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

Discontinuous Phase Transitions

Yesterday, Demeter Kiss from Cambridge gave a seminar in Oxford about a model for self-destructive percolation on $\mathbb{Z}^2$ that had implications for the (non-)existence of an infinite-parameter forest fire model on the same lattice. I enjoyed talking about this and his recent work on the related model of frozen percolation on $\mathbb{Z}^2$. Considering these models in the lattice setting present a whole range of interesting geometric challenges that are not present in the mean-field case that has mainly occupied my research direction so far.

The afternoon’s discussion included lots of open problems about percolation. Several of these are based around continuity of the phase transition, so I thought I would write a quite post about some simple examples of this, and one example where it does not hold.

A helpful base example is bond percolation on the lattice $\mathbb{Z}^2$. Here, we specify some probability p in [0,1], and we declare edges of the lattice open with probability p, independently of each other. We then consider the graph induced by the open edges. We say that percolation occurs if the origin is contained in an infinite open component. The terminology arises from the interpretation as fluid being added at the origin and flowing down open edges. We define $\theta(p)$ to be the probability that the origin is in an infinite component when the parameter is p. By translation-invariance, we can get some sort of 0-1 law, to conclude that there is an infinite component somewhere in the system with probability either 0 or 1, depending on whether $\theta(p)$ is positive or zero. Indeed, we can further show that if it is positive, then with probability 1 there is a unique infinite component.

We define the critical probability $p_c:= \inf\{\theta(p)>0\}$. A question worth asking is then, what is $\theta(p_c)$? In some examples, we can find $p_c$, but we cannot prove that $\theta(p)$ is continuous around $p_c$. In the case of $\mathbb{Z}^2$ this is known, and it is known from work of Kesten that $p_c=1/2$. See below for a plot of $\theta(p)$ in this setting (obtained from this blog, though possibly originating elsewhere).

The aim is to find an example where we do not have such a continuous phase transition. The original work on frozen percolation took place on trees, and one of Kiss’s results is confirms that these show qualitatively different phenomena to the same process on the lattice. In some sense, trees lie halfway between a lattice and a mean-field model, since there is often some independence when we look down the tree from a given generation, if it is well-defined to use such language.

Anyway, first we consider percolation on an infinite regular rooted k-ary tree. This means we have a root, which has k children, each of which in turn has k children, and so on. As before we consider bond percolation with parameter p. In this setting, we have a language to describe the resulting open component of the root. The offspring distribution of any vertex in the open component is given by Bin(k,p) independently of everything else, so we can view this component as the realisation of a Galton-Watson tree with this offspring distribution. This distribution has finite mean kp, and so we can state explicitly when the survival probability is positive. This happens when the mean is greater than 1, ie p>1/k.

For our actual example, we will consider the survival probability, but the technicalities are easier to explain if we look at the extinction probability, now using the language of branching processes. Suppose the offspring distribution has pgf given by

$f(x)=p_0+p_1x+p_2x^2+\ldots.$

Then the extinction probability q satisfies f(q)=q. I want to pause to consider what happens if this equation has multiple solutions. Indeed, in most interesting cases it will have multiple solutions, since f(1) will always be 1 if it is a non-defective offspring distribution. It is typically cited that: the extinction probability q is the smallest solution to this equation. I want to discuss why that is the case.

To approach this, we have to consider what extinction means. It is the limit in the event sense of the events {we are extinct after n generations}. Let the probabilities of these events be $q_n$, so $q_0=0$. Then by a straightforward coupling argument, we must have

$0=q_0\le q_1\le q_2 \le\ldots\le q:= \lim q_n \le 1.$

But, by the same generating function argument as before, $q_{n+1}=f(q_n)\ge q_n$. So if we split [0,1] into regions A where $f(x)\ge x$ and B where $f(x), all the $(q_n)$s must occur in the former, and so since it is closed, their limit must be in A also. Note that if f(x) intersects x lots of times, then region A is not necessarily connected. In the diagram below, in moving from $q_n$ to $q_{n+1}$ we might jump across part of B.

This is bad, as we are trying to prove that q is the right boundary of the connected component of A containing 0. But this cannot happen, as f is monotonic. So if one of the roots of f(x)=x in between the hypothesised $q_n is called z, then $f(q_n)< f(z)=z < q_{n+1}$, a contradiction.

Ok, so now we are ready to consider our counterexample to continuity over the percolation threshold. See references for a link to the original source of this example. We have to choose a slightly more complicated event than mere survival or extinction. We consider bond percolation as before on the infinite ternary tree, where every vertex has precisely 3 offspring. Our percolation event is now that the root is the root of an infinite binary tree. That is, the root has at least two children, each of which have at least two children, each of which, and so on.

If we set this probability equal to q, and the probability of an edge being open equal to p, then we have the recurrence:

$q=3p^2(1-p)q^2+p^3[3q^2(1-q)+q^3].$

The first term corresponds to the root having two open edges to offspring, and the second to the root having all three open edges to offspring. After manipulating, we end up with

$q\left[2p^3q^2-3p^2q+1\right]=0.$

We are therefore interested in roots of the quadratic lying between 0 and 1. The discriminant can be evaluated as

$\Delta=p^3(9p-8),$

and so there are no real roots where p<8/9. But when p=8/9, we have a repeated root at q=27/32, which is obviously not zero!

This equation is qualitatively different to the previous one for the extinction probability of a Galton-Watson tree. There, we had a quadratic, with one root at 1. As we varied p, the other root moved continuously from greater than one to less than one, so it passed through 1, giving continuity at the critical probability. Here, we have a cubic, again with one root at 1. But now the other roots are complex for small p, meaning that the local minimum of the cubic lies above the x-axis. As p gets to the critical value, it the local minimum passes below the x-axis, and suddenly we have a repeated root, not at zero.

I would like to have a neat probabilistic heuristic for this result, without having to make reference to generating functions. At the moment, the best I can come up with is to say that the original problem is simple, in the sense that the critical probability is as small as it could be while still making sense in expectation. To be concrete, when the mean of the offspring generation is less than 1, the expected size of the nth generation tends to zero, so there certainly could not be positive probability of having an infinite component.

Whereas in the binary tree example, we only require p=2/3 to have, in expectation, the right number of open edges to theoretically allow an infinite binary tree. If we think of percolation as a dynamic process by coupling in p, essentially as we move from p=2/3 to p=8/9 we need to add enough edges near the origin to be able to take advantage of the high density of edges available far from the origin. The probability of this working given you start from n vertices grows much faster (as n grows) than in the original problem, so you might expect a faster transition.

This is so content-free I’m reluctant even to call it a heuristic. I would be very interested to hear of any more convincing argument for this phenomenon!

REFERENCES

Dekking, Pakes – On family trees and subtrees of simple branching processes (link)

Branching Processes and Dwass’s Theorem

This is something I had to think about when writing my Part III essay, and it turns out to be relevant to some of the literature I’ve been reading this week. The main result is hugely helpful for reducing a potentially complicated combinatorial object to a finite sum of i.i.d. random variables, which in general we do know quite a lot about. I was very pleased with the proof I came up with while writing the essay, even if in the end it turned out to have appeared elsewhere before. (Citation at end)

Galton-Watson processes

A Galton-Watson process is a stochastic process describing a simple model for evolution of a population. At each stage of the evolution, a new generation is created as every member of the current generation produces some number of `offspring’ with identical and independent (both across all generations and within generations) distributions. Such processes were introduced by Galton and Watson to examine the evolution of surnames through history.

More precisely, we specify an offspring distribution, a probability distribution supported on $\mathbb{N}_0$. Then define a sequence of random variables $(Z_n,n\in\mathbb{N})$ by:

$Z_{n+1}=Y_1^n+\ldots+Y_{Z_n}^n,$

where $(Y_k^n,k\geq 1,n\geq 0)$ is a family of i.i.d. random variables with the offspring distribution $Y$. We say $Z_n$ is the size of the $n$th generation. From now on, assume $Z_0=1$ and then we call $(Z_n,n\geq 0)$ a Galton-Watson process. We also define the total population size to be

$X:=Z_0+Z_1+Z_2+\ldots,$

noting that this might be infinite. We refer to the situation where $X<\infty$ finite as extinction, and can show that extinction occurs almost surely when $\mathbb{E}Y\leq 1$, excepting the trivial case $Y=\delta_1$. The strict inequality parts are as you would expect. We say the process is critical if $\mathbb{E}Y=1$, and this is less obvious to visualise, but works equally well in the proof, which is usually driven using generating functions.

Total Population Size and Dwass’s Theorem

Of particular interest is $X$, the total population size, and its distribution. The following result gives us a precise and useful result linking the probability of the population having size $n$ and the distribution of the sum of $n$ RVs with the relevant offspring distribution. Among the consequences are that we can conclude immediately, by CLT and Cramer’s Large Deviations Theorem, that the total population size distribution has power-law decay in the critical case, and exponential decay otherwise.

Theorem (Dwass (1)): For a general branching process with a single time-0 ancestor and offspring distribution $Y$ and total population size $X$:

$\mathbb{P}(X=k)=\frac{1}{k}\mathbb{P}(Y^1+\ldots+ Y^k=k-1),\quad k\geq 1$

where $Y^1,\ldots,Y^k$ are independent copies of $Y$.

We now give a proof via a combinatorial argument. The approach is similar to that given in (2). Much of the literature gives a proof using generating functions.

Proof: For motivation, consider the following. It is natural to consider a branching process as a tree, with the time-0 ancestor as the root. Suppose the event $\{X=k\}$ in holds, which means that the tree has $k$ vertices. Now consider the numbers of offspring of each vertex in the tree. Since every vertex except the root has exactly one parent, and there are no vertices outside the tree, we must have $Y^1+\ldots+Y^k=k-1$ where $Y^1,\ldots,Y^k$ are the offspring numbers in some order. However, observe that this is not sufficient. For example, if $Y^1$ is the number of offspring of the root, and $k\geq 2$, then we must have $Y^1\geq 1$. Continue reading