An army of gamblers and a martingale

coin toss
Start flipping a coin. How long should you wait (on average) before observing the pattern {HHTHHTT} !? Easy you might say, that’s simply the hitting time of the state {H} starting from state {A} of the following Markov chain.

Indeed, one can solve this kind of problem using a first step analysis. In other words, if {a} is the expected number of steps before reaching {H} starting from {A}, {b} is the expected number of steps before reaching {H} starting from {B}, {c} is the expected number of steps before reaching {H} starting from {C}, etc… one can readily see that {a,b, \ldots, h} verify the system of equations

\displaystyle \begin{array}{rcl} a &=& 1 + \frac12(a+b), \quad b = 1 + \frac12(a+c), \quad c = 1 + \frac12(c+d) \\ d &=& 1 + \frac12(a+e), \quad e = 1 + \frac12(a+f), \quad f = 1 + \frac12(c+g) \\ g &=& 1 + \frac12(e+h), \quad h = 0. \end{array}

After a finite amount of time and coffee one then come up with the answer {a=128}. That was a little bit tedious and I don’t really want to know how long one should wait on average before observing the sequence {HHTHTTTHTHHHHTTHTHTTH}.

It turns out that there is a very neat way to approach this kind of problems without solving any linear system. Apparently, this trick was discovered in the 80s by Shuo-Yen Robert Li. As often in probability, this boils down to introducing a clever martingale. Suppose that one wants to know how long it takes on average before observing the magic coin pattern {x_1x_2 \ldots x_n} of length {n \geq 1} with {x_i \in \{H,T\}} for all {1 \leq i \leq n}. To this end, one will introduce an army of gamblers, a casino, a game of chance and a martingale. The martingale {M} describes the amount of money earned by the casino. At each round (starting at round {t=1}) the casino tosses a coin and the gamblers can make bets. If a gambler starts betting just before round {k}, he wins if the coin shows {x_1}. If he decides to bet again at round {k+1}, he wins if the coin shows {x_2}, etc… Suppose that there are infinitely many gamblers {G_0, G_1, G_2, \ldots} and that gambler {G_k} starts betting just before round {k+1}. The game is fair in the sense that if a gambler bets an amount {A}, with probability {1/2} he gets {2A} and with probability {1/2} he losses his bet. One supposes that gambler {G_k} starts betting {1} dollar just before round {k+1} and keeps reinvesting his total capital until he either losses everything (total loss equals {1} dollar) or observes the magic coin pattern {x_1x_2 \ldots x_n} (total earning equals {2^n-1}). Indeed, the amount of money earned by the casino is a martingale and {M_0=0}. For example, {M_1=1} if the first gambler {G_0} loses his bet and {M_1=-1} if the wins. Consider the time {\tau} when the magic coin pattern first appears. This is a stopping time and the optional stopping theorem for martingales shows that {\mathbb{E}[M_{\tau}] = 0}. Indeed, one can also compute this quantity another way. By time {\tau}, all the losers have given {1} dollar to the casino. Also, if the gambler {G_k} is still in the game he has won a total amount {2^{\tau-k}-1}. in other words,

\displaystyle M_{\tau} = \tau - \sum_{j=0}^{\tau-1} 2^{\tau-k} \, \delta_j

where {\delta_j=1} if the gambler {G_j} is still in the game after the round {\tau} and {\delta_j=0} otherwise. The fundamental remark is that the quantity {\sum_{j=1}^{\tau} 2^{\tau-k+1} \, \delta_j} is not random so that {\mathbb{E}[\tau] = \sum_{j=0}^{\tau-1} 2^{\tau-k} \, \delta_j}. Indeed, right after round {\tau} one can know with certainty whether {G_{k}} is still in the game or not. Gambler {G_{\tau-k}} is in the game only if the {k} last letters of the magic word {x_1x_2 \ldots x_n} are the same as the first {k} letters. For example, if the magic word is {HHTHHTH} then the gamblers who have seen the pattern {H},{HHTH} and {HHTHHTH} at time {\tau} are still in the game when the magic word first appear. In this case this shows that {\mathbb{E}[\tau] = 2^1+2^4+2^7}.

gambler

For example, this immediately shows that we have to wait on average {2^{21}+2^1} coin tosses before observing the pattern {HHTHTTTHTHHHHTTHTHTTH}.

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Conditioned Brownian motion, 1st Laplacian eigenfunction, etc…

Brownian Motion

Brownian Motion

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

Consider a bounded domain {D \subset \mathbb{R}^d} and a Brownian motion conditioned to stay inside {D}. 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 {D}. To make everything simple, and because this does not change anything, it suffices to study the situation where {D=[0,1] \subset \mathbb{R}}. In other words, what does a real Brownian motion conditioned to stay inside the segment {[0,1]} 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 {\delta>0} and a standard random walk {X_k} with increment {\pm \sqrt{\delta}}: with probability {\frac12} the random walk goes up {+\sqrt{\delta}} and with probability {\frac12} it goes down {-\sqrt{\delta}}. For clarity, assume that there exists an integer {m_{\delta} \in \mathbb{N}} such that {m_{\delta} \sqrt{\delta} = 1}. Suppose as well that it is conditioned on the event {X_k \in [0,1]} for {k=1, \ldots, N}. Later, we will consider the limiting case {N \rightarrow \infty} and {\delta \rightarrow 0}, in this order, to recover the Brownian motion case.

Conditioning

First, let us compute the probability transitions of the random walk {\{X_k\}_{k=0}^N} conditioned on the event {X_k \in [0,1]} for {k=1, \ldots, N}. For clarity, let us denote the conditioned random walk by {\hat{X} = \big\{ \hat{X}_k \big\}_{k \geq 0}}. 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 {\hat{P}(k,x,y) = \mathbb{P}[\hat{X}_{k+1}=y \,|\hat{X}_k=x]} with {x,y \in D_{\delta}} where {D_{\delta} = \sqrt{\delta} \mathbb{N} \; \cap \; [0,1]}. One can compute the probability that the conditioned Markov chain follows a given trajectory {(x_0, \ldots, x_N)} of {D_{\delta}},

\displaystyle  \begin{array}{rcl}  \mathbb{P}[\hat{X}_0 = x_0, \ldots, \hat{X}_N=x_N] = \frac{1}{Z(N,\delta)} P(x_0,x_1) \times \ldots \times P(x_{N-1}, x_N) \end{array}

where {P(x,y)} is the transition kernel of the unconditioned Markov chain and {Z(N,\delta) = \mathbb{P}[X_k \in D_{\delta} \; \text{for} \; k=0, \ldots, N]} is a normalisation constant. The Doob h-transform simply consists in noticing that this also reads

\displaystyle  \begin{array}{rcl}  \mathbb{P}[\hat{X}_0 = x_0, \ldots, \hat{X}_N=x_N] = \hat{P}(0,x_0,x_1) \times \ldots \times \hat{P}(N-1,x_{N-1}, x_N) \end{array}

where the conditioned Markov kernel is {\hat{P}(k,x_k,x_{k+1}) = \frac{P(x,y) \, h(k+1,y)}{h(k,x)}} and the function {h(\cdot, \cdot)} is defined by

\displaystyle  \begin{array}{rcl}  h(k,x) = \mathbb{P}[X_{j} \in D_{\delta} \; \text{for} \; k+1 \leq j \leq N \, |X_k=x]. \end{array}

Of course we have {h(N,x)=1} for all {x \in D_{\delta}}. The quantity {h(x,k)} is the probability that a random walk starting at {x \in D_{\delta}} at time {k} remains inside {D_{\delta}} for time {k,k+1, \ldots, N}. Consequently, in order to find the transition probabilities of the conditioned kernel, it suffices to compute the quantities {h(k, x)} for all {x \in D_{\delta}}. Since we are interested in the limiting case {N \rightarrow \infty}, it actually suffices to consider the case {k=0}. It can be computed recursively since {h(k,x) = \frac12 \, h(k+1, x+\sqrt{\delta}) + \frac12 \, h(k+1, x-\sqrt{\delta})} with the appropriate boundary conditions. In other words, adopting the obvious matrix notations, the vector {h(k,\cdot)} satisfies

\displaystyle  \begin{array}{rcl}  h(k,\cdot) = A h(k+1, \cdot) \end{array}

where {\big( A_{i,j} \big)_{0 \leq i,j \leq m_{\delta}}} is the usual tridiagonal matrix given by {A_{i,j} = 1} if, and only if, {|i-j|=1} and {A_{i,j} = 0} otherwise. It is related to the discrete Laplacian operator. Indeed, because all the eigenvalues of {A} are real with modulus strictly inferior to {1}, it follows that {h(0,\cdot) = A^{N} \textbf{1} \approx \lambda^N \, \varphi_{\delta}(\cdot)} where {\lambda_{\delta}} is the highest eigenvalue of {A} and {\varphi_{\delta}(\cdot)} the associated eigenfunction. The eigenvalues of {A} are well-known, and as {\delta \rightarrow 0}, the highest eigenvalue converges to {1} and the associated eigenfunction converges to the first eigenfunction of the Laplacian on the domain {D=[0,1]} with Dirichlet boundary. In our case it is {\varphi(u) = \sin(\pi \, u)} and

\displaystyle  \begin{array}{rcl}  \lim_{N \rightarrow \infty} \frac{ h(1,x \pm \sqrt{\delta})}{h(0,x)} = \lambda_{\delta}^{-1} \frac{ \varphi_{\delta}(x \pm \sqrt{\delta})}{\varphi_{\delta}(x)}. \end{array}

In other words, the random walk with increments {\pm \sqrt{\delta}} conditioned to stay inside {D_{\delta}} has probability transitions given by

\displaystyle  \begin{array}{rcl}  \hat{P}(x, x \pm \sqrt{\delta}) = \frac{{\lambda_{\delta}}^{-1}}{2} \frac{ \varphi_{\delta}(x \pm \sqrt{\delta})}{\varphi_{\delta}(x)}. \end{array}

Next section investigates the limiting case {\delta \rightarrow 0}.

Conclusion

We have computed the dynamics of the conditioned random walk with space-increments {\sqrt{\delta}}. To obtain the dynamics of the conditioned Brownian motion it suffices to consider the limiting case {\delta \rightarrow 0}. The drift of the resulting diffusion is given by

\displaystyle  \begin{array}{rcl}  \text{(drift)} &=& \lim_{\delta \rightarrow 0} \frac{1}{\delta} \Big\{ \hat{P}(0,x,x+\sqrt{\delta})\sqrt{\delta} - \hat{P}(0,x,x-\sqrt{\delta})\sqrt{\delta} \Big\} \\ &=& \lim_{\delta \rightarrow 0} \frac{{\lambda_{\delta}}^{-1}}{2 \sqrt{\delta}} \Big\{ \frac{ \varphi_{\delta}(x + \sqrt{\delta})}{\varphi_{\delta}(x)} - \frac{ \varphi_{\delta}(x - \sqrt{\delta})}{\varphi_{\delta}(x)} \Big\} = \frac{1}{2} \frac{\varphi'(x)}{\varphi(x)}. \end{array}

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

\displaystyle  \begin{array}{rcl}  \text{(volatility)} &=& \lim_{\delta \rightarrow 0} \frac{1}{\delta} \Big\{ \hat{P}(0,x,x+\sqrt{\delta})\delta + \hat{P}(0,x,x-\sqrt{\delta})\delta \Big\} \\ &=& \lim_{\delta \rightarrow 0} \frac{{\lambda_{\delta}}^{-1}}{2} \Big\{ \frac{ \varphi_{\delta}(x + \sqrt{\delta})}{\varphi_{\delta}(x)} + \frac{ \varphi_{\delta}(x - \sqrt{\delta})}{\varphi_{\delta}(x)} \Big\} = 1. \end{array}

As the consequence, this shows that a Brownian motion conditioned to stay inside {D=[0,1]} follows the stochastic differential equation {dX = \frac{1}{2} \frac{\varphi'(X)}{\varphi(X)} \, dt + dW_t} where {\varphi(\cdot)} is the first eigenfunction of the Laplacian on {D} with Dirichlet boundary conditions. More generaly, the same argument would show that a Brownian motion in {\mathbb{R}^d} conditioned to stay inside a nice bounded domain {D \subset \mathbb{R}^d} evolves according to the stochastic differential equation

\displaystyle  \begin{array}{rcl}  dX = \frac{1}{2} \frac{\nabla \varphi(X)}{\varphi(X)} \, dt + dW_t \end{array}

where {\varphi:D \rightarrow \mathbb{R}} is the first eigenfunction of the Laplacian on {D}. This is a Langevin diffusion and one can immediately see that the invariant distribution of this diffusion is given by

\displaystyle  \begin{array}{rcl}  \pi_{\infty}(x) \, \propto \varphi(x). \end{array}

For example, the following plot depicts the first eigenfunction of the Laplacian on the domain {D=[0,1]^2 \subset \mathbb{R}^2}.

First Eigenfunction

Curvature for Markov Chains

Recently, Yann Ollivier developed a nice theory of Ricci curvature for Markov chains. In many ways, this can be seen as a geometric language giving another view on the notion of path coupling, developed at the end of the {90}‘s by Martin Dyer and co-workers. It has to be noted that this new notion of curvature is very general and does not need the state space where the Markov chain evolves to have any differential structure, as can be expected at first sight. Any state space endowed with a metric suffices.

Let {P} be a Markov kernel on a metric state space {(S,d)}. We would like to quantify how long it takes for two different particles evolving according to the Markovian dynamic given by {P} to meet. If the first particle starts at {x \in S} and the second at {y \in S}, the initial distance between them is {d(x,y)}. At time {t>0}, what is the average distance between these two particles. For example, if {W^x} and {W^y} are two Brownian motions in {{\mathbb R}^n} started from {x} and {y} respectively, there is no reason why {W^x_t} and {W^y_t} should be closer from each other than {x=W^x_0} and {y=W^y_0}. Indeed, one can even show that whatever the coupling of these two Brownian motions we have {\mathop{\mathbb E}[d(W^x_t, W^y_t)] \geq d(x,y)}: this is roughly speaking because the Euclidean space {{\mathbb R}^n} 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 {(S,d)}. The usual total variation distance {d(\mu,\nu)} defined by

\displaystyle  d(\mu,\nu) \;=\; \sup_{A \subset S} \; |\mu(A)-\nu(A)| \ \ \ \ \ (1)

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 {d(\cdot,\cdot)} of the space {E} and develop a notion of curvature, we use the Wasserstein distance {W(\mu,\nu)} between probability measures. It is defined as

\displaystyle  W(\mu,\nu) \;=\; \sup\Big\{ \mu(f) - \nu(f) \;:\; \text{Lip}(f) \leq 1\Big\}. \ \ \ \ \ (2)

The distance {d(\cdot,\cdot)} is crucial to this definition: a change of distance implies a change of the class of {1}-Lipschitz functions. Since {\mu(f) - \nu(f) = \mathop{\mathbb E}[f(X) - f(Y)]} for any coupling {(X,Y)} of {\mu} and {\nu}, and since the function {f} is {1}-Lipschitz, it follows that {\mathop{\mathbb E}[f(X) - f(Y)] \leq \mathop{\mathbb E}[d(X,Y)]}. Consequently, for any coupling {(X,Y)} we have {W(\mu,\nu) \leq \mathop{\mathbb E}[d(X,Y)]}. Taking the infimum over all the couplings {(X,Y)} leads to the inequality

\displaystyle  W(\mu,\nu) \;=\; \sup_{\text{Lip}(f) \leq 1} \; |\mu(f) - \nu(f)| \;\leq\; \inf_{(X,Y)} \; \mathop{\mathbb E}[d(X,Y)]. \ \ \ \ \ (3)

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

\displaystyle   W(\mu,\nu) \;=\; \sup_{\text{Lip}(f) \leq 1} \; |\mu(f) - \nu(f)| \;=\; \inf_{(X,Y)} \; \mathop{\mathbb E}[d(X,Y)]. \ \ \ \ \ (4)

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

2. Notion of Curvature

Denoting by {m_x = \delta_x P} the one step distribution of the Markov chain started from {x} in the sense that {m_x(A) = \mathop{\mathbb P}[X_1 \in A \;| X_0 = x ]}, we define the local (Ricci) curvature {\kappa(x,y) \in {\mathbb R}} between {x} and {y} as

\displaystyle   W(m_x, m_y) = d(x,y) \cdot (1-\kappa(x,y)). \ \ \ \ \ (5)

The closer to {1} is {\kappa(x,y)}, the more the trajectories started at {x} tend to meet the trajectories started at {y}.

Trajectories tend to coalesce

The interesting case is when the infimum {\inf_{x,y} \, \kappa(x,y)} is strictly positive,

\displaystyle   \inf_{x,y \in E} \kappa(x,y) \;=\; \kappa > 0. \ \ \ \ \ (6)

In this case we say that the Markov kernel {P} is positively curved on {(S,d)}. It should be noted that in many natural spaces it suffices to ensure that {\kappa(x,y) \;\geq\; \kappa} for all neighbouring states {x} and {y} to ensure that {\kappa(x,y) \;\geq\; \kappa} for any pair {x,y \in S}. This can be proved thanks to the so called Gluing Lemma. A space without curvature correspond to the case {\kappa=0}: for example, a symmetric random walk on {\mathbb{Z}^d} and a Brownian motion on {{\mathbb R}^d} have both zero curvature. The curvature {\kappa} is a property of both the metric space {(S,d)} and the Markov kernel {P}: indeed, different Markov chain on the same metric space {(S,d)} have generally different associated curvature. Given a metric space {(S,d)} carrying a probability distribution {\pi}, this is an interesting problem to construct a {\pi}-invariant Markov chain with the highest possible curvature {\kappa}.

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 {\{X_t\}_{t \geq 0}} as the largest real number {\kappa} such that for any {x,y \in (S,d)} and {\kappa' < \kappa} we have

\displaystyle  W(m_x^{\delta}, m_y^{\delta}) \;\leq\; (1-\delta \kappa') \; d(x,y) \ \ \ \ \ (7)

for every {\delta} small enough. The quantity {m_x^{\delta}} is the distribution of {X_{\delta}} when started from {x} in the sense that {m_x^{\delta}(A) = \mathop{\mathbb P}[X_{\delta} \in A \; |X_0=x]}.

3. Contraction property

We now show that a positive curvature implies a contraction property. Equation (5) shows that {W(\delta_x P, \delta_y P) \leq W(\delta_x,\delta_y) \cdot (1-\kappa)} for any {x,y \in S}. A simple argument shows that one can indeed generalize the situation to any two distributions {\mu,\nu} in the sense that

\displaystyle   W(\mu P, \nu P) \leq W(\mu,\nu) \cdot (1-\kappa). \ \ \ \ \ (8)

Proof: For any pair {x,y \in S} consider a coupling {(U_{x,y}, V_{x,y})} of {m_x} and {m_y} such that {W(m_x,m_y)=\mathop{\mathbb E}[d(U_{x,y}, V_{x,y})]}. Now, choose an optimal coupling {(X,Y)} of {\mu} and {\nu}. This is straightforward to check that {(U_{X,Y}, V_{X,Y})} is a coupling (in general not optimal) of {\mu P} and {\nu P} so that

\displaystyle  \begin{array}{rcl}  W(\mu P, \nu P) &\leq& \mathop{\mathbb E}[d(U_{X,Y}, V_{X,Y})] = \mathop{\mathbb E}[\; \mathop{\mathbb E}[d(U_{x,y}, V_{x,y}) \;|X=x, Y=y] ] \\ &=& \mathop{\mathbb E}[ W(m_X, m_Y) ] = \mathop{\mathbb E}[ d(X,Y) \cdot (1-\kappa(X,Y)) ]\\ &\leq& (1-\kappa) \; \mathop{\mathbb E}[ d(X,Y) ] = (1-\kappa) \; W(\mu,\nu). \end{array}

\Box

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

\displaystyle  W(\mu P^t, \pi) \leq (1-\kappa)^{t} \; W(\mu,\pi). \ \ \ \ \ (9)

In other words, there is exponential convergence (in the Wasserstein metric) to the invariance distribution {\pi} at rate {(1-\kappa)^t}. In continuous time, this reads

\displaystyle  W(\mu P^t, \pi) \;\leq\; e^{-\kappa t} \; W(\mu,\pi). \ \ \ \ \ (10)

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 {\Psi:{\mathbb R} \rightarrow {\mathbb R}} that is uniformly elliptic in the sense {\Psi^{''}(x) \geq \lambda > 0}. The Langevin diffusion {dz = -\frac{1}{2} \Psi'(z) \, dt + dW} has invariant distribution {\pi} with density proportional to {e^{-\Psi(x)}}. Given a time step {\delta}, the Euler discretization of this diffusion reads
    \displaystyle  x^{k+1} = x^k - \frac{1}{2} \Psi'(x^k) \, \delta + \sqrt{\delta} \; \xi \ \ \ \ \ (11) 

    where {\xi \sim {\mathcal N}(0,1)}. Given two starting points {x^0=x} and {y^0=y}, using the same noise {\xi} to define {x^1} and {y^1} it immediately follows that

    \displaystyle  \begin{array}{rcl}  W(x^1, y^1) &\leq& (x-y) \; \Big(1 - \frac{\delta}{2} \frac{\Psi'(x)-\Psi'(y)}{x-y} \Big)\\ &\leq& (x-y) \; (1-\frac{\lambda}{2} \delta). \end{array}

    In other words, the Langevin diffusion {\{z_t\}_{t \geq 0}} is positively curved with curvature (at least) equal to {\kappa = \frac{\lambda}{2}}.

     

  2. Brownian motion on a sphere: consider a Brownian motion on the unit sphere of {{\mathbb R}^n}. Consider two points {X,Y} on this unit sphere: by symmetry, one can always rotate the coordinates so that that {X=(\sqrt{1-h^2},0,h)} and {X=(\sqrt{1-h^2},0,-h)} for some {h \in [0,1]}. For {h \ll 1} the (geodesic) distance {d(X,Y)} is approximated by {d(X,Y) \approx 2h}. One can couple two Brownian motions {W^X} and {W^Y}, one started at {X} and the other one started at {Y}, by the usual symmetry with respect to the plane {\mathcal{P} = \{(x,y,z): z=0\}}: in other words, {W^Y} is the reflexion of {W^X} with respect to {\mathcal{P}}. One can check (good exercise!) that the diffusion followed by the {z}-coordinate of a Brownian motion on the unit sphere of {{\mathbb R}^n} is simply given by
    \displaystyle  dz = -\frac{1}{2}(n-1)z \, dt + \sqrt{1-z^2} \, dW. \ \ \ \ \ (12) 

    With this coupling, for small time {\delta \ll 1}, it follows that

    \displaystyle  \begin{array}{rcl}  z^X_{\delta} &\approx& h - \frac{1}{2} (n-1) h \, \delta + \sqrt{1-h^2} \sqrt{\delta} \; \xi\\ z^Y_{\delta} &\approx& -h + \frac{1}{2} (n-1) h \, \delta - \sqrt{1-h^2} \sqrt{\delta} \; \xi \end{array}

    where {\xi \sim {\mathcal N}(0,1)} is used as the same source of randomness for {z^X_{\delta}} and {z^Y_{\delta}} since {W^Y} is the reflexion of {W^X}. Since {d(W^X_{\delta}, W^X_{\delta}) \approx |z^X_{\delta} - z^Y_{\delta}|} it readily follows that

    \displaystyle  \begin{array}{rcl}  d(W^X_{\delta}, W^X_{\delta}) \; \leq \; \big(1- \frac{1}{2}(n-1)\delta \big)\; d(x,y). \end{array}

    In other words, the curvature of a Brownian motion on the unit sphere of {{\mathbb R}^n} is equal to {\frac{1}{2}(n-1)}. 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 {\mathop{\mathbb E} \|W_{t+\delta}-W_t\|^2 \approx n \delta}.

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

Sampling conditioned Markov chains, and diffusions

In many situations it might be useful to know how to sample a Markov chain (or a diffusion process) {X_k} between time {0} and {T}, conditioned on the knowledge that {X_0=a} and {X_T=b}. This conditioned Markov chain {X^{\star}} is still a Markov chain but in general is not time homogeneous. Moreover, it is generally very difficult to compute the transition probabilities of this conditioned Markov chain since they depend on the knowledge of the transition probabilities {p(t,x,y) = \mathop{\mathbb P}(X_t=y | X_0=x)} of the unconditioned Markov chain, which are usually not available: this has been discussed in this previous post on Doob h-transforms. Perhaps surprisingly, this article by Michael Sorensen and Mogens Bladt shows how this is sometimes quite easy to sample good approximations of such a conditioned Markov chain, or diffusion.

1. Reversible Markov chains

Remember that a Markov chain {X_k} on the state space {S} with transition operator {p(x,y) = \mathop{\mathbb P}(X_{k+1}=y \,|X_k=x)} is reversible with respect to the probability {\pi} if for any {x,y \in S}

\displaystyle  \pi(x) p(x,y) = p(y,x) \pi(y). \ \ \ \ \ (1)

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 {y_1, y_2, \ldots, y_n} is equal to {\pi(y_1)p(y_1, y_2) \ldots p(y_{n-1}, y_n)}, which is also equal to {\pi(y_n)p(y_n,y_{n-1}) \ldots p(y_2, y_1)} 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 {a} to {b}, namely the segment {[a,b]}. Indeed, the situation is completely different in higher dimensions.

First, let us remark that any Markov chain on {\mathbb{Z}}, that has {\pi} as invariant distribution and that can only make jumps of size {+1} or {-1} 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 {u_k-d_{k+1}=0} for any {k \in \mathbb{Z}} where {u_k = \pi(k) p(k,k+1)} is the upward flux at level {k} and {d_k = \pi(k)p(k,k-1)} is the downward flux. Because {\pi} is invariant it satisfies the usual balance equations,

\displaystyle  \begin{array}{rcl}  u_k+d_k &=& \pi(k) = \pi(k-1)p(k-1,k)+\pi(k+1)p(k+1,k)\\ &=& u_{k-1} + d_{k+1} \end{array}

so that {u_{k}-d_{k+1} = u_{k-1}-d_{k}}. Interating we get that for any {n \in \mathbb{Z}} we have {u_k-d_{k+1} = u_{k-n}-d_{k-n+1}}: the conclusion follows since {\lim_{m \rightarrow \pm \infty} u(m) = \lim_{m \rightarrow \pm \infty} d(m) = 0}.

This simple result on skip-free Markov chains gives also the result for many one dimensional diffusions {dX_t = \alpha(X_t) \, dt + \sigma(X_t) \, dW_t} since under regularity assumptions on {\alpha} and {\sigma} they can be seen as limit of skip-free one dimensional Markov chains on {\epsilon \mathbb{Z}}. 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 {X} is reversible Markov chain and that we would like to sample a path conditioned on the event {X_0=a} and {X_T=b}:

  1. sample the path {X^{(a)}} between {t=0} and {t=T}, starting from {X^{(a)}_0=a}
  2. sample the path {X^{(b)}} between {t=0} and {t=T}, starting from {X^{(b)}_0=b}
  3. if there exists {t \in [0,T]} such that {X^{(a)}_t = X^{(b)}_{T-t}} then define the path {X^{\star}} by\displaystyle  X^{\star}_s = \left\{ \begin{array}{ll} X^{(a)}_s	& \text{ for } s \in [0,t]\\ X^{(b)}_s	& \text{ for } s \in [t,T], \end{array} \right. \ \ \ \ \ (2)and otherwise go back to step {1}.

Indeed, the resulting path {X^{\star}} 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 {X_0=0} and {X_{1000}=50}, with parameter {\mathop{\mathbb P}(X_{t+1}=k+1 | X_t=k) = 0.4 = 1-\mathop{\mathbb P}(X_{t+1}=k-1 | X_t=k)}.

Conditioned Birth-Death process

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 {t} such that {X^{(a)}_t = X^{(b)}_{T-t}} is extremely small, and in many cases equal to {0} 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.

Potts model and Monte Carlo Slow Down

A simple model of interacting particles

The mean field Potts model is extremely simple: there are {N} interacting particles {x_1, \ldots, x_N} and each one of them can be in {q} different states {1,2, \ldots, q}. Define the Hamiltonian

\displaystyle  H_N(x) = -\frac{1}{N} \sum_{i,j} \delta(x_i, x_j)

where {x=(x_1, \ldots, x_N)} and {\delta} is the Kronecker symbol. The normalization {\frac{1}{N}} ensures that the energy is an extensive quantity so that the mean energy per particle {h_N(x) = \frac{1}{N} H_N(x)} does no degenerate to {0} or {+\infty} for large values of {N}. The sign minus is here to favorize configurations that have a lot of particles in the same state. The Boltzman distribution at inverse temperature {\beta} on {\{1, \ldots, q\}^N} is given by

\displaystyle P_{N,\beta} = \frac{1}{Z_N(\beta)} e^{-\beta H_N(x)}

where {Z_N(\beta)} is a normalization constant. Notice that if we choose a configuration uniformly at random in {\{1, \ldots, q\}^N}, with overwhelming probability the ratio of particles in state {k} will be close to {\frac{1}{q}}. Also it is obvious that if we define

\displaystyle  L^{(N)}_k(x) = \frac{1}{N} \, \Big( \textrm{Number of particles in state }k \Big)

then {L=(L^{(N)}_1, \ldots, L^{(N)}_q)} will be close to {(\frac{1}{q}, \ldots, \frac{1}{q})} for a configuration taken uniformly at random. Stirling formula even says that the probability that {L} is close to {\nu = (\nu_1, \ldots, \nu_q)} is close to {e^{-N \, R(\nu)}} where

\displaystyle R(\nu) = \nu_1 \ln(q\nu_1) + \ldots + \nu_q \ln(q\nu_q).

Indeed {(\frac{1}{q}, \ldots, \frac{1}{q}) = \textrm{argmin} \, R(\nu)}. 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 {H_N(x)} 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 {(\frac{1}{q}, \ldots, \frac{1}{q})}) 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 {\beta_c} such that:

  • for {\beta < \beta_c} the entropy wins the battle: the most probable configurations are close to the ratio {(\frac{1}{q}, \ldots, \frac{1}{q})}
  • for {\beta > \beta_c} the energy effect shows up: there are {q} most probable configurations that are the permutations of {(a_{\beta},b_{\beta},b_{\beta}, \ldots, b_{\beta})} where {a_{\beta}} and {b_{\beta}} are computable quantities.

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

\displaystyle  h(\beta) = \lim \frac{1}{N} H_N(\textrm{most probable state})

then this function has a discontinuity at {\beta=\beta_c}. 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 {L=(L_1, \ldots, L_q)} to have an accurate picture of the system. For example, a typical Markov Chain Monte Carlo approach would run as follows:

  • choose a particle {x_i} uniformly at random in {\{1,2, \ldots, N\}}
  • try to switch its value uniformly in {\{1,2, \ldots, q\} \setminus \{x_i\}}
  • compute the Metropolis ratio
  • update accordingly.

If we do that {10^5} times for {q=3} states at inverse temperature {\beta=1.5} and for {100} particles (which is fine since we only need to keep track of the {3}-dimensional ratio vector) and plot the result in barycentric coordinates we get a picture that looks like:
Potts Model MCMC

Here I started with a configuration where all the particles were in the same states i.e ratio vector equal to {(1,0,0)}. We can see that even with {10^5} steps, the algorithm struggles to go from one most probable position {(a,b,b)} to the other two {(b,a,b)} and {(b,b,a)} – 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.

Random permutations and giant cycles

The other day, Nathanael Berestycki presented some very nice results on random permutations. Start with the identity permutation of {\mathfrak{S}_N} and compose with random transpositions: in short, at time {k} we consider the random permutation {\sigma_k = \tau_k \circ \ldots \tau_{2} \circ \tau_1} where {\tau_i} is a transposition chosen uniformly at random among the {\frac{N(N-1)}{2}} transpositions of {\mathfrak{S}_N}. How long do we have to wait before seeing a permutation that has a cycle of length greater than {\epsilon N}, where {\epsilon>0} is a fixed constant ?

The answer has been known for quite a long time (O. Schramm,2005), but no simple proof was available: the highlight of Nathanael’s talk was a short and elegant proof of the fact that we only have to wait a little bit more than {N/2} steps to see this giant cycle! The first half of the proof is such a beauty that I cannot resist in sharing it here!

1. Visual representation

First, this is more visual to represent a permutation by its cycle decomposition. Consider for example the permutation {\sigma = (1,3,4)(5,6,8,9,10,11)(2)(7) \in \mathfrak{S}_{10}} that sends {1 \rightarrow 3 \rightarrow 4 \rightarrow 1} etc… It can be graphically represented by the following pictures:

cycle structure - visual representation

cycle structure - visual representation

If this permutation is composed with the transposition {(4,6)} , this gives the following cycle structure:

two cycles merge

two cycles merge

If  it is composed with {(8,11)} this gives,

a cycles breaks down into two

a cycles breaks down into two

It is thus pretty clear that the cycle structure of {\sigma_{k+1}} can be obtained from the cycle structure of {\sigma_k} in only 2 ways:

  • or two cycles of {\sigma_k} merge into one,
  • or one cycle of {\sigma_k} breaks into two cycles.

In short, the process {\{\sigma_k\}_k} behaves like a fragmentation-coalescence processes. At the beginning, the cycle structure of {\sigma_0=\text{Id}} is trivial: there are {N} trivial cycles of length one! Then, as times goes by, they begin to merge, and sometimes they also break down. At each step two integers {i} and {j} are chosen uniformly at random (with {i \neq j}): if {i} and {j} belong to the same cycle, this cycle breaks into two parts. Otherwise the cycle that contains {i} merges with the cycle that contains {j}. Of course, since at the beginning all the cycles are extremely small, the probability that a cycles breaks into two part is extremely small: a cycle of size {k} breaks into two part with probability roughly equal to {\Big(\frac{k}{N}\Big)^2}.

2. Comparison with the Erdos-Renyi random Graph

Since the fragmentation part of the process can be neglected at the beginning, this is natural to compare the cycle structure at time {k} with the rangom graph {G(N,p_k)} where {\frac{N(N-1)}{2}p_k = k}: this graph has {N} vertices and roughly {k} edges since each vertex is open with probability {p_k}. A coupling argument makes this idea more precise:

  1. start with {\sigma_0=\textrm{Id}} and the graph {G_0} that has {N} disconnected vertices {\{1,2, \ldots, N\}}
  2. choose two different integers {i \neq j}
  3. define {\sigma_{k+1}=(i,j) \circ \sigma_k}
  4. to define {G_{k+1}}, take {G_k} and join the vertices {i} and {j} (never mind if they were already joined)
  5. go to step {2}.

The fundamental remark is that each cycle of {\sigma_k} is included in a connected component of {G_k}: this is true at the beginning, and this property obviously remains true at each step. The Erdos-Renyi random graph {G(n,p)} is one of the most studied object: it is well-known that for {c<1}, with probability tending to {1}, the size of the largest connected component of {G(N,\frac{c}{N})} is of order {\ln(N)}. Since {p_k \sim \frac{2k}{N^2}}, this immediately shows that for {k \approx \alpha N}, with {\alpha < \frac{1}{2}} the length of the longuest cycle in {\sigma_k} is of order {\ln(N)} or less. This is one half of the result. It remains to work a little bit to show that if we wait a little bit more than {\frac{N}{2}}, the first cycle of order {N} appears.

For a permutation {\sigma \in \mathfrak{S}_N} with cycle decomposition {\sigma = \prod_{j=1}^r c_i} (cycles of length {1} included), one can consider the quantity

\displaystyle  T(\sigma) = \sum_{j=1}^R |c_j|^2.

This is a quantity between {N} (for the identity) and {N^2} that measures in some extends the sizes of the cycles. In order to normalize everything and to take into account that the cycle structure abruptly changes near the critical value {\frac{N}{2}}, one can also observe this quantity on a logarithmic scale:

\displaystyle  s(\sigma) = \frac{ \ln\Big( T(\sigma)/N^2 \Big) }{\ln(N)}.

This quantity {s(\sigma)} lies between {0} and {1}, and evolves quite smoothly with time: look at what it looks like, once the time has been rescaled by a factor {N}:

fluid limit ?

fluid limit ?

The plot represented the evolution of the quantity {s\Big(\sigma(\alpha N)\Big)} between time k=0 and k=\frac{N}{2} for dimensions N=10000,20000,30000.

Question: does the random quantity s(\alpha,N)={s\Big(\sigma(\alpha N)\Big)} converge (in probability) towards a non trivial constant {S(\alpha) \in (0,1)} ? It must be the analogue quantity for the random graph {G(N,\frac{2\alpha}{N})}, so I bet it is a known result …