Lecture 2 – Connectivity threshold

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

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

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

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

Immediate remarks

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

Heuristics – Isolated vertices

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

Method 1: second-moment method

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

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

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

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

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

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

Method 2: first-moment method

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

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

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

It turns out that it is annoying to count components of size k because when we take an expectation, we can’t study all connected components on k vertices equally, since the probability of a given component structure depends on the exact number of edges it contains (which must be at least k-1, but will be larger if it includes lots of cycles). It is much easier to study the number of spanning trees of components of size k. So if the component itself is a tree, it has exactly one spanning tree, and if it has more cyclic structure then it will have several spanning trees.

The point is that we expect the expected number of such components to be very very small when k\ge 2, so it doesn’t really matter if we overcount by some factor. Somewhat abusing notation, we define X_k to be the number of such spanning trees in G(n,p), so

\mathbb{E}[X_k]=\binom{n}{k} k^{k-2}p^{k-1}(1-p)^{k(n-k)}.

Here the terms determine, respectively:

  • the number of ways to choose the vertex set of the component;
  • the number of trees on that vertex set (Cayley’s formula);
  • the probability of E(G(n,p)) including the edges of the tree in question;
  • the probability that G(n,p) includes no edges between the vertex of the tree and the rest of the graph (so it is actually a component).

Applying Stirling’s formula, and bounding helpfully, we obtain

\mathbb{E}[X_k]\le n(npe)^k \exp(-k(n-k)p).

We have already seen (modulo the abuse of notation) that \mathbb{E}[X_1]\rightarrow 0. Now we need to show that

\mathbb{E}\left[ \sum_{k=2}^{\lfloor n/2\rfloor} X_k\right]\rightarrow 0.

At this point, we should optimise our approximations based on how easy they are to sum, rather than how strong they are. One weak option is

\mathbb{E}[X_k] \le n(npe^{1-np/2})^k,

but it turns out that this is perfectly strong enough, and gives a geometric series, which is easy to sum. The only issue is that we get

\mathbb{E}\left[\sum_{k=\ell}^{\lfloor n/2\rfloor} X_k\right] < n \left(\frac{\log n}{\sqrt{n}}\right)^\ell,

so the sum is only bounded as we want when \ell\ge 3. Fortunately, almost any argument will work to show \mathbb{E}[X_2]\rightarrow 0 separately.

Once we know that \mathbb{E}[\sum X_k] \rightarrow 0, we can use the fact that for non-negative-valued RVs X,

\mathbb{P}(X\ne 0)=\mathbb{P}(X\ge 1)\le \mathbb{E}[X],

which you can view as an example of Markov’s inequality if you want, but really is just one of the fundamental properties of integer-valued RVs that allows useful interchange of probabilities and expectations.

In any case, combining this shows that with high probability there is no cut into classes of size k and n-k for any k, which is the same as being connected.

Remarks

  • The similarity to the famous coupon collector problem is unsurprising, modulo the heuristic about isolated vertices. Under the right setup (continuous-time, for example), whether the coupons have been collected is independent between different coupons. That’s not the case here, as can be seen in the second-moment calculation. But under any version of the coupon collector problem, the second-moment calculation setup is essentially the same as here.
  • It becomes clear in the proof that the threshold for the existence of cuts into k and n-k with high probability is p(n)\sim \frac{\log n}{kn} for fixed k. In particular for every k>1, we are working with a range of p much larger than this threshold in the calculation above, so it’s unsurprising that we could make weak assumptions and approximations, and still show that the expected number of such cuts vanishes.
  • Once one has a coupling of G(n,p) for all p, one can ask whether the graph becomes connected at exactly the same time / value of p as the last isolated vertex receives an incident edge. I will not discuss this more here, as it’s one of the extended exercises, but this is just the start of a family of questions concerning whether closely-related properties occur simultaneously or not.

Renyi’s study of G(n,m)

Renyi’s original argument concerned the related model G(n,m), that is, a uniform choice amongst the graphs on vertex set [n] with exactly m edges. There’s no right or wrong answer to the question of which is the better choice as the canonical model for a random graph. However, for this question, I guess it is natural to pose the question for G(n,m) since there’s an obvious extremal answer in terms of edges. That is, we require at least n-1 edges before there’s any chance that the graph is connected (cf the exercise about trees), so it’s natural to ask how many edges are required before there’s a strong chance that the G(n,m) is connected.

Clearly, conditioning on m edges being present in G(n,p) gives G(n,m). We can view G(n,p) as a mixture of distributions G(n,m) across values of m from 0 to \binom{n}{2}. Some graph properties are clearly badly focused under this mixture, for example the property A={the graph has an even number of edges}.

However, for increasing events (which are preserved under the addition of edges), proving that an event holds whp for G(n,p) given it holds whp for G(n,m), and vice versa, is normally straightforward.

For example, given a sequence p(n), take m(n) to be a sequence such that E(G(n,p))\ge m(n) with high probability. This is just about the tail of the binomial distribution, so for example taking m(n)=\binom{n}{2}p - \omega(\sqrt{n^2p}) suffices. Then if increasing property A holds for G(n,m) with high probability, it holds for G(n,p) with high probability too.

One can argue similarly in the other direction.

An interesting adjustment to this argument is required in the final section of my paper with James Martin in ALEA concerning a model of random forests which does not have the natural coupling property enjoyed by G(n,p) and G(n,q). This may be the subject of an upcoming post, time permitting.

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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_js 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_is 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.

Hall’s Marriage Theorem

Hall’s Marriage Theorem gives conditions on when the vertices of a bipartite graph can be split into pairs of vertices corresponding to disjoint edges such that every vertex in the smaller class is accounted for. Such a set of edges is called a matching. If the sizes of the vertex classes are equal, then the matching naturally induces a bijection between the classes, and such a matching is called a perfect matching.

The number of possible perfect matchings of K_{n,n} is n!, which is a lot to check. Since it’s useful to have bijections, it’s useful to have matchings, so we would like a simple way to check whether a bipartite graph has a matching. Hall’s Marriage Theorem gives a way to reduce the number of things to check to 2^n, which is still large. However, much more importantly, the condition for the existence of a matching has a form which is much easier to check in many applications. The statement is as follows:

Given bipartite graph G with vertex classes X and Y, there is a matching of X into Y precisely when for every subset A\subset X, |\Gamma(A)|\ge |A|, where \Gamma(A) is the set of vertices joined to some vertex in A, called the neighbourhood of A.

Taking A=X, it is clear that |Y|\ge |X| is a necessary condition for the result to hold, unsurprisingly. Perhaps the most elementary standard proof proceeds by induction on the size of X, taking the smallest A to give a contradiction, then using the induction hypothesis to lift smaller matchings up to the original graph. This lifting is based on the idea that a subset relation between sets induces a subset relation between their neighbourhoods.

In this post, I want to consider this theorem as a special case of the Max-Flow-Min-Cut Theorem, as this will support useful generalisations much more easily. The latter theorem is a bit complicated notationally to set up, and I don’t want it to turn into the main point of this post, so I will summarise. The Wikipedia article, and lots of sets of lecture notes are excellent sources of more detailed definitions.

The setting is a weakly-connected directed graph, with two identified vertices, the source, with zero indegree, and the sink, with zero outdegree. Every other vertex lies on a path (not necessarily unique) between the source and the sink. Each edge has a positive capacity, which should be thought of as the maximum volume allowed to flow down the edge. A flow is a way of assigning values to each edge so that they do not exceed the capacity, and there is volume conservation at each interior vertex. That is, the flow into the vertex is equal to flow out of the vertex. The value of the flow is the sum of the flows out of the source, which is necessarily equal to the sum of flows into the sink.

A cut is a partition of the vertices into two classes, with the source in one and the sink in the other. The value of a cut is then the sum of the capacities of any edges going from the class containing the source to the class containing the sink. In most examples, the classes will be increasing, in the sense that any path from source to sink changes class exactly once.

The Max-Flow-Min-Cut Theorem asserts that the maximum value of a flow through the system is equal to the minimum value of a cut. The proof is elementary, though it relies on defining a sensible algorithm to construct a minimal cut from a maximal flow that is not going to be interesting to explain without more precise notation available.

First we explain why Hall’s Marriage Theorem is a special case of this result. Suppose we are given the setup of HMT, with edges directed from X to Y with infinite capacity. We add edges of capacity 1 from some new vertex x_0 to each vertex of X, and from each vertex of Y to a new vertex y_0. The aim is to give necessary and sufficient conditions for the existence of a flow of value |X|. Note that one direction of HMT is genuinely trivial: if there is a matching, then the neighbourhood size condition must hold. We focus on the other direction. If the maximum flow is less than |X|, then there should be a cut of this size as well. We can parameterise a cut by the class of vertices containing the source, say S. Let A=SnX. If \Gamma(A)\not\subset S, then there will be an infinite capacity edge in the cut. So if we are looking for a minimal cut, this should not happen, hence \Gamma(A)\subset S if S is minimal. Similarly, there cannot be any edges from X\A to \Gamma(A). The value of the cut can then be given by

|\Gamma(A)|+|X|-|A|, which is at least |X| if the neighbourhood size assumption is given. Note we can use the same method with the original edges having capacity one, but we have to track slightly more quantities.

This topic came up because I’ve been thinking about fragmentation chains over this holiday. I have a specific example concerning forests of unrooted trees in mind, but won’t go into details right now. The idea is that we often have distributions governing random partitions of some kind, let’s say of [n]. Conditioning on having a given number of classes might give a family of distributions P_{n,k} for the partitions of [n] into k parts. We would be interested to know how easy it is to couple these distributions in a nice way. One way would be via a coalescence or fragmentation process. In the latter, we start with [n] itself, then at each step, split one of the parts into two according to some (random, Markovian) rule. We are interested in finding out whether such a fragmentation process exists for a given distribution.

It suffices to split the problem into single steps. Can we get from P_{n,k} to P_{n,k+1}?

The point I want to make is that this is just a version of Hall’s Marriage Theorem again, at least in terms of proof method. We can take X to be the set of partitions of [n] into k parts, and Y the set of partitions into (k+1) parts. Then we add a directed edge with infinite capacity between x\in X and y\in Y if y can be constructed from x by breaking a part into two. Finally, we connect a fresh vertex x_0 to each edge in X, only now we insist that the capacity is equal to P_{n,k}(x), and similarly an edge from y to y_0 with capacity equal to P_{n,k+1}(y). The existence of a fragmentation chain over this step is then equivalent to the existence of a flow of value 1 in the directed graph network.

Although in many cases this remains challenging to work with, which I will explore in a future post perhaps, this is nonetheless a useful idea to have in mind when it comes to deciding on whether such a construction is possible for specific examples.

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.