Given two (non-neighboring) cities on the map, how long does take to travel between them?

To make the problem more tractable, mathematicians usually assume that the time it takes to travel between neighboring cities are independent from each other and identically distributed according to a distribution . In other words, each edge of the lattice carries a random variable . The random variables are and distributed according to . It takes

to travel on the optimal route between city and city . The optimal route is usually called a geodesic. If one starts from the origin, one may wonder what are the cities that can be reached in less that hours, say. On the next picture, each red dot represents a city that can be reached in less than hours when the travel time between any two neighboring city is exponentially distributed with mean hour.

One may also wonder how the optimal routes look like. On the next picture I have plotted cities (the origin is the blue dotted city) and highlighted the optimal routes between the origin and each one of the cities on the border of the map.

Is it very different from the highway system around a typical city?

For those who want to play with this model, you can modify the quick and dirty Python code that can be found here. The study of these models is an active area of research.

Brownian Motion

A few days ago, a student asked the following question:

Consider a bounded domain and a Brownian motion conditioned to stay inside . How does it look like ? What is the invariant distribution ?

This is very simple, and surprisingly interesting. This involves the first eigenfunction of the Laplacian on . To make everything simple, and because this does not change anything, it suffices to study the situation where . In other words, what does a real Brownian motion conditioned to stay inside the segment look like.

Discretisation

One could directly do the computations in a continuous setting but, as it is often the case, it is simpler to consider the usual random walk discretisation of a Brownian motion. For this purpose, consider a time discretisation and a standard random walk with increment : with probability the random walk goes up and with probability it goes down . For clarity, assume that there exists an integer such that . Suppose as well that it is conditioned on the event for . Later, we will consider the limiting case and , in this order, to recover the Brownian motion case.

Conditioning

First, let us compute the probability transitions of the random walk conditioned on the event for . For clarity, let us denote the conditioned random walk by . This is a Doob h-transform, and the resulting process is a non-homogenous Markov chain. In this simple case the computations are straightforward. The conditioned Markov chain has transition probabilities given by with where . One can compute the probability that the conditioned Markov chain follows a given trajectory of ,

where is the transition kernel of the unconditioned Markov chain and is a normalisation constant. The Doob h-transform simply consists in noticing that this also reads

where the conditioned Markov kernel is and the function is defined by

Of course we have for all . The quantity is the probability that a random walk starting at at time remains inside for time . Consequently, in order to find the transition probabilities of the conditioned kernel, it suffices to compute the quantities for all . Since we are interested in the limiting case , it actually suffices to consider the case . It can be computed recursively since with the appropriate boundary conditions. In other words, adopting the obvious matrix notations, the vector satisfies

where is the usual tridiagonal matrix given by if, and only if, and otherwise. It is related to the discrete Laplacian operator. Indeed, because all the eigenvalues of are real with modulus strictly inferior to , it follows that where is the highest eigenvalue of and the associated eigenfunction. The eigenvalues of are well-known, and as , the highest eigenvalue converges to and the associated eigenfunction converges to the first eigenfunction of the Laplacian on the domain with Dirichlet boundary. In our case it is and

In other words, the random walk with increments conditioned to stay inside has probability transitions given by

Next section investigates the limiting case .

Conclusion

We have computed the dynamics of the conditioned random walk with space-increments . To obtain the dynamics of the conditioned Brownian motion it suffices to consider the limiting case . The drift of the resulting diffusion is given by

The same computation gives the volatility of the resulting diffusion. It is given by

As the consequence, this shows that a Brownian motion conditioned to stay inside follows the stochastic differential equation where is the first eigenfunction of the Laplacian on with Dirichlet boundary conditions. More generaly, the same argument would show that a Brownian motion in conditioned to stay inside a nice bounded domain evolves according to the stochastic differential equation

where is the first eigenfunction of the Laplacian on . This is a Langevin diffusion and one can immediately see that the invariant distribution of this diffusion is given by

For example, the following plot depicts the first eigenfunction of the Laplacian on the domain .

First Eigenfunction

1. Heisenberg Uncertainty Principle

The Heisenberg Uncertainty principle states that a signal cannot be both highly concentrated in time and highly concentrated in frequency. For example, consider a square-integrable function normalised so that . In this case, defines a probability distributions on the real line. The Fourier isometry shows that its Fourier transform also defines a probability distribution . In order to measure how spread-out these two probability distributions are, one can use their variance. The uncertainty principle states that

where designates the variance of the distribution . This shows for example that and cannot be simultaneously less than .

We can play the same with a vector and its (discrete) Fourier transform . The Fourier matrix is defined by

where the normalisation coefficient ensures that . The Fourier inversion formula reads . To measure how spread-out the coefficients of a vector are, one can look at the size of its support

If an uncertainty principle holds, one should be able to bound from below. There is no universal constant such that

for any and . Indeed, if and one readily checks that and so that . Nevertheless, in this case this gives . Indeed, a beautiful result of L. Donoho and B. Stark shows that

Maybe more surprisingly, and crucial to applications, they even show that this principle can be made robust by taking noise and approximations into account. This is described in the next section.

2. Robust Uncertainty Principle

Consider a subset of indices and the orthogonal projection on . In other words, if , then

We say that a vector is -concentrated on if . The -robust uncertainty principle states that if is concentrated on and is -concentrated on then

Indeed, the case gives Equation (5). The proof is surprisingly easy. On introduces the reconstruction operator defined by

In words: take a vector, delete the coordinate outside , take the Fourier transform, delete the coordinate outside , finally take the inverse Fourier transform. The special case simply gives . The proof consists in bounding the operator norm of from above and below. The existence of a vector that is -concentrated such that is -concentrated gives a lower bound. In details:

• Upper bound: it is obvious that is a contraction in the sense that since are isometries and are orthogonal projections. Moreover, the smaller and , the smaller the norm of . Since is an isometry we have , where the supremum is over all that are supported by . Cauchy-Schwarz shows that

  

  In other words, the reconstruction operator satisfies . This is a non-trivial bound only if .

• Lower bound: consider that satisfies and . The Fourier transform is an isometry so that the triangle inequality easily shows that in the sense that

  

  This proves the lower bound .

In summary, the reconstruction operator always satisfies . Moreover, the existence of satisfying and implies the lower bound . Therefore, the sets and must verify the uncertainty principle

Notice that we have only use the fact that the entries of the Fourier matrix are bounded in absolute value by . We could have use for example any other unitary matrix . The bound for all gives the uncertainty principle . In general since is an isometry.

One can work out an upper bound and a lower bound for the reconstruction operator using the -norm instead of the -norm. Doing so, one can prove that if and then

3. Stable Reconstruction

Let us see how the uncertainty principle can be used to reconstruct a corrupted signal. For example, consider a discrete signal corrupted by some noise . In top of that, let us suppose that on the set the receiver does not receive any information. In other words, the receiver can only observe

In general, it is impossible to recover the original , or an approximation of it, since the data for are lost forever. In general, even if the received signal is very weak, the deleted data might be huge. Nevertheless, under the assumption that the frequencies of are supported by a set satisfying , the reconstruction becomes possible. It is not very surprising since the hypothesis implies that the signal is sufficiently smooth so that the knowledge of on is enough to construct an approximation of for . We assume that the set is known to the observer. Since the components on are not observed, one can suppose without loss of generality that there is no noise on these components ( we have for ): the observation can be described as

It is possible to have a stable reconstruction of the original signal if the operator can be inverted: there exists a linear operator such that for every . If this is the case, the reconstruction

satisfies so that . A sufficient condition for to exist is that for every . This is equivalent to

Since for we have it follows that where is the reconstruction operator. The condition thus follows from the bound . Consequently the operator is invertible since satisfies

Since we have the bound . Therefore

Notice also that can easily and quickly be approximated using the expansion

Nevertheless, there are two things that are not very satisfying:

• the bound is extremely restrictive. For example, if of the component are corrupted, this imposes that the signal has only (and not ) non-zero Fourier coefficients.
• in general, the sets and are not known by the receiver.

The theory of compressed sensing addresses these two questions. More on that in forthcoming posts, hopefully … Emmanuel Candes recently gave extremely interesting lectures on compressed sensing.

4. Exact Reconstruction: Logan Phenomenon

Let us consider a slightly different situation. A small fraction of the components of a signal are corrupted. The noise is not assumed to be small. More precisely, suppose that one observes

where is the set of corrupted components and is an arbitrary noise that can potentially have very high intensity. Moreover, the set is unknown to the receiver. In general, it is indeed impossible to recover the original signal . Surprisingly, if one assumes that the signal is band-limited the Fourier transform of is supported by a known set that satisfies

it is possible to recover exactly the original . The main argument is that the condition (20) implies that the signal has more that of its energy on in the sense that . Consequently, since is the set where the signal is perfectly known, enough information is available to recover the original signal . To prove the inequality , it suffices to check that

In this case we have , which prove that has more that of its energy on . Indeed, the same conclusion holds with the -norm if the condition holds. The magic of the -norm is that condition implies that the signal is solution of the optimisation problem

This would not be true if one were using the norm instead. This idea was first discovered in a slightly different context in Logan’s thesis (1965). More details here. To prove (22), notice that any can be written as with . Therefore, since , it suffices to show that

This is equivalent to proving that for any non-zero we have

It is now that the -norm is crucial. Indeed, we have

which gives the conclusion. This would not work for any -norm with since in general : the inequality is in the wrong sense. In summary, as soon as the condition is satisfied, one can recover the original signal by solving the minimisation problem (22).

\noindent Because it is not hard to check that in general we always have

this shows that the condition implies that exact recovery is possible! Nevertheless, the condition is far from satisfying. The inequality is tight but in practice it happens very often that even if is much bigger that .

The take away message might be that perfect recovery is often possible if the signal has a sparse representation in an appropriate basis (in the example above, the usual Fourier Basis): the signal can then be recovered by -minimisation. For this to be possible, the representation basis (here the Fourier basis) must be incoherent with the observation basis in the sense that the coefficients of the change of basis matrix (here, Fourier matrix) must be small if is the observation basis and is the representation basis then must be small for every . For example, for the Fourier basis we have for every .

black: observation basis, green:representation basis

Let be a Markov kernel on a metric state space . We would like to quantify how long it takes for two different particles evolving according to the Markovian dynamic given by to meet. If the first particle starts at and the second at , the initial distance between them is . At time , what is the average distance between these two particles. For example, if and are two Brownian motions in started from and respectively, there is no reason why and should be closer from each other than and . Indeed, one can even show that whatever the coupling of these two Brownian motions we have : this is roughly speaking because the Euclidean space has no curvature. The situation is quite different if we were instead considering Brownian motions on a sphere: in this case, trajectories tend to coalesce.

1. Wasserstein distance

In the sequel, we will need to use a notion of distance between probability distributions on the metric space . The usual total variation distance defined by

is not adapted to our purpose since the metric structure of the space is not exploited. Instead, in order to take into account the distance of the space and develop a notion of curvature, we use the Wasserstein distance between probability measures. It is defined as

The distance is crucial to this definition: a change of distance implies a change of the class of -Lipschitz functions. Since for any coupling of and , and since the function is -Lipschitz, it follows that . Consequently, for any coupling we have . Taking the infimum over all the couplings leads to the inequality

This is a deep result that on any reasonable space the inequality is in fact an equality. Indeed, Kantorovich duality states that on any Radon space we have

It is interesting to note that under mild conditions on the state space one can always find a coupling that achieves the infimum of (4): this is an easy compactness argument.

2. Notion of Curvature

Denoting by the one step distribution of the Markov chain started from in the sense that , we define the local (Ricci) curvature between and as

The closer to is , the more the trajectories started at tend to meet the trajectories started at .

Trajectories tend to coalesce

The interesting case is when the infimum is strictly positive,

In this case we say that the Markov kernel is positively curved on . It should be noted that in many natural spaces it suffices to ensure that for all neighbouring states and to ensure that for any pair . This can be proved thanks to the so called Gluing Lemma. A space without curvature correspond to the case : for example, a symmetric random walk on and a Brownian motion on have both zero curvature. The curvature is a property of both the metric space and the Markov kernel : indeed, different Markov chain on the same metric space have generally different associated curvature. Given a metric space carrying a probability distribution , this is an interesting problem to construct a -invariant Markov chain with the highest possible curvature .

Indeed, the notion of curvature readily generalizes to continuous time Markov processes by taking a limiting case of (5). For example, one can define the curvature of the continuous time Markov process as the largest real number such that for any and we have

for every small enough. The quantity is the distribution of when started from in the sense that .

3. Contraction property

We now show that a positive curvature implies a contraction property. Equation (5) shows that for any . A simple argument shows that one can indeed generalize the situation to any two distributions in the sense that

Proof: For any pair consider a coupling of and such that . Now, choose an optimal coupling of and . This is straightforward to check that is a coupling (in general not optimal) of and so that

Equation (8) is extremely powerful since it immediately shows that

In other words, there is exponential convergence (in the Wasserstein metric) to the invariance distribution at rate . In continuous time, this reads

In other words, the higher the curvature, the faster the convergence to equilibrium.

4. Examples

Let us give examples of positively curved Markov chains.

1. Langevin diffusion with convex potential: consider a convex potential that is uniformly elliptic in the sense . The Langevin diffusion has invariant distribution with density proportional to . Given a time step , the Euler discretization of this diffusion reads
    

   where . Given two starting points and , using the same noise to define and it immediately follows that

   

   In other words, the Langevin diffusion is positively curved with curvature (at least) equal to .

    

2. Brownian motion on a sphere: consider a Brownian motion on the unit sphere of . Consider two points on this unit sphere: by symmetry, one can always rotate the coordinates so that that and for some . For the (geodesic) distance is approximated by . One can couple two Brownian motions and , one started at and the other one started at , by the usual symmetry with respect to the plane : in other words, is the reflexion of with respect to . One can check (good exercise!) that the diffusion followed by the -coordinate of a Brownian motion on the unit sphere of is simply given by
    

   With this coupling, for small time , it follows that

   

   where is used as the same source of randomness for and since is the reflexion of . Since it readily follows that

   

   In other words, the curvature of a Brownian motion on the unit sphere of is equal to . Maybe surprisingly, the higher the dimension, the faster the convergence to equilibrium. This is not so unreal if one notices that the Brownian increment satisfies .

3. Other examples: see the original text for many other examples.

Lazy, I asked myself if it were possible to write lines long Python code to solve this innocent looking enigma. The whole pyramid is entirely determined by the numbers lying at the bottom, and each one of them is an integer between and : these numbers must be different so that there are at most possibilities to test! Brute force won’t work my friend!

When stupid brute force does not work, one can still try annealing/probabilist methods: this works pretty well for Sudoku (which is NP-hard) as this is brilliantly described here and there. The principle is simple: if one can find a good energy function such that a solution to the problem corresponds to a low energy configuration, one can do MCMC-simulating annealing-etc on the target distribution

The issue is that it might be very difficult to choose a sensible energy function . Foolishly, I first tried the following energy function, and then ran a random walk Metropolis algorithm with as target probability:

where is the numbers of levels that one can fill, starting from the bottom, without encountering any problem no repetition and no number greater than . With different values of and letting run the algorithm for a few millions iterations ( min on my crappy laptop), one can easily produce configurations that are -levels high: but the algorithm never found any real solution a configuration with height equal to .

Now I am curious wether this is possible to produce a non-stupid energy function so that this riddle is solvable in a reasonable amount of time by standard MCMC – annealing methods.

As a conclusion, I should mention that with a pen and a cup of coffee, one can easily find a solution: I will not spoil the fun, but just say that the configuration space is not that big if one think more carefully about it…

1. Reversible Markov chains

Remember that a Markov chain on the state space with transition operator is reversible with respect to the probability if for any

In words, this means that looking at a trajectory of this Markov chain, this is impossible to say if the time is running forward or backward: indeed, the probability of observing is equal to , which is also equal to if the chain is reversible.

This is precisely this property of invariance by time reversal that allows to sample from conditioned reversible Markov chain. Since under mild conditions a one dimensional diffusion is also reversible, this also shows that (ergodic) one dimensional conditioned diffusions are sometimes quite easy to sample from!

2. One dimensional diffusions are reversible!

The other someone told me that almost any one dimensional diffusion is reversible! I did not know that, and I must admit that I still find this result rather surprising. Indeed, this is not true for multidimensional diffusions, and this is very easy to construct counter-examples. What makes the result works for real diffusions is that there is only one way to go from to , namely the segment . Indeed, the situation is completely different in higher dimensions.

First, let us remark that any Markov chain on , that has as invariant distribution and that can only make jumps of size or is reversible: such a Markov chain is usually called skip-free in the litterature. This is extremely easy to prove, and since skip-free Markov chains have been studied a lot, I am sure that this result is somewhere in the litterature (any reference for that?). To show the result, it suffices to show that for any where is the upward flux at level and is the downward flux. Because is invariant it satisfies the usual balance equations,

so that . Interating we get that for any we have : the conclusion follows since .

This simple result on skip-free Markov chains gives also the result for many one dimensional diffusions since under regularity assumptions on and they can be seen as limit of skip-free one dimensional Markov chains on . Indeed, I guess that the usual proof of this result goes through introducing the scale function and the speed measure of the diffusion, but I would be very glad if anyone had another pedestrian approach that gives more intuition into this.

3. How to sample conditioned reversible Markov chains

From what has been said before, I am sure that this becomes quite clear how a conditioned reversible Markov chain can be sampled from. Suppose that is reversible Markov chain and that we would like to sample a path conditioned on the event and :

1. sample the path between and , starting from
2. sample the path between and , starting from
3. if there exists such that then define the path byand otherwise go back to step .

Indeed, the resulting path is an approximation of the realisation of the conditioned Markov chain: this is not hard to prove it, playing around with the definition of time reversibility. It is not hard at all to adapt this idea to reversible diffusions, though the result is indeed again an approximation. The interesting question is to discuss how good this approximation is (see the paper by Michael Sorensen and Mogens Bladt )

For example, here is a sample from a Birth-Death process, conditioned on the event and , with parameter .

Conditioned Birth-Death process

4. Final remark

It might be interesting to notice that this method is especially inefficent for multidimensional processes: the probability of finding an instant such that is extremely small, and in many cases equal to for diffusions! This works pretty well for one dimensional diffusion thanks to the continuity of the path and the intermediate value property. Nevertheless, even for one dimensional diffusion this method does not work well at all when trying to sample from conditioned paths between two meta-stable position: this is precisely this situation that is interesting in many physics when one wants to study the evolution of a particle in a double well potential, for example. In short, sampling conditioned (multidimensional) diffusions is still a very difficult problem.

1. Erdos Renyi Random Graph model

Remember that the Erdos Renyi random graph is a graph with vertices such that any two of them are connected with probability , independently.

Erdos-Renyi random graph model

This model of random graphs has been extensively studied and this has long been known that undergoes a phase transition at : for , the largest component of has a size of order while for the largest connected component , also called the giant component, has size of order . In another blog entry, another consequence of this phase transition has been used.

One can even show that for there exists a constant such that for any ,

Simply put, the largest connected component of has size approximatively equal to . Maybe surprisingly, this is quite simple to compute the value of through fluid limit techniques.

2. Exploration of the graph

Let be a configuration of . This configuration is explored through a classical depth-first algorithm. To keep track of what is doing, a process is introduced. This process captures essentially all the information needed to study the size of the connected component: it is then shown that behaves essentially deterministically when looked under the appropriate scale.

1. initilisation: t=0: Color all the vertices in green, the current vertex is and define .

2. t=t+1: Color the current vertex in red (=visited): if a vertex is connected to and is green, color it in grey ( this vertex has still to be visited, but its ‘parent’ has already been visited): these are the children of .
3. update the following way, , and
   • if has children, choose one of them and make it the new current vertex. Go to .
   • If has no children then the new current vertex is its parent. Go to .
   • if has no children and no parent, the new current vertex is one of the remaining (if any) green vertex. Go to
   • else: we are finished.

A moment of though reveals that and reaches precisely when the connected component of the vertex has all been explored. Then reaches precisely when the second connected component has all been explored, etc … This is why if and then the connected components have size exactly for . The largest component has size .

Z process

3. Fluid Limit

It is quite easy to show why the rescaled process should behave like a differential equation (ie: fluid limit). While exploring the first component, represents the number of gray vertices at time so that represents the number of green vertices: this is why

so that where . Also, the variance statisfies so that all the ingredients for a fluit limit are present: converges to the differential equation

whose solution is . This is why is implicitly defined as the solution of

As expected, and .

The mean field Potts model is extremely simple: there are interacting particles and each one of them can be in different states . Define the Hamiltonian

where and is the Kronecker symbol. The normalization ensures that the energy is an extensive quantity so that the mean energy per particle does no degenerate to or for large values of . The sign minus is here to favorize configurations that have a lot of particles in the same state. The Boltzman distribution at inverse temperature on is given by

where is a normalization constant. Notice that if we choose a configuration uniformly at random in , with overwhelming probability the ratio of particles in state will be close to . Also it is obvious that if we define

then will be close to for a configuration taken uniformly at random. Stirling formula even says that the probability that is close to is close to where

Indeed . The situation is quite different under the Boltzman distribution since it favorizes the configurations that have a lot of particles in the same state: this is because the Hamiltonian is minimized for configurations that have all the particles in the same state. In short there is a competition between the entropy (there are a lot of configurations close to the ratio ) and the energy that favorizes the configurations where all the particles are in the same state.

With a little more work, one can show that there is a critical inverse temperature such that:

• for the entropy wins the battle: the most probable configurations are close to the ratio
• for the energy effect shows up: there are most probable configurations that are the permutations of where and are computable quantities.

The point is that above the system has more than one stable equilibrium point. Maybe more important, if we compute the energy of these most probable states

then this function has a discontinuity at . I will try to show in the weeks to come how this behaviour can dramatically slow down usual Monte-Carlo approach to the study of these kind of models.

Hugo Touchette has a very nice review of statistical physics that I like a lot and a good survey of the Potts model. Also T. Tao has a very nice exposition of related models. The blog of Georg von Hippel is dedicated to similar models on lattices, which are far more complex that this mean field approximation presented here.

MCMC Simulations

These is extremely easy to simulate this mean field Potts model since we only need to keep track of the ratio to have an accurate picture of the system. For example, a typical Markov Chain Monte Carlo approach would run as follows:

• choose a particle uniformly at random in
• try to switch its value uniformly in
• compute the Metropolis ratio
• update accordingly.

If we do that times for states at inverse temperature and for particles (which is fine since we only need to keep track of the -dimensional ratio vector) and plot the result in barycentric coordinates we get a picture that looks like:

Here I started with a configuration where all the particles were in the same states i.e ratio vector equal to . We can see that even with steps, the algorithm struggles to go from one most probable position to the other two and – in this simulation, one of the most probable state has even not been visited! Indeed, this approach was extremely naive, and this is quite interesting to try to come up with better algorithms. Btw, Christian Robert’s blog has tons of interesting stuffs related to MCMC and how to boost up the naive approach presented here.

]]> http://linbaba.wordpress.com/2010/06/22/potts-model-and-monte-carlo-slow-down/feed/ 0 linbaba potts-mcmc