# BMO1 2018

The first round of the British Mathematical Olympiad was sat yesterday. The paper can be found here, and video solutions here. Copyright for the questions is held by BMOS. They are reproduced here with permission.

I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following commentaries on the problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so. I’ve written quite a lot about Q5 because I found it hard (or at least time-consuming) and somewhat atypical, and I’ve written a lot about Q6 because there was a lot to say. I hope at least some of this is interesting to some readers of all levels of olympiad experience.

Question 1

A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of 3, and each digit {0,1,…,9} appears exactly once on the blackboard. In how many ways can this be done? (Note that a two-digit number cannot begin with zero.)

It’s a trope of BMO1 that the first question must be doable by some sort of exhaustive calculation or listing exercise. Of course, that is rarely the most efficient solution.

However, there is normally a trade-off between eliminating all listing, and reducing to a manageable task.

The key observation here is that writing the integers in increasing order is really just a way to indicate that order of the choices doesn’t matter. Even if that seems counter-intuitive. The question wants to know how many ways to choose these five numbers. The order of choice doesn’t matter since we’re going to put them in ascending order on the blackboard anyway.

You want to make your choices with as much independence as possible. So it would, for example, be a bad idea to choose the smallest number first. How many possibilities are there where the smallest number is 24? What about 42? What about 69? These are all different, and some are zero, so will make the computation very taxing.

However, you might notice that the digits {0,3,6,9} have to go together to form two numbers, and the rest have to pair up with one digit from {1,4,7} and one from {2,5,8}. You might know that an integer is divisible by 3 precisely if its digit sum is divisible by 3, but in this context you wouldn’t lose too much time by simply listing everything! These tasks are now completely separate, so you can take the number of ways to pair up {0,3,6,9} and multiply by the number of ways to pair up {1,4,7} and {2,5,8}. You need to take care over the ordering. It does (obviously) matter which is the first digit and which is the second digit in a number!

# BMO2 2018

The second round of the British Mathematical Olympiad was taken yesterday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.

I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length (and in particular not just presenting direct solutions) is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly.

All these questions could be solved by able children. In fact, each could be solved by able children in less than an hour. You definitely count as an able child if you qualified or if your teacher allowed you to make an open entry! Others count too naturally. But most candidates won’t in fact solve all the questions, and many won’t solve any. And I think candidates often come up with the wrong reasons why they didn’t solve problems. “I didn’t know the right theorems” is very very rarely the reason. Olympiad problems have standard themes and recurring tropes, but the task is not to look at the problem and decide that it is an example of Olympiad technique #371. The task is actually to have as many ideas as possible, and eliminate the ones that don’t work as quickly as possible.

The best way to realise that an idea works is to solve the problem immediately. For the majority of occasions when we’re not lucky enough for that to happen, the second-best way to realise that an idea works is to see that it makes the problem look a bit more like something familiar. Conversely, the best way to realise that an idea doesn’t work is to observe that if it worked it would solve a stronger but false problem too. (Eg Fermat’s Last Theorem *does* have solutions over the reals…) The second-best way to realise that an idea doesn’t work is to have the confidence that you’ve tried it enough and you’ve only made the problem harder, or less familiar.

Both of these second-best ideas do require a bit of experience, but I will try to explain why none of the ideas I needed for various solutions this year required any knowledge beyond the school syllabus, some similarities to recent BMOs, and a small bit of creativity.

As usual, the caveat that these are not really solutions, and certainly not official solutions, but they are close enough to spoil the problems for anyone who hasn’t tried them by themselves already. Of course, the copyright for the problems is held by BMOS, and reproduced here with permission.

Question One

I wrote this question. Perhaps as a focal point of the renaissance of my interest in geometry, or at least my interest in teaching geometry, I have quite a lot to say about the problem, its solutions, its origin story, the use of directed angles, the non-use of coordinate methods and so on. In an ideal world I would write a book about this sort of thing, but for now, a long and separate post is the answer.

This will be available once I’ve successfully de-flooded my apartment.

Question Two

I also wrote this problem, though I feel it’s only fair to show the version I submitted to the BMO committee. All the credit for the magical statement that appears above lies with them. There is a less magical origin story as well, but hopefully with some interesting combinatorial probability, which is postponed until the end of this post.One quick observation is that in my version Joe / Hatter gets to keep going forever. As we shall see, all the business happens in the first N steps, but a priori one doesn’t know that, and in my version it forces you to strategise slightly differently for Neel / Alice. In the competition version, we know Alice is done as soon as she visits a place for a second time, but not in the original. So in the original we only have to consider ‘avoid one place’ rather than the multiple possibilities now of ‘avoid one place’ or ‘visit a place again’.

But I think the best idea is to get Alice to avoid one particular place $c\not\equiv 0$ whenever possible. At all times she has two possible options for where to go next, lets say $b_k+a_k, b_k-a_k$ in the language of the original statement. We lose nothing by assuming $-N/2 < a_k\le N/2$, and certainly it would be ridiculous for Joe / Hatter ever to choose $a_k=0$. The only time Alice’s strategy doesn’t work is when both of these are congruent to $c$, which implies $N\,|\, 2a_k$, and thus we must have $N= 2a_k$. In other words, Alice’s strategy will always work if N is odd.

I think it’s really worth noticing that the previous argument is weak. We certainly did not show that N must be odd for Alice to win. We showed that Alice can avoid a congruence class modulo an odd integer. We didn’t really need that odd integer to be N for this to work. In particular, if N has an odd factor p (say a prime), then the same argument works to show that we can avoid visiting any site with label congruent to 1 modulo p.

It’s actually very slightly more complicated. In the original argument, we didn’t need to use any property of $b_k$. But obviously here, if $b_k\equiv 1$ modulo p and $p\,|\,a_k$, then certainly $b_{k+1}\equiv 1$ modulo p. So we have to prove instead that Alice can ensure she never ‘visits 1 modulo p for the first time’. Which is fine, by the same argument.

So, we’ve shown that Neel / Alice wins if N is odd, or has an odd factor. The only values that remain are powers of 2. I should confess that I was genuinely a little surprised that Joe / Hatter wins in the power of 2 case. You can find a construction fairly easily for N=2 and N=4, but I suspected that might be a facet of small numbers. Why? Because it still felt we could avoid a particular site. In order for Alice’s strategy to fail, we have to end up exactly opposite the particular site at exactly the time when the next $a_k=N/2$, and so maybe we could try to avoid that second site as well, and so on backwards?

But that turned out to be a good example of something that got very complicated quite quickly with little insight. And, as discussed at the beginning, that’s often a sign in a competition problem that your idea isn’t so good. (Obviously, when composing a problem, that’s no guarantee at all. Sometimes things are true but no good ideas work.) So we want other ideas. Note that for N=4, the sequence (2,1,2) works for Joe / Hatter, because that forces Alice / Neel to visit either (0,2,1,3) or (0,2,3,1). In particular, this strategy gave Alice no control on the first step nor the last step, and the consequence is that we force her to visit the evens first, then transfer to an odd, and then force her to visit the other odd.

We might play around with N=8, or we might proceed directly to a general extension. If we have a Joe / Hatter strategy for N, then by doubling all the $a_k$s, we have a strategy for 2N which visits all the even sites in the first N steps. But then we can move to an odd site eg by taking $a_N=1$. Just as in the N=4 case, it doesn’t matter which odd site we start from, since if we again double all the $a_k$s, we will visit all the other odd sites. This gives us an inductive construction of a strategy for powers of two. To check it’s understood, the sequence for N=8 is (4,2,4,1,4,2,4).

Although we don’t use it, note that this strategy takes Alice on a tour of sites described by decreasing order of largest power of two dividing the label of the site.

# DGFF 4 – Properties of the Green’s function

I’m at UBC this month for the PIMS probability summer school. One of the long courses is being given by Marek Biskup about the Discrete Gaussian Free Field (notes and outline here) so this seems like a good moment to revive the sequence of posts about the DGFF. Here’s DGFF1, DGFF2, DGFF3 from November.

The first draft of this post was about the maximum of the DGFF in a large box $V_N$, and also about the Green’s function $G^{V_N}(x,y)$, which specifies the covariance structure of the DGFF. This first draft also became too long, so I’m splitting it into two somewhat shorter ones. As we’ll see, some understanding and standard estimates of the Green’s function is enough to say quite a bit about the maximum. In this first post, we’ll explore some ‘low-hanging fruit’ concerning the Green’s function, as defined through a simple random walk, which are useful, but rarely explained in the DGFF literature.

Symmetry of Green’s function

We start with one of these low-hanging fruit. If $G^{V_N}$ is to be a covariance matrix, it has to be symmetric. In the first post, showing that the definition of the DGFF as a random field with given Hamiltonian is equivalent to $\mathcal{N}(0,G^{V_N})$ certainly can be viewed as a proof of symmetry. However, it would be satisfying if there was a direct argument in the language of the definition of the Green’s function.

To make this self-contained, recall the random walk definition of $G^{V_N}(x,y)$. Let $(S_m)_{m\ge 0}$ be simple random walk on $V_N$, and $\mathbb{P}_x,\,\mathbb{E}_x$ denote starting the random walk at $x\in V_N$. As usual, let $\tau_y,\,\tau_A$ denote the hitting time of a vertex y or a set A respectively. Then

$G^{V_N}(x,y):= \mathbb{E}_x \left[ \sum_{m=0}^{\tau_{\partial V_N}}1_{(S_m=y) }\right].$

That is, $G^{V_N}(x,y)$ is the expected number of visits to y by a random walk from x, before it exits $V_N$.

Let’s drop the superscript for now, as everything should hold for a more general subset of the lattice. I don’t think it’s immediately obvious at the level of Markov chains why G(x,y)=G(y,x). In particular, it’s not the case that

$\mathbb{P}_x(\tau_y < \tau_{D^c}) = \mathbb{P}_y(\tau_x <\tau_{D^c}),$

and it feels that we can’t map between paths $x \to \partial D$ and $y\to \partial D$ in a way that preserves the number of visits to y and x, respectively. However, we can argue that for any m

$\mathbb{P}_x(S_m=y, \tau_{D^c}>m) = \mathbb{P}_y(S_m=x, \tau_{D^c}>m),$

by looking at the suitable paths of $(S_m)$. That is, if we have a path $x=S_0,S_1,\ldots,S_m=y$ that stays within D, then the probability of seeing this path starting from x and its reverse direction starting from y are equal. Why? Because

$\mathbb{P}_x(S_0=x,S_1=v_1,\ldots,S_{m-1}=v_{m-1},S_m=y) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_\ell)},$

and

$\mathbb{P}_y(S_0=y,S_1=v_{m-1},\ldots,S_{m-1}=v_1, S_m=x) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_{m-\ell})} = \prod_{\ell=1}^m \frac{1}{\mathrm{deg}(v_\ell)}.$

Since $D\subset \mathbb{Z}^d$ and x,y are in the interior of D, we must have $\mathrm{deg}(x)=\mathrm{deg}(y)$, and so these two expressions are equal. Summing over all such two-way paths, and then all m gives the result.

Fixing one argument

We now focus on $G^D(\cdot,y)$, where the second argument is fixed. This is the solution to the Poisson equation

$\Delta G^D(\cdot,y) = -\delta_y(\cdot),\quad G^D(x,y)=0,\; \forall x\in \partial D.$

To see this, can use a standard hitting probability argument (as here) with the Markov property. This is harmonic in $D\backslash \{y\}$, and since we know

$G^D(y,y)= \frac{1}{\mathbb{P}_y(\text{RW hits }\partial D\text{ before returning to }y)},$

this uniquely specifies $G^D(\cdot,y)$. Anyway, since harmonic functions achieve their maxima at the boundary, we have $G(y,y)\ge G(x,y)$ for all $x\in D$. We can also see this from the SRW definition as

$G(x,y)=G(y,x) = \mathbb{P}_y (\tau_x < \tau_{\partial D} ) G(x,x) \le G(x,x).$

Changing the domain

Now we want to consider nested domains $D\subset E$, and compare $G^D(\cdot,\cdot)$ and $G^E(\cdot,\cdot)$ on DxD. The idea is that for SRW started from $x\in D$, we have $\tau_{\partial D}\le \tau_{\partial E}$, since one boundary is contained within the other. From this, we get

$G^D(x,y)\le G^E(x,y),\quad \forall x,y\in D,$

and we will use the particular case y=x.

For example, if $x\in V_N$, the box with width N, then the box with width 2N centred on x contains the whole of $V_N$. So, if we set $\bar {V}_{2N}:= [-N,N]^d$, then with reference to the diagram, we have

$G^{V_N}(x,x)\le G^{\bar{V}_{2N}}(0,0),\quad x\in V_N.$

As we’ll see when we study the maximum of the DGFF on $V_N$, uniform control over the pointwise variance will be a useful tool.

Maximising the Green’s function

The idea of bounding $G^{V_N}(x,x)$ by $G^{\bar V_{2N}}(0,0)$ for any $x\in V_N$ is clever and useful. But a more direct approach would be to find the value of x that maximises $G^{V_N}(x,x)$. We would conjecture that when $V_N$ has a central vertex, then this is the maximiser.

We can prove this directly from the definition of the Green’s function in terms of random walk occupation times. Let’s assume we are working with $\bar{V}_N$ for even N, so that 0 is the central vertex. Again, since

$G^D(x,x)=\frac{1}{\mathbb{P}_x(\text{RW hits }\partial D\text{ before returning to }x)},$ (*)

it would suffice to show that this probability is minimised when x=0. This feels right, since 0 is furthest from the boundary. Other points are closer to the boundary in some directions but further in others, so we can’t condition on the maximum distance from its start point achieved by an excursion of SRW (we’re vertex-transitive, so these look the same from all starting points), as even allowing for the four possible rotations, for an excursion of diameter slightly larger than N, starting at the centre is maximally bad.

However, intuitively it does feel as if being closer to the boundary makes you more likely to escape earlier. In fact, with a bit more care, we can couple the SRW started from 0 and the SRW started from $r=(r^x,r^y)\ne 0$ such that the latter always exits first. For convenience we’ll assume also that $r^x,r^y$ are both even.

I couldn’t find any reference to this, so I don’t know whether it’s well-known or not. The following argument involves projecting into each axis, and doing separate couplings for transitions in the x-direction and transitions in the y-direction. We assume WLOG that x is in the upper-right quadrant as shown. Then, let $0=S_0,S_1,S_2,\ldots$ be SRW started from 0, and we will construct $r=R_0,R_1,R_2,\ldots$ on the same probability space as $(S_m)_{m\ge 0}$ as follows. For every m, we set the increment $R_{m+1}-R_m$ to be $\pm(S_{m+1}-S_m)$. It remains to specify the sign, which will be determined by the direction of the S-increment, and a pair of stopping times. The marginal is therefore again an SRW, started from r. Temporarily, we use the unusual notation $S_m= (S^x_m,S^y_m)$ for the coordinates of $S_m$.

So, if $S_{m+1}-S_m=(1,0), (-1,0)$, ie S moves left or right, then we set

$R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m) &\quad \text{if }mT^x.\end{cases}$ (*)

where $T^x:= \min\{m\,:\, R^x_m=S^x_m\}$. That is, $R^x$ moves in the opposing direction to $S^x$ until the first time when they are equal (hence the parity requirement), and then they move together. WLOG assume that $r^x>0$. Then suppose $S^x_m=\pm N$ and such m is minimal. Then by construction, if $m\ge T^x$, then $R^x_m=\pm N$ also. If $m, then we must have $S^x_m=-N$, and so since $R^x$‘s trajectory is a mirror image of $S^x$‘s, in fact $R^x_m = N+r^x>N$, so $R^x$ hit +N first. In both cases, we see that $R^x$ hits $\pm N$ at the same time or before $S^x$.

In other words, when $S^x_m$ has non-negative x coordinate, the lazy random walk $R^x$ follows the same trajectory as $S^x$, and when it has negative x coordinate, the $R^x$ mirrors $S^x$. At some time, it may happen that $S^x_m= R^x_m=0$ (recall the parity condition on r). Call this time $T^x$. We then adjust the description of the coupling so that (*) is the mechanism for $m, and then for $m\ge T^x$, we take $S^x_m=R^x_m$.

Similarly, if $S_{m+1}-S_m =(0,1), (0,-1)$, ie S moves up or down, then we set

$R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m)&\quad \text{ if }m

with corresponding definition of the stopping time $T^y$.

This completes the coupling, and by considering $T^x\wedge T^y$, we have shown what that the exit time for the walk started from zero dominates the exit time for walk started from r. Recall that so far we are in the case where the box has even width and $r=(r^x,r^y)$ has even coordinates.

This exit time comparison isn’t exactly what we need to compare $G^N(0,0)$ and $G^N(x,x)$. It’s worth remarking at this stage that if all we cared about was the Green’s function on the integer line [-N,N], we would have an easier argument, as by the harmonic property of $G(\cdot,y)$

$G^{[-N,N]}(0,r)=\frac{N-r}{N}G^{[-N,N]}(0,0),$

$G^{[-N,N]}(r,0) = \frac{N}{N+r}G^{[-N,N]}(r,r),$

and so $G(0,0)>G(r,r)$ follows by symmetry. To lift from 1D to 2D directly, we need a bit more than this. It’s possible that S returns in both x- and y- coordinates more often than R, but never at the same time. Fortunately, the coupling we defined slightly earlier does give us a bit more control.

Let $\tau^x(S), \tau^x(R)$ be the first times that $S^x, R^x$ hit $\pm N$. Under this coupling, for any $m\ge 0$

$\mathbb{P}(S^x_m=0, m

since these events are literally equal. Since we showed that $\tau^x(R)\le \tau^x(S)$ almost surely, we can further deduce

$\mathbb{P}(S^x_m=0,m

$=\mathbb{P}(R^x_m=r^x, m

To address the corresponding events for which $m\ge T^x$, we apply the strong Markov property at $T^x$, to obtain SRW $Z_m$ started from r/2, and let $\tau_{-N},\tau_{+N}$ be the hitting times of $-N,+N$ respectively and $\tau_{\pm N}=\tau_{-N}\wedge \tau_{+N}$. It will now suffice to prove that

$\mathbb{P}(Z_m=0, m< \tau_{\pm N}) \ge \mathbb{P}(Z_m=r,m<\tau_{\pm N}),$ (**)

as then we can apply the law of total probability and sum over values of $T^x$ and $m\ge 0$.

To prove this result, we consider the following bijection between trajectories of length m from r/2 to {0,r}. We decompose the trajectories into excursions away from r/2, and then a final meander from r/2 to {0,r} that stays on the same side of r/2. We construct the new trajectory by preserving all the initial excursions, but reversing all the steps of the final meander. So if the original trajectory ended up at 0, the image ends up at r. Trivially, the initial excursions in the image only hit $\pm N$ if the excursions in the original trajectory did this too. But it’s also easy to see, by a similar argument to the coupling at the start of this section, that if the original trajectory ends at r and does not hit $\pm N$, then so does the image. However, the converse is not true. So we conclude (**), and thus

$\mathbb{P}(S_m^x=0) \ge \mathbb{P}(R_m^x=0)$

for all m by combining everything we have seen so far. And so we can now lift to a statement about $S_m$ itself, that is considering both coordinates separately.

The remaining cases for r require a little more care over the definition of $T^x$, though the same projection argument works, for fundamentally the same reason. (Note that in the above argument, if $S^x_m=-N$ and $m, then in fact $R^x_m\ge N+2$, and so it’s not hard to convince yourself that a sensible adjustment to the stopping time will allow a corresponding result with $R^x_m\ge N+1$ in the odd $r^x$ case.) The case for N odd is harder, since in one dimension there are two median sites, and it’s clear by symmetry that we can’t couple them such that RW from one always exits at least as early as RW from the other. However, the distributions of exit times started from these two sites are the same (by symmetry), and so although we can’t find a coupling, we can use similar stopping times to obtain a result in probability.

In the next post, we’ll see how to apply this uniform bound on $G^{V_N}(x,x)$ to control the maximum of the DGFF on $V_N$. In particular, we address how the positive correlations of DGFF influence the behaviour of the maximum by comparison with independent Gaussians at each site.

# Random walks conditioned to stay positive

In this post, I’m going to discuss some of the literature concerning the question of conditioning a simple random walk to lie above a line with fixed gradient. A special case of this situation is conditioning to stay non-negative. Some notation first. Let $(S_n)_{n\ge 0}$ be a random walk with IID increments, with distribution X. Take $\mu$ to be the expectation of these increments, and we’ll assume that the variance $\sigma^2$ is finite, though at times we may need to enforce slightly stronger regularity conditions.

(Although simple symmetric random walk is a good example for asymptotic heuristics, in general we also assume that if the increments are discrete they don’t have parity-based support, or any other arithmetic property that prevents local limit theorems holding.)

We will investigate the probability that $S_n\ge 0$ for n=0,1,…,N, particularly for large N. For ease of notation we write $T=\inf\{n\ge 0\,:\, S_n<0\}$ for the hitting time of the negative half-plane. Thus we are interested in $S_n$ conditioned on T>N, or T=N, mindful that these might not be the same. We will also discuss briefly to what extent we can condition on $T=\infty$.

In the first paragraph, I said that this is a special case of conditioning SRW to lie above a line with fixed gradient. Fortunately, all the content of the general case is contained in the special case. We can repose the question of $S_n$ conditioned to stay above $n\alpha$ until step N by the question of $S_n-n\alpha$ (which, naturally, has drift $\mu-\alpha$) conditioned to stay non-negative until step N, by a direct coupling.

Applications

Simple random walk is a perfectly interesting object to study in its own right, and this is a perfectly natural question to ask about it. But lots of probabilistic models can be studied via naturally embedded SRWs, and it’s worth pointing out a couple of applications to other probabilistic settings (one of which is the reason I was investigating this literature).

In many circumstances, we can desribe random trees and random graphs by an embedded random walk, such as an exploration process, as described in several posts during my PhD, such as here and here. The exploration process of a Galton-Watson branching tree is a particularly good example, since the exploration process really is simple random walk, unlike in, for example, the Erdos-Renyi random graph G(N,p), where the increments are only approximately IID. In this setting, the increments are given by the offspring distribution minus one, and the hitting time of -1 is the total population size of the branching process. So if the expectation of the offspring distribution is at most 1, then the event that the size of the tree is large is an atypical event, corresponding to delayed extinction. Whereas if the expectation is greater than one, then it is an event with limiting positive probability. Indeed, with positive probability the exploration process never hits -1, corresponding to survival of the branching tree. There are plenty of interesting questions about the structure of a branching process tree conditional on having atypically large size, including the spine decomposition of Kesten [KS], but the methods described in this post can be used to quantify the probability, or at least the scale of the probability of this atypical event.

In my current research, I’m studying a random walk embedded in a construction of the infinite-volume DGFF pinned at zero, as introduced by Biskup and Louidor [BL]. The random walk controls the gross behaviour of the field on annuli with dyadically-growing radii. Anyway, in this setting the random walk has Gaussian increments. (In fact, there is a complication because the increments aren’t exactly IID, but that’s definitely not a problem at this level of exposition.) The overall field is decomposed as a sum of the random walk, plus independent DGFFs with Dirichlet boundary conditions on each of the annuli, plus asymptotically negligible corrections from a ‘binding field’. Conditioning that this pinned field be non-negative up to the Kth annulus corresponds to conditioning the random walk to stay above the magnitude of the minimum of each successive annular DGFF. (These minima are random, but tightly concentrated around their expectations.)

Conditioning on $\{T > N\}$

When we condition on $\{T>N\}$, obviously the resulting distribution (of the process) is a mixture of the distributions we obtain by conditioning on each of $\{T=N+1\}, \{T=N+2\},\ldots$. Shortly, we’ll condition on $\{T=N\}$ itself, but first it’s worth establishing how to relate the two options. That is, conditional on $\{T>N\}$, what is the distribution of T?

Firstly, when $\mu>0$, this event always has positive probability, since $\mathbb{P}(T=\infty)>0$. So as $N\rightarrow\infty$, the distribution of the process conditional on $\{T>N\}$ converges to the distribution of the process conditional on survival. So we’ll ignore this for now.

In the case $\mu\le 0$, everything is encapsulated in the tail of the probabilities $\mathbb{P}(T=N)$, and these tails are qualitatively different in the cases $\mu=0$ and $\mu<0$.

When $\mu=0$, then $\mathbb{P}(T=N)$ decays polynomially in N. In the special case where $S_n$ is simple symmetric random walk (and N has the correct parity), we can check this just by an application of Stirling’s formula to count paths with this property. By contrast, when $\mu<0$, even demanding $S_N=-1$ is a large deviations event in the sense of Cramer’s theorem, and so the probability decays exponentially with N. Mogulskii’s theorem gives a large deviation principle for random walks to lie above a line defined on the scale N. The crucial fact here is that the probabilistic cost of staying positive until N has the same exponent as the probabilistic cost of being positive at N. Heuristically, we think of spreading the non-expected behaviour of the increments uniformly through the process, at only polynomial cost once we’ve specified the multiset of values taken by the increments. So, when $\mu<0$, we have

$\mathbb{P}(T\ge(1+\epsilon)N) \ll \mathbb{P}(T= N).$

Therefore, conditioning on $\{T\ge N\}$ in fact concentrates T on N+o(N). Whereas by contrast, when $\mu=0$, conditioning on $\{T\ge N\}$ gives a nontrivial limit in distribution for T/N, supported on $[1,\infty)$.

A related problem is the value taken by $S_N$, conditional on {T>N}. It’s a related problem because the event {T>N} depends only on the process up to time N, and so given the value of $S_N$, even with the conditioning, after time N, the process is just an unconditioned RW. This is a classic application of the Markov property, beloved in several guises by undergraduate probability exam designers.

Anyway, Iglehart [Ig2] shows an invariance principle for $S_N | T>N$ when $\mu<0$, without scaling. That is $S_N=\Theta(1)$, though the limiting distribution depends on the increment distribution in a sense that is best described through Laplace transforms. If we start a RW with negative drift from height O(1), then it hits zero in time O(1), so in fact this shows that conditonal on $\{T\ge N\}$, we have T= N +O(1) with high probability. When $\mu=0$, we have fluctuations on a scale $\sqrt{N}$, as shown earlier by Iglehart [Ig1]. Again, thinking about the central limit theorem, this fits the asymptotic description of T conditioned on T>N.

Conditioning on $T=N$

In the case $\mu=0$, conditioning on T=N gives

$\left[\frac{1}{\sqrt{N}}S(\lfloor Nt\rfloor ) ,t\in[0,1] \right] \Rightarrow W^+(t),$ (*)

where $W^+$ is a standard Brownian excursion on [0,1]. This is shown roughly simultaneously in [Ka] and [DIM]. This is similar to Donsker’s theorem for the unconditioned random walk, which converges after rescaling to Brownian motion in this sense, or Brownian bridge if you condition on $S_N=0$. Skorohod’s proof for Brownian bridge [Sk] approximates the event $\{S_N=0\}$ by $\{S_N\in[-\epsilon \sqrt{N},+\epsilon \sqrt{N}]\}$, since the probability of this event is bounded away from zero. Similarly, but with more technicalities, a proof of convergence conditional on T=N can approximate by $\{S_m\ge 0, m\in[\delta N,(1-\delta)N], S_N\in [-\epsilon \sqrt{N},+\epsilon\sqrt{N}]\}$. The technicalities here emerge since T, the first return time to zero, is not continuous as a function of continuous functions. (Imagine a sequence of processes $f^N$ for which $f^N(x)\ge 0$ on [0,1] and $f^N(\frac12)=\frac{1}{N}$.)

Once you condition on $T=N$, the mean $\mu$ doesn’t really matter for this scaling limit. That is, so long as variance is finite, for any $\mu\in\mathbb{R}$, the same result (*) holds, although a different proof is in general necessary. See [BD] and references for details. However, this is particularly clear in the case where the increments are Gaussian. In this setting, we don’t actually need to take a scaling limit. The distribution of Gaussian *random walk bridge* doesn’t depend on the mean of the increments. This is related to the fact that a linear transformation of a Gaussian is Gaussian, and can be seen by examining the joint density function directly.

Conditioning on $T=\infty$

When $\mu>0$, the event $\{T=\infty\}$ occurs with positive probability, so it is well-defined to condition on it. When $\mu\le 0$, this is not the case, and so we have to be more careful.

First, an observation. Just for clarity, let’s take $\mu<0$, and condition on $\{T>N\}$, and look at the distribution of $S_{\epsilon N}$, where $\epsilon>0$ is small. This is approximately given by

$\frac{S_{\epsilon N}}{\sqrt{N}}\stackrel{d}{\approx}W^+(\epsilon).$

Now take $\epsilon\rightarrow\infty$ and consider the RHS. If instead of the Brownian excursion $W^+$, we instead had Brownian motion, we could specify the distribution exactly. But in fact, we can construct Brownian excursion as the solution to an SDE:

$\mathrm{d}W^+(t) = \left[\frac{1}{W^+(t)} - \frac{W^+(t)}{1-t}\right] \mathrm{d}t + \mathrm{d}B(t),\quad t\in(0,1)$ (**)

for B a standard Brownian motion. I might return in the next post to why this is valid. For now, note that the first drift term pushes the excursion away from zero, while the second term brings it back to zero as $t\rightarrow 1$.

From this, the second drift term is essentially negligible if we care about scaling $W^+(\epsilon)$ as $\epsilon\rightarrow 0$, and we can say that $W^+(\epsilon)=\Theta(\sqrt{\epsilon})$.

So, returning to the random walk, we have

$\frac{S_{\epsilon N}}{\sqrt{\epsilon N}}\stackrel{d}{\approx} \frac{W^+(\epsilon)}{\sqrt{\epsilon}} = \Theta(1).$

At a heuristic level, it’s tempting to try ‘taking $N\rightarrow\infty$ while fixing $\epsilon N$‘, to conclude that there is a well-defined scaling limit for the RW conditioned to stay positive forever. But we came up with this estimate by taking $N\rightarrow\infty$ and then $\epsilon\rightarrow 0$ in that order. So while the heuristic might be convincing, this is not the outline of a valid argument in any way. However, the SDE representation of $W^+$ in the $\epsilon\rightarrow 0$ regime is useful. If we drop the second drift term in (**), we define the three-dimensional Bessel process, which (again, possibly the subject of a new post) is the correct scaling limit we should be aiming for.

Finally, it’s worth observing that the limit $\{T=\infty\}=\lim_{N\rightarrow\infty} \{T>N\}$ is a monotone limit, and so further tools are available. In particular, if we know that the trajectories of the random walk satisfy the FKG property, then we can define this limit directly. It feels intuitively clear that random walks should satisfy the FKG inequality (in the sense that if a RW is large somewhere, it’s more likely to be large somewhere else). You can do a covariance calculation easily, but a standard way to show the FKG inequality applies is by verifying the FKG lattice condition, and unless I’m missing something, this is clear (though a bit annoying to check) when the increments are Gaussian, but not in general. Even so, defining this monotone limit does not tell you that it is non-degenerate (ie almost-surely finite), for which some separate estimates would be required.

A final remark: in a recent post, I talked about the Skorohod embedding, as a way to construct any centered random walk where the increments have finite variance as a stopped Brownian motion. One approach to conditioning a random walk to lie above some discrete function is to condition the corresponding Brownian motion to lie above some continuous extension of that function. This is a slightly stronger conditioning, and so any approach of this kind must quantify how much stronger. In Section 4 of [BL], the authors do this for the random walk associated with the DGFF conditioned to lie above a polylogarithmic curve.

References

[BD] – Bertoin, Doney – 1994 – On conditioning a random walk to stay nonnegative

[BL] – Biskup, Louidor – 2016 – Full extremal process, cluster law and freezing for two-dimensional discrete Gaussian free field

[DIM] – Durrett, Iglehart, Miller – 1977 – Weak convergence to Brownian meander and Brownian excursion

[Ig1] – Iglehart – 1974 – Functional central limit theorems for random walks conditioned to stay positive

[Ig2] – Iglehart – 1974 – Random walks with negative drift conditioned to stay positive

[Ka] – Kaigh – 1976 – An invariance principle for random walk conditioned by a late return to zero

[KS] – Kesten, Stigum – 1966 – A limit theorem for multidimensional Galton-Watson processes

[Sk] – Skorohod – 1955 – Limit theorems for stochastic processes with independent increments

# 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.

# DGFF 1 – The discrete Gaussian free field from scratch

I’ve moved to Haifa in northern Israel to start a post-doc in the probability group at the Technion, and now that my thesis is finished I want to start blogging again. The past couple of weeks have been occupied with finding an apartment and learning about the Discrete Gaussian Free Field. All questions about the apartment are solved, but fortunately lots remain open about the DGFF, so I thought I’d write some background about this object and methods which have been used to study it.

Background – Random walk bridge

When we think of a random walk, we usually think of the index as time, normally going forwards. So for a random walk bridge, we might assume $Z_0=0$, and then condition on $Z_N=0$, thinking of this as a demand that the process has returned to zero at the future time. In some applications, this is the ideal intuition, but in others, it is more useful to think of the random walk bridge

$(0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0),$

as a random height function indexed by [0,N], where the probability of a given path decomposes naturally into a product depending on the N increments, up to a normalising constant.

Naturally, we are interested in the asymptotic behaviour of such a random walk bridge when $N\rightarrow\infty$. So long as the step distribution has finite variance, a conditioned version of Donsker’s theorem shows that the rescaled random walk bridge converges in distribution to Brownian bridge. Note that Brownian bridge

$(B^{\mathrm{br}}_t, t\in[0,1])$

can be constructed either by conditioning a standard Brownian motion B to return to zero at time one (modulo some technicalities – this event has zero probability), or by applying an appropriate (random) linear shift

$B^{\mathrm{br}}(t):= B(t) - tB(1).$ (*)

It is not too hard to calculate the distribution of $B^{\mathrm{br}}(t)$ for each $t\in[0,1]$, and with a bit more work, one can calculate the joint distribution of $(B^{\mathrm{br}}(s),B^{\mathrm{br}}(t))$. In particular, the joint distribution is multivariate Gaussian, and so everything depends on the covariance ‘matrix’ (which here is indexed by [0,1]).

So if we return to a random walk bridge what should the step distribution be? Simple symmetric RW is a natural choice, as then lots of the quantities we might want to consider boil down to combinatorial calculations. Cleverness and Stirling’s formula can often get us useful asymptotics. But there are lots of inconveniences, not least the requirement to be careful about parity (N has to be even for a start unless you make the walk lazy, in which case the combinatorics becomes harder), and even if these can be overcome in a given calculation, it would be better not to have this.

The claim is that the random walk with Gaussian increments is by far the easiest to analyse asymptotically. As a further heuristic, think about the statement of the central limit theorem in the case where the underlying distribution is normal: it’s true but obvious. [Indeed, it’s my favourite piece of advice to anyone taking second year probability exams to check that your proposed statement of CLT does actually work for $N(\mu,\sigma^2)$…] More concretely, if a RW has Gaussian increments, then the path $(Z_1,\ldots,Z_N)$ is a multivariate normal, or a Gaussian process with finite index set. In particular, covariances define the distribution. It remains a Gaussian process after conditioning on $Z_N=0$, and the linear tilting argument at (*) remains true here, and can indeed be applied to turn any boundary conditions into any other boundary conditions.

The discrete Gaussian free field

We know how to generalise the domain of a random walk to higher dimensions. But what generalising the index to higher dimension? So now there is definitely no arrow of time, and the notion of a random height function above $\mathbb{Z}^2$ (or a subset of it) is helpful, for which a scaling limit might be a random surface rather than Brownian motion.

Because we can’t well-order $\mathbb{Z}^d$, it’s harder to define any such random object on the entire lattice immediately, so we start with compact connected subsets, with zero boundary conditions, as in the one-dimensional case of random walk bridge. Formally, let D be a finite subset of $\mathbb{Z}^d$, and the boundary $\partial D$ those elements of $D^c$ which are adjacent to an element of D, and let $\bar D:= D\cup \partial D$.

Then, the discrete Gaussian free field on D is a random real vector $h^D=(h^D_x: x\in \bar D)$, with probability density proportional to

$\mathbf{1}\{h^D_x=0, x\in\partial D\}\exp\left ( - \frac{1}{4d} \sum_{x\sim y}(h^D_x - h^D_y)^2 \right),$ (1)

where we write $x\sim y$ if that x,y are adjacent in $\bar D$. We won’t at any stage worry much about the partition function which normalises this pdf. Note also that $\frac{1}{4d}$ is just a convenient choice of constant, which corresponds to one of the canonical choices for the discrete Laplacian. Adjusting this constant is the same as uniformly rescaling the values taken by the field.

The immediate interpretation of (1) is that the values taken by the field at vertices which are close to each other are positively correlated. Furthermore, the form of the density is Gaussian. Concretely, if the values of $h^D$ are fixed everywhere except one vertex $x\in D$, then the conditional distribution of $h^D_x$ is Gaussian. Later, or in subsequent posts, we will heavily develop this idea. Alternatively, we could if we really wanted describe the model in terms of independent Gaussians describing the ‘increment’ along each edge in D (which we should direct), subject to a very large number of conditions, namely that the sum of increments along any directed cycle is zero. This latter description might be more useful if you wanted to define a DGFF on a more sparse graph, but won’t be useful in what follows.

Note that we can rearrange the Laplacian in (1) in terms of the transition kernel p( ) of the simple random walk of D to obtain

$\exp\left( -\frac12 (h^D)^T (\mathbf{P}-\mathbf{1})h^D \right),$

where $P_{x,y}=p(y-x)$ is the transition matrix of SRW on D. In particular, this means that the free field is Gaussian, and we can extract the covariances via

$\mathrm{Cov}(h^D_x,h^D_y) = \left[ (\mathbf{1}-\mathbf{P})^{-1}\right]_{x,y}$

$= \left[\sum_{n\ge 0} \mathbf{P}^n\right]_{x,y} = \sum_{n\ge 0} \mathbb{P}_x\left[X_n=y,\tau_{\partial D}>n\right],$

where, under $\mathbb{P}_x$, $(X_0,X_1,\ldots)$ is simple random walk started from x.

This final quantity records the expected number of visits to y before leaving the domain D, for a random walk started at x, and is called the Green’s function.

In summary, the DGFF on D is the centred Gaussian random vector indexed by $\bar D$ with covariance given by the Green’s function $G_D(x,y)$.

How many of these equivalences carries over to more general D-indexed random fields is discussed in the survey paper by Velenik. But it’s worth emphasising that having the covariance given by the Green’s function as in the definition we’ve just given is a very nice property, as there are lots of pre-existing tools for calculating these. By contrast, it’s hard to think of a natural model for an integer-valued surface of this kind, as an analogue to SRW.

[Though definitely not impossible. The nicest example I’ve heard of is for height functions of large uniform domino tilings within their ‘arctic circle’, which have GFF asymptotics. See this paper by Kenyon.]

A continuous limit?

We motivated the discussion of random walk bridge by the limit object, namely Brownian bridge. Part of the reason why the DGFF is more interesting than Gaussian random walk bridge, is that the limit object, the (continuum) Gaussian free field is hard to define classically in two dimensions.

We might suppose that the DGFF in $V_N$, the square box of width N has some scaling limit as $N\rightarrow\infty$. However, for fixed $x,y\in [0,1]^2$, (and taking integer parts component-wise), well-known asymptotics for SRW in a large square lattice (more on this soon hopefully) assert that

$\mathrm{Cov}(h^{V_N}_{\lfloor Nx \rfloor},h^{V_N}_{\lfloor Ny\rfloor}) \sim \log |x-y|,$ (2)

and so any scaling limit will rescale only the square domain, not the height (since there is no N on the RHS of (2)). However, then the variance of the proposed limit is infinite everywhere.

So the GFF does not exist as a random height function on $[0,1]^2$, with the consequence that a) more care is needed over its abstract definition; b) the DGFF in 2D on a large square is an interesting object, since it does exist in this sense.

What makes it ‘free’?

This seemed like a natural question to ask, but I’ve received various answers. Some sources seem to suggest that having zero boundary condition is free. Other sources refer to the Hamiltonian (that is the term inside the exponential function at (1) ) as free since it depends only on the increments between values. If the Hamiltonian also depends on the heights themselves, for example via the addition of a $\sum_{x} \Psi(h^D_x)$ term, then for suitable choice of function $\Psi$, this is interpreted as a model where the particles have mass. The physical interpretation of these more general Gibbs measures is discussed widely, and I’m not very comfortable with it all at the moment, but aim to come back to it later, when hopefully I will be more comfortable.

# Fair games and the martingale strategy III

Gambler’s Ruin

Continuing directly from the previous post, the nicest example of the optional stopping theorem we developed there is to example a simple random walk constrained between two values, say 0 and N. This represents an idealised gambling situation, where the gambler stops playing either when they reach some pre-agreed profit, or when they go bankrupt. We assume that we start at level k, for k = 1,2,…,N-1.

Naturally, we want to know the probabilities of winning (ie getting to N) and losing (ie going bankrupt). We could set this up by conditioning on the first step. Let $p_k$ be the probability of winning starting from level k. Then we must have

$p_k= \frac12 p_{k+1}+\frac12 p_{k-1},\quad k=1,\ldots,N-1,$ (*)

with the obvious boundary conditions $p_0=0, p_N=1$. In an ideal world, we just know how to solve second order difference equations like (*). Well, actually it isn’t too hard, because we can see from (*) directly that

$p_{k+1}-p_k = p_k-p_{k-1},$

and so $p_k$ is a linear function of k, and so $p_k = k/N$ follows pretty much immediately.

But, we can also use OST profitably. Let T be the time at which we first hit 0 or N. It’s intuitively clear that this should have finite expectation, since the problems you might encounter with just the hitting time of a single level shouldn’t apply. Or you can consider the expected number of steps before you see N ups or downs in a row, which certainly provides an upper bound on T. This random number of steps is sort of geometric (at least, can be upper bounded by a geometric RV) and so has finite expectation. So can apply OST to X at T, and we have

$\mathbb{E}[X_T] = N\cdot \mathbb{P}(X_T=N) + 0 \cdot \mathbb{P}(X_T=0) = \mathbb{E}[X_0]=k,$

from which we also derive $p_k=k/N$.

The reason we talk about gambler’s ruin is by considering the limit $N\rightarrow\infty$ with k fixed. After a moment’s thought, it’s clear we can’t really talk about stopping the process when we hit infinity, since that won’t happen at any finite time. But we can ask what’s the probability that we eventually hit zero. Then, if we imagine a barrier at level N, the probability that we hit 0 at some point is bounded below by the probability that we hit 0 before we hit level N (given that we know we hit either zero or level N with probability one), and this is $\frac{N-k}{N}$, and by choosing N large enough, we can make this as close to 1 as we want. So the only consistent option is that the probability of hitting 0 at some point is one. Hence gambler’s ruin. With probability one, ruin will occur. There’s probably a moral lesson hiding there not especially subtly.

So the deal here seems to be that if you just care about your average, it doesn’t matter how to choose to play a sequence of fair games. But what if you care about something other than your average? In any real setting, we maybe care about slightly more than this. Suppose I offer you a bet on a coin toss: you get £3 if it comes up heads, and I get £1 if it comes up tails. Sounds like a good bet, since on average you gain a pound. But what about if you get £10,003 if it comes up heads and I get £10,001 if it comes up tails? I’m guessing you’re probably not quite so keen now.

But if you were an international bank, you might have fewer reservations about the second option. My intention is not to discuss whether our valuation of money is linear here, but merely to offer motivation for the financial option I’m about to propose. The point is that we are generally risk-averse (well, most of us, most of the time) and so we are scared of possible large losses, even when there is the possibility of large profits to balance it out.

Let’s assume we have our simple random walk, and for definiteness let’s say it starts at £1. Suppose (eg as a very niche birthday present) we have the following opportunity: at any point between now and time t=5, we have the right to buy one unit of the stock for £2.

We want to work out how much this opportunity, which from now on I’m going to call an option, is worth on average. Note that now it does seem that when we choose to cash in the option will have an effect on our return, and so we will have to include this in the analysis.

Note that, once we’ve bought a unit of the stock, we have an asset which is following a simple random walk (ie sequential fair games) and so from this point on its expected value remains unchanged. So in terms of expectation, we might as well sell the stock at the same moment we buy it. So if we cash in the option when the stock is currently worth £X, we will on average have a return of £(X-2). This means that we’ll only ever consider exercising our option if the current value of the stock is greater than £2. This narrows down our strategy slightly.

This sort of option minimises the risk of a large loss, since the worst thing that happens is that you never choose to exercise your option. So if you actually paid for the right to have this option, that cost is the largest amount you can lose. In the trading world, this type of opportunity is called an American option.

The trick here is to work backwards in time, thinking about strategies. If at time t=4, the stock is worth £1, then the best that can happen is that it’s worth £2 at time t=5, and this still gains you no wealth overall. Similarly if it’s worth £0 at time t=3. So we’ve identified a region where, if the stock value enters this region, we might as well rip up our contract, because we definitely aren’t going to gain anything. Remember now that we’ve also said you won’t ever cash in if the stock’s value is at most £2, because you don’t gain anything on average.

Now suppose that the stock has value £3 at time t=4. There’s no danger of it ever getting back below £2 during the lifetime of the option, so from now on your potential return is following the trajectory of a simple random walk, ie a fair game. So on average, it makes no difference whether you cash in now, or wait until t=5, or some combination of the two. The same argument holds if the stock has value £4 at time t=3 or time t=4, and so we can identify a region where you might as well cash in.

What about the final region? If the stock value is greater than £2, but not yet in the definitely-cash-in area, what should you do? Well, if you think about it, the value of the stock is a fair game. But your return should be better than that, because the stock price doesn’t take account of the fact that you wouldn’t buy in (and make a loss overall) if the value drops below £2. So at this stage, your future options are better than playing a fair game, and so it doesn’t make sense (in terms of maximising your *average*) to cash in.

Now we can actually work backwards in time to establish how much any starting value is worth under this optimal strategy. We can fill in the values in the ‘doomed’ area (ie all zeros) and on the ‘cash in now’ area (ie current value minus 2), and construct backwards using the fact that we have a random walk.

The final answer ends up being 7/16 if the stock had value £1 at time 0. Note that the main point here is that working out the qualitative form of the strategy was the non-trivial part. Once we’d done that, everything was fairly straightforward. I claim that this was a reasonably fun adjustment to the original problem, but have minimal idea whether pricing options is in general an interesting thing to do.

Anyway, I hope that provided an interesting overview to some of the topics of interest within the question of how to choose strategies for games based on random processes.

# Fair games and the martingale strategy I

I went back to my school a couple of weeks ago and gave a talk. I felt I’d given various incarnations of a talk on card-shuffling too many times, so it was time for a new topic. The following post (and time allowing, one or two more) is pretty much what I said.

The Martingale Strategy

Suppose we bet repeatedly on the outcome of tossing a fair coin. Since it’s November, my heart is set on buying an ice cream that costs £1, so my aim is to win this amount from our game. My strategy is this:

First, I bet £1. If I win, then that’s great, because I now have made exactly enough profit to buy the ice cream. If I lose, then I play again, and this time I bet £2. Again, if I win, then my total profit is £2-£1 = £1, so I stop playing and buy the ice cream. If I lose, then I play a third time, again doubling my stake. So if I win for the first time on the seventh go, my overall profit will be

£64 – (£1+£2+£4+£8+£16+£32) = £1,

and it’s clear that this can be continued and I will eventually win a round, and at this point my total profit will be £1. So I will always eventually be able to buy my ice cream.

But, there’s nothing special about the value £1, so I could replace the words ‘ice cream’ with ‘private tropical island’, so why am I still here in the UK on a wet Monday when I could be on my beach lounger?

There are some fairly obvious reasons why the strategy I’ve described is not actually a fail-safe way to make a profit. For a start, although with probability one a head will come up eventually, there is a small positive chance that the first 200 rolls will all be tails. At this point, I would have accrued a debt of roughly $2^{200}$ pounds, and this is slightly more than the number of atoms in the universe. All this for an ice cream?

So there are major problems carrying out this strategy in a finite world. And of course, it’s no good if we stop after a very large but finite number of turns, because then there’s always this very small chance that we’ve made a very large loss, which is bad, partly because we can’t have the ice cream, but also because it exactly cancels out the chance of making our £1 profit, and so our overall average profit is exactly zero.

Though I’ve set this up in an intentionally glib fashion, as so often is the case, we might have stumbled across an interesting mathematical idea. That is, if we play a fair game a finite number of times, we have a fair game overall, meaning our overall average profit is zero. But if we are allowed to play a potentially infinite number of times, then it’s not clear how to define our overall ‘average’ profit, since we feel it ought to be zero, as an extension of the finite case, but also might be positive, because it ends up being £1 with probability one.

It’s tempting at this stage to start writing statements like

$1 \times 1 + (-\infty) \times 0=0 ,$

to justify why this might have come about, where we consider the infinitely unlikely event that is infinitely costly. But this is only convincing at the most superficial level, and so it makes more sense to think a bit more carefully about under exactly what circumstances we can extend our observation about the overall fairness of a finite sequence of individual fair games.

A second example

The previous example was based upon a series of coin tosses, and we can use exactly the same source of randomness to produce a simple random walk. This is a process that goes up or down by 1 in each time step, where each option happens with probability ½, independently of the history.

We could avoid the requirement to deal with very large bets by always staking £1, and then cashing in the first time we have a profit of £1. Then, if we start the random walk at zero, it models our profit, and we stop the first time it gets to 1. It’s not obvious whether we hit 1 with probability one. Let’s show this.

In order to hit some positive value k, the random walk must pass through 1, 2, and so on, up to (k-1) and then finally k. So $\mathbb{P}(\text{hit k}) = [\mathbb{P}(\text{hit 1})]^k$. And similarly for negative values. Also, the probability that we return to zero is the same as the probability that we ever hit 1, since after one time-step they are literally the same problem (after symmetry). So, if the probability of hitting 1 is p<1, then the number of visits to zero is geometric (supported on 1,2,3,…) with parameter p, and so

$\mathbb{E}[\text{visits to k}] = \mathbb{E}[\text{visits to zero}] \times \mathbb{P}(\text{hit k})=(1+1/p) \times p^{|k|} = (p+1)p^{|k|-1}.$

Thus, when we sum over all values of k, we are summing a pair of geometric series with exponent <1, and so we get a finite answer. But if the expected number of visits to anywhere (ie the sum across all places) is finite, this is clearly ridiculous, since we are running the process for an infinite time, and at each time-step we must be somewhere! So we must in fact have p=1, and thus another potential counter-example to the claim that a sequence of fair games can sometimes be unfair.

We might have exactly the same set of practical objections, such as this method requiring arbitrarily large liquidity (even though it doesn’t grow exponentially fast so doesn’t seem so bad).

What will actually turn out to be useful is that although the bets are now small, the average time until we hit 1 is actually infinite. Remember that, even though most things we see in real life don’t have this property, it is completely possible for a random variable to take finite values yet have infinite expectation.

Notes on the Martingale Strategy

There’s no reason why the originally proposed strategy had to be based upon fair coin tosses. This strategy might work in a more general setting, where the chance of winning on a given turn is not ½, or is not even constant. So long as at each stage you bet exactly enough that, if you win, you recoup all your losses so far, and one extra pound, this has the same overall effect.

Of course, we need to check that we do eventually win a round, which is not guaranteed if the probability of winning (conditional on not having yet won) decays sufficiently fast. If we let $p_k$ be the probability of winning on turn k, given that we haven’t previously won, then we require that the probability of never winning $\prod_{k\ge 1}(1-p_k)=0$. By taking logs and taking care of the approximations, it can be seen that the divergence or otherwise of $\sum p_k$ determines which way this falls.

In the next post, we’ll talk about how the two problems encountered here, namely allowing large increments, and considering a stopping time with infinite expectation are exactly the two cases where something can go wrong. We’ll also talk about a slightly different setting, where the choice of when to stop playing becomes a bit more dynamic and complicated.

# Ornstein-Uhlenbeck Process

A large part of my summer has been spent proving some technical results pertaining to the convergence of some functionals of a critical Frozen Percolation process. This has been worthwhile, but hasn’t involved a large amount of reading around anything in particular, which has probably contributed to the lack of posts in recent months. Perhaps a mixture of that and general laziness?

Anyway, it turns out that the limit of the discrete processes under consideration is the Ornstein-Uhlenbeck process. The sense in which this limit holds (or at least, for now, is conjectured to hold) is something for another article. However, I thought it would be worth writing a bit about this particular process and why it is interesting.

The O-U process is described by the SDE

$dX_t=-\beta (X_t-\mu)dt+\sigma dW_t,$

where W is a standard Brownian motion. We think of $\mu$ as the ‘mean’. The extent to which this behaves as a mean will be discussed shortly. The process is then mean-reverting, in the sense that the drift is directed against deviations of the process away from this mean. The parameter $\beta$ measures the extent of this mean reversion, while as usual $\sigma$ controls the magnitude of the Brownian noise.

The motivation for considering mean-reverting processes is considerable. One measure of this is how many equations with articles on Wikipedia turn out to be precisely this Ornstein-Uhlenbeck process with different context or notation. In most cases, the motivation arises because Brownian motion is for some reason unsuitable to take as a canonical random process. We will see why the O-U process is somehow the next most canonical choice for a random process.

In physics, it is sometimes unsatisfactory to model the trajectory of a particle with Brownian motion (even though this motivated the name…) as the velocities are undefined (see this post from ages ago), or infinite, depending on your definition of velocity. Using the Ornstein-Uhlenbeck process to model the velocity of a particle is often a satisfactory alternative. It is not unreasonable that there should be a mean velocity, presumably zero. The mean reversion models a frictional force from the underlying medium, while the Brownian noise describes random collisions with similar particles.

In financial applications, the Ornstein-Uhlenbeck model has been applied, apparently under the title of the Vasicek model since the 70s to describe quantities such as interest rates where there is some underlying reason to ban indefinite growth, and require mean reversion. Another setting might be a commodity which, because of external driving factors, has over the relevant time-scale well-defined mean value, around which mean-reverting fluctuations on the observed time-scale can be described. As with other financial models, it is undesirable for a process to take negative values. This can be fixed by taking a positive mean, then setting the volatility to be state dependent, decaying to zero as the state tends to zero, so for small values, the positive drift dominates. I don’t fully understand why patching this aspect is significantly more important than patching any other non-realistic properties of the model, but the resulting SDE is, at least in one particular case where the volatility is $\sqrt{X_t}$, called the Cox-Ingersoll-Ross model.

Anyway, a mathematical reason to pay particular attention to this Ornstein-Uhlenbeck process is the following. It is the unique family of continuous Markov processes to have a stationary Gaussian distribution. It is the mean-reverting property that is key. There is no chance of Brownian motion having any stationary distribution, let alone a Gaussian one. If this isn’t clear, you can convince yourself by thinking of the stationary distribution of SRW on $\mathbb{Z}$. Since the process is space-homogeneous, the only stationary measure is the uniform measure.

I want to focus on one particular property of the O-U process, through which some other aspects will be illuminated. If we take $\sigma=\beta$ and let $\beta\rightarrow\infty$, then the stationary processes converge to white noise.

First though, we should note this is perhaps the easiest SDE to solve explicitly. We consider $X_t e^{\theta t}$, and applying Ito’s lemma rapidly gives

$X_t=\mu + (X_0-\mu)e^{-\beta t}+\sigma\int_0^t e^{-\beta(t-s)}dW_s.$

W is Gaussian so the distribution of $X_t$ conditional on $X_0=x_0$ is also Gaussian, and since W is centred we can read off the expectation. Applying the Ito isometry then gives the variance. In conclusion:

$X_t\stackrel{d}{=}\mathcal{N}(\mu+(x_0-\mu)e^{-\beta t}, \frac{\sigma^2}{\beta}(1-e^{-2\beta t})).$

In particular, note that the variation has no dependence on $x_0$. So as t grows to infinity, this converges to $\mathcal{N}(\mu, \frac{\sigma^2}{\beta})$. This is, unsurprisingly, the stationary distribution of the process.

To address the white noise convergence, we need to consider $\text{Cov}(X_0,X_t)$ in stationarity. Let’s assume WLOG that $\mu=0$ so most of the expectations will vanish. We obtain

$\text{Cov}(X_0,X_t)=\mathbb{E}[X_0X_t]=\mathbb{E}_{x_0}\left[\mathbb{E}[X_t| X_0=x_0]\right]=\mathbb{E}[X_0^2 e^{-\beta t}]= \frac{sigma^2}{2\beta}2^{-\beta t}.$

If we want, the Chapman-Kolmogorov equations work particularly nicely here, and we are able to derive a PDE for the evolution of the density function, though obviously this is very related to the result above. This PDE is known as the Fokker-Planck equation.

So, in particular, when $\sigma=\beta\rightarrow \infty$, this covariance tends to 0. I’m not purporting that this constitutes a proof that the Ornstein-Uhlenbeck processes converge as processes to white noise. It’s not obvious how to define process convergence, not least because there’s flexibility about how to view white noise as a process. One doesn’t really want to define the value of white noise at a particular time, but you can consider the covariance of integrals of white noise over disjoint intervals as a limit, in similar way to convergence of finite dimensional distributions.

The fact that taking $\beta=0$ gives Brownian motion, and this case gives white noise, intermediate versions of the Ornstein-Uhlenbeck process are sometimes referred to as coloured noise.

Finally, the Ornstein-Uhlenbeck process emerges as the scaling limit of mean-reverting discrete Markov chains, analogous to Brownian motion as the scaling limit of simple random walk. One particularly nice example is the Ehrenfest Urn model. We have two urns, and 2N balls. In each time step one of the 2N balls is chosen uniformly at random, and it is moved to the other urn. So a ball is more likely to be removed from an urn with more than N balls. We can view this as a model for molecules in, say a room, with a slightly porous division between them, eg a small hole. More complicated interface models in higher dimensions lead to fascinating PDEs, such as the famous KPZ equation, which are the subject of much ongoing interest in this area.

This result can be an application of the theory of convergence of Markov chains to SDEs pioneered by Stroock and Varadhan, about which more may follow very soon. In any case, it turns out that the fluctuations in the Ehrenfest Urn model are on the scale of $\sqrt{n}$, unsurprisingly, and are given by a centred Ornstein-Uhlenbeck process.

Investigating this has reminded me how much I’ve forgotten, or perhaps how little I ever knew, about the technicalities of stochastic processes are their convergence results, so next up will probably be a summary of all the useful definitions and properties for this sort of analysis.

# Hitting Probabilities for Markov Chains

This continues my previous post on popular questions in second year exams. In the interest of keeping it under 2,500 words I’m starting a new article.

In a previous post I’ve spoken about the two types of Markov chain convergence, in particular, considering when they apply. Normally the ergodic theorem can be used to treat the case where the chain is periodic, so the transition probabilities do not converge to a stationary distribution, but do have limit points – one at zero corresponding to the off-period transitions, and one non-zero. With equal care, the case where the chain is not irreducible can also be treated.

A favourite question for examiners concerns hitting probabilities and expected hitting times of a set A. Note these are unlikely to come up simultaneously. Unless the hitting probability is 1, the expected hitting time is infinite! In both cases, we use the law of total probability to derive a family of equations satisfied by the probabilities/times. The only difference is that for hitting times, we add +1 on the right hand side, as we advance one time-step to use the law of total probability.

The case of hitting probabilities is perhaps more interesting. We have:

$h_i^A = 1,\; i\in A, \quad h_i^A=\sum_{j\in S}p_{ij}h_j^A,\; i\not\in A.$

There are two main cases of interest: where the chain is finite but has multiple closed communicating classes, and where the chain is infinite, so even though it is irreducible, a trajectory might diverge before hitting 0.

For the case of a finite non-irreducible Markov chain, this is fairly manageable, by solving backwards from states where we know the values. Although of course you could ask about the hitting probability of an open state, the most natural question is to consider the probability of ending up in a particular closed class. Then we know that the hitting probability starting from site in the closed class A is 1, and the probability starting from any site in a different closed class is 0. To find the remaining values, we can work backwards one step at a time if the set of possible transitions is sparse enough, or just solve the simultaneous equations for $\{h_i^A: i\text{ open}\}$.

We therefore care mainly about an infinite state-space that might be transient. Typically this might be some sort of birth-and-death chain on the positive integers. In many cases, the hitting probability equations can be reduced to a quadratic recurrence relation which can be solved, normally ending up with the form

$h_i=A+B\lambda^i$,

where $\lambda$ might well be q/p or similar if the chain is symmetric. If the chain is bounded, typically you might know $h_0=1, h_N=0$ or similar, and so you can solve two simultaneous equations to find A and B. For the unbounded case you might often only have one condition, so you have to rely instead on the result that the hitting probabilities are the minimal solution to the family of equations. Note that you will always have $h^i_i=1$, but with no conditions, $h^i_j\equiv 1$ is always a family of solutions.

It is not clear a priori what it means to be a minimal solution. Certainly it is not clear why one solution might be pointwise smaller than another, but in the case given above, it makes sense. Supposing that $\lambda<1$, and A+B=1 say, then as we vary the parameters, the resulting set of ‘probabilities’ does indeed vary monotically pointwise.

Why is this true? Why should the minimum solution give the true hitting probability values? To see this, take the equations, and every time an $h_i^A$ appears on the right-hand side, substitute in using the equations. So we obtain, for $i\not\in A$,

$h_i^A=\sum_{j\in A}p_{ij}+\sum_{j\not\in A} p_{ij}h_j^A,$

and after a further iteration

$h_i^A=\sum_{j_1\in A}p_{ij_1}+\sum_{j_1\not\in A, j_2\in A}p_{ij_1}p_{j_1j_2}+\sum_{j_1,j_2\not\in A}p_{ij_1}p_{j_1j_2}h_{j_2}^A.$

So we see on the RHS the probability of getting from i to A in one step, and in two steps, and if keep iterating, we will get a large sum corresponding to the probability of getting from i to A in 1 or 2 or … or N steps, plus an extra term. Note that the extra term does not have to correspond to the probability of not hitting A by time N. After all, we do not yet know that $(h_{i}^A)$ as defined by the equations gives the hitting probabilities. However, we know that the probability of hitting A within N steps converges to the probability of hitting A at all, since the sequence is increasing and bounded, so if we take a limit of both sides, we get $h_i^A$ on the left, and something at least as large as the hitting probability starting from i on the right, because of the extra positive term. The result therefore follows.

It is worth looking out for related problems that look like a hitting probability calculation. There was a nice example on one of the past papers. Consider a simple symmetric random walk on the integers modulo n, arranged clockwise in a circle. Given that you start at state 0, what is the probability that your first return to state 0 involves a clockwise journey round the circle?

Because the system is finite and irreducible, it is not particularly interesting to consider the actual hitting probabilities. Also, note that if it is convenient to do so, we can immediately reduce the problem when n is even. In two steps, the chain moves from j to j+2 and j-2 with probability ¼ each, and stays at j with probability ½. So the two step chain is exactly equivalent to the lazy version of the same dynamics on n/2.

Anyway, even though the structure is different, our approach should be the same as for the hitting probability question, which is to look one step into the future. For example, to stand a chance of working, our first two moves must both be clockwise. Thereafter, we are allowed to move anticlockwise. There is nothing special about starting at 0 in defining the original probability. We could equally well ask for the probability that starting from j, the first time we hit 0 we have moved clockwise round the circle.

The only thing that is now not obvious is how to define moving clockwise round the circle, since it is not the case that all the moves have to be clockwise to have experienced a generally clockwise journey round the circle, but we definitely don’t want to get into anything complicated like winding numbers! In fact, the easiest way to make the definition is that given the hitting time of 0 is T, we demand that the chain was at state n at time T-1.

For convenience (ie to make the equations consistent) we take $h_0=0, h_n=1$ in an obvious abuse of notation, and then

$h_j=\frac12h_{j-1}+\frac12 h_{j+1},$

from which we get

$h_j=a+bj \Rightarrow h_j=\frac{j}{n}.$

Of course, once we have this in mind, we realise that we could have cut the circle at 0 (also known as n) and unfolded it to reduce the problem precisely to symmetric gambler’s ruin. In particular, the answer to the original problem is 1/2n, which is perhaps just a little surprising – maybe by thinking about the BM approximation to simple random walk, and that BM started from zero almost certainly crosses zero infinitely many times near we might have expected this probability to decay faster. But once it is unfolded into gambler’s ruin, we have the optimal stopping martingale motivation to reassure us that this indeed looks correct.