frac{l\sqrt{n}}{n}).$

If we view this as a function of l and differentiate, we get

$d(LHS)=\sqrt{n}dl \log (1-\frac{l}{\sqrt{n}})\approx \sqrt{n}dl \left[-\frac{l}{\sqrt{n}}-\frac{l^2}{2n}\right]\approx -ldl.$

When l is zero, the LHS should be zero, so we can obtain the desired result (*) by integrating then taking an exponential.

# The Chinese Restaurant Process

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

REFERENCES

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

# Urns and the Dirichlet Distribution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Polya’s Urn

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

# Increments of Random Partitions

The following is problem 2.1.4. from Combinatorial Stochastic Processes:

Let $X_i$ be the indicator of the event that i the least element of some block of an exchangeable random partition $\Pi_n$ of [n]. Show that the joint law of the $(X_i,1\leq i\leq n)$ determines the law of $\Pi_n$.

As Pitman says, this is a result by Serban Nacu, the paper for which can be found here. In this post I’m going to explain what an exchangeable random partition is, how to prove the result, and a couple of consequences.

The starting point is the question ‘what is an exchangeable random partition?’ The most confusing aspect is that there are multiple definitions depending on whether the blocks of the partition are sets or just integers corresponding to a size. Eg, {1,2,4} u {3} is a partition of [4], corresponding to the partition 3+1 of 4. Obviously one induces the other, and in an exchangeable setting the laws of one may determine the laws of the other.

In the second case, we assume 3+1 is the same partition as 1+3. If order does matter then we call it a composition instead. This gets a bit annoying for set partitions, as we don’t want these to be ordered either. But if we want actually to talk about the sets in question we have to give them labels, which becomes an ordering, so we need some canonical way to assign these labels. Typically we will say $\Pi_n=\{A_1,\ldots,A_k\}$, where the curly brackets indicate that we don’t care about order, and we choose the labels by order of appearance, so by increasing order of least elements.

We say that a random partition $\Pi_n$ of [n] is exchangeable if its distribution is invariant the action on partitions induced by the symmetric group. That is, relabelling doesn’t change probabilities. We can express this functionally by saying

$\mathbb{P}(\Pi_n=\{A_1,\ldots,A_k\})=p(|A_1|,\ldots,|A_k|),$

for p a symmetric function. This function is then called the exchangeable partition probability function (EPPF) by Pitman.

Consider a partition of 4 into sets of sizes 3 and 1. There is a danger that this definition looks like it might be saying that the probability that A_1 is the set of size 3 is the same as the probability that A_1 is the set of size 1. This would be a problem because we expect to see some size-biasing to the labelling. Larger sets are more likely to contain small elements, merely because they contain more elements. Fortunately the definition is not broken after all. The statement above makes no reference to the probabilities of seeing various sizes for A_1 etc. For that, we would have to sum over all partitions with that property. It merely says that the partitions:

$\{1,2,3\}\cup\{4\},\quad \{1,2,4\}\cup\{3\},\quad\{1,3,4\}\cup\{2\},\quad \{2,3,4\}\cup\{1\}$

have respective probabilities:

$p(3,1),\quad p(3,1),\quad p(3,1),\quad p(1,3),$

and furthermore these are equal.

Anyway, now let’s turn to the problem. The key idea is that we want to be looking at strings of 0s and 1s that can only arise in one way. For example, the string 10…01 can only arise corresponding to the partitions {1,2,…,n-1} u {n} and {1,2,…,n-2,n} u {n-1}. So now we know p(n-1,1) and so also p(1,n-1). Furthermore, note that 10…0 and 11…1 give the probabilities of 1 block of size n and n blocks of size 1 respectively at once.

So then the string 10…010 can only arise from partitions {1,2,…,n-2,n} u {n-1} or {1,2,…,n-2} u {n-1,n}. We can calculate the probability that it came from the former using the previously found value of p(n-1,1) and a combinatorial weighting, so the remaining probability is given by p(2,n-2). Keep going. It is clear what ‘keep going’ means in the case of p(a,b) but for partitions with more than two blocks it seems a bit more complicated.

Let’s fix k the number of blocks in partitions under consideration, and start talking about compositions, that is $a_1+\ldots+a_k=n$. The problem we might face in trying to generalise the previous argument is that potentially lots of compositions might generate the same sequence of 0s and 1s, so the ‘first time’ we consider a composition might be the same for more than one composition. Trying it out in the case k=3 makes it clear that this is not going to happen, but we need some partial ordering structure to explain why this is the case.

Recall that a composition with k blocks is a sequence $a=(a_1,\ldots,a_k)$ which sums to n. Let’s say a majorizes b if all its partial sums are at least as large. That is $a_1+\ldots+a_l\geq b_1+\ldots+b_l$ for all $1\leq l \leq k$. We say this is strict if at least one of the inequalities is strict. It is not hard to see that if a majorizes b then this is strict unless a = b.

Since we don’t care about ordering, we assume for now that all compositions are arranged in non-increasing order. So we find a partition corresponding to some such composition $a_1,\ldots,a_k$. The partition is:

$\{1,\ldots,a_1\}\cup\{a_1+1,\ldots,a_1+a_2\}\cup\{a_1+a_2+1,\ldots,a_1+a_2+a_3\}\cup\ldots\cup\{n-a_k,\ldots,n\}.$

This generates a sequence of 0s and 1s as describe above, with $a_i-1$ 0s between the i’th 1 and the (i+1)th 1. The claim is that given some composition which admits a partition with this same corresponding sequence, that composition must majorize a. Proof by induction on l. So in fact we can prove Nacu’s result inductively down the partial ordering described. We know the probability of the sequence of 0s and 1s corresponding to the partition of [n] described by assumption. We know the probability of any partition corresponding to a composition which majorizes a by induction, and we know how many partitions with this sequence each such composition generates. Combining all of this, we can find the probability corresponding to a.

Actually I’m not going to say much about consequences of this except to paraphrase very briefly what Nacu says in the paper. One of the neat consequences of this result is that it allows us to prove in a fairly straightforward way that the only infinite family of exchangeable random partitions with independent increments is the so-called Chinese Restaurant process.

Instead of attempting to prove this, I will explain what all the bits mean. First, the Chinese Restaurant process is the main topic of the next chapter of the book, so I won’t say any more about it right now, except that its definition is almost exact what is required to make this particular result true.

We can’t extend the definition of exchangeable to infinite partitions immediately, because considering invariance under the symmetric group on the integers is not very nice, in particular because there’s a danger all the probabilities will end up being zero. Instead, we consider restrictions of the partition to $[n]\subset\mathbb{N}$, and demand that these nest appropriately, and are exchangeable.

Independent increments is a meaningful thing to consider since one way to construct a partition, infinite or otherwise, is to consider elements one at a time in the standard ordering, either adding the new element to an already present block, or starting block. Since 0 or 1 in the increment sequence corresponds precisely to these events, it is meaningful to talk about independent increments.

# Bell Polynomials

Trees with a single cycle

When counting combinatorial objects, it is often the case that we have two types of structure present at different levels. The aim of this post is to introduce the Bell polynomials, which provides the most natural notation for describing this sort of situation, and to mention some of the results that become easier to derive in this framework. This post is based on material and exercises from Chapter 1 of Jim Pitman’s book Combinatorial Stochastic Processes, which is great, and also available online here.

The structures that Bell polynomials enumerate are called composite structures in this account. Rather than give a definition right away, I shall give an example. An object I have been thinking about in the past few weeks are graphs on n vertices containing precisely one cycle. Some of the background for this has been explained in recent posts.

In a recent post on Prufer codes, I gave the classical argument showing that the number of trees on n vertices is $n^{n-2}$. We might consider a unicyclic graph to be a tree with an extra edge. But if we consider the number of ways to add a further vertex to a tree, we get

$n^{n-2}\left[\binom{n}{2}-(n-1)\right]=n^{n-2}\binom{n-1}{2}.$

Obviously, we have overcounted. If the single cycle in a graph has length k, then the graph has been counted exactly k times in this enumeration. But it is not obvious how many graphs have a single cycle of length k.

Instead, we stop worrying about exactly how many of these there are, as there might not be a simple expression anyway. As soon as we start using them in any actual argument, it will be useful to know various properties about the graphs, but probably not exactly how many there are.

Let’s focus on this single cycle of length k say. If we remove the edges of the cycle, we are left with a collection of trees. Why? Well if there was a cycle in the remaining graph, then the original graph would have had at least two cycles. So we have a collection of trees, unsurprisingly called a forest. Remembering that some of the trees may in fact be a single vertex (on the cycle), it is clear that there is a bijection between these trees and the vertices of the cycle in the obvious way. We can think of the graph as a k-cycle, dressed with trees.

Alternatively, once we have specified its size, we can forget about the k-cycle altogether. The graph is precisely defined by a forest of k trees on n vertices, with a specified root in each tree indicating which vertex lies on the cycle, and a permutation specifying the cyclic ordering of the trees. We can write this as

$N_{n,k}=(k-1)!\sum_{(A_1,\ldots,A_k)\in\mathcal{P}^k(n)}a_1^{a_1-1}\cdot\ldots\cdot a_k^{a_k-1},\quad \text{for }a_i=|A_i|,$

where $\mathcal{P}^k(n)$ is the number of partitions of [n] with k blocks. Remember that the blocks in a partition are necessarily unordered. This makes sense in this setting as the cyclic permutation chosen from the (k-1)! possibilities specifies the order on the cycle.

Bell Polynomials

The key point about this description is that there are two types of combinatorial structure present. We have the rooted trees, and also a cyclic ordering of the rooted trees. Bell polynomials generalise this idea. It is helpful to be less specific and think of partitions of [n] into blocks. There are $w_j$ arrangements of any block of size j, and there are $v_k$ ways to arrange the blocks, if there are k of them. Note that we assume $v_k$ is independent of the arrangements within the collection of blocks. So in the previous example, $w_j=j^{j-2}$, and $v_k=(k-1)!$. Pitman denotes these sequences by $v_\bullet,w_\bullet$. Then the (n,k)th partial Bell polynomial, $B_{n,k}(w_\bullet)$ gives the number of divisions into k blocks:

$B_{n,k}(w_\bullet):=\sum_{(A_1,\ldots,A_k)\in\mathcal{P}^k(n)}\prod_{i=1}^k w_{a_i}.$

The total number of arrangements is given by the Bell polynomial

$B_n(v_\bullet,w_\bullet):=\sum_{k=1}^n v_k B_{n,k}(w_\bullet).$

Here are some other examples of Bell polynomials. The Stirling numbers of the first kind $c_{n,k}$ give the number of permutations of [n] with k cycles. Since we don’t want to impose any combinatorial structure on the set of cycles, we don’t need to consider $v_\bullet$, and the number of ways to make a j-cycle from a j-block is $w_j=(j-1)!$, so $c_{n,k}:=B_{n,k}((\bullet-1)!)$. Similarly, the Stirling numbers of the second kind $S_{n,k}$ give the number of permutations of [n] into k blocks. Almost by definition, $S_{n,k}:=B_{n,k}(1^\bullet)$, where $1^\bullet$ is defined to be the sequence containing all 1s.

Applications

So far, this is just a definition that gives an abbreviated description for the sizes of several interesting sets of discrete objects. Having clean notation is always important, but there are further advantages of using Bell polynomials. I don’t want to reproduce the entirety of the chapter I’ve read, so my aim for this final section is to give a very vague outline of why this is a useful formulation.

Bell polynomials can be treated rather nicely via generating functions. The key to this is to take a sum not over partitions, but rather over ordered partitions, which are exactly the same, except now we also care about the order of the blocks. This has the advantage that there is a correspondence between ordered partitions with k blocks and compositions with k terms. If the composition is $n_1+\ldots+n_k=n$, it is clear why there are $\binom{n}{n_1,\ldots,n_k}$ ordered partitions encoding this structure. This multinomial coefficient can be written as a product of factorials of $n_i$s over i, and so we can write:

$B_{n,k}(w_\bullet)=\frac{n!}{k!}\sum_{(n_1,\ldots,n_k)}\prod_{i=1}^k \frac{w_{n_i}}{n_i!}.$

This motivates considering the exponential generating function given by

$w(\xi)=\sum_{j=1}^\infty w_j\frac{\xi_j}{j!},$

as this leads to the neat expressions:

$B_{n,k}(w_\bullet)=n![\xi^n]\frac{w(\xi)^k}{k!},\quad B_n(v_\bullet,w_\bullet)=n![\xi^n]v(w(\xi)).$

The Bell polynomial $B_n(v_\bullet,w_\bullet)$ counts the number of partitions of [n] subject to some extra structure. If we choose uniformly from this set, we get a distribution on this combinatorial object, for which the Bell polynomial provides the normalising constant. If we then ignore the extra structure, the sequences $v_\bullet,w_\bullet$ induce a probability distribution on the set of partitions of n. This distribution is known as a Gibbs partition. It is interesting to consider when and whether it is possible to define a splitting mechanism such that the Gibbs partitions can be coupled to form a fragmentation process. This is the opposite of a coalescence process. Here, we have a sequence of masses, and at each integer time we have rules to determine which mass to pick, and a rule for how to break it into two pieces. It is certainly not the case that for an arbitrary splitting rule and sequences $v_\bullet,w_\bullet$, the one-step fragmentation of the Gibbs partition on n gives the corresponding Gibbs partition on (n-1).

CLT for random permutations

For the final demonstration of the use of Bell polynomials, I am going to sketch the outline of a solution to exercise 1.5.4. which shows that the number of cycles in a uniformly chosen permutation has a CLT. This is not at all obvious, since the number of permutations of [n] with k cycles is given by $B_{n,k}((\bullet-1)!)$ and there is certainly no simple form for this, so the possibility of doing a technical limiting argument seems slim.

For ease of notation, we copy Pitman and write $c_{n,k}:=B_{n,k}((\bullet-1)!)$ as before. First we show exercise 1.2.3. which asserts that

$x(x+1)\ldots(x+(n-1))=\sum_{k=1}^n c_{n,k}x^k.$

We argue combinatorially. The RHS is the number of ways to choose $\sigma\in S_n$ and a colouring of [n] with k colours such that the orbits of $\sigma$ are monochromatic. We prove that the LHS also has this property by induction on the number of vertices. We claim there is a 1-to-(x+n) map from configurations on n vertices to configurations on (n+1) vertices. Given $\sigma\in S_n$ and colouring, for any $a\in[n]$, we construct $\sigma_a\in S_{n+1}$ by $\sigma_a(a)=n+1$, $\sigma_a(n)=\sigma(a)$ and for all other x, $\sigma_a(x)=\sigma(x)$. We give n+1 the same colour as a. This gives us n possibilities. Alternatively, we can map (n+1) to itself and give it any colour we want. This gives us x possibilities. A slightly more careful argument shows that this is indeed a 1-to-(x+n) map, which is exactly what we require.

So the polynomial

$A_n(z)=\sum_{k=0}^nc_{n,k}z^k,$

has n real zeros, which allows us to write

$\frac{c_{n,k}}{A_n(1)}=\mathbb{P}(X_1+\ldots+X_n=k),$

where the Xs are independent but not identically distributed Bernoulli trials. The number of cycles is then given by this sum, and so becomes a simple matter to verify the CLT by checking a that the variances grows appropriately. As both mean and variance are asymptotically log n, we can conclude that:

$\frac{K_n - \log n}{\sqrt{\log n}}\stackrel{d}{\rightarrow} N(0,1).$

In a future post, I want to give a quick outline of section 1.3. which details how the Bell polynomials can be surprisingly useful to find the moments of infinitely divisible distributions.