# 100k Views

When I started this blog, I implicitly made a promise to myself that I would be aiming for quality of posts rather than quantity of posts. Evaluating quality is hard, whereas one always feels some vague sense of purpose by seeing some measure of output increase. Nonetheless, I feel I have mostly kept this promise, and haven’t written that much solely for the sake of getting out another post. This post is something of an exception, since I noticed in the office on Friday that this blog was closing in on 100,000 page views. Some of these were not actually just me pressing F5 repeatedly. Obviously this is no more relevant an integer to be a threshold as any other, and one shouldn’t feel constrained by base 10, but it still feels like a good moment for a quick review.

Here are some interesting statistics over the $(3+\epsilon)$ years of this website’s existence.

• 175 posts, not including this one, or the three which are currently in draft status. This works out at about 4.7 posts a month. By some margin the most prolific period was May 2012, when I was revising for my Part III exams in Cambridge, and a series of posts about the fiddliest parts of stochastic calculus and network theory seemed a good way to consolidate this work. I’ve learned recently that PhDs are hard, and in fact it’s been a year since I last produced at least five posts in a month, if you discount the series of olympiad reports, which though enjoyable, don’t exactly require a huge amount of mathematical engagement.
• By at least one order of magnitude, the most viewed day was 17th August 2014, when several sources simultaneously linked to the third part of my report on IMO 2014 in Cape Town. An interesting point to note is that WordPress counts image views separately to page views, so the rare posts which have a gallery attached count well in a high risk / high return sense. In any case, the analytics consider that this day resulted in 2,366 views by 345 viewers. During a typical period, the number of views per visitor fluctuates between roughly 1.5 and 1.8, so clearly uploading as many photos of maths competitions as possible is the cheap route to lots of hits, at least by the WordPress metric.

• One might well expect the distributions involved in such a setup to follow a power-law. It’s not particularly clear from the above data about views per month since late 2012 whether this holds. One anomalously large data point (corresponding to the interest in IMO 2014) does not indicate a priori a power law tail… In addition, there is a general upward trend. Since a substantial proportion of traffic arrives from Google, one might naively assume that traffic rate might be proportion to amount of content, which naturally will grow in time, though it seems impractical to test this. One might also expect more recent posts to be more popular, though in practice this seems not to have been the case.
• The variance in popularity of the various posts has been surprisingly large. At some level, I guess I thought there would be more viewers who browse through lots of things, but such people would probably do so from the home page, so it doesn’t show up as viewing lots of different articles. There is some internal linking between some pages, but not enough to be a major effect.
• At either end of the spectrum, a post about coupling and the FKG inequality has received only 16 views in 2.5 years, while a guide to understanding the Levy-Khintchine formula has, in slightly less time, racked up 2,182 hits. There are direct reasons for this. Try googling Levy-Khintchine formula and see what comes up. In a sense, this is not enough, since you also need people to be googling the term in question, and picking topics that are hard but not super-hard at a masters level is probably maximising interest. But I don’t have a good underlying reason for why some posts should end up being more Google-friendly than others.
• In particular, quality of article seems at best loosely correlated with number of views. This is perhaps worrying, both for my reputation, and for the future of written media, but we will see. Indeed, two articles on the Dubins-Schwarz theorem and a short crib sheet for convergence of random variables, both get a regular readership, despite seeming to have been written (in as much as a blog post can be) on the back of an envelope. I also find it amusing that the Dubins-Schwarz theorem is always viewed at the same time of the year, roughly mid-February, as presumably it comes up in the second term of masters courses, just like it did in my own.
• By contrast, I remain quite pleased with the Levy-Khintchine article. It’s the sort of topic which is perfectly suited to this medium, since most books on Levy processes seem to assume implicit that their readers will already be familiar with this statement. So it seemed like a worthwhile enterprise to summarise this derivation, and it’s nice to see that others clearly feel the same, and indeed I still find some posts of this flavour useful as revision for myself.

• This seemed like a particularly good data set in which to go hunting for power-laws. I appreciate that taking a print-screen of an Excel chart will horrify many viewers, but never mind. The above plot shows the log of page view values for those mathematical articles with at least 200 hits. You can see the Levy-Khintchine as a mild anomaly at the far left. While I haven’t done any serious analysis, this looks fairly convincing.
• I haven’t engaged particularly seriously in interaction with other blogs and other websites. Perhaps I should have done? I enjoy reading such things, but networking in this fashion seems like a zero-sum game overall except for a few particularly engaged people, even if one gets a pleasing spike in views from a reciprocal tweet somewhere. As a result, the numbers of comments and out-going clicks here is essentially negligible.
• Views from the UK narrowly outnumber views from the US, but at present rates this will be reversed very soon. I imagine if I discounted the olympiad posts, which are sometimes linked from UK social media, this would have happened already.
• From purely book-keeping curiosity, WordPress currently thinks the following countries (and territories – I’m not sure how the division occurs…) have recorded exactly one view: Aaland Islands, Afghanistan, Belize, Cuba, Djibouti, El Salvador, Fiji, Guatemala, Guernsey, Laos, Martinique, Mozambique, New Caledonia, Tajikistan, US Virgin Islands, and Zambia. Visiting all of those would be a fun post-viva trip…

Conclusion

As I said, we all know that 100,000 is just a number, but taking half an hour to write this has been a good chance to reflect on what I’ve written here in the past three years. People often ask whether I would recommend that they start something similar. My answer is ‘probably yes’, so long as the writer is getting something out of most posts they produce in real time. When writing about anything hard and technical, you have to accept that until you become very famous, interest in what you produce is always going to be quite low, so the satisfaction has to be gained from the process of understanding and digesting the mathematics itself. None of us will be selling the musical rights any time soon.

I have two pieces of advice to anyone in such a position. 1) Wait until you’ve written five posts before publishing any of them. This is a good test of whether you actually want to do it, and you’ll feel much more plausible linking to a website with more than two articles on it. 2) Don’t set monthly post count targets. Tempting though it is to try this to prevent your blog dying, it doesn’t achieve anything in the long term. If you have lots to say, say lots; if you don’t, occasionally saying something worthwhile feels a lot better when you look back on it than producing your target number of articles which later feel underwhelming.

I don’t know whether this will make it to $6+2\epsilon$ years, but for now, I’m still enjoying the journey through mathematics.

# 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.

# Subordinators and the Arcsine rule

After the general discussion of Levy processes in the previous post, we now discuss a particular class of such processes. The majority of content and notation below is taken from chapters 1-3 of Jean Bertoin’s Saint-Flour notes.

We say $X_t$ is a subordinator if:

• It is a right-continuous adapted stochastic process, started from 0.
• It has stationary, independent increments.
• It is increasing.

Note that the first two conditions are precisely those required for a Levy process. We could also allow the process to take the value $\infty$, where the hitting time of infinity represents ‘killing’ the subordinator in some sense. If this hitting time is almost surely infinite, we say it is a strict subordinator. There is little to be gained right now from considering anything other than strict subordinators.

Examples

• A compound Poisson process, with finite jump measure supported on $[0,\infty)$. Hereafter we exclude this case, as it is better dealt with in other languages.
• A so-called stable Levy process, where $\Phi(\lambda)=\lambda^\alpha$, for some $\alpha\in(0,1)$. (I’ll define $\Phi$ very soon.) Note that checking that the sample paths are increasing requires only that $X_1\geq 0$ almost surely.
• The hitting time process for Brownian Motion. Note that this does indeed have jumps as we would need. (This has $\Phi(\lambda)=\sqrt{2\lambda}$.)

Properties

• In general, we describe Levy processes by their characteristic exponent. As a subordinator takes values in $[0,\infty)$, we can use the Laplace exponent instead:

$\mathbb{E}\exp(-\lambda X_t)=:\exp(-t\Phi(\lambda)).$

• We can refine the Levy-Khintchine formula;

$\Phi(\lambda)=k+d\lambda+\int_{[0,\infty)}(1-e^{-\lambda x})\Pi(dx),$

• where k is the kill rate (in the non-strict case). Because the process is increasing, it must have bounded variation, and so the quadratic part vanishes, and we have a stronger condition on the Levy measure: $\int(1\wedge x)\Pi(dx)<\infty$.
• The expression $\bar{\Pi}(x):=k+\Pi((x,\infty))$ for the tail of the Levy measure is often more useful in this setting.
• We can think of this decomposition as the sum of a drift, and a PPP with characteristic measure $\Pi+k\delta_\infty$. As we said above, we do not want to consider the case that X is a step process, so either d>0 or $\Pi((0,\infty))=\infty$ is enough to ensure this.

Analytic Methods

We give a snapshot of a couple of observations which make these nice to work with. Define the renewal measure U(dx) by:

$\int_{[0,\infty)}f(x)U(dx)=\mathbb{E}\left(\int_0^\infty f(X_t)dt\right).$

If we want to know the distribution function of this U, it will suffice to consider the indicator function $f(x)=1_{X_t\leq x}$ in the above.

The reason to exclude step processes specifically is to ensure that X has a continuous inverse:

$L_x=\sup\{t\geq 0:X_t\leq x\}$ so $U(x)=\mathbb{E}L_x$ is continuous.

In fact, this renewal measure characterises the subordinator uniquely, as we see by taking the Laplace transform:

$\mathcal{L}U(\lambda)=\int_{[0,\infty)}e^{-\lambda x}U(dx)=\mathbb{E}\int e^{-\lambda X_t}dt$

$=\int \mathbb{E}e^{-\lambda X_t}dt=\int\exp(-t\Phi(\lambda))dt=\frac{1}{\Phi(\lambda)}.$

The Arcsine Law

X is Markov, which induces a so-called regenerative property on the range of X, $\mathcal{R}$. Formally, given s, we do not always have $s\in\mathcal{R}$ (as the process might jump over s), but we can define $D_s=\inf\{t>s:t\in\mathcal{R}\}$. Then

$\{v\geq 0:v+D_s\in\mathcal{R}\}\stackrel{d}{=}\mathcal{R}.$

In fact, the converse holds as well. Any random set with this regenerative property is the range of some subordinator. Note that $D_s$ is some kind of dual to X, since it is increasing, and the regenerative property induces some Markovian properties.

In particular, we consider the last passage time $g_t=\sup\{s, in the case of a stable subordinator with $\Phi(\lambda)=\lambda^\alpha$. Here, $\mathcal{R}$ is self-similar with scaling exponent $\alpha$. The distribution of $\frac{g_t}{t}$ is thus independent of t. In this situation, we can derive the generalised arcsine rule for the distribution of $g_1$:

$\mathbb{R}(g_1\in ds)=\frac{\sin \alpha\pi}{\pi}s^{\alpha-1}(1-s)^{-\alpha}ds.$

The most natural application of this is to the hitting time process of Brownian Motion, which is stable with $\alpha=\frac12$. Then $g_1=S_1-B_1$, in the usual notation for the supremum process. Furthermore, we have equality in distribution of the processes (see previous posts on excursion theory and the short aside which follows):

$(S_t-B_t)_{t\geq 0}\stackrel{d}{=}(|B_t|)_{t\geq 0}.$

So $g_1$ gives the time of the last zero of BM before time 1, and the arcsine law shows that its distribution is given by:

$\mathbb{P}(g_1\leq t)=\frac{2}{\pi}\text{arcsin}\sqrt{t}.$

# The Levy-Khintchine Formula

Because of a string of coincidences involving my choice of courses for Part III and various lecturers’ choices about course content, I didn’t learn what a Levy process until a few weeks’ ago. Trying to get my head around the Levy-Khintchine formula took a little while, so the following is what I would have liked to have been able to find back then.

A Levy process is an adapted stochastic process started from 0 at time zero, and with stationary, independent increments. This is reminiscent, indeed a generalisation, of the definition of Brownian motion. In that case, we were able to give a concrete description of the distribution of $X_1$. For a general Levy process, we have

$X_1=X_{1/n}+(X_{2/n}-X_{1/n})+\ldots+(X_1-X_{1-1/n}).$

So the distribution of $X_1$ is infinitely divisible, that is, can be expressed as the distribution of the sum n iid random variables for all n. Viewing this definition in terms of convolutions of distributions may be more helpful, especially as we will subsequently consider characteristic functions. If this is the first time you have seen this property, note that it is not a universal property. For example, it is not clear how to write a U[0,1] random variable as a convolution of two iid RVs. Note that exactly the same argument suffices to show that the distribution of $X_t$ is infinitely divisible.

It will be most convenient to work with the characteristic functions

$\mathbb{E}\exp(i\langle \lambda,X_t\rangle).$

By stationarity of increments, we can show that this is equal to

$\exp(-\Psi(\lambda)t)\quad\text{where}\quad \mathbb{E}\exp(i\langle \lambda,X_1\rangle)=:\exp(-\Psi(\lambda)).$

This function $\Psi(\lambda)$ is called the characteristic exponent. The argument resembles that used for Cauchy’s functional equations, by dealing first with the rationals using stationarity of increments, then lifting to the reals by the (right-)continuity of

$t\mapsto \mathbb{E}\exp(i\langle \lambda,X_t\rangle).$

As ever, $\Psi(\lambda)$ uniquely determines the distribution of $X_1$, and so it also uniquely determines the distribution of Levy process. The only condition on $\Psi$ is that it be the characteristic function of an infinitely divisible distribution. This condition is given explicitly by the Levy-Khintchine formula.

Levy-Khintchine

$\Psi(\lambda)$ is the characteristic function of an infinitely divisible distribution iff

$\Psi(\lambda)=i\langle a,\lambda\rangle +\frac12 Q(\lambda)+\int_{\mathbb{R}^d}(1-e^{i\langle \lambda,x\rangle}+i\langle \lambda,x\rangle 1_{|x|<1})\Pi(dx).$

for $a\in\mathbb{R}^d$, Q a quadratic form on $\mathbb{R}^d$, and $\Pi$ a so-called Levy measure satisfying $\int (1\wedge |x|^2)\Pi(dx)<\infty$.

This looks a bit arbitrary, so first let’s explain what each of these terms ‘means’.

• $i\langle a,\lambda\rangle$ comes from a drift of $-a$. Note that a deterministic linear function is a (not especially interesting) Levy process.
• $\frac12Q(\lambda)$ comes from a Brownian part $\sqrt{Q}B_t$.

The rest corresponds to the jump part of the process. Note that a Poisson process is an example of a Levy process, hence why we might consider thinking about jumps in the first place. The reason why there is an indicator function floating around is that we have to think about two regimes separately, namely large and small jumps. Jumps of size bounded below cannot happen too often as otherwise the process might explode off to infinity in finite time with positive probability. On the other hand, infinitesimally small jumps can happen very often (say on a dense set) so long as everything is controlled to prevent an explosion on the macroscopic scale.

There is no canonical choice for where the divide between these regimes happens, but conventionally this is taken to be at $|x|=1$. The restriction on the Levy measure near 0 ensures that the sum of the squares all jumps up some finite time converges absolutely.

• $\Pi\cdot 1_{|x|\geq 1}$ gives the intensity of a standard compound Poisson process. The jumps are well-spaced, and so it is a relatively simple calculation to see that the characteristic function is

$\int_{\mathbb{R}^d}(1-e^{i\langle \lambda,x\rangle})1_{|x|\geq 1}\Pi(dx).$

The intensity $\Pi\cdot 1_{|x|<1}$ gives infinitely many hits in finite time, so if the expectation of this measure is not 0, we explode immediately. We compensate by drifting away from this at rate

$\int_{\mathbb{R}^d}x1_{|x|<1}\Pi(dx).$

To make this more rigorous, we should really consider $1_{\epsilon<|x|<1}$ then take a limit, but this at least explains where all the terms come from. Linearity allows us to interchange integrals and inner products, to get the term

$\int_{\mathbb{R}^d}(1-e^{-i\langle \lambda,x\rangle}+i\langle\lambda,x\rangle 1_{|x|<1})\Pi(dx).$

If the process has bounded variation, then we must have Q=0, and also

$\int (1\wedge |x|)\Pi(dx)<\infty,$

that is, not too many jumps on an |x| scale. In this case, then this drift component is well-defined and linear $\lambda$, so can be incorporated with the drift term at the beginning of the Levy-Khintchine expression. If not, then there are some $\lambda$ for which it does not exist.

There are some other things to be said about Levy processes, including

• Stable Levy processes, where $\Psi(k\lambda)=k^\alpha \Psi(\lambda)$, which induces the rescaling-invariance property: $k^{-1/\alpha}X_{kt}\stackrel{d}{=}X$. The distribution of each $X_t$ is then also a stable distribution.
• Resolvents, where instead of working with the process itself, we work with the distribution of the process at a random exponential time.