# The reflection principle and conditioned RWs

I haven’t published a post about probability for far too long. Several are queued, so perhaps this will be the start of a deluge.

Anyway, with my advisor at Technion, I’m still working on some problems concerning Gaussian random walk subject to some conditioning which is complicated, but in practice (we hope) only mildly different to conditioning the walk to stay positive. Our conditioning at step n depends on some external randomness, but also on the future trajectory of the walk (related to the embedding of the walk in a 2D DGFF), thus ruining the possibility of applying the Markov property in any proof without significant preliminary work.

It seemed worth writing briefly about some of these results in a slightly simpler setting. The goal is to assemble many of the ingredients required to prove a local limit for Gaussian random walk conditioned to stay positive, in a sense which will be clarified towards the end. This is not the best way to go about getting scaling limits (as discussed briefly here, and for which see references [Ig74] and [Bo76]), and it’s probably not the best way to get local limits in the simplest setting, but it’s the method we are currently working to generalise, and follows the outline of [B-JD05], but in much less technical detail.

Probabilities via the reflection principle

We start with Brownian motion. The reflection principle, as described briefly in this post from the depths of history, is a classical technique for studying the maximum of Brownian motion. Roughly speaking, we exploit the fact that $(W_t,t\ge 0)\stackrel{d}=(-W_t,t\ge 0)$, but we then apply this at the hitting time of a particular positive value, using the Strong Markov Property.

Let $S_t=\max_{0\le s\le t}W_s$ be the running maximum of the Brownian motion $W_t$, and $\tau_b$ the hitting time of b. Then

$\mathbb{P}(S_t\ge b, B_t\le a)=\mathbb{P}(\tau_b

which, by SMP at $\tau_b$ and the reflection invariance of a standard BM, is equal to

$\mathbb{P}(\tau_b

This obviously assumed $b\ge a$, but if we set $b=a$, we find

$\mathbb{P}(S_t\ge b)=\mathbb{P}(B_t>b)+\mathbb{P}(S_t\ge b,B_t\le b)=2\mathbb{P}(B_t\ge b).$

Or, in other words, $S_t\stackrel{d}=|B_t|$.

While we can’t derive such nice equalities in distribution, the reflection principle is robust enough to treat more complicated settings, such as Brownian bridge.

We might want to ask about the maximum of a standard Brownian bridge, but we can be more general, and ask about the maximum of a Brownian bridge with drift (let’s say general bridge here). It’s important to remember that a general Brownian bridge has the same distribution as a linear transformation of a standard Brownian bridge. Everything is Gaussian, after all. So asking whether the maximum of a general Brownian bridge is less than a particular value is equivalent to asking whether a standard Brownian bridge lies below a fixed line. Wherever possible, we make such a transformation at the start and perform the simplest version of the required calculation.

So, suppose we have a bridge B from (0,0) to (t,a), and we want to study $\max_{s\in[0,t]} B_s$. Fix some $b>a$, and work with a standard Brownian motion $W_s$. By a similar argument to before,

$\mathbb{P}(\tau_b\le t, W_t\in[a,a+\mathrm{d}x]) = \mathbb{P}(W_t\in [2b-a-\mathrm{d}x,2b-a]) = \frac{\mathrm{d}x}{\sqrt{2\pi t}}e^{-(2b-a)^2/2t},$

and

$\mathbb{P}(W_t\in[a,a+\mathrm{d}x])=\frac{\mathrm{d}x}{\sqrt{2\pi t}}e^{-a^2/2t}.$

So

$\mathbb{P}(\max_{s\in[0,t]}B_t\ge b) = \exp\left(\frac{-(2b-a)^2 + a^2}{2t}\right).$

Random walk conditioned to stay positive

Our main concern is conditioning to stay above zero. Let $\mathbb{P}_{0,x}^{t,y}$ be some complete if cumbersome notation for a Brownian bridge B from (0,x) to (t,y). Then another simple transformation of the previous result gives

$\mathbb{P}_{0,x}^{t,y}(B_s\ge 0,\,s\in[0,t])=1-\exp\left( \frac{-(x+y)^2 + (x-y)^2}{2t} \right)= 1-\exp\left(-\frac{2xy}{t}\right).$

Then, if $xy\ll t$, we can approximate this by $\frac{2xy}{t}$. (*)

Extend the notation so $\mathbb{P}_{0,x}$ describes Brownian motion started from (0,x). Then integrating over y, gives

$\mathbb{P}_{0,x}(B_s\ge 0,\, s\in[0,t] ) = \frac{x}{t}\mathbb{E}[B_t\vee 0] = \sqrt{\frac{2}{\pi}} \frac{x}{\sqrt{t}}.$

(It might appear that we have integrated the approximation (*) over parts of the range where it is not a valid approximation, but the density of $B_t=\Theta(t)$ vanishes exponentially fast, and so actually it’s not a problem.)

We now want to extend this to random walks. Some remarks:

• We used the Gaussian property of Brownian motion fairly heavily throughout this derivation. In general random walks are not Gaussian, though we can make life easier by focusing on this case.
• We also used the continuity property of Brownian motion when we applied the reflection principle. For a general random walk, it’s hopeless to consider the hitting times of individual values. We have to consider instead the hitting times of regions $\tau_{(-\infty,b]}$, and so on. One can still apply SMP and a reflection principle, but this gives bounds rather than equalities. (The exception is simple random walk, for which other more combinatorial methods may be available anyway.)
• On the flip side, if we are interested in Brownian motion/bridge staying positive, we can’t start it from zero, as then the probability of this event is zero, by Blumenthal’s 0-1 law. By contrast, we can certainly ask about random walk staying positive when started from zero without taking a limit.

A useful technique will be the viewpoint of random walk as the values taken by Brownian motion at a sequence of stopping times. This Skorohod embedding is slightly less obvious when considering a general random walk bridge inside a general Brownian bridge, but is achievable. We want to study quantities like

$\mathbb{P}(S_k\ge 0,\, k=1,\ldots,n \big| S_0=x,S_n=y),$

where for simplicity let’s just take $(S_k,k\ge 0)$ to be a random walk with standard Gaussian increments. It’s possible we might want to take a scaling limit in x and y as functions of n. But first if we take x,y fixed, and embed the random walk bridge with these endpoints into the corresponding Brownian bridge with $t\approx n$, we are then faced with the question:

What’s the probability that the Brownian bridge goes below zero, but the embedded RW with n steps does not?

If the Brownian bridge conditioned to go below zero spends time $\Theta_p(n)$ below zero, then for large n it’s asymptotically very unlikely that the n places at which we embed the random walk avoids this set of intervals.

Several technical estimates are required to make this analysis rigorous. The conclusion is that there exists a function $f(x)$ for which $f(x)=x(1+o(1))$ as $x\rightarrow\infty$, such that

$q_n(x,y):=\mathbb{P}(S_k\ge 0,\, k=0,1,\ldots,n \,\big|\, S_0=x,S_n=y) \sim \frac{2f(x)f(y)}{n},$

$\text{and}\quad q_n(x):=\mathbb{P}(S_k\ge 0,\,k=0,1,\ldots,n\,\big|\, S_0=x)\sim \sqrt{\frac{2}{\pi}}\frac{f(x)}{\sqrt{n}}.$

As earlier, the second is obtained from the first by integrating over suitable y. This function $f$ has to account for the extra complications when either end-point is near zero, for which the event where the Brownian motion goes negative without the random walk going negative requires additional care.

Limits for the conditioned random walk

In the previous post on this topic, we addressed scaling limits in space and time for conditioned random walks. But we don’t have to look at the classical Donsker scaling to see the effects of conditioning to stay positive. In our setting, we are interested in studying the distribution of $S_m$ conditional on the event $(S_1\ge 0,S_2\ge 0,\ldots, S_n\ge 0)$, with limits taken in the order $n\rightarrow\infty$ and then $m\rightarrow\infty$.

(At a more general level, it is meaningful to describe the random walk conditioned on staying positive forever. Although this would a priori require conditioning on an event of probability zero, it can be handled formally as an example of an h-transform.)

As explained in that previous post, the scaling invariance of the Bessel process $W^+$ (which it’s not unreasonable to think of as ‘Brownian motion conditioned to stay non-negative’) suggests that this limit should exist, and be given by the entrance law of $W^+$. But this is hard to extract from a scaling limit.

However, we can use the previous estimate to start a direct calculation.

$\mathbb{P}(S_m\in \mathrm{d}y \,\big|\, S_k\ge 0,\, k=1,\ldots,n) = \frac{q_m(0,y) q_{n-m}(y) \mathbb{P}(S_m\in\mathrm{d}y)}{q_n(0)}.$

Here, we used the Markov property at time m to split the event that $S_m=y$ and the walk stays positive into two time-intervals. We will later take m large, so we may approximate as

$\frac{2f(0)f(y)/m \times \sqrt{\frac{2}{\pi}}f(y)/\sqrt{n-m}\times \mathbb{P}(S_m\in\mathrm{d}y) } { \sqrt{\frac{2}{\pi}}f(0)/\sqrt{n}}\stackrel{n\rightarrow\infty}=\frac{2f(y)^2}{m}\mathbb{P}(S_m\in\mathrm{d}y).$

This final probability emphasises that as $m\rightarrow\infty$ we only really have to consider $y=\Theta(\sqrt{m})$, so set $y=z\sqrt{m}$, and we obtain

$\lim_{n\rightarrow\infty}\mathbb{P}(\frac{S_m}{\sqrt{m}}\in\mathrm{d}z\,\big|\, S_k\ge 0,\,k=1,\ldots,n)$

$\sim \sqrt{m}\cdot\frac{2z^2m}{m}\cdot \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{m}}e^{-z^2/2} = \sqrt{\frac{2}{\pi}}z^2 e^{-z^2/2}.$

This is precisely the entrance law of the 3-dimensional Bessel process, usually denoted $R$. This process is invariant under time-rescaling in the same fashion as Brownian motion. Indeed, one representation of R is as the radial part of a three-dimensional Brownian motion, given by independent BMs in each coordinate. (See [Pi75] for explanation of the relation to ‘BM conditioned to stay non-negative’.) We could complete the analogy by showing that $q_n(x,y)$ converges to the transition density of R as well. (Cf the prelude to Theorem 2.2 of [B-JD05].)

Final remarks

The order of taking limits is truly crucial. We can also obtain a distributional scaling limit at time n under conditioning to stay non-negative up to time n. But then this is the size-biased normal distribution $\sim ze^{-z^2/2}$ (the Rayleigh distribution), rather than the square-size-biased normal distribution we say in this setting. And we can exactly see why. Relative to the normal distribution which applies in the absence of conditioning, we require size-biasing to account for the walk staying positive up to time m, and then also size-biasing to account for the walk staying positive for the rest of time (or up to n in the $n\rightarrow\infty$ limit if you prefer).

The asymptotics for $q_n(x,y)$ were the crucial step, for which only heuristics are present in this post. It remains the case that estimates of this kind form the crucial step in other more exotic conditioning scenarios. This is immediately visible (even if the random walk notation is rather exotic) in, for example, Proposition 2.2 of [CHL17], of which we currently require a further level of generalisation.

References

[Bo76] – Bolthausen – On a functional central limit theorem for random walks conditioned to stay positive

[B-JD05] – Bryn-Jones, Doney – A functional limit theorem for random walk conditioned to stay non-negative

[CHL17] – Cortines, Hartung, Louidor – The structure of extreme level sets in branching Brownian motion

[Ig74] – Iglehart – Functional central limit theorems for random walks conditioned to stay positive

[Pi75] – Pitman – One-dimensional Brownian motion and the three-dimensional Bessel process

# 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.

# Sticky Brownian Motion

This follows on pretty much directly from the previous post about reflected Brownian motion. Recall that this is a process defined on the non-negative reals which looks like Brownian motion away from 0. We consider whether RBM is the only such process, and how any alternative might be constructed as a limit of discrete-time Markov processes.

One of the alternatives is called Sticky Brownian motion. This process spends more time at 0 than reflected Brownian motion. In fact it spends some positive proportion of time at 0. My main aim here is to explain why some intuitive ideas I had about how this might arise are wrong.

The first thought was to ensure that each visit to 0 last some positive measure of time. This could be achieved by staying at 0 for an Exp(1) duration, every time the process visited it. It doesn’t seem unreasonable that this might appear as the limit of a standard SRW adjusted so that on each visit to 0 the walker waits for a time given by independent geometric distributions. These distributions are memoryless, so that is fine, but by Blumenthal’s 0-1 Law, starting from 0 a Brownian motion hits zero infinitely many times before any small time t. So in fact the process described above would be identically zero as before it gets anywhere it would have to spend some amount of time at 0 given by an infinite sum of Exp(1) RVs.

We will return later to the question of why the proposed discrete-time model will still converge to reflected BM rather than anything more exotic. First though, we should discount the possibility of any intermediate level of stickiness, where the set of times spent at 0 still has measure zero, but the local time at 0 grows faster than for standard reflected BM. We can define the local time at 0 through a limit

$L_t=\lim_{\epsilon\downarrow 0}\frac{1}{2\epsilon}\text{Leb}(\{0\le s \le t: |B_t|<\epsilon\})$

of the measure of time spent very near 0, rescaled appropriately. So if the measure of the times when the process is at 0 is zero, then the local time is determined by behaviour near zero rather than by behaviour at zero. More precisely, on the interval $[-\epsilon,\epsilon]$, the process behaves like Brownian motion, except on a set of measure zero, so the local time process should look the same as that of BM itself. Note I don’t claim this as a formal proof, but I hope it is a helpful heuristic for why you can’t alter the local time process without altering the whole process.

At this stage, it seems sensible to define Sticky Brownian motion. For motivation, note that we are looking for a process which spends a positive measure of time at 0. So let’s track this as a process, say $C_t$. Then the set of times when C is increasing is sparse, as it coincides with the process being 0, but we know we cannot wait around at 0 for some interval of time without losing the Markov property. So C shares properties with the local time of a reflected BM. The only difference is that the measure of times when C is increasing is positive here, but zero for the local time.

So it makes sense to construct the extra time spent at zero from the local time of a standard reflected BM. The heuristic is that we slow down the process whenever it is at 0, so that local time becomes real time. We can also control the factor by which this slowing-down happens, so define

$\sigma(s)=\rho L(s)+s,$

where L is the local time process of an underlying reflected BM, and $\rho>0$ is a constant. So $\sigma$ is a map giving a random time-change. Unsurprisingly, we now define Sticky BM as the reflected BM with respect to this time-change. To do this formally, it is easiest to define a family of stopping times $\{\tau_t\}$, such that $\sigma(\tau_t)=t, \tau_{\sigma(s)}=s$, then if X is the reflected BM, define $Y_t=X_{\tau_t}$ for the sticky BM.

It is worth thinking about what the generator of this process should be. In particular, why should it be different to reflected BM? The key observation is that the drift of the underlying reflected BM is essentially infinite at 0. By slowing down the process at 0, this drift becomes finite. So the martingale associated with sticky BM is precisely a time-changed version of the martingale associated with the underlying reflected BM, but this time-change is precisely what is required to give a generator. We get:

$(\mathcal{L}f)(x)=\begin{cases}\frac12f''(x)&\quad x>0\\ \rho^{-1}f'(0) &\quad x=0.\end{cases}$

Now that we have the generator, it starts to become apparent how sticky BM might appear as a limit of discrete-time walks. The process must look like mean-zero, unit-variance RW everywhere except near 0, where the limiting drift should be $\rho^{-1}$. Note that when considering the limiting drift near zero, we are taking a joint limit in x and h. The order of this matters. As explained at the end of the previous article, we only need to worry about the limiting drift along sequences of $x,h$ such that $a_h(x)\rightarrow 0$. If no such sequences exist, or the limiting drift along any of these is infinite, then we actually have a reflected boundary condition.

This highlights one confusing matter about convergence of reflected processes. The boundary of the discrete-time process should converge to the boundary of the reflected process, but we also have to consider where reflective behaviour happens. Can we get sticky BM with reflection only at the boundary in the discrete-time processes? The answer turns out to be no. At the start of this article, I proposed a model of SRW with geometric waiting times whenever the origin was visiting. What is the limit of this?

The trick is to consider how long the discrete process spends near 0, after rescaling. It will spend a multiple 1/p more time at 0 itself, where p is the parameter of the geometric distribution, but no more time than expected at any point $x\in(0,\epsilon)$. But time spent in $(0,\epsilon)$ dominates time spent at 0 before this adjustment, so must also dominate it after the adjustment, so in the limit, the proportion of time spent at 0 is unchanged, and so in particular it cannot be positive.

Because of all of this, in practice it seems that most random walks we might be interested in converge (if they converge to a process at all) to a reflected SDE/diffusion etc, rather than one with sticky boundary conditions. I feel I’ve been talking a lot about Markov processes converging, so perhaps next, or at least soon, I’ll write some more technical things about exactly what conditions and methods are required to prove this.

REFERENCES

S. Varadhan – Chapter 16 from a Lecture Course at NYU can be found here.

# Reflected Brownian Motion

A standard Brownian motion is space-homogeneous, meaning that the behaviour of $B_{T+t}-B_T$ does not depend on the value of $B_T$. By Donsker’s Theorem, such a Brownian motion is also the limit in a process space of any homogeneous random walk with zero-drift and constant variance, after suitable rescaling.

In many applications, however, we are interested in real-valued continuous-time Markov processes that are defined not on the whole of the real line, but on the half-line $\mathbb{R}_{\ge 0}$. So as BM is the fundamental real-valued continuous-time Markov process, we should ask how we might adjust it so that it stays non-negative. In particular, we want to clarify uniqueness, or at least be sure we have found all the sensible ways to make this adjustment, and also to consider how Donsker’s Theorem might work in this setting.

We should consider what properties we want this non-negative BM to have. Obviously, it should be non-negative, but it is also reasonable to demand that it looks exactly like BM everywhere except near 0. But since BM has a scale-invariance property, it is essentially meaningful to say ‘near 0’, so we instead demand that it looks exactly like BM everywhere except at 0. Apart from this, the only properties we want are that it is Markov and has continuous sample paths.

A starting point is so-called reflected Brownian motion, defined by $X_t:=|B_t|$. This is very natural and very convenient for analysis, but there are some problems. Firstly, this has the property that it looks like Brownian motion everywhere except 0 only because BM is space-homogeneous but also symmetric, in the sense that $B_t\stackrel{d}{=}-B_t$. This will be untrue for essentially any other process, so as a general method for how to keep stochastic processes positive, this will be useless. My second objection is a bit more subtle. If we consider this as an SDE, we get

$dX_t=\text{sign}(B_t)dB_t.$

This is a perfectly reasonable SDE but it is undesirable, because we have a function of B as coefficient on the RHS. Ideally, increments of X would be a function of X, and the increments of B, rather than the values of B. That is, we would expect $X_{t+\delta t}-X_t$ to depend on $X_t$ and on $(B_{t+s}-B_t, 0\le s\le \delta t)$, but not on $B_t$ itself, as that means we have to keep track of extra information while constructing X.

So we need an alternative method. One idea might be to add some non-negative process to the BM so that the sum stays non-negative. If this process is deterministic and finite, there there is some positive probability that the sum will eventually be negative, so this won’t do. We are looking therefore so a process which depends on the BM. Obviously we could take $\max(-B_t,0)$, but this sum would then spend macroscopic intervals of time at 0, and these intervals would have the Raleigh distribution (for Brownian excursions) rather than the exponential distribution, hence the process given by the sum would not be memoryless and Markov.

The natural alternative is to look for an increasing process $A_t$, and then it makes sense to talk about the minimal increasing process that has the desired property. A moment’s thought suggests that $A_t=-min_{s\le t}B_t$ satisfies this. So we have the decomposition

$B_t=-A_t+S_t,$

where $S_t$ is the height of B above its running minimum. So S is an ideal alternative definition of reflecting BM. In particular, when B is away from its minimum, $dB_t=dS_t$, so this has the property that it evolves exactly as the driving Brownian motion.

What we have done is to decompose a general continuous process into the sum of a decreasing continuous process and a non-negative process. This is known as the Skorohod problem, and was the subject of much interest, even in the deterministic case. Note that process A has the property that it is locally constant almost everywhere, and is continuous, yet non-constant. Unsurprisingly, since A only changes when the underlying BM is 0, A is continuous with respect to the local time process at 0. In fact, A is the local time process of the underlying Brownian motion, by comparison with the construction by direct reflection.

One alternative approach is to look instead at the generator of the process. Recall that the generator of a process is an operator on some space of functions, with $\mathcal{L}f$ giving the infinitissimal drift of $f(X_t)$. In the case of Brownian motion, the generator $(\mathcal{L}f)(x)=\frac12 f''(x)$ for bounded smooth functions f. This is equivalent to saying that

$f(X_t)-f(X_0)-\int_0^t \frac12 f''(X_s)ds$ (*)

is a martingale. This must hold also for reflected Brownian motion, whenever x is greater than 0. Alternatively, if the function f is zero in a small neighbourhood of 0, it should have the same generator with respect to reflected BM. Indeed, for a general smooth bounded function f, we can still consider the expression (*) with respect to reflected BM. We know this expression behaves as a martingale except when X is zero. If f'(0)>0, and T is some hitting time of 0, then $f(X_{T+\delta T})-f(X_T)\ge 0$, hence the expression (*) is a submartingale. So if we restrict attention to functions with f'(0)=0, the generator remains the same. Indeed, by patching together all such intervals, it can be argued that even if f'(0) is not zero,

$f(X_t)-f(X_0)-\int_0^t \frac12 f''(X_s)ds - f'(0)A_t$

is a martingale, where A is the local time process at zero.

I was aware when I started reading about this that there was another family of processes called ‘Sticky Brownian Motion’ that shared properties with Reflected BM, in that it behaves like standard BM away from zero, but is also constrained to the non-negative reals. I think this will get too long if I also talk about that here, so that can be postponed, and for now we consider reflected BM as a limit of reflected (or other) random walks, bearing in mind that there is at least one other candidate to be the limit.

Unsurprisingly, if we have a family of random walks constrained to the non-negative reals, that are zero-drift unit-variance away from 0, then if they converge as processes, the limit is Brownian away from zero, and non-negative. Note that “away from 0” means after rescaling. So the key aspect is behaviour near zero.

What is the drift of reflected BM at 0? We might suspect it is infinite because of the form of the generator, but we can calculate it directly. Given $X_0=0$, we have:

$\frac{\mathbb{E}X_t}{t}=\frac{\mathbb{E}|B_t|}{t}=\frac{\sqrt{t}\mathbb{E}|B_1|}{t},$

so letting $t\rightarrow 0$, we see indeed that the drift is infinite at 0.

For convergence of discrete processes, we really need the generators to converge. Typically we index the discrete-time processes by the time unit h, which tends to 0, and $b_h(x),a_h(x)$ are the rescaled drift and square-drift from x. We assume that we don’t see macroscopic jumps in the limit. For the case of simple random walk reflected at 0, it doesn’t matter exactly how we construct the joint limit in h and x, as the drift is uniform on x>0, but in general this does matter. I don’t want to discuss sticky BM right now, so it’s probably easiest to be vague and say that the discrete Markov processes converge to reflected BM so long they don’t spend more time than expected near 0 in the limit, as the title ‘sticky’ might suggest.

The two ways in which this can happen is if the volatility term $a_h(x)$ is too small, in which case the process looks almost deterministic near 0, or if the drift doesn’t increase fast enough. And indeed, this leads to two conditions. The first is straightforward, if $a_h(x)$ is bounded below, in the sense that $\liminf_{h,x\rightarrow 0} a_h(x)\ge C>0$, then we have convergence to reflected BM. Alternatively, the only danger can arise down those subsequences where $a_h(x)\rightarrow 0$, so if we have that $b_h(x)\rightarrow +\infty$ whenever $h,x,a_h(x)\rightarrow 0$, then this convergence also holds.

Next time I’ll discuss what sticky BM means, what it doesn’t mean, why it isn’t easy to double the local time, and how to obtain sticky BM as a limit of discrete random walks in a similar way to the above.

REFERENCES

S. Varadhan – Chapter 16 from a Lecture Course at NYU can be found here.

# Supremum of Brownian Motion

We define the supremum process of Brownian Motion by:

$S_t:=\sup_{0\leq s\leq t}B_s.$

Here are two facts about Brownian Motion. Firstly, the Reflection Principle:

$\mathbb{P}(S_t\geq b,B_t\leq a)=\mathbb{P}(B_t\geq 2b-a),$

which we motivate by ‘stopping’ at time $S_t$, and using the SMP for Brownian Motion, even though it isn’t a stopping time. By setting a=b, we get:

$\mathbb{P}(S_t\geq b)=\mathbb{P}(S_t\geq b,B_t\leq b)+\mathbb{P}(B_t\geq b)=2\mathbb{P}(B_t\geq b)=\mathbb{P}(|B|\geq b),$

and conclude that

$S_t\stackrel{d}{=}|B_t|\quad\text{for each }t\geq 0.$

The second fact comes from the decomposition of BM into local times and excursions:

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

where L is the local time process at 0, and this equality in distribution holds for the processes. See the previous post on excursion theory for explanation of what local times mean.

In particular, combining these two facts gives:

$S_t\stackrel{d}{=}S_t-B_t\quad\text{for every }t\geq 0.$

I thought that was rather surprising, and wanted to think of a straightforward reason why this should be true. I think the following works:

Brownian motion is time-reversible. In particular, as processes, we have

$(B_s)_{s\geq 0}\stackrel{d}{=}(B_{t-s}-B_t)_{s\geq 0}$

$\Rightarrow \sup_{0\leq r\leq t}B_r\stackrel{d}{=}\sup_{0\leq r\leq t}(B_{t-r}-B_t)$

$\Rightarrow S_t\stackrel{d}{=}S_t-B_t.$

# Subordinators and the Arcsine rule

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

We say $X_t$ is a subordinator if:

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

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

Examples

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

Properties

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

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

• We can refine the Levy-Khintchine formula;

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

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

Analytic Methods

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

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

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

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

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

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

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

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

The Arcsine Law

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

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

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

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

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

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

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

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

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

# Brownian Excursions and Local Time

I’ve been spending a fair bit of time this week reading and thinking about the limits of various combinatorial objects, in particular letting the number of vertices tend to $\infty$ in models of random graphs with various constraints. Perhaps predictably, like so many continuous stochastic objects, yet again the limiting ‘things’ turn out to be closely linked to Brownian Motion. As a result, I’ve ended up reading a bit about the notion of local time, and thought it was sufficiently elegant even by itself to justify a quick post.

Local Time

In general, we might be interested in calculating a stochastic integral like

$\int_0^t f(B_s)ds$.

Note that, except in some highly non-interesting cases, this is a random variable. Our high school understanding of Riemannian integration encourages thinking of this as a ‘pathwise’ integral along the path evolving in time. But of course, that’s orthogonal to the approach we start thinking about when we are introduced to the Lebesgue integral. There we think about potential values of the integrand, and weight their contribution by the (Lebesgue) measure of the subset of the domain in which they appear.

Can we do the same for the stochastic integral? That is, can we find a measure which records how long the Brownian Motion spends at a point x? This measure will not be deterministic – effectively the stochastic behaviour of BM will be encoded through the measure rather than the argument of the function.

The answer is yes, and the measure in question is referred to as local time. More formally, we want

$\int_0^t f(B_s)ds=\int_\mathbb{R}f(x)L(t,x)dx.$ (*)

where the local time L(t,x) is a random process, increasing for fixed x. Informally, one could take

$\partial_t L(t,x) \propto 1(B_t=x)$

but clearly in practice that won’t do at all for a definition, and so instead we use (*). In the usual way, if we want (*) to hold for all reasonably nice functions f, it suffices to check it for the indicator functions of Borel sets. L(t,.) is therefore often referred to as occupation density, while L(.,A) is local time.

Local Time as natural index for Excursions

An excursion, for example of Brownian Motion, is a segment of the path that has zero value only at its endpoints. Alternatively, it is a maximal open interval of time such that the path is away from 0. We want to specify the measure on these excursions. Here are some obvious difficulties.

By Blumenthal’s 0-1 law, BM started from zero hits zero infinitely often in any time interval [0,e], so in the same way that there is no first positive rational, there is no first excursion. We could pick the excursion occurring in progress at a fixed time t, but this is little better. Firstly, the resultant measure is size-biased by the length of the excursion, and more importantly, the proximity of t to the origin may be significant unless we know of some memorylessness type of property to excursions.

Local time allows us to solve these problems. We restrict attention to $L_t:=L(t,0)$, the occupation density of 0. Let’s think about some advantages of indexing excursions by local time rather than by the start time:

• The key observation is that local time remains constant on excursions. That is, if we are avoiding 0, the local time at 0 cannot grow because the BM spends no time there!
• If we use start time, then we have a countably infinite number of small excursions accumulating close to 0, ie with very small start time. However, local time increases rapidly when there are lots of small excursions. Remember, lots of small excursions means that the BM hits 0 lots of times. So local time grows quickly through the annoying bits, and effectively provides a size-biasing for excursions that allows us to ignore the effects of the ‘Blumenthal excursions’ near time 0.
• When indexed by time, excursions might be Markovian, in the sense that subsequent excursions (and in particular their lengths) are independent of past excursions.This is certainly not the case if you index by start time! If an excursion starts at time t and has length u, then the ‘next’ excursions, in as much as that makes sense, must surely start at time t+u.

We know there are only countably many excursions, hence there are only countably many local times which pertain to an excursion. This motivates considering the set of excursions as a Poisson Point Process on local time. Once you’ve had this idea, everything follows quite nicely. Working out the distribution of the constant rate (which is a measure on the set of excursions) remains, but essentially we now have a sensible framework for tracking the process of excursions, and from this we can reconstruct the original Brownian Motion.

# Dubins-Schwarz Theorem

In developing the stochastic integral, much of our motivation has come from considering integrals with respect to Brownian Motion. In this section, we develop some results which justify that Brownian Motion is the canonical stochastic process with non-zero quadratic variation (which is related, but not directly equivalent to the property of infinite total variation). In particular, we shall observe the Dubins-Schwarz theorem, which shows that martingales with unbounded (as time $\rightarrow\infty$) quadratic variation ARE Brownian Motion, up to a (stochastic) time change.

Recall Levy’s characterisation of a d-dimensional BM, which allows us to avoid considering independent normal increments. Given $X^1,\ldots,X^d\in\mathcal{M}_{c,loc}$:

$X=(X^1,\ldots,X^d)$ a BM iff $[X^i,X^j]_t=\delta_{ij}t$

Obviously, one direction has been shown as part of the construction and properties of quadratic variation. For the other direction,, because laws are precisely defined by characteristic functions, it suffices to show that

$\mathbb{E}\left[\exp(i\langle \theta,X_t-X_s\rangle)|\mathcal{F}_s\right]=\exp(-\frac12||\theta||^2(t-s))$

We set $Y_t:=\langle \theta,X_t\rangle$, and deduce $[Y]=t||\theta||^2$ and $Z=\mathcal{E}(iY)=\exp(iY_t+\frac12[Y]_t)\in\mathcal{M}_{c,loc}$, and furthermore is bounded on compact [0,t], hence is a true martingale. So $\mathbb{E}\left(\frac{Z_t}{Z_s}|\mathcal{F}_s\right)=1$ which is pretty much what was required.

Now, Dubins-Schwarz states

Theorem: Given $M\in\mathcal{M}_{c,loc}, M_0=0, [M]_\infty=\infty$ almost surely, if we set $\tau_s:=\inf\{t:[M]_t>s\}$, then $B_s:=M_{\tau_s}$ is a $(\mathcal{F}_{\tau_s})$-BM, with $M_t=B_{[M]_t}$.

This final result is clear if $[M]_t$ is almost surely strictly increasing in t: just take $s=[M]_t$ in the definition.

We know B is cadlag: we first show B as defined is almost surely continuous. It remains to show $B_{s-}=B_s\,\forall s>0\iff M_{\tau_{s-}}=M_{\tau_s}$, noting that $\tau_{s-}=\inf\{t\geq 0:[M]_t=s\}$ (by continuity) is a stopping time also.

The only interesting case is if $\tau_{s-}<\tau_s$, for which need to show [M] is constant. This is intuitively obvious, but formally, we must appeal to $(M^2-[M])^{\tau_s}$ which is UI, since $\mathbb{E}[M^{\tau_s}]_\infty<\infty$. Now may apply OST to obtain $\mathbb{E}[M_{\tau_s}^2-M_{\tau_{s-}}^2|\mathcal{F}_{\tau_{s-}}]=\mathbb{E}[(M_{\tau_s}-M_{\tau_{s-}})^2|\mathcal{F}_{\tau_{s-}}]=0$ which implies M is almost surely constant on $[\tau_{s-},\tau_s]$. We need to lift this to the case where it holds for all s simultaneously almost surely. Note that cadlag almost surely plus almost surely continuous at each point does not implies almost surely continuous everywhere (eg consider H(U(0,1)) with H the Heaviside function and U a uniform distribution). Instead, we record intervals of constancy of both $M_t,[M]_t$. That is, we set

$T_r=\inf\{t>r:M_t\neq M_r\},\quad S_r=\inf\{t>r:[M]_t\neq [M]_r\}$

Then these are cadlag, and by above $T_r=S_r\,\forall r\in\mathbb{Q}^+$ a.s. therefore $T_r=S_r\,\forall r$ almost surely. Thus M, [M] are constant on the same intervals.

We also check B is adapted to $\mathcal{G}_t=\mathcal{F}_{\tau_t}$. STP $X_T1_{\{T<\infty\}}$ is $\mathcal{F}_T$-measurable for X cadlag adapted. Approximating T discretely from above gives the result, exploiting that the result is clear if T has countable support. Now, obtain $M^{\tau_s}\in\mathcal{M}_c^2$, so $M_{t\wedge \tau_s}$ UI by Doob, so by OST, get $\mathbb{E}[M_{\tau_s}|\mathcal{F}_{\tau_s}]=M_{\tau_r}$, to get B a martingale. The finally:

$\mathbb{E}[B_s^2-s|\mathcal{G}_r]=\mathbb{E}[(M^2-[M])_{\tau_s}|\mathcal{F}_{\tau_s}]=M_{\tau_r}^2-[M]_{\tau_r}=B_r^2-r$

And so we can apply Levy’s characterisation to finish the result.

# Brownian Motion is not finite variation

There is a natural definition of ‘pathwise’ stochastic integrals of a certain type of ‘simple’ process with respect to cadlag non-decreasing processes. It can be a shown that a function is of finite variation iff it can be expressed as the difference of two such functions. Hence, these finite variation processes can be used as variable of integration via an obvious linear extension. One direction of this result is obvious; the other is fiddly. To proceed, we show that the total valuation process is cadlag (and, obviously, increasing), and then check that $a'=\frac12(v+a),a''=\frac12(v-a)$ are processes satisfying the conditions of the result.

Our overall aim is to define integrals with respect to Brownian Motion since that is (in a sense to be made precise through the Dubins-Schwarz theorem later) the canonical non-trivial stochastic process with non-zero quadratic variation. The result we demonstrate shows that it is not possible to define the integral with respect to BM through pathwise finite variation integrals.

Theorem: $M\in\mathcal{M}_{c,loc},M_0=0$ a.s. is of finite variation. Then M is indistinguishable from 0.

We will show this for M a bounded martingale with bounded variation. Why does this suffice? In general, set $S_n:=\inf\{t,V_t\leq n\}$, noting that V is continuous adapted non-decreasing. If $M^{S_n}\equiv 0\,\forall n$, then we are done, as the $S_n$s are increasing. But this is a bounded martingale with bounded variation.

To prove this, we make use of the orthogonality relation which is a key trick for this sort of result: If M is a martingale, with $M_s,M_t\in L^2$, for s<t, then just by multiplying out:

$\mathbb{E}[(M_t-M_s)^2|\mathcal{F}_s]=\mathbb{E}[M_t^2-M_s^2|\mathcal{F}_s]$ a.s.

Now, for this particular result, we decompose

$\mathbb{E}[M_t^2]=\mathbb{E}\left[\sum_{k=0}^{2^n-1}(M_{(k+1)2^{-n}t}^2-M_{k2^{-n}t}^2)\right]=\mathbb{E}[\sum (M_{(k+1)2^{-n}t}-M_{k2^{-n}t})^2]$

and then we bound this last term as

$\leq \mathbb{E}\left[\sup_k [M_+-M_-]\sum_k |M_+-M_-|\right]$

Now, as $n\uparrow\infty$, we have $\sum_k |M_+-M_-|\uparrow V_t\leq N$ by the boundedness assumption. Furthermore, M is almost surely continuous on [0,t] and so it is in fact uniformly continuous, which allows us to conclude that

$\sup_k |M_+-M_-|\downarrow 0$

By bounded convergence, this limit applies equally under the expectation, and so conclude that $\mathbb{E}M_t^2=0$ for each time t, and so for each time t the martingale is almost surely equal to 0. In the usual, can lift this to rational points by countability, then to all points by continuity.

# Strong Markov Property for BM

The Strong Markov Property is the most important result to demonstrate for any Markov process, such as Brownian Motion. It is also probably the most widely requested item of bookwork on the Part III Advanced Probability exam. I feel it is therefore worth practising writing as quickly as possible.

Theorem (SMP): Take $(B_t)$ a standard $(\mathcal{F}_t)$-BM, and T an a.s. finite stopping time. Then $(B_{T+t}-B_T,t\geq 0)$ is a standard BM independent of $\mathcal{F}_T$.

Proof: We write $B_t^{(T)}=B_{T+t}-B_T$ for ease of notation. We will show that for any $A\in\mathcal{F}_T$ and F bounded, measurable:

$\mathbb{E}[1_AF(B_{T+t_1}-B_T,\ldots,B_{T+t_n}-B_T)]=\mathbb{P}(A)\mathbb{E}F(B_{t_1},\ldots,B_{t_n})$

This will suffice to establish independence, and taking $A=\Omega\in\mathcal{F}_t$ shows that $B_t^T$ is a standard BM since (Levy), BM is uniquely characterised by its finite joint distributions.

To prove the result, we approximate discretely, and apply the Markov property.

$\mathbb{E}[1_AF(B_{t_1}^{(T)},\ldots)]=\lim_{m\rightarrow\infty}\sum_{k=1}^\infty \mathbb{E}[1_{A\cap\{T\in((k-1)2^{-m},k2^{-m}]\}}F(B_{t_1}^{(k2^{-m})},\ldots)]$

by bounded convergence, using continuity of F, right-continuity of B, and that $T<\infty$ a.s. (so that $1_A=\sum 1_{A\cap \{T\in(-,-]\}}$)

$\stackrel{\text{WMP}}{=}\lim_{m\rightarrow\infty}\sum_{k=1}^\infty \mathbb{P}[A\cap\{T\in((k-1)2^{-m},k2^{-m}]\}]\mathbb{E}F(B_{t_1},\ldots,B_{t_n})$

$\stackrel{\text{DOM}}{=}\mathbb{P}(A)\mathbb{E}F(B_{t_1},\ldots,B_{t_n})$

which is exactly what we required.

Remarks: 1) We only used right-continuity of the process, and characterisation by joint marginals, so the proof works equally well for Levy processes.

2) We can in fact show that it is independent of $\mathcal{F}_T^+$, by considering $T+\frac{1}{n}$ which is still a stopping time, then taking a limit in this as well in the above proof. For details of a similar result, see my post on Blumenthal’s 0-1 Law.