# 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.

# CLT and Stable Distributions

One of the questions I posed at the end of the previous post about the Central Limit Theorem was this: what is special about the normal distribution?

More precisely, for a large class of variables (those with finite variance) the limit in distribution of $S_n$ after a natural rescaling is distributed as N(0,1). As a starting point for investigating similar results for a more general class of underlying distributions, it is worth considering what properties we might require of a distribution if it is to appear as a limit in distribution of sums of IID RVs, rescaled if necessary.

The property required is that the distribution is stable. In the rest of the post I am going to give an informal precis of the content of the relevant chapter of Feller.

Throughout, we assume a collection of IID RVs, $X,X_1,X_2,\ldots$, with the initial sums $S_n:=X_1+\ldots+X_n$. Then we say $X$ is stable in the broad sense if

$S_n\stackrel{d}{=}c_nX+\gamma_n,$

for some deterministic parameters $c_n,\gamma_n$ for every n. If in fact $\gamma_n=0$ then we say $X$ is stable in the strict sense. I’m not sure if this division into strict and broad is still widely drawn, but anyway. One interpretation might be that a collection of distributions is stable if they form a non-trivial subspace of the vector space of random variables and also form a subgroup under the operation of adding independent RVs. I’m not sure that this is hugely useful either though. One observation is that if $\mathbb{E}X$ exists and is 0, then so are all the $\gamma_n$s.

The key result to be shown is that

$c_n=n^{1/\alpha}$ for some $0<\alpha\leq 2$.

Relevant though the observation about means is, a more useful one is this. The stability property is retained if we replace the distribution of $X$ with the distribution of $X_1-X-2$ (independent copies naturally!). The behaviour of $c_n$ is also preserved. Now we can work with an underlying distribution that is symmetric about 0, rather than merely centred. The deduction that $\gamma_n=0$ still holds now, whether or not X has a mean.

Now we proceed with the proof. All equalities are taken to be in distribution unless otherwise specified. By splitting into two smaller sums, we deduce that

$c_{m+n}X=S_{m+n}=c_mX_1+c_nX_2.$

Extending this idea, we have

$c_{kr}X=S_{kr}=S_k^{(1)}+\ldots+S_k^{(r)}=c_kX_1+\ldots+c_kX_r=c_kS_r=c_kc_rX.$

Note that it is not even obvious yet that the $c_n$s are increasing. To get a bit more control, we proceed as follows. Set $v=m+n$, and express

$X=\frac{c_m}{c_v}X_1+\frac{c_n}{c_v}X_2,$

from which we can make the deduction

$\mathbb{P}(X>t)\geq \mathbb{P}(X_1>0,X_2>t\frac{c_v}{c_n})=\frac12\mathbb{P}(X_2>t\frac{c_v}{c_n}).$ (*)

So most importantly, by taking $t>>0$ in the above, and using that X is symmetric, we can obtain an upper bound

$\mathbb{P}(X_2>t\frac{c_v}{c_n})\leq \delta<\frac12,$

in fact for any $\delta<\frac12$ if we take $t$ large enough. But since

$\mathbb{P}(X_2>0)=\frac12(1-\mathbb{P}(X_2=0)),$

(which should in most cases be $\frac12$), this implies that $\frac{c_v}{c_n}$ cannot be very close to 0. In other words, $\frac{c_n}{c_v}$ is bounded above. This is in fact regularity enough to deduce that $c_n=n^{1/\alpha}$ from the Cauchy-type functional equation (*).

It remains to check that $\alpha\leq 2$. Note that this equality case $\alpha=2$ corresponds exactly to the $\frac{1}{\sqrt{n}}$ scaling we saw for the normal distribution, in the context of the CLT. This motivates the proof. If $\alpha>2$, we will show that the variance of X is finite, so CLT applies. This gives some control over $c_n$ in an $n\rightarrow\infty$ limit, which is plenty to ensure a contradiction.

To show the variance is finite, we use the definition of stable to check that there is a value of t such that

$\mathbb{P}(S_n>tc_n)<\frac14\,\forall n.$

Now consider the event that the maximum of the $X_i$s is $>tc_n$ and that the sum of the rest is non-negative. This has, by independence, exactly half the probability of the event demanding just that the maximum be bounded below, and furthermore is contained within the event with probability $<\frac14$ shown above. So if we set

$z(n)=n\mathbb{P}(X>tc_n)$

we then have

$\frac14>\mathbb{P}(S_n>tc_n)\geq\frac12\mathbb{P}(\max X_i>tc_n)=\frac12[1-(1-\frac{z}{n})^n]$

$\iff 1-e^{-z(n)}\leq \frac12\text{ for large }n.$

So, $z(n)=n(1-F(tc_n))$ is bounded as $n$ varies. Rescaling suitably, this gives that

$x^\alpha(1-R(x))

This is exactly what we need to control the variance, as:

$\mathbb{E}X^2=\int_0^\infty \mathbb{P}(X^2>t)dt=\int_0^\infty \mathbb{P}(X^2>u^2)2udu$

$=\int_0^\infty 4u\mathbb{P}(X>u)du\leq \int_0^\infty 1\wedge\frac{4M}{u^{-(\alpha-1)}}du<\infty,$

using that X is symmetric and that $\alpha>2$ for the final equalities. But we know from CLT that if the variance is finite, we must have $\alpha=2$.

All that remains is to mention how stable distributions fit into the context of limits in distribution of RVs. This is little more than a definition.

We say F is in the domain of attraction of a broadly stable distribution R if

$\exists a_n>0,b_n,\quad\text{s.t.}\quad \frac{S_n-b_n}{a_n}\stackrel{d}{\rightarrow}R.$

The role of $b_n$ is not hugely important, as a broadly stable distribution is in the domain of attraction of the corresponding strictly stable distribution.

The natural question to ask is: do the domains of attraction of stable distributions (for $0<\alpha\leq 2$) partition the space of probability distributions, or is some extra condition required?

Next time I will talk about stable distributions in a more analytic context, and in particular how a discussion of their properties is motivated by the construction of Levy processes.