# Skorohod embedding

Background

Suppose we are given a standard Brownian motion $(B_t)$, and a stopping time T. Then, so long as T satisfies one of the regularity conditions under which the Optional Stopping Theorem applies, we know that $\mathbb{E}[B_T]=0$. (See here for a less formal introduction to OST.) Furthermore, since $B_t^2-t$ is a martingale, $\mathbb{E}[B_T^2]=\mathbb{E}[T]$, so if the latter is finite, so is the former.

Now, using the strong Markov property of Brownian motion, we can come up with a sequence of stopping times $0=T_0, T_1, T_2,\ldots$ such that the increments $T_k-T_{k-1}$ are IID with the same distribution as T. Then $0,B_{T_1},B_{T_2},\ldots$ is a centered random walk. By taking T to be the hitting time of $\{-1,+1\}$, it is easy to see that we can embed simple random walk in a Brownian motion using this approach.

Embedding simple random walk in Brownian motion.

The Skorohod embedding question asks: can all centered random walks be constructed in this fashion, by stopping Brownian motion at a sequence of stopping time? With the strong Markov property, it immediately reduces the question of whether all centered finite-variance distributions X can be expressed as $B_T$ for some integrable stopping time T.

The answer to this question is yes, and much of what follows is drawn from, or at least prompted by Obloj’s survey paper which details the problem and rich history of the many approaches to its solution over the past seventy years.

Applications and related things

The relationship between random walks and Brownian motion is a rich one. Donsker’s invariance principle asserts that Brownian motion appears as the scaling limit of a random walk. Indeed, one can construct Brownian motion itself as the limit of a sequence of consistent random walks with normal increments on an increasingly dense set of times. Furthermore, random walks are martingales, and we know that continuous, local martingales can be expressed as a (stochastically) time-changed Brownian motion, from the Dubins-Schwarz theorem.

The Skorohod embedding theorem can be used to prove results about random walks with general distribution by proving the corresponding result for Brownian motion, and checking that the construction of the sequence of stopping times has the right properties to allow the result to be carried back to the original setting. It obviously also gives a coupling between a individual random walk and a Brownian motion which may be useful in some contexts, as well as a coupling between any pair of random walks. This is useful in proving results for random walks which are much easier for special cases of the distribution. For example, when the increments are Gaussian, or when there are combinatorial approaches to a problem about simple random walk. At the moment no aspect of this blog schedule is guaranteed, but I plan to talk about the law of the iterated logarithm shortly, whose proof is approachable in both of these settings, as well as for Brownian motion, and Skorohod embedding provides the route to the general proof.

At the end, we will briefly compare some other ways to couple a random walk and a Brownian motion.

One thing we could do is sample a copy of X independently from the Brownian motion, then declare $T= \tau_{X}:= \inf\{t\ge 0: B_t=X\}$, the hitting time of (random value) X. But recall that unfortunately $\tau_x$ has infinite expectation for all non-zero x, so this doesn’t fit the conditions required to use OST.

Skorohod’s original method is described in Section 3.1 of Obloj’s notes linked above. The method is roughly to pair up positive values taken by X appropriately with negative values taken by X in a clever way. If we have a positive value b and a negative value a, then $\tau_{a,b}$, the first hitting time of $\mathbb{R}\backslash (a,b)$ is integrable. Then we choose one of these positive-negative pairs according to the projection of the distribution of X onto the pairings, and let T be the hitting time of this pair of values. The probability of hitting b conditional on hitting {a,b} is easy to compute (it’s $\frac{-a}{b-a}$) so we need to have chosen our pairs so that the ‘probability’ of hitting b (ie the density) comes out right. In particular, this method has to start from continuous distributions X, and treat atoms in the distribution of X separately.

The case where the distribution X is symmetric (that is $X\stackrel{d}=-X$) is particularly clear, as then the pairs should be $(-x,x)$.

However, it feels like there is enough randomness in Brownian motion already, and subsequent authors showed that indeed it wasn’t necessary to introduce extra randomness to provide a solution.

One might ask whether it’s possible to generate the distribution on the set of pairs (as above) out of the Brownian motion itself, but independently from all the hitting times. It feels like it might be possible to make the distribution on the pairs measurable with respect to

$\mathcal{F}_{0+} = \bigcap\limits_{t>0} \mathcal{F}_t,$

the sigma-algebra of events determined by limiting behaviour as $t\rightarrow 0$ (which is independent of hitting times). But of course, unfortunately $\mathcal{F}_{0+}$ has a zero-one law, so it’s not possible to embed non-trivial distributions there.

Dubins solution

The exemplar for solutions without extra randomness is due to Dubins, shortly after Skorohod’s original argument. The idea is to express the distribution X as the almost sure limit of a martingale. We first use the hitting time of a pair of points to ‘decide’ whether we will end up positive or negative, and then given this information look at the hitting time (after this first time) of two subsequent points to ‘decide’ which of four regions of the real interval we end up in.

I’m going to use different notation to Obloj, corresponding more closely with how I ended up thinking about this method. We let

$a_+:= \mathbb{E}[X \,|\, X>0], \quad a_- := \mathbb{E}[X\,|\, X<0],$ (*)

and take $T_1 = \tau_{\{a_-,a_+\}}$. We need to check that

$\mathbb{P}\left( B_{T_1}=a_+\right) = \mathbb{P}\left(X>0\right),$

for this to have a chance of working. But we know that

$\mathbb{P}\left( B_{T_1}=a_+\right) = \frac{a_+}{a_+-a_-},$

and we can also attack the other side using (*) and the fact that $\mathbb{E}[X]=0$, using the law of total expectation:

$0=\mathbb{E}[X]=\mathbb{E}[X\,|\, X>0] \mathbb{P}(X>0) + \mathbb{E}[X\,|\,X<0]\mathbb{P}(X<0) = a_+ \mathbb{P}(X>0) + a_- \left(1-\mathbb{P}(X>0) \right),$

$\Rightarrow\quad \mathbb{P}(X>0)=\frac{a_+}{a_+-a_-}.$

Now we define

$a_{++}=\mathbb{E}[X \,|\, X>a_+],\quad a_{+-}=\mathbb{E}[X\,|\, 0

and similarly $a_{-+},a_{--}$. So then, conditional on $B_{T_1}=a_+$, we take

$T_2:= \inf_{t\ge T_1}\left\{ B_t\not\in (a_{+-},a_{++}) \right\},$

and similarly conditional on $B_{T_1}=a_-$. By an identical argument to the one we have just deployed, we have $\mathbb{E}\left[B_{T_2} \,|\,\mathcal{F}_{T_1} \right] = B_{T_1}$ almost surely. So, although the $a_{+-+}$ notation now starts to get very unwieldy, it’s clear we can keep going in this way to get a sequence of stopping times $0=T_0,T_1,T_2,\ldots$ where $B_{T_n}$ determines which of the $2^n$ regions of the real line any limit $\lim_{m\rightarrow\infty} B_{T_m}$ should lie in.

A bit of work is required to check that the almost sure limit $T_n\rightarrow T$ is almost surely finite, but once we have this, it is clear that $B_{T_n}\rightarrow B_T$ almost surely, and $B_T$ has the distribution required.

We want to know how close we can make this coupling between a centered random walk with variance 1, and a standard Brownian motion. Here, ‘close’ means uniformly close in probability. For large times, the typical difference between one of the stopping times $0,T_1,T_2,\ldots$ in the Skorohod embedding and its expectation (recall $\mathbb{E}[T_k]=k$) is $\sqrt{n}$. So, constructing the random walk $S_0,S_1,S_2,\ldots$ from the Brownian motion via Skorohod embedding leads to

$\left |S_k - B_k \right| = \omega(n^{1/4}),$

for most values of $k\le n$. Strassen (1966) shows that the true scale of the maximum

$\max_{k\le n} \left| S_k - B_k \right|$

is slightly larger than this, with some extra powers of $\log n$ and $\log\log n$ as one would expect.

The Komlos-Major-Tusnady coupling is a way to do a lot better than this, in the setting where the distribution of the increments has a finite MGF near 0. Then, there exists a coupling of the random walk and the Brownian motion such that

$\max_{k\le n}\left|S_k- B_k\right| = O(\log n).$

That is, there exists C such that

$\left[\max_{k\le n} \left |S_k-B_k\right| - C\log n\right] \vee 0$

is a tight family of distributions, indeed with uniform exponential tail. To avoid digressing infinitely far from my original plan to discuss the proof of the law of iterated logarithm for general distributions, I’ll stop here. I found it hard to find much coverage of the KMT result apart from the challenging original paper, and many versions expressed in the language of empirical processes, which are similar to random walks in many ways relevant to convergence and this coupling, but not for Skorohod embedding. So, here is a link to some slides from a talk by Chatterjee which I found helpful in getting a sense of the history, and some of the modern approaches to this type of normal approximation problem.

# Birthday Coincidences and Poisson Approximations

This morning, Facebook was extremely keen to remind me via every available medium that four of my friends celebrate their birthday today. My first thought was that I hope they all enjoy their day, and my second thought was to ask what the chance of this was. I have about 200 Facebook friends, and so this struck me as an unlikely occurrence. But this problem has form, and it felt worthwhile to try some calculations to see if my intuition was well-founded.

Siméon Denis Poisson celebrated his 234th birthday on 21st June this year.

The classical birthday problem

The starting point is the question: how many friends do you have to have before you expect to start seeing anyone sharing a birthday? There are a ridiculous number of articles about this on the web already, so I will say little, except that I don’t want to call this the ‘birthday paradox’, because it’s not a paradox at all. At best it might be counter-intuitive, but then the moral should be to change our intuition for this type of problem.

Throughout, let’s discount February 29th, as this doesn’t add much. So then, to guarantee having a shared pair of birthdays, you need to have 366 friends. But if you have a mere 23 friends, then the probability of having some pair that share a birthday is slightly greater than a half. The disparity between these two numbers leads to the counter-intuition. Some people might find it helpful to think that instead of counting friends, we should instead be counting pairs of friends, but I don’t personally find this especially helpful.

For me, thinking about the calculation in very slightly more generality is helpful. Here, and throughout, let’s instead take N to be the number of days in a year, and K the number of friends, or kids in the class if you prefer. Then, as usual, it is easier to calculate the probability that no two share a birthday (that is, that all the birthdays are distinct) than the probability that some two share a birthday. We could think of the number of ways to pick the set of birthdays, or we could look at the kids one-at-a-time, and demand that their birthday is not one of those we’ve already seen. Naturally, we get the same answer, that is

$\frac{^N P_K}{N^K} = 1\cdot \frac{N-1}{N}\cdot\ldots \frac{N-K+1}{N}.$

We’ve assumed here that all birthdates are equally likely. We’ll come back to this assumption right at the end. For now, let’s assume that both N and K are large, and we’ll try to decide roughly how large K has to be in relation to N for this answer to be away from 0 and 1. If we pair opposite terms up, we might approximate this by

$(\frac{N-\frac{K}{2}}{N})^K = (1-\frac{K}{2N})^K\approx e^{-K^2/2N}.$

In fact, AM-GM says that this is an overestimate, and a bit more care can be used to show that this is a good-approximation to first order. So we see that if $K=\Theta(\sqrt{N})$ for large N, we get a non-trivial limit.

Challenges for four-way shared birthdays

So the original problem I posed is harder, because there isn’t (unless I’m missing something) a natural way to choose birthdays one-at-a-time, or describe the set of suitable birthday sets. There are two major obstacles in a calculation such as this. Firstly, the overlap of people, that is we might have five or more birthdays overlapping; secondly, the overlap of days, that is we might have several days with four (or more) birthdays. We’ll end up worrying more about the second situation.

We start by eliminating both problems, by asking for the probability that exactly four friends are born on January 1st. The general form of this probability is $\frac{\binom{K}{4} }{N^4} \cdot (\frac{N-1}{N})^{K-4}$. Now, if $K\ll N$, this final term should not be significant. Removing this is not exactly the same as specifying the probability that at least four birthdays are on January 1st. But in fact this removal turns a lower bound (because {exactly four}<{at least four}) into an upper (in fact a union) bound. So if the factor being removed is very close to one, we can use whichever expression is more convenient.

In the real life case of N=365, K=200, this term is not negligible. But accounting for this, we get that the probability of exactly four birthdays on 1st January is ~0.0021. Our upper bound on the probability of at least four is ~0.0036.

But now that we know the probability for a given day, can we calculate $(1-0.0021)^{365}$ to estimate the probability that we never have four-overlap? When we did our previous iterative calculation, we were using independence of the different kids’ birthdays. But the event that we have four-overlap on January 1st is not quite independent of the event that we have four-overlap on January 2nd. Why? Well if we know at least four people were born on January 1st, there are fewer people left (potentially) to be born on January 2nd. But maybe this dependence is mild enough that we can ignore it?

We can, however, use some moment estimates. The expected number of days with four-overlap is $365\cdot 0.0021 \approx 0.77$. So the probability that there is at least one day with four-overlap is at most ~0.77.

But we really want a lower bound. So, maybe we can borrow exactly the second-moment argument we tried (there for isolated vertices in the random graph) in the previous post? Here, the probability that both January 1st and January 2nd are four-overlapping is

$\frac{\binom{K}{4}\binom{K-4}{4}}{N^8}\cdot (\frac{N-2}{N})^{K-8}\approx 4.3\times 10^{-6}.$

From this, we can evaluate the expectation of the square of the number of days with four-overlap, and thus find that the variance is ~0.74. So we use Chebyshev, calling this number of days #D for now:

$\mathbb{P}(\# D=0)\le \mathbb{P}(|\#D - \mathbb{E}\# D|^2 \ge (\mathbb{E}\# D)^2 ) \le \frac{\mathrm{Var} \# D}{(\mathbb{E} \#D)^2}.$

In our case, this unfortunately gives us an upper bound greater than 1 on this probability, and thus a lower bound of zero on the probability that there is at least one day with four-overlap. Which isn’t especially interesting…

Fairly recently, I spoke about the Lovasz Local Lemma, which can be used to find lower bounds on the probabilities of intersections of events, many of which are independent (in a particular precise sense). Perhaps this might be useful here? The natural choice of ‘bad event’ is that particular 4-sets of people share a birthday. There are $\binom{K}{4}$ such events, and each is independent of the collection of $\binom{K-4}{4}$ disjoint events. Thus we can consider using LLL if $e\cdot (\binom{K}{4}-\binom{K-4}{4})\cdot 0.0021 \le 1$. Unfortunately, this difference of binomial coefficients is large in our example, and so in fact the LHS has order $10^3$.

Random number of friends – coupling to a Poisson Process

All of these methods failed because without independence we had to use estimates which were really not tight at all. But we can re-introduce independence if we remove the constraints on the model. Suppose instead of demanding I have K friends, I instead demand that I have a random number of friends, with distribution Poisson(K). Now it is reasonable to assume that for each day, I have a Poisson(K/365) friends with that birthday, independently for each day.

If we end up having exactly K friends with this random choice, then the distribution of the number of 4-overlap days is exactly the same as in the original setup. However, crucially, if we end up having at most K friends with this random choice, the distribution of the number of 4-overlap days is stochastically dominated by the original distribution. So instead let’s assume we have Poisson(L) friends, where L<K, and see how well we can do. For definiteness, we’ll go back to N=365, K=200 now. Let’s say X is the distribution of birthdays in the original model, and $\Xi$ for the distribution of birthdays in the model with a random number of friends

Then

$\mathbb{P}(\exists \ge 4\text{-overlap in }\Xi) = 1- \mathbb{P}(\mathrm{Po}(L/365)\le 3)^365.$ (*)

Now we can write the crucial domination relation as

$\mathbb{P}(\exists \ge 4\text{-overlap in }X)\ge \mathbb{P}( \exists \ge 4\text{-overlap in }\Xi \,|\, |\Xi|\le 200),$

and then use an inequality version of the law of total probability to bound further as

$\ge \frac{ \mathbb{P}(\exists \ge 4\text{-overlap in }\Xi) - \mathbb{P}(|\Xi|>200)}{\mathbb{P}(|\Xi|\le 200)}.$

This is a function of L, and in principle we could find its maximum, perhaps as $N\rightarrow\infty$. Here, though, let’s just take L=365/2 and see what happens. For (*) we get ~0.472.

To estimate $\mathbb{P}(\mathrm{Po}(365/2)>200)$, observe that this event corresponds to 1.4 standard deviations above the mean, so we can approximate using quantiles of the normal distribution, via the CLT. (Obviously this isn’t completely precise, but it could be made precise if we really wanted.) I looked up a table, and this probability is, conveniently for calculations, roughly 0.1. Thus we obtain a lower bound of $\frac{0.472-0.1}{0.9}$. Allowing for the fairly weak estimates at various points, we still get a lower bound of around 0.4. Which is good, because it shows that my intuition wasn’t right, but that I was in the right ball-park for it being a ‘middle-probability event’.

Remarks and References

– The reason for doing the upper bound for the probability of exact 4-overlap is that the same argument for at-least-4-overlap would have given an upper bound of 1. However, this Poisson Process coupling is also a much better method for obtaining an upper bound on either event.

– Birthdays are not uniformly distributed through the year. The deviation is strong enough that even from the set of birth frequencies (rather than the sequence of birth frequencies), we can reject a null hypothesis of uniformity. Early September is pretty close to the maximum. Two comments: 1) this is the time of year where small variations in birth date have a big effect on education, especially in primary school; 2) we are 37 weeks into the year…

– It is known that 187 friends is the first time the probability of having at-least-4-overlap is greater than ½. You can find the full sequence on OEIS as A014088. I used to have about 650 Facebook friends, before I decided that I’d prefer instead the pleasant surprise of finding out what old acquaintances were up to when I next spoke to them. In this case, the median of the distribution of the largest number sharing a birthday would be seven.

– Eric Weisstein’s article on Mathworld is, in my opinion, the best resource for a mathematician on the first two pages of Google hits by at least two orders of magnitude. In the notation of this article, we were calculating $P_4(n=200,d=365)$. There are also some good general asymptotics, or at least recipes for asymptotics, in equations (17) and (18).

– The paper Methods for Studying Coincidences by Diaconis and Mosteller is, as one might expect, extremely readable, and summarises many results and applications, including several generalisations.

# Sharpness of Phase Transition in Percolation

On the flight to Romania a couple of weeks ago, I read this very nice paper by Duminil-Copin and Tassion in which, as a corollary to a more general result in a companion paper, they provide a short proof of a result about percolation on $\mathbb{Z}^d$. Mitch talked about this paper at our final Junior Probability Seminar of this term yesterday, so it seemed appropriate to write up something about this nice argument. I must confess I know relatively little about these problems, and in particular know nothing about how the original authors, Aizenmann + Barsky (1987), and Menshikov (1986) approached this problem, but experts have said that this is substantially easier to digest.

Rather than reference the first paper repeatedly, I remark now that everything which follows comes from there.

We consider conventional bond percolation on the edges of $\mathbb{Z}^d$, for $d\ge 2$, and are interested in whether the the origin percolates with positive probability. That is, that zero is contained in an infinite component. As usual we define $p_c=\sup\{p: \mathbb{P}_p(0\leftrightarrow\infty)=0\}$ to be the critical probability above which percolation happens with positive probability. Defining $\theta(p)=\mathbb{P}_p(0\leftrightarrow\infty)$, we do not know whether $\theta(p_c)=0$ for some values of d, notably d=3.

If the origin is connected to infinity, it is by definition connected to the boundary of every $\Lambda_n:=[-n,n]^d$. The number of distinct paths from the origin to $\partial \Lambda_n$ is bounded by the number of self-avoiding walks on the lattice of length n starting from 0, which grows at most as fast as $(2d-1)^n$. In particular, we know that $p_c\ge \frac{1}{2d-1}$, but also, for any $p<\frac{1}{2d-1}$, the probability $\mathbb{P}_p[0\leftrightarrow \partial\Lambda_n]$ decays exponentially in n. We would expect this in fact to hold for all $p, and this is something that the authors prove, called Item 1. They also show that the percolation probability grows at least linearly beyond $p_c$, specifically (called Item 2)

$\theta(p)\ge \frac{p-p_c}{p(1-p_c)}.$

The proof here proceeds by considering the function of subsets S which contain the origin:

$\varphi_p(S):= p\sum_{(x,y)\in\Delta S} \mathbb{P}_p[0\stackrel{S}{\leftrightarrow} x],\quad S\subset \mathbb{Z}^d$.

In words, this gives the expected number of edges across the boundary which are connected to the origin by a path within S. So this gives a measure of how likely we are to escape S, and in particular, an upper bound on the probability that an open path exists from 0 to outside S. The authors then define the alternative critical probability

$\tilde p_c := \sup_\{p\text{ s.t. }\exists\text{ finite }0\in S\text{ with }\varphi_p(S)<1\}.$

They will show that $\tilde p_c$ satisfies the statements of both Item 1 and Item 2. Item 2 for $\tilde p_c$ implies $\tilde p_c\le p_c$, and Item 1 for $\tilde p_c$ implies $p_c\le \tilde p_c$, so this is exactly what we need.

They show Item 1 first. We consider this set S for which $\varphi_p(S)<1$, and take some box $\Lambda_L$ which strictly contains S. Now we consider the probability of escaping from a box of size kL. The reason why considering this definition of S works really nicely is that it makes it possible to split this event of escaping from $\Lambda_{kL}$ into an event involving subjects of various disjoint sets of edges being open, so we can use independence.

We decompose the path from 0 to $\partial\Lambda_{kL}$ based on the first time it leaves S. We are mindful that there might be lots of paths from from 0 to this boundary. The way we are bounding means it doesn’t matter if we have lots of suitable paths, but they should all spend a maximal number of steps in S, in the sense that whenever the path re-enters S, say to vertex z, there is no open path from 0 to z contained in S. Let the vertex on $\partial S$ we leave from for the first time be x. Then, for all vertices y later in the path, $0\stackrel{S}{\not\leftrightarrow}y$.

So under any suitable path, now take y to be the vertex directly following x, hence $(x,y)\in\Delta S$. If we take $\mathcal{C}$ to be the set of vertices z for which $0\stackrel{S}{\leftrightarrow}z$, we can split the expression based on S to obtain:

$\mathbb{P}_p[0\leftrightarrow \partial \Lambda_{kL}]\le p \sum_{(x,y)\in\Delta S}\sum_{C\subset S} \mathbb{P}_p[0\stackrel{S}{\leftrightarrow} x,\mathcal{C}=C] \mathbb{P}_p[y\stackrel{\mathbb{Z}^d\backslash C}{\leftrightarrow}\partial\Lambda_{kL}].$

Splitting based on C gives us independence between all of these sets of edges, but then we immediately forget about it. Irrespective of choice of y (recall, $y\in S\subset \Lambda_L$), this final probability is definitely bounded by $\mathbb{P}_p[0\leftrightarrow \partial \Lambda_{(k-1)L}]$, while the p and the first term can be summed over C to give $\varphi_p(S)$. They obtain:

$\mathbb{P}_p[0\leftrightarrow \partial \Lambda_{kL}] \le \varphi_p(S)\mathbb{P}_p[y\leftrightarrow \partial \Lambda_{(k-1)L}] \le \varphi_p(S)^{k-1},$

where the final relation holds by induction, and clearly gives exponential decay as required.

For Item 2 we use Russo’s formula. Here we have a slightly simpler example than the most general version, since the event under consideration $A_n=\{0\leftrightarrow \partial\Lambda_n\}$ is increasing with respect to adding edges. It is also a function of a finite number of edges. Then we can consider $\frac{d}{dp}\mathbb{P}_p[A]$ under the coupling which adds each edge independently as a Poisson process with (locally) rate $\frac{1}{1-t}$. (We take this to be the rate to avoid having to reparameterise exponentially between time and probability. Here t=p.)

Just for ease, we only consider the right-derivative at p. Then with $\mathbb{P}$ as the law of the coupled process:

$\frac{d}{dp}\mathbb{P}_p[A] \approx \frac{1}{\epsilon} \mathbb{P}[A\text{ holds at }p+\epsilon,\text{ but not at }p]$

$= \frac{1}{\epsilon} \sum_{e\in E}\mathbb{P}[e\text{ added between }p,p+\epsilon\text{ and }e\text{ completes event }A]$

$+ \frac{1}{\epsilon}\mathbb{P}[\text{two or more edges relevant to }A\text{ added}].$

Since the number of edges whose states determine A is finite, this second term vanishes as $\epsilon\rightarrow 0$. So

$=\frac{1}{\epsilon}\sum \frac{\epsilon}{1-p} \mathbb{P}(A \text{ doesn't hold at p, and edge }e\text{ pivotal at p}).$

Taking the limit in $\epsilon$ in this example gives

$\frac{d}{dp}\mathbb{P}_p[0\leftrightarrow \partial\Lambda_n] = \frac{1}{1-p} \sum_{e\in\Lambda_n} \mathbb{P}_p[e\text{ pivotal, }0\not\leftrightarrow \partial \Lambda_n].$

The argument then proceeds in a similar way to Item 1, decomposing $\{0\not\leftrightarrow \partial \Lambda_n\}$ by conditioning on the set of vertices $\mathcal{S}$ from which it is not possible to get to $\partial \Lambda_n$. In particular, this set is an excellent candidate to view as S, since on this event it contains 0 by definition. Once we have specified $\mathcal{S}$ we know which edges might be pivotal, namely those across the boundary of $\mathcal{S}$. Crucially, the event $\{\mathcal{S}=S\}$ only depends on those edges between the boundary of S and $\partial \Lambda_n$, so is independent of the event $\{0\stackrel{S}{\leftrightarrow}x\}$ whenever $x\in\partial \mathcal{S}$. So applying this version of Russo gives

$\frac{d}{dp}\mathbb{P}_p[0\leftrightarrow \partial\Lambda_n] = \frac{1}{1-p}\sum_{0\in S\subset \Lambda_n} \sum_{(x,y)\in\Delta S} \mathbb{P}_p[0\stackrel{S}{\leftrightarrow} x]\mathbb{P}_p[\mathcal{S}=S].$

It is clear where $\varphi_p(S)$ might turn up within the sum (after removing a factor of p), so for a bound we can take $\inf_S \varphi_p(S)$ outside the sum, and arrive at

$\frac{d}{dp}\mathbb{P}_p[0\leftrightarrow \partial\Lambda_n] \ge \frac{1}{p(1-p)}\inf_{0\in S\subset \Lambda_n} (1-\mathbb{P}_p[0\leftrightarrow \partial \Lambda_n].$

It wasn’t immediately clear to me immediately that this implied the given form of Item 2 (though it certainly is consistent). I think perhaps I was trying to be too complicated and thinking about Gronwall’s lemma when in fact everything really follows from bounding $\inf \varphi_p(S)$ below by 1 (since we have assumed $p>\tilde p_c$ here), then integrating the differential inequality

$\frac{d}{dp}\left[ \frac{p}{1-p}f(p) \right] = \frac{p}{1-p}f'(p)+\frac{1}{(1-p)^2}f(p) \ge \frac{1}{(1-p)^2}.$

I include this not because it’s an interesting part of the argument (I don’t think it is really) but because I struggled to digest it first time round.

What is interesting is how well this choice to consider $\varphi_p(S)$ works out. In both parts of the argument, sets which work well for splitting the crossing probabilities into disjoint edge events mesh nicely with considering this function after conditioning on sub- or super-criticality with respect to $\tilde p_c$.

# Real Trees – Root Growth and Regrafting

Two weeks ago in our reading group meeting, Raphael told us about Chapter Five which introduces root growth and regrafting. One of the points of establishing the Gromov-Hausdorff topology in this book was to provide a more natural setting for a discussion of tree-valued processes. Indeed in what follows, one can imagine how to start the construction of a similar process for the excursions which can be used to encode real trees, involving cutting off sub-excursions above one-sided local minima, then glueing them back in elsewhere. But taking account of the equivalence structure will be challenging, and it is much nicer to be able to describe cutting a tree in two by removing a single point without having to worry about quotient maps.

We have seen in Chapter Two an example of a process defined on the family of rooted trees with n labelled vertices which has the uniform rooted tree as an invariant distribution. Given a rooted tree with root p, we choose uniformly at random a vertex p’ in [n] to be the new root. Then if p’=p we do nothing, otherwise we remove the unique first edge in the path from p’ to p, giving two trees. Adding an edge from p to p’ completes the step and gives a new tree with p’ as root. We might want to take a metric limit of these processes as n grows and see whether we end up with a stationary real tree-valued process whose marginals are the BCRT.

To see non-trivial limiting behaviour, it is most interesting to consider the evolution of a particular subtree (which includes the root) through this process. If the vertex chosen for cutting lies in our observed subtree, then the subtree undergoes a prune and regraft operation. On the other hand, if the vertex chosen for cutting does not lie in the subtree, then we do not see any effect of the pruning, except the addition of a new vertex below the original root, which becomes the new root. So essentially, from the point of view of our observed subtree, the root is growing.

Now we can think about interpreting the dynamics of a natural limit process acting on real trees. The key idea is that we don’t change the set on which the tree is supported much, but instead just change the metric. In particular, we will keep the original tree, and add on length at unit rate. Of course, where this length gets added on entirely determines the metric structure of the tree, but that doesn’t stop us giving a simple ‘name’ for the extra length. If we consider a process $X^T$ starting from a particular finite subtree T, then at time t, the tree $X^T_t$ is has vertex set $T \coprod (0,t]$. (Finite subtree here means that it has finite total length.)

Root regrafting should happen at a rate proportional to the total length of the current observed tree. This is reasonable since after all it is supported within a larger tree, so in the discrete case the probability of a prune-regrafting event happening within a given observed subtree is proportional to the number of vertices in that subtree, which scales naturally as length in the real tree limit. It turns out that to get unit rate root growth with $\Theta(1)$ rate prune-regrafting, we should consider subtrees of size $\sqrt{n}$ within a host tree of size n as $n\rightarrow\infty$. We also rescale the lengths by $\frac{1}{\sqrt{n}}$, and time by $\sqrt{n}$ so we actually see prune-regraft events.

Furthermore, if the subtree is pruned, the location of the pruning is chosen uniformly by length of the current observed subtree. So we can view the pruning process as being driven by a Poisson point process with intensity given by the instantaneous length measure of the tree, which at time t has vertex set $T\coprod (0,t]$. It will turn out to be consistent that there is a ‘piecewise isometry’ for want of a better phrase between the metric (and thus length measure) on $X^T_t$ and the canonical induced measure on $T\coprod (0,t]$, so we can describe the instances and locations of the pruning events via a pair of PPPs. The first is supported on $T \times [0,\infty)$, and the second on $\{(t,x): 0 \le x \le t\}$, since we only ‘notice’ pruning at the point labelled x if the pruning happens at some time t after x was created.

If we start from a compact tree T, then the total intensity of this pair is finite up to some time t, and so we have a countable sequence $\tau_0=0<\tau_1<\tau_2<\ldots$ of times for pruning events. It is easy to describe (but a bit messy to notate) the evolution of the metric between these pruning times. Essentially the distance between any pair of points in the observed tree at time $\tau_m$ with root $\rho_{\tau_m}$ is constant between times $\tau_m,\tau_{m+1}$, and new points are added so that the distance between $\rho_{\tau_m}$ and any new point $a\in(\tau_m,\tau_{m+1}]$ is $a-\tau_m$, and everything thing else follows from straightforward consideration of geodesics.

When a pruning event happens at point $x_m$ at time $\tau_m$, distances are preserved within the subtree above $x_m$ in $X^T_{\tau_m -}$, and within the rest of the tree. Again, an expression for the cross distances is straightforward but requires a volume of notation not ideally suited to this medium.

The natural thing to consider is the coupled processes started from different subtrees (again both must contain the original root) of the same host tree. Say $T^1,T^2\le T$, then it is relatively easy to check that $X^{T^1}_t,X^{T^2}_t \le X^T_t \,\forall t$, when we drive the processes by consistent coupled Poisson processes. Furthermore, it is genuinely obvious that the Hausdorff distance between $X^{T^1}_t,X^{T^2}_t$, here viewed as compact subsets of $(X^T_t, d^T_t)$ remains constant during root growth phase.

Less obvious but more important is that the Hausdorff distance decreases during regrafting events. Suppose that just before a regrafting event, the two subtrees are T’ and T”, and the Hausdorff distance between them is $\epsilon$. This Hausdorff distance is with respect to the metric on the whole tree T. [Actually this is a mild abuse of notation – I’m now taking T to be the whole tree just before the regraft, rather than the tree at time 0.]

So for any $a\in T'$, we can choose $b\in T''$ such that $d_T(a,b)\le \epsilon$. This is preserved under the regraft unless the pruning point lies on the geodesic segment (in T) between a and b. But in that case, the distance between a and the pruning point is again at most $\epsilon$, and so after the regrafting, a is at most $\epsilon$ away from the new root, which is in both subtrees, and in particular the regrafted version of T”.

This is obviously a useful first step on the path to proving any kind of convergence result. There are some technicalities which we have skipped over. It is fairly natural that this leads to a Markov process when the original tree is finite, but it is less clear how to define these dynamics when the total tree length is infinite, as we don’t want regrafting events to be happening continuously unless we can bound their net effect in some sense.

Last week, Franz showed us how to introduce the BCRT into matters. Specifically, that BCRT is the unique stationary distribution for this process. After a bit more work, the previous result says that for convergence properties it doesn’t matter too much what tree we start from, so it is fine to start from a single point. Then, the cut points and growth mechanism corresponds very well to the Poisson line-breaking construction of the BCRT. With another ‘grand coupling’ we can indeed construct them simultaneously. Furthermore, we can show weak convergence of the discrete-world Markov chain tree algorithm to the process with these RG with RG dynamics.

It does seem slightly counter-intuitive that a process defined on the whole of the discrete tree converges to a process defined through subtrees. Evans remarks in the introduction to the chapter that this is a consequence of having limits described as compact real trees. Then limitingly almost all vertices are close to leaves, so in a Hausdorff sense, considering only $\sqrt{n}$ of the vertices (ie a subtree) doesn’t really make any difference after rescaling edge lengths. I feel I don’t understand exactly why it’s ok to take the limits in this order, but I can see why this might work after more checking.

Tomorrow, we will have our last session, probably discussing subtree prune-and-regraft, where the regrafting does not necessarily happen at the root.

# The Yule Process

The second problem sheet for classes on the Applied Probability course this term features a long question about the Yule process. This is probably the simplest example of a birth process. It’s named for the British statistician George Udny Yule, though some sources prefer to call it the Yule-Furry process for the American physicist Wendell Furry who used it as a model of a radioactive reaction.

The model is straightforward. At any time there is some number of individuals in the population, and each individual gives birth to an offspring at constant rate $\lambda$, independently from the rest of the population. After a birth has happened, the parent and child evolve independently. In the notation of general birth processes, the birth rate when there are n individuals is $\lambda_n=\lambda n$.

Note that if we start with two or more individuals, the sizes of the two or more families of descendents evolve as a continuous-time Polya’s urn. The arrivals process speeds up with time, but the jump chain is exactly Polya’s urn. Unsurprisingly, the Yule process can be found embedded in preferential attachment models, and other processes which are based around Polya’s urn with extra information.

This is a discrete, random version of exponential growth. Since the geometric distribution is the discrete analogue of the exponential distribution, we probably shouldn’t be surprised to learn that this is indeed the distribution of the process at some fixed time t, when it is started from a single original ancestor. This is all we care about, since the numbers of descendents from each different original ancestors are independent. In general, the distribution of the population size at some fixed time will be negative binomial, that is, a sum of IID geometric distributions.

The standard method here is to proceed using generating functions. Conditioning on the first splitting time gives two independent copies of the original process over a shorter time-scale. One derives an ODE in time for the generating function evaluated at any particular value z. This can be solved uniquely for each z, and patching together gives the generating function of the distribution at any specific time t, which can be seen to coincide with the corresponding generating function of the geometric distribution with parameter $e^{-\lambda t}$.

So we were trying to decide whether there might be a more heuristic argument for this geometric distribution. The method we came up with is not immediate, but does justify the geometric distribution in a couple of steps. First, we say that the birth times are $T_2,T_3,\ldots$, so between times $[T_n,T_{n+1})$ there are n individuals, with $T_1:=0$ for concreteness. Then by construction of the birth process, $T_{n+1}-T_n\stackrel{d}{=}\mathrm{Exp}(\lambda n)$.

We now look at these ‘inter-birth times’ backwards, starting from $T_{n+1}$. Note that $\mathrm{Exp}(\lambda n)$ is the distribution of the time for the first of n IID $\mathrm{Exp}(\lambda)$ clocks to ring. But then, looking backwards, the next inter-birth time is thus the distribution of the time for one of (n-1) IID $\mathrm{Exp}(\lambda)$ clocks to ring. So by memorylessness of the exponential distribution (discussed at great length on the first problem sheet), we can actually take these (n-1) clocks to be exactly those of the original n clocks which did not ring first. Continuing this argument, we can show that the first (in the original time direction) inter-birth time corresponds to the time spent waiting for the final clock to ring. Rewriting this observation formally:

$T_{n+1}\stackrel{d}{=}\max\{X_i : X_1,\ldots,X_n\stackrel{\text{iid}}{\sim}\mathrm{Exp}(\lambda)\}.$ (*)

To return to justifying the geometric form of the distribution, we need to clarify the easiest relationship between the population size at a fixed size and these birth times. As we are aiming for the geometric distribution, the probability of the event $\{X_t>n\}$ will be most useful. Clearly this event is the same as $\{T_{n+1}, and from the description involving maxima of IID exponentials, this is easy to compute as $(1-e^{-\lambda t})^n$, which is exactly what we want.

There are two interesting couplings hidden in these constructions. On closer inspection they turn out to be essentially the same from two different perspectives.

We have specified the distribution of $T_n$ at (*). Look at this distribution on the right hand side. There is a very natural way to couple these distributions for all n, namely to take some infinite sequence $X_1,X_2,\ldots$ of IID $\mathrm{Exp}(\lambda)$ random variables, then use initial sequences of these to generate each of the $T_n$s as described in (*).

Does this coupling correspond to the use of these IID RVs in the birth process? Well, in fact it doesn’t. Examining the argument, we can see that $X_1$ gives a different inter-birth time for each value of t in the correspondence proposed. Even more concretely, in the birth process, almost surely $T_{n+1}>T_n$ for each n. This is not true if we take the canonical coupling of (*). Here, if $X_n<\max\{X_1,\ldots,X_{n-1}\}$, which happens with high probability for large n, we have $T_{n+1}=T_n$ in the process of running maxima.

Perhaps more interestingly, we might observe that this birth process gives a coupling of the geometric distributions. If we want to recover the standard parameterisation of the geometric distribution, we should reparameterise time. [And thus generate an essentially inevitable temptation to make some joke about now having a Yule Log process.]

Let’s consider what the standard coupling might be. For a binomial random variable, either on [n] or some more exotic set, as in percolation, we can couple across all values of the parameter by constructing a family independent uniform random variables, and returning a 1 if $U_i>1-p$ and so on, where p is the parameter of a specific binomial realisation.

We can do exactly the same here. A geometric distribution can be justified as the first success in a sequence of Bernoulli trials, so again we can replace the relevant Bernoulli distribution with a uniform distribution. Take $U_1,U_2,\ldots$ to be IID U[0,1] random variables. Then, we have:

$X_t=\stackrel{d}{=}\bar X_t:= \max\{n: U_1,\ldots,U_{n-1}\ge e^{-\lambda t}\}.$

The equality in distribution holds for any particular value of t by constructing. But it certainly doesn’t hold uniformly in t. Note that if we define $\bar X_t$ as a process, then typically the jumps of this process will be greater than 1, which is forbidden in the Yule process.

So, we have seen that this Yule process, even though its distribution at a fixed time has a standard form, provides a coupling of such distributions that is perhaps slightly surprising.

# Skorohod Representation Theorem

Continuing the theme of revising theory in the convergence of random processes that I shouldn’t have forgotten so rapidly, today we consider the Skorohod Representation Theorem. Recall from the standard discussion of the different modes of convergence of random variables that almost sure convergence is among the strongest since it implies convergence in probability and thus convergence in distribution. (But not convergence in $L_1$. For example, take U uniform on [0,1], and $X_n=n\mathbf{1}_{\{U<\frac{1}{n}\}}$.)

Almost sure convergence is therefore in some sense the most useful form of convergence to have. However, it comes with a strong prerequisite, that the random variables be defined on the same probability space, which is not required for convergence in distribution. Indeed, one can set up weak versions of convergence in distribution which do not even require the convergents to be random variables. The Skorohod representation theorem gives a partial converse to this result. It states some conditions under which random variables which converge in distribution can be coupled on some larger probability space to obtain almost sure convergence.

Skorohod’s original proof dealt with convergence of distributions defined on complete, separable metric spaces (Polish spaces). The version discussed here is from Chapter 5 of Billingsley [1], and assumes the limiting distribution has separable support. More recent authors have considered stronger convergence conditions (convergence in total variation or Wasserstein distance, for example) with weaker topological requirements, and convergence of random variables defined in non-metrizable spaces.

Theorem (Skorohod representation theorem): Suppose that distributions $P_n\Rightarrow P$, where P is a distribution with separable support. Then we can define a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and random variables $X,(X_n)_{n\ge 1}$ on this space such that the laws of $X,X_n$ are $P,P_n$ respectively and $X_n(\omega)\rightarrow X(\omega)$ for all $\omega\in\Omega$.

NB. We are proving ‘sure convergence’ rather than merely almost sure convergence! It is not surprising that this is possible, since changing the value of all the $X_n$s on a set with measure zero doesn’t affect the conditions for convergence in distribution.

Applications: Before going through the Billingsley proof, we consider one simple application of this result. Let S be a separable metric space containing the support of X, and g a continuous function $S\rightarrow S'$. Then

$X_n\stackrel{a.s.}{\rightarrow}X\quad\iff\quad g(X_n)\stackrel{a.s.}{\rightarrow}g(X).$

So, by applying the Skorohod representation theorem once, and the result that almost sure convergence implies convergence in distribution, we have shown that

$X_n\stackrel{d}{\rightarrow}X\quad\iff \quad g(X_n)\stackrel{d}{\rightarrow}g(X),$

subject to these conditions on the space supporting X.

Proof (from [1]): Unsurprisingly, the idea is to construct realisations of the $(X_n)$ from a realisation of X. We take X, and a partition of the support of X into small measurable sets, chosen so that the probability of lying in a particular set is almost the same for $X_n$ as for X, for large n. Then, the $X_n$ are constructed so that for large n, with limitingly high probability $X_n$ lies in the same small set as X.

Constructing the partition is the first step. For each $x\in S:=\mathrm{supp}(X)$, there must be some radius $\frac{\epsilon}{4} such that $P(\partial B(x,r_x)=0$. This is where we use separability. Since every point in the space is within $\frac{\epsilon}{4}$ of some element of a countable sequence of elements of the space, we can take a countable subset of these open balls $B(x,r_x)$ which cover the space. Furthermore, we can take a finite subset of the balls which cover all of the space apart from a set of measure at most $\epsilon$. We want the sets to be disjoint, and we can achieve this by removing the intersections inductively in the obvious way. We end up with a collection $B_0,B_1,\ldots,B_k$, where $B_0$ is the leftover space, such that

• $P(B_0)<\epsilon$
• $P(\partial B_i)=0,\quad i=0,1,\ldots,k$
• $\mathrm{diam}(B_i)<\epsilon,\quad i=1\ldots,k$.

Now suppose for each m, we take such a partition $B^m_0,B^m_1,\ldots,B^m_{k_m}$, for which $\epsilon_m=\frac{1}{2^m}$. Unsurprisingly, this scaling of $\epsilon$ is chosen so as to use Borel-Cantelli at the end. Then, from convergence in distribution, there exists an integer $N_m$ such that for $n\ge N_m$, we have

$P_n(B^m_i)\ge (1-\epsilon_m)P(B^m_i),\quad i=0,1,\ldots,k_m.$ (*)

Now, for $N_m\le n , for each $B^m_i$ with non-zero probability under P, take $Y_{n,i}$ to be independent random variables with law $P_n(\cdot | B^m_i)$ equal to the restriction onto the set. Now take $\xi\sim U[0,1]$ independent of everything so far. Now we make concrete the heuristic for constructing $X_n$ from X. We define:

$X_n=\sum_{i=0}^{k_m}\mathbf{1}_{\{\xi\le 1-\epsilon_m, X\in B^m_i\}} Y_{n,i} + \mathbf{1}_{\{\xi>1-\epsilon_m\}}Z_n.$

We haven’t defined $Z_n$ yet. But, from (*), there is a unique distribution such that taking $Z_n$ to be independent of everything so far, with this distribution, we have $\mathcal{L}(X_n)=P_n$. Note that by iteratively defining random variables which are independent of everything previously defined, our resulting probability space $\Omega$ will be a large product space.

Note that $\xi$ controls whether the $X_n$ follow the law we have good control over, and we also want to avoid the set $B^m_0$. So define $E_m:=\{X\not \in B^m_0, \xi\le 1-\epsilon_m\}$. Then, $P(E_m)<2\epsilon_m=2^{-(m-1)}$, and so by Borel-Cantelli, with probability 1, $E_m$ holds for all m larger than some threshold. Let us call this $\liminf_m E_m=: E$, and on this event E, we have by definition $X_n \rightarrow X$. So we have almost sure convergence. But we can easily convert this to sure convergence by removing all $\omega\in\Omega$ for which $\xi(\omega)=1$ and setting $X_n\equiv X$ on $E^c$, as this does not affect the distributions.

Omissions:

• Obviously, I have omitted the exact construction of the distribution of $Z_n$. This can be reverse reconstructed very easily, but requires more notation than is ideal for this medium.
• It is necessary to remove any sets $B^m_i$ with zero measure under P for the conditioning to make sense. These can be added to $B^m_0$ without changing any of the required conditions.
• We haven’t dealt with any $X_n$ for $n.

The natural question to ask is what happens if we remove the restriction that the space be separable. There are indeed counterexamples to the existence of a Skorohod representation. The clearest example I’ve found so far is supported on (0,1) with a metric inducing the discrete topology. If time allows, I will explain this construction in a post shortly.

References

[1] – Billingsley – Convergence of Probability Measures, 2nd edition (1999)

# Coupling from the Past

In a long series of previous posts I have talked about mixing times for Markov chains. We consider how long it takes for the distribution of a particular Markov chain to approach equilibrium. We are particularly interested in the asymptotics when some parameter of the model grows, such as the size of the state space, grows to infinity.

But why are we interested in the underlying problem? The idea of Markov Chain Monte Carlo methods is to sample from an intractable distribution by instead sampling from a Markov chain which approximates the distribution well at large times. A distribution might be intractable because it is computationally demanding to work out the normalising constant, or it might be distributed uniformly on a complicated combinatorial set. If, however, the distribution is the equilibrium distribution of some Markov chain, then we know how to at least sample from a distribution which is close to the one we want. But we need to know how long to run the process. We will typically tolerate some small error in approximating the distribution (whether we measure this in terms of total variation distance or some other metric doesn’t really matter at this heuristic level), but we need to know how it scale. If we double the size of the system, do we need to double the number of iterations of the chain, or square it. This is really important if we are going to use this for large real-world models with finite computing power!

Sometimes though, an approximation is not enough. If we want an exact sample from the equilibrium distribution, Markov chains typically will not help us as it is only in very artificial examples that the distribution after some finite time is actually the equilibrium distribution. One thing that we might use is a stationary time, which is a stopping time T, for which $X_T\stackrel{d}{=}\pi$. Note that there is one trivial way to do this. We can sample Y from distribution $\pi$ before starting the process, then stop X at the first time T for which $X_T=Y$. But this is no help really, as we need to have Y in the first place!

So we are really interested in less trivial stationary times. Perhaps the best example is the top-to-random shuffle. Here we are given a pack of labelled cards, WLOG initially in descending order at each step we move the top card in the pile to a randomly-chosen location in the pile (which includes back onto the top). Then it turns out that the first time we move the card originally at the bottom from the top to somewhere is a strong stationary time. This is fairly natural, as by this time, every card has been involved in at least one randomising event.

Anyway, so this gives a somewhat artificial way to sample from the uniform distribution on a pack of cards. This strong stationary time is almost surely finite, with distribution given by the coupon collector problem, for which the expectation grows as $n\log n$, where n is the number of cards.

The problem with this method is that it is not easy in general to come up with a non-contrived stationary time such as this one. The idea of coupling from the past, discussed by some previous authors but introduced in this context by Propp and Wilson in the mid ’90s, is another method to achieve perfect sampling from the equilibrium distribution of a Markov chain. The idea here is to work backwards rather than forwards. The rest of this post, which discusses this idea, is based on the talk given at the Junior Probability Seminar by Irene, and on the chapter in the Levin, Peres, Wilmer book.

The key to the construction is a coupling of the transitions of a Markov chain. In the setting of a simple random walk, we have by construction a coupling of the transitions. It doesn’t matter which state we are at: we toss a coin to decide whether to move up or down, and we can do this without reference to our current position. Levin, Peres and WIlmer call this a random mapping representation in general, and it is yet another concept that is less scary than its definition might suggest.

Given a transition matrix P on state space S, such a representation is a function

$\phi: S\times[0,1]\rightarrow S,\text{ s.t. }\mathbb{P}(\phi(i,U)=j)=p_{ij},$

where U is a U(0,1) random variable independent of choice of i. In particular, once we have the random value of u, we can consider $\phi(i,u)$ as i varies, to obtain a random map $S\rightarrow S$. Crucially, this map is not necessarily a bijection.

Note first that there are many possibilities for constructing the representation $\phi$. For some chains, and some representations, in particular random walks on vertex-transitive graphs (such as SRW – only for now we are restricting attention to finite state spaces) it is possible to choose $\phi$ so that it always gives a bijection, but it is also always possible to choose it so that there is some probability it doesn’t give a bijection.

Let $U_1,U_2,\ldots$ be an IID sequence of U[0,1] random variables, and write $\phi_i$ for the random map induced by $U_i$. Then consider the sequence of iterated maps:

$\phi_1, \phi_1\circ \phi_2, \ldots, \phi_1\circ\ldots\circ\phi_n,$

and let T be the (random) smallest time such that the image of $\phi_1\circ\ldots\circ \phi_T$ is a single state. Ie, as we go backwards in time through the maps $\phi_i$, we are gradually losing various states, corresponding to the maps not being bijections. Since the state space is finite, and the probability of not being a bijection is positive, it can be shown that T is almost surely finite. The claim then is that

$Y=\text{Im}(\phi_1\circ\ldots\circ \phi_T)$

is distributed as the equilibrium distribution of the chain. We finish by proving this.

Proof: Since the algorithm terminates after finite time almost surely, given any $\epsilon>0$, we can choose N such that the probability the algorithm stops in at most N steps is greater than $1-\epsilon$.

Now run the Markov chain from time -N, started in the equilibrium distribution, with the transition from time -t to -(t-1) given by the random mapping driven by $U_t$. Thus at time 0, the distribution of the chain is still the equilibrium distribution. But if we condition on the event that $T\le N$, then $X_0=\phi_1\circ \ldots \circ\phi_n(X_{-N})=Y$ regardless of the initial value. So $\mathbb{P}(X_0\ne Y)<\epsilon$, and hence the result follows, since $\epsilon>0$ was arbitrary.

What makes this easier than strong stationary times is that we don’t have to be clever to come up with the stopping time. It is however still important to know how long on average it takes to run the algorithm. At the end of her talk, Irene showed how to adapt this algorithm to deal with Probabilistic Cellular Automata. Roughly speaking, these are a sequence of infinite strings of 0s and 1s. The value of some element is determined randomly as a function of the values in the row underneath, say the element directly underneath and the two either side. In that setting, if you start with a finite subsequence and couple from the past by looking down to lower rows, each time you drop down a row you consider one further element, so in fact the coupling from the past algorithm has to eliminate possibilities fast enough to make up for this, if we want to terminate almost surely in finite time.

Here’s a link to the paper which discusses this in fuller detail.

# Mixing of the Noisy Voter Model

I’ve been spending a bit of time recently thinking about some interacting particle systems and mixing time problems, so it was interesting to find a short paper due to Harishchandra Ramadas on arXiv a couple of weeks ago combining these two areas. I enjoyed reading it, so I thought I would summarise its arguments here, in preparation for perhaps talking about it at this week’s junior probability seminar.

The voter model is a graph-based interacting particle system. The underlying motivation is to describe the spread of opinions through a population. Each vertex represents an agent who has an opinion which might vary over time. At exponential rates, an agent will replace their opinion with the opinion of one of their neighbours, chosen uniformly at random. Note that, obviously, this might not actually involve a change of opinion.

One problem with this model is that we might be interested in convergence to equilibrium, but this might be hard to specify, as the state where everything is 0 and the state where everything is 1 are absorbing, and similarly any binomial distribution on the set of vertices is invariant. It is also not entirely realistic to assume that future updates to the system are entirely based on the current configuration. In any actual application, we should allow for some, possibly very small, influence from the outside, whether due to random errors in the update rule, or some external factors which haven’t been explicitly accounted for.

So we consider instead the noisy voter model where, in addition to the dynamics described above, each vertex changes state at rate $\delta$, independently of the configuration or the history. [Note that we are now considering the set of opinions {0,1} ]

We are interested in the mixing time of this system. It is not immediately clear that this should have an invariant distribution, but it seems reasonable since the noise, which converts 0 to 1 and 1 to 0 at equal rates if the proportion of 0s and 1s is equal, moves the system towards something like Bin(V(G),1/2), although the correlations need to take account of the local geometry.

Anyway, we assume $\delta>0$ is fixed, and n tends to infinity. Ramadas proves that the dynamics have the property of optimal temporal mixing, which states that given copies of the process X, Y starting from different initial conditions

$||X_t-Y_t||\le C n e^{-ct},$

for suitable choice of constants. Note that it doesn’t matter whether X and Y are coupled or independent. Total variation distance is a function of distribution, not of probability space. I don’t have a great intuition for exactly what this means, except to make the obvious comment that the distance from uniformity grows at most uniformly as the size of the system grows. Note that this immediately implies that the mixing time is O(log n). The rest of the paper proves that the noisy voter model has this property.

The method is to consider the relation:

$||X-Y||\le \mathbb{P}(X\ne Y),$

which says that the total variation distance gives the minimum of the RHS probability over couplings of X and Y. Anyway, as discussed in the previous post on duality in such interacting particle systems, the key step is to consider the process as a deterministic walk through a random environment. Ie, we apply the same moves to configurations started from different initial conditions.

The first step is to break down the noise mechanism to make it easier to couple everything. Instead of saying that a vertex changes state at rate $\delta$, we instead have two noise processes. One sets a vertex’s state to 0 at rate $\delta$, and another sets a vertex’s state to 1 at rate $\delta$ independently. Since at any configuration, for each vertex, precisely one of these processes has no effect.

So for the coupling, we run the same noise processes, and we choose the neighbours from which to update our opinions the same in both processes. The only difference is the initial configuration. With these two processes, X and Y defined on the same probability space, Ramadas defines $Z_t(v)=\mathbf{1}\{X_t(v)\neq Y_t(v)\}$, which counts how many vertices are different between the two processes. Almost surely, eventually all the Zs will be 0, and this state is absorbing because of how we’ve defined this coupling. We are interested to find out how long it takes for this event, which we could think of like a strong stationary time, to happen.

To control this, we track the evolution in the probabilities of $Z_t(v)=0$ jointly. We note that after a noise event at vertex v, we definitely have $Z_t(v)=0$. After an opinion-change event, whether $Z_t(v)$ changes depends on whether $Z_t(u)$ is 1, where u is the uniformly-chosen vertex from which v inherits its opinion in both processes. We obtain the following ODEs:

$\frac{d}{dt}\mathbb{P}(Z_t(v)=1)=\frac{1}{d(v)}\left(\sum_{u:u\sim v}\mathbb{P}(Z_t(v)=0,Z_t(u)=1)\right)$

$-\frac{1}{d(v)}\left(\sum_{u:u\sim v}\mathbb{P}(Z_t(v)=1,Z_t(u)=0)\right)-2\delta \mathbb{P}(Z_t(v)=1).$

We interpret this sum as a negative bit plus two terms which we would hope would cancel each other out. It seems impractical to attempt to formalise this heuristic. However, we can instead choose to consider at time t to consider the ODE for whichever vertex v maximises $Z_t(v)$. It’s then clear that the RHS is less than $-2\delta \mathbb{P}(Z_t(v)=1)$.

Right-differentiabililty of $Z_t(v_{max})$ follows from differentiability of $Z_t(v)$. The number of vertices is finite, and almost surely the vertex attaining this maximum can’t change too often. In particular, from the ODE, we must have $Z_t(v_{max})$ bounded above by $Ce^{-2\delta t}$. From which it follows that

$\mathbb{P}(X\ne Y)\le nCe^{-2\delta t },$

and we’ve shown this optimal temporal mixing property, and hence the mixing time has order log n.

What I’ve discussed so far is taken entirely from Section 3 of Ramadas’s paper. My first thought was whether this bound is actually correct in scale, as we have used only the size of the graph rather than any finer information about its graph structure. The author discusses that it is not possible to use a general result due to Hayes and Sinclair on this topic in this setting, as it is not reversible.

I certainly can’t answer this question, but I will make a comment on the extreme examples. If the graph is the empty graph on n vertices, then the voter model dynamics are null, so we essentially have just a coupon collector problem going on. It doesn’t lose any generality to start from entirely 0s, then at time t, the distribution of the system will be Bin(n,p), with p a function of time, but not n. Now note that

$||\mathrm{Bin}(n,p)-\mathrm{Bin}(n,q)||_{TV}=\Theta(\sqrt{n}|p-q|),$

so in this example, we need to choose a time so that $p=\frac12+\Theta(\sqrt{n})$. This is really a statement about the central limit theorem. It’s clear that Bin(n,1/2) should genuinely be the equilibrium distribution here. The evolution of p(t) is exponential, and hence we get the O(log n) scaling emerging. Note that the presence of $\sqrt{n}$ rather than any other power of n makes no difference, though it would be relevant if we actually wanted to know the constant.

At the other end of the connectivity spectrum, we consider the dynamics on the complete graph. Here, let’s count the number of 1s, and call it W(t) say. Then W goes from k to k+1 at rate $\frac{(n-k)\delta}{2}+\frac{k(n-k)}{n}$, and from k to k-1 at rate $\frac{k\delta}{2}+\frac{k(n-k)}{n}$. In particular, note in expectation, the contributions from the voter model dynamics cancel. This fits with our intuition that it is solely the noise driving the system towards equilibrium.

If we take an ODE approximation, we get:

$\frac{dX}{dt}=\frac{n}{2}-\delta x,$

from which it follows that

$t(x)=\alpha-\frac{1}{\delta}\log(C-\delta x),$

which gives precisely the correct scaling, when we take x=n/2.

I would be very interested to think about or hear about arguments for a general lower bound across all graphs.

REFERENCES

Hayes, Sinclair (2007) – A general lower bound for mixing of single-site dynamics on graphs (arXiv)

Ramadas (2014) – Mixing of the Noisy Voter Model (arXiv link at top of article)

# 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.

# Mixing Times 6 – Aldous-Broder Algorithm and Cover Times

In several previous posts, I’ve discussed the Uniform Spanning Tree. The definition is straightforward: we choose uniformly at random from the set of trees which span a fixed underlying graph. But for a dense underlying graph, there are a very large number spanning trees. Cayley’s formula says that the complete graph K_n has $n^{n-2}$ spanning trees, so to select from this list is impractical.

We seek a better algorithm. In a post about a year ago, I presented the result that the path between two fixed points x and y in the UST is distributed as the path generated by Loop-Erased Random Walk, for which we start at x and delete cycles as they appear. An initial problem might be that this only gives us a single path, which might be enough in some contexts, but in general we will want to specify the whole tree. Wilson’s Algorithm is an unsurprising but useful extension to this equivalence which does just that. You start by constructing the LERW between two vertices, then you add the LERW which connects some other vertex to the path you already have. Then you take a further vertex not currently explored and start LERW there, continuing until you hit the tree that you already have. Iterate this process, which must terminate after at most n steps when there are no vertices which to start from. The tree thus obtained is the UST. The tricky part is proving that the method for selecting which unused vertices to start from has no effect on the distribution of paths between two fixed points.

I want to consider a different algorithm, discovered roughly simultaneously by Aldous and Broder. Start a random walk on the underlying graph at some particular vertex. Every time we traverse an edge which takes us to a vertex we haven’t yet explored, add this edge to the tree. For now I don’t want to give a proof that this algorithm works, but rather to talk about how fast it works, because it ties in nicely with something from the Mixing Times book we’ve been reading recently. It is clear that the algorithm terminates at the first time the random walk has visited every vertex. This is a stopping time, called the cover time of the Markov chain. If we are working with an underlying complete, then we notice that this is annoying, because it means that the cover time will increase like n.log n. That is, it will take an increasingly long time to gather the final few vertices into the tree. Perhaps some combination of Aldous-Broder initially then Wilson’s method for the final o(n) vertices might be preferable?

I want to discuss how to treat this cover time. Often we have information about the hitting times of states from other states $\mathbb{E}_x T_y$. A relationship between S, the hitting time, defined to be the maximum of the previous display over x and y, and the expected cover time would be useful, especially for a highly symmetric graph like the complete graph where the expected hitting times are all the same.

Matthews’ Method relates these two for an irreducible finite Markov chain on n states. It says:

$t_{cov}\leq t_{hit}\left(1+\frac12+\ldots+\frac 1 n\right).$

We first remark that this agrees with what we should get for the random walk on the complete graph. There, the hitting time of x from y is a geometric random variable with success probability 1/n, hence expectation is n. The cover time is the standard coupon collector problem, giving expectation n log n, and the sum of reciprocals factor is asymptotically a good approximation.

The intuition is that if we continue until we hit state 1, then reset and continue until we hit state 2, and so on, by the time we hit state n after (n-1) iterations, this is a very poor overestimate of the cover time, because we are actually likely to have hit most states many times. What we want to do really is say that after we’ve hit state 1, we continue until we hit state 2, unless we’ve already done so, in which case we choose a different state to aim for, one which we haven’t already visited. But this becomes complicated because we then need to know the precise conditional probabilities of visiting any site on the way between two other states, which will depend rather strongly on the exact structure of the chain.

Peres et al give a coupling proof in Chapter 11 of their book which I think can be made a bit shorter, at least informally. The key step is that we still consider hitting the sites in order, only now in a random order.

That is, we choose a permutation $\sigma\in S_n$ uniformly at random, and we let $T_k$ be the first time that states $\sigma(1),\ldots,\sigma(k)$ have all been visited. This is a random time that is measurable in the product space, and for each $\sigma$ it is a stopping time.

The key observation is that $\mathbb{P}(T_{k+1}=T_k)=1-\frac{1}{k+1}$. This holds conditional on any path of the Markov chain because the requirement for the event is that $\sigma(k+1)$ is visited after $\{\sigma(1),\ldots,\sigma(k)\}$. The statement therefore holds as stated as well as just pathwise. Then, by the SMP, conditional on $\{T_{k+1}>T_k\}$, we have

$T_{k+1}-T_k \leq_{st} t_{hit}.$

Note that by the definition of $t_{hit}$, this bound on the hitting time $T_{k+1}$ is unaffected by concerns about where the chain actually is at $T_k$ (since it is not necessarily at $\sigma(k)$).

So, removing the conditioning, we have:

$\mathbb{E}\Big[T_{k+1}-T_k\Big]\leq\frac{1}{k+1}t_{hit},$

and so the telescoping sum gives us Matthews’ result.

One example is the cover time of random walk on the n x n torus, which turns out to be

$O(n^2(\log n)^2).$

If anyone remembers that Microsoft screensaver from many years ago which started with a black screen and a snake leaving a trail of white pixels as it negotiated the screen, this will be familiar. The last few black bits take a frustratingly long while to disappear. Obviously that isn’t quite a random walk, but it perhaps diminishes the surprise that it should take this long to find the cover time.

There are a couple of interesting things I wanted to say about electrical networks for Markov chains and analytic methods for mixing times, but the moment may have passed, so this is probably the last post about Mixing Times. Plans are in motion for a similar reading group next term, possible on Random Matrices.