# 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.

# DGFF 3 – Gibbs-Markov property for entropic repulsion

In the previous post, we saw that it isn’t much extra effort to define the DGFF with non-zero boundary conditions, by adding onto the zero-BC DGFF the unique (deterministic) harmonic function which extends the boundary values into the domain. We also saw how a Gibbs-Markov property applies, whereby the values taken by the field on some sub-region $A\subset D$ depend on the values taken on $D\backslash A$ only through values taken on $\partial A$.

In this post, we look at how this property and some other methods are applied by Deuschel [1] to study the probability that the DGFF on a large box in $\mathbb{Z}^d$ is positive ‘everywhere’. This event can be interpreted in a couple of ways, all of which are referred to there as entropic repulsion. Everything which follows is either taken directly or paraphrased directly from [1]. I have tried to phrase this in a way which avoids repeating most of the calculations, instead focusing on the methods and the motivation for using them.

Fix dimension $d\ge 2$ throughout. We let $P^0_N$ be the law of the DGFF on $V_N:=[-N,N]^d\subset \mathbb{Z}^d$ with zero boundary conditions. Then for any subset $A\subset \mathbb{Z}^d$, in an intuitively-clear abuse of notation, we let

$\Omega^+(A):= \{ h_x\ge 0, x\in A\},$

be the event that some random field h takes only non-negative values on A. The goal is to determine $P^0_N ( \Omega^+(V_N))$. But for the purposes of this post, we will focus on showing bounds on the probability that the field is non-negative on a thin annulus near the boundary of $V_N$, since this is a self-contained step in the argument which contains a blog-friendly number of ideas.

We set $(L_N)$ to be a sequence of integers greater than one (to avoid dividing by zero in the statement), for which $\frac{L_N}{N}\rightarrow 0$. We now define for each N, the annulus

$W_N = \{v\in V_N: L_N\le d_{\mathbb{Z}^d}(v, V_N^c)\le 2L_N \}$

with radius $L_N$ set a distance $L_N$ inside the box $V_N$. We aim to control $P^N_0 (\Omega^+(W_N))$. This forms middle steps of Deuschel’s Propositions 2.5 and 2.9, which discuss $P^N_0(\Omega^+(V_{N-L_N}))$. Clearly there is the upper bound

$P^N_0(\Omega^+(V_{N-L_N})) \le P^N_0(\Omega^+(W_N))$ (1)

and a lower bound on $P^N_0(\Omega^+(V_{N-L_N}))$ is obtained in the second proposition by considering the box as a union of annuli then combining the bounds on each annulus using the FKG inequality.

Upper bound via odds and evens

After removing step (1), this is Proposition 2.5:

$\limsup_{N\rightarrow \infty} \frac{L_N}{N^{d-1} \log L_N} \log P^N_0(\Omega^+(W_N)) < 0.$ (2)

This is giving a limiting upper bound on the probability of the form $L_N^{-CN^{d-1}/L_N}$, though as with all LDP estimates, the form given at (2) is more instructive.

Morally, the reason why it is unlikely that the field should be non-negative everywhere within the annulus is that the distribution at each location is centred, and even though any pair of values are positively correlated, this correlation is not strong enough to avoid this event being unlikely. But this is hard to corral into an upper bound argument directly. In many circumstances, we want to prove upper bounds for complicated multivariate systems by projecting to get an unlikely event for a one-dimensional random variable, or a family of independent variables, even if we have to throw away some probability. We have plenty of tools for tail probabilities in both of these settings. Since the DGFF is normal, a one-dimensional RV that is a linear combination (eg the sum) of all the field heights is a natural candidate. But in this case we would have thrown away too much probability, since the only way we could dominate is to demand that the sum $\sum_{x\in W_N}h^N_x\ge 0$, which obviously has probability 1/2 by symmetry. (3)

So Deuschel splits $W_N$ into $W_N^o,W_N^e$, where the former includes all vertices with odd total parity in $W_N$ and the latter includes all the vertices with even total parity in the interior of $W_N$. (Recall that $\mathbb{Z}^d$ is bipartite in exactly this fashion). The idea is to condition on $h^N\big|_{W^o_N}$. But obviously each even vertex is exactly surrounded by odd vertices. So by the Gibbs-Markov property, conditional on the odd vertices, the values of the field at the even vertices are independent. Indeed, if for each $v\in W_N^e$ we define $\bar h_v$ to be the average of its neighbours (which is measurable w.r.t to the sigma-algebra generated by the odd vertices), then

$\{h_v: v\in W_N^e \,\big|\, \sigma(h_w: w\in W_N^o)\},$

is a collection of independent normals with variance one, and where the mean of $h_v$ is $\bar h_v$.

To start finding bounds, we fix some threshold $m=m_N\gg 1$ to be determined later, and consider the odd-measurable event $A_N$ that at most half of the even vertices v have $\bar h_v\ge m$. So $A_N^c\cap \Omega^+(W_N)$ says that all the odd vertices are non-negative and many are quite large. This certainly feels like a low-probability event, and unlike at (3), we might be able to obtain good tail bounds by projection into one dimension.

In the other case, conditional on $A_N$, there are a large number of even vertices with conditional mean at most m, and so we can control the probability that at least one is negative as a product

$(1-\varphi(m))^{\frac12 |W_N^e|}$. (4)

Note that for this upper bound, we can completely ignore the other even vertices (those with conditional mean greater than m).

So we’ll go back to $A_N^c \cap \Omega^+(W_N)$. For computations, the easiest one-dimensional variable to work with is probably the mean of the $\bar h_v$s across $v\in W_N^e$, since on $A_N^c\cap \Omega^+(W_N)$ this is at least $\frac{m}{2}$. Rather than focus on the calculations themselves involving

$\bar S^e_N:= \frac{1}{|W_N^e|} \sum\limits_{v\in W_N^e} \bar h_v,$

let us remark that it is certainly normal and centered, and so there are many methods to bound its tail, for example

$P^0_N \left( \bar S^e_N \ge \frac{m}{2} \right) \le \exp\left( \frac{-m^2}{8\mathrm{Var}(\bar S^e_N)} \right),$ (5)

as used by Deuschel just follows from an easy comparison argument within the integral of the pdf. We can tackle the variance using the Green’s function for the random walk (recall the first post in this set). But before that, it’s worth making an observation which is general and useful, namely that $\bar S^e_N$ is the expectation of

$S^e_N:= \sum{1}{|W_N^e|}\sum\limits_{v\in W_N^e} h_v$

conditional on the odds. Directly from the law of total variance, the variance of any random variable X is always larger than the variance of $\mathbb{E}[X|Y]$.

So in this case, we can replace $\mathrm{Var}(\bar S^e_N)$ in (5) with $\mathrm{Var}(S^e_N)$, which can be controlled via the Green’s function calculation.

Finally, we choose $m_N$ so that the probability at (4) matches the probability at (5) in scale, and this choice leads directly to (2).

In summary, we decomposed the event that everything is non-negative into two parts: either there are lots of unlikely local events in the field between an even vertex and its odd neighbours, or the field has to be atypically large at the odd sites. Tuning the parameter $m_N$ allows us to control both of these probabilities in the sense required.

Lower bound via a sparse sub-lattice

To get a lower bound on the probability that the field is non-negative on the annulus, we need to exploit the positive correlations in the field. We use a similar idea to the upper bound. If we know the field is positive and fairly large in many places, then it is increasingly likely that it is positive everywhere. The question is how many places to choose?

We are going to consider a sub-lattice that lives in a slightly larger region than $W_N$ itself, and condition the field to be larger than $m=m_N$ everywhere on this lattice. We want the lattice to be sparse enough that even if we ignore positive correlations, the chance of this happening is not too small. But we also want the lattice to be dense enough that, conditional on this event, the chance that the field is actually non-negative everywhere in $W_N$ is not too small either.

To achieve this, Deuschel chooses a sub-lattice of width $\lfloor\epsilon L_N^{2/d}\rfloor$, and sets $\Lambda_N(\epsilon)$ to be the intersection of this with the annulus with radii $[N-\frac{5}{2}L_N, N-\frac{1}{2}L_N]$, to ensure it lives in a slightly larger region than $W_N$ itself. The scaling of this sub-lattice density is such that when a random walk is started at any $v\in W_N$, the probability that the RW hits $\Lambda_N(\epsilon)$ before $\partial V_N$ is asymptotically in (0,1). (Ie, not asymptotically zero or one – this requires some definitely non-trivial calculations.) In particular, for appropriate (ie large enough) choice of $\epsilon$, this probability is at least 1/2 for all $v\in W_N$. This means that after conditioning on event $B_N:=\{h_v\ge m : v\in \Lambda_N(\epsilon)\}$, the conditional expectation of $h_w$ is at least $\frac{m}{2}$ for all $w\in W_N\backslash \Lambda_N(\epsilon)$. Again this uses the Gibbs-Markov property and the Gaussian nature of the field. In particular, this conditioning means we are left with the DGFF on $V_N\backslash \Lambda_N(\epsilon)$, ie with boundary $\partial V_N\cup \Lambda_N(\epsilon)$, and then by linearity, the mean at non-boundary points is given by the harmonic extension, which is linear (and so increasing) in the boundary values.

At this point, the route through the calculations is fairly clear. Since we are aiming for a lower bound on the probability of the event $\Omega^+(W_N)$, it’s enough to find a lower bound on $P^0_N(\Omega^+(W_N)\cap B)$.

Now, by positive correlation (or, formally, the FKG inequality) we can control $P^0_N(B)$ just as a product of the probabilities that the field exceeds the threshold at each individual site in $\Lambda_N(\epsilon)$. Since the value of the field at each site is normal with variance at least 1 (by definition), this is straightforward.

Finally, we treat $P^0_N(\Omega^+(W_N) \,\big|\, B)$. We’ve established that, conditional on B, the mean at each point of $W_N\backslash \Lambda_N(\epsilon)$ is at least $\frac{m}{2}$, and we can bound the variance above too. Again, this is a conditional variance, and so is at most the corresponding original variance, which is bounded above by $\sigma_N^2:=\mathrm{Var}(h^N_0)$. (This fact that the variance is maximised at the centre is intuitively clear when phrased in terms of occupation times, but the proof is non-obvious, or at least non-obvious to me.)

Since each of the event $h_v^N\ge 0$ for $v\in W_N\backslash \Lambda_N(\epsilon)$ is positively correlated with B, we can bound the probability it holds for all v by the product of the probabilities that it holds for each v. But having established that the conditional mean is at least $\frac{m_N}{2}$ for each v, and the variance is uniformly bounded above (including in N), this gives an easy tail bound of the form we require.

Again it just remains to choose the sequence of thresholds $m_N$ to maximise the lower bound on the probability that we’ve found in this way. In both cases, it turns out that taking $m_N= \sqrt{C\log N}$ is sensible, and this turns out to be linked to the scaling of the maximum of the DGFF, which we will explore in the future.

References

[1] – J-D Deuschel, Entropic Repulsion of the Lattice Free Field, II. The 0-Boundary Case. Available at ProjectEuclid.

# DGFF 2 – Boundary conditions and Gibbs-Markov property

In the previous post, we defined the Discrete Gaussian Free Field, and offered some motivation via the discrete random walk bridge. In particular, when the increments of the random walk are chosen to be Gaussian, many natural calculations are straightforward, since Gaussian processes are well-behaved under conditioning and under linear transformations.

Non-zero boundary conditions

In the definition of the DGFF given last time, we demanded that $h\equiv 0$ on $\partial D$. But the model is perfectly well-defined under more general boundary conditions.

It’s helpful to recall again the situation with random walk and Brownian bridge. If we want a Brownian motion which passes through (0,0) and (1,s), we could repeat one construction for Brownian bridge, by taking a standard Brownian motion and conditioning (modulo probability zero technicalities) on passing through level s at time 1. But alternatively, we could set

$B^{\mathrm{drift-br}}(t) = B(t)+ t(s-B(1)),\quad t\in[0,1],$

or equivalently

$B^{\mathrm{drift-br}}(t)=B^{\mathrm{br}}(t)+ st, \quad t\in[0,1].$

That is, a Brownian bridge with drift can be obtain from a centered Brownian bridge by a linear transformation, and so certainly remains a Gaussian process. And exactly the same holds for a discrete Gaussian bridge: if we want non-zero values at the endpoints, we can obtain this distribution by taking the standard centred bridge and applying a linear transformation.

We can see how this works directly at the level of density functions. If we take $0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0$ a centred Gaussian bridge, then the density of $Z=\mathbf{z}\in \mathbb{R}^{N+1}$ is proportional to

$\mathbf{1}\{z_0=z_N=0\}\exp\left( -\frac12 \sum_{i=1}^N (z_i-z_{i-1})^2 \right).$ (3)

So rewriting $z_i= y_i- ki$ (where we might want $k=s/N$ to fit the previous example), the sum within the exponent rearranges as

$-\frac12 \sum_{i=1}^N (y_i-y_{i-1} - k)^2 = -\frac12 \sum_{i=1}^N (y_i-y_{i-1})^2 - 2k(y_N-y_0)+ Nk^2.$

So when the values at the endpoints $z_0,z_n,y_0,y_N$ are fixed, this middle term is a constant, as is the final term, and thus the density of the linearly transformed bridge has exactly the same form as the original one.

In two or more dimensions, the analogue of adding a linear function is to add a harmonic function. First, some notation. Let $\varphi$ be any function on $\partial D$. Then there is a unique harmonic extension of $\varphi$, for which $\nabla \varphi=0$ everywhere on D, the interior of the domain. Recall that $\nabla$ is the discrete graph Laplacian defined up to a constant by

$(\nabla \varphi) _x = \sum\limits_{x\sim y} \varphi_x - \varphi_y.$

If we want $h^D$ instead to have boundary values $\varphi$, it’s enough to replace $h^D$ with $h^D+\varphi$. Then, in the density for the DGFF ( (1) in the previous post), the term in the exponential becomes (ignoring the $\frac{1}{4d}$ )

$-\sum\limits_{x\sim y} \left[ (h^D_x-h^D_y)^2 + (\varphi_x-\varphi_y)^2 +2(h^D_x - h^D_y)(\varphi_x-\varphi_y)\right].$

For each $x\in D$, on taking this sum over its neighbours $y\in \bar D$, the final term vanishes (since $\varphi$ is harmonic), while the second term is just a constant. So the density of the transformed field, which we’ll call $h^{D,\varphi}$ is proportional to (after removing the constant arising from the second term above)

$\mathbf{1}\left\{h^{D,\varphi}_x = \varphi_x,\, x\in\partial D\right\} \exp\left( -\frac{1}{4d} \sum\limits_{x\sim y} \left( h^{D,\varphi}_x - h^{D,\varphi}_y \right)^2 \right).$

So $h^{D,\varphi}:= h^D + \varphi$ satisfies the conditions for the DGFF on D with non-zero boundary conditions $\varphi$.

Harmonic functions and RW – a quick review

Like the covariances in DGFF, harmonic functions on D are related to simple random walk on D stopped on $\partial D$. (I’m not claiming a direct connection right now.) We can define the harmonic extension $\varphi$ to an interior point x by taking $\mathbb{P}_x$ to be the law of SRW $x=Z_0,Z_1,Z_2,\ldots$ started from x, and then setting

$\varphi(x):= \mathbb{E}\left[ \varphi_{\tau_{\partial d}} \right],$

where $\tau_{\partial D}$ is the first time that the random walk hits the boundary.

Inverse temperature – a quick remark

In the original definition of the density of the DGFF, there is the option to add a constant $\beta>0$ within the exponential term so the density is proportional to

$\exp\left(-\beta \sum\limits_{x\sim y} (h_x-h_y)^2 \right).$

With zero boundary conditions, the effect of this is straightforward, as varying $\beta$ just rescales the values taken by the field. But with non-zero boundary conditions, the effect is instead to vary the magnitude of the fluctuations of the values of the field around the (unique) harmonic function on the domain with those BCs. In particular, when $\beta\rightarrow \infty$, the field is ‘reluctant to be far from harmonic’, and so $h^D \Rightarrow \varphi$.

This parameter $\beta$ is called inverse temperature. So low temperature corresponds to high $\beta$, and high stability, which fits some physical intuition.

A Markov property

For a discrete (Gaussian) random walk, the Markov property says that conditional on a given value at a given time, the trajectory of the process before this time is independent of the trajectory afterwards. The discrete Gaussian bridge is similar. Suppose we have as before $0=Z_0,Z_1,\ldots, Z_N=0$ a centred Gaussian bridge, and condition that $Z_k=y$, for $k\in\{1,\ldots,N-1\}$, and $y\in\mathbb{R}$. With this conditioning, the density (3) splits as a product

$\mathbf{1}\{z_0=z_N=0, z_k=y\}\exp\left(-\frac12 \sum\limits_{i=1}^N (z_i-z_{i-1})^2 \right) =$

$\mathbf{1}\{z_0=0,z_k=y\} \exp\left(-\frac12 \sum\limits_{i=1}^k (z_i-z_{i-1})^2 \right) \cdot \mathbf{1}\{z_k=y,z_N=0\} \exp\left(-\frac12 \sum\limits_{i=k+1}^N (z_i-z_{i-1})^2 \right).$

Therefore, with this conditioning, the discrete Gaussian bridge splits into a pair of independent discrete Gaussian bridges with drift. (The same would hold if the original process had drift too.)

The situation for the DGFF is similar, though rather than focusing on the condition, it makes sense to start by focusing on the sub-domain of interest. Let $A\subset D$, and take $B=\bar D\backslash A$. So in particular $\partial A\subset B$.

Then we have that conditional on $h^D\big|_{\partial A}$, the restricted fields $h^D\big|_{B\backslash \partial A}$ and $h^D\big|_A$ are independent. Furthermore, $h^D\big|_A$ has the distribution of the DGFF on A, with boundary condition given by $h^D\big|_{\partial A}$. As in the discrete bridge, this follows just by splitting the density. Every gradient term corresponds to an edge in the underlying graph that lies either entirely inside $\bar A$ or entirely inside B. This holds for a general class of Gibbs models where the Hamiltonian depends only on the sum of some function of the heights (taken to be constant in this ‘free’ model) and the sum of some function of their nearest-neighbour gradients.

One additional and useful interpretation is that if we only care about the field on the restricted region A, the dependence of $h^D\big|_A$ on $h^D\big|_{D\backslash A}$ comes only through $h^D\big|_{\partial A}$. But more than that, it comes only through the (random) harmonic function which extends the (random) values taken on the boundary of A to the whole of A. So, if $h^A$ is an independent DGFF on A with zero boundary conditions, we can construct the DGFF $h^D$ from its value on $D\backslash A$ via

$h^D_x \stackrel{d}= h^A_x + \varphi^{h^D\big|_{\partial A}},$

where $\varphi^{h^D\big|_{\partial A}}$ is the unique harmonic extension of the (random) values taken by $h^D$ on $\partial A$ to $\bar A$.

This Markov property is crucial to much of the analysis to come. There are several choices of the restricted domain which come up repeatedly. In the next post we’ll look at how much one can deduce by taking A to be the even vertices in D (recalling that every integer lattice $\mathbb{Z}^d$ is bipartite), and then taking A to be a finer sublattice within D. We’ll use this to get some good bounds on the probability that the DGFF is positive on the whole of D. Perhaps later we’ll look at a ring decomposition of $\mathbb{Z}^d$ consisting of annuli spreading out from a fixed origin. Then the distribution of the field at this origin can be considered, via the final idea discussed above, as the limit of an infinite sequence of random harmonic functions given by the values taken by the field at increasingly large radius from the origin. Defining the DGFF on the whole lattice depends on the existence or otherwise of this local limit.