Markov process on a torus with prescribed invariant distribution

Question

In Euclidean space, $\mathbb R^d$, the Langevin diffusion $${\rm d}X_t=b(X_t){\rm d}t+\sigma(X_t){\rm d}W_t\tag1,$$ where $\sigma:\mathbb R^d\to\mathbb R^{d\times k}$, $$b:=\frac{\Sigma+U}2\nabla\ln p+\frac12\nabla\cdot\Sigma,\tag2$$ $\Sigma:=\sigma\sigma^\ast$ and $(W_t)_{t\ge0}$, $U\in\mathbb R^{d\times d}$ is any anti-symmetric matrix is a $\mathbb R^k$-valued standard Wiener process, has an unique invariant measure with density $p$ with respect to the Lebesgue measure on $\mathbb R^d$.

Now, in my application, I'm working on $[0,1)^d$ instead. So I have a given density $p:[0,1)^d\to[0,\infty)$. In my application, it works nicely to consider $[0,1)^d$ with toroidal boundary and the standard MCMC sampling approach is Metropolis-Hastings with proposal kernel given by the wrapped normal distribution.

What I would like to do now is defining a (continuous-time) Markov process, with values in $[0,1)^d$, with unique invariant measure with density $p$ with respect to the restriction of the Lebesgue measure on $\mathbb R^d$ to $[0,1)^d$. Is this possible?

One might think that we can simply use $(1)$ with toroidal wrapping, but as this paper shows, the resulting process is not even Markov anymore.

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EDIT1

I think the question is trickier than one might think at first glance. It seems like the Markov property of $(1)$ will not be maintained under the transformation by $$\iota:\mathbb R^d\to[0,1)^d\;,\;\;\;x\mapsto x-\lfloor x\rfloor\tag3.$$ A counterxample is given here.

However, we got the following general result: Assume $(E,\mathcal E)$ is a measurable space and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$. Now, given another measurable space $(\tilde E,\tilde{\mathcal E})$ and an $(\mathcal E,\tilde{\mathcal E})$-measurable $\varphi:E\to\tilde E$ satisfying

1. $\varphi(E)=\tilde E$;
2. If $t\ge0$ and $\tilde f:\tilde E\to\mathbb R$ is bounded and $\tilde E$-measurable, then $$\kappa_t(\tilde f\circ\varphi)=\tilde g\circ\varphi\tag4$$ for some bounded $\tilde{\mathcal E}$-measurable $\tilde g:\tilde E\to\mathbb R$,

then $\tilde g$ is uniquely determined and hence $$\tilde\kappa_tf:=\tilde g\tag5$$ is well-defined. If now $(X_t)_{t\ge0}$ is a Markov process, with respect to some filtration $(\mathcal F_t)_{t\ge0}$, with transiton semigroup $(\kappa_t)_{t\ge0}$, then $$\tilde X_t:=\varphi\circ X_t\;\;\;\text{for }t\ge0$$ is a Markov process with respect to the same filtration $(\mathcal F_t)_{t\ge0}$ with transition semigroup $(\tilde\kappa_t)_{t\ge0}$.

We clearly would want to apply this for $E=\mathbb R^d$, $\tilde E=[0,1)^d$, $\varphi=\iota$ and $(X_t)_{t\ge0}$ given by $(1)$. However, with that I end up with $$\operatorname E\left[f\left(\tilde X_{s+t}\right)\mid\tilde X_s\right]=\operatorname E\left[\left(\kappa_t(f\circ\iota)\right)(X_s)\mid\tilde X_s\right]\tag6.$$ But this will not be possible in general.

EDIT 2:

If $X$ is a Lévy process, we can actually show that $\tilde X$ is again a time-homogeneous Markov process. The fact that the increments of $X$ are independent in that case, is crucial for the proof. Now, a Langevin-like process $X$ - at least in the form $(1)$ + $(2)$ - will not be a Lévy process. So, the currently given answer does actually not answer my question. We still don't verified that there is a Markov process with prescribed distribution. While the paper referenced in the answer seems to claim the existence, it is unfortunately not accessible to me due to the heavy amount of differential geometry I'm not aware of. Also, while it's nice to have an expression for the generator, it is still not sufficient to simulate it on a computer, for which I would need to know the precise shape of the transition semigroup. I'm sure that in the present, way simpler setting, a direct construction is possible. Any advice and or answer on that would be highly appreciated.

EDIT 3

Meanwhile I was able to prove that the toroidal wrapping of any Lévy process is a Markov process and I could identify its transition semigroup in terms of the transition semigroup of the original process. However, the open question is whether the same can be shown for the toroidal wrapping of a general SDE (under the condition of a periodic drift and a periodic diffusion coefficient); at least for the one with Langevin drift and constant diffusion coefficient.

Answer 1

The main result of Barp et al. (2021) contains as special case the answer to your question. In a nutshell, given a (possibly non-normalized) target density on the $d$-torus $p: \mathbb{T}^d \to \mathbb{R}$, the general diffusion process that leaves it invariant has infinitesimal generator defined by: $$ \mathcal{L} f(x) = \color{blue}{b(x) \cdot \nabla f(x)} + \color{red}{(\Sigma(x) \nabla p(x) + \operatorname{div}(\Sigma(x)) )\cdot \nabla f(x) } + \color{red}{\operatorname{trace}\left(\Sigma(x) D^2 f(x)\right)} $$ where blue highlights the non-reversible part involving the vector field $b: \mathbb{T}^d \to \mathbb{R}^d$ which is divergence-free wrt to the target distribution (i.e., $\operatorname{div}(p b) \equiv 0$) and red highlights the reversible part of the diffusion (that is, reversible wrt the target).

An earlier work that derives a general SDE with specified target on $\mathbb{R}^d$ is Ma, Chen, Fox (2015).

Addendum

Re Markov property and ergodicity of BM on a flat torus $\mathbb{T}$ identified with $[0,1)$ (this question came up in one of the comments below), let $(B_t)_{t \ge 0}$ be standard BM on $\mathbb{R}$. Then as the OP writes, the process $(X_t)_{t \ge 0}$ defined by $X_t = B_t - \lfloor B_t \rfloor \in [0,1)$ is a BM on a flat torus. Analogously to the wrapped normal distribution, the transition probability density function of $X_t$ given $X_0 = x \in [0,1)$ is explicitly $$ p_t(x,y) = \frac{1}{\sqrt{2 \pi t}} \sum_{k \in \mathbb{Z}} e^{-|x-y-k|^2/(2t)} \mathbb{1}_{[0,1)}(x) \mathbb{1}_{[0,1)}(y) $$ for any $t>0$. As a byproduct, clearly the process $(X_t)_{t \ge 0}$ is Markovian and irreducible.

Now, for any $t>0$, owing to the regularity of $p_t$ (it is smooth with bounded derivatives), the corresponding transition kernel is strong Feller (as in standard BM on $\mathbb{R}$), in the sense that $x \mapsto \int_{[0,1)} p_t(x,y) f(y) dy $ is a continuous and bounded function for all bounded and measurable functions $f : [0,1) \to \mathbb{R}$. The process also leaves invariant the uniform distribution on $[0,1)$ in the sense that $$ \int_{[0,1)} \int_{(a,b)} p_t(x,y) dy dx = \int_a^b \left( \frac{1}{\sqrt{2 \pi t}} \int_{\mathbb{R}} e^{-|x-y|^2/(2t)} dy \right) dx = b-a $$ for all $a,b \in [0,1)$ such that $b>a$. (In fact, the transition density is reversible w.r.t. the uniform distribution.) Combining this with the strong Feller property, we obtain uniqueness of the invariant measure. (See, e.g., Corollary 2.7.)

Answer 2

Meanwhile, I can give a canonical answer. I will omit proofs in the following, but every theorem below can be proven without further assumptions by standard arguments.

Let

• $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $\mathbb R^d$ and $$\tilde\kappa_t(x,\;\cdot\;):=\kappa_t(x,\;\cdot\;)\circ\iota^{-1}\;\;\;\text{for }x\in[0,1)^d\text{ and }t\ge0;$$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
• $(X_t)_{t\ge0}$ be an $\mathbb R^d$-valued process on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname E\left[f(X_{s+t})\mid X_s\right]=(\kappa_tf)(X_s)\tag a$$ for all bounded Borel measurable $g:\mathbb R^d\to\mathbb R$ and $s,t\ge0$ and $$W:=\iota\circ X.$$

Remark: $(\tilde\kappa_t)_{t\ge0}$ is not necessarily a semigroup and $(a)$ does not imply that $X$ is a Markov process.

Theorem 1 Let $s,t\ge0$ and $f:[0,1)^d\to\mathbb R$ be bounded and Borel measurable $$\operatorname E\left[g(W_{s+t})\mid W_s\right]=(\tilde\kappa_tg)(W_s).$$

That is, if $W$ would be a Markov process (which it is not in general) and $(\tilde\kappa_t)_{t\ge0}$ is a semigroup, then the transition semigroup of $W$ would be $(\tilde\kappa_t)_{t\ge0}$. While $W$ is not a Markov process in general, it always holds that $W_{s+t}$ and $\mathcal F^W_s$ are conditionally independent given $X_s$; i.e. $$\operatorname E\left[f(W_{s+t}\mid\mathcal F^W_s\right]=\operatorname E\left[f(W_{s+t})\mid X_s\right]\tag b.$$

Now, let

• $\tilde b:[0,1)^d\to\mathbb R^d$ be Borel measurable and $$b:=\tilde b\circ\iota;$$
• $k\in\mathbb N$;
• $\sigma:[0,1)^d\to\mathbb R^{d\times k}$ be Borel measurable and $$\sigma:=\tilde\sigma\circ\iota;$$
• $B$ be an $\mathbb R^k$-valued Wiener process on $(\Omega,\mathcal A,\operatorname P)$ with covariance operator $\operatorname{id}_{\mathbb R^k}$.

Assume there is a continuous process $(X^x_t)_{(t,\;x)\in[0,\;\infty)\times\mathbb R^d}$ on $(\Omega,\mathcal A,\operatorname P)$ with $$X^x_t=x+\int_0^tb(X^x_s)\;{\rm d}s+\int_0^t\sigma(X^x_s)\;{\rm d}B_s\;\;\;\text{for all }t\ge0\tag c$$ almost surely for all $x\in\mathbb R^d$. We can easily prove that $$(X_{s+t})_{t\ge0}\sim X^{X_s}\tag d$$ for all $s\ge0$ and $$X^{x+k}=k+X^x\;\;\;\text{for all }k\in\mathbb Z^d\text{ almost surely}\tag e$$ for all $x\in\mathbb R^d$. Assuming that $X_0$ is $\mathcal F^B$-measurable (or $B$ is actually a Brownian motion with respect to a larger filtration $\mathcal F$ such that $X_0$ is $\mathcal F_0$-measurable), we et the following result:

Theorem 2 Let $s,t\ge0$ and $f:[0,1)^d\to\mathbb R$ be bounded and Borel measurable. Then, $$\operatorname E\left[f(W_{s+t})\mid\mathcal F^W_s\right]=\operatorname E\left[W_{s+t}\mid W_s\right].\tag f$$

For the proof we are using that $(e)$ implies $$X^{X_s}=X^{W_s}\;\;\;\text{almost surely}\tag g.$$ Moreover, we are using $(d)$ and $$\operatorname E\left[g(X^{W_s}_t)\mid X_s\right]=\left.\operatorname E\left[g(X^x_t)\right]\right|_{x\;=\;W_s}\;\;\;\text{almost surely}\tag h$$ for all bounded Borel measurable $g:\mathbb R^d\to\mathbb R$. Given that, we obtain \begin{equation}\begin{split}\operatorname E\left[f(W_{s+t})\mid W_s\right]&=\operatorname E\left[\operatorname E\left[f(W_{s+t})\mid X_s\right]\mid W_s\right]\\&=\operatorname E\left[\operatorname E\left[(f\circ\iota)(X^{W_s}_t)\mid X_s\right]\mid W_s\right]\\&=\operatorname E\left[\left.\operatorname E\left[(f\circ\iota)(X^x_t)\right]\right|_{x\;=\;W_s}\mid W_s\right]\\&=\left.\operatorname E\left[(f\circ\iota)(X^x_t)\right]\right|_{x\;=\;W_s}\\&=\operatorname E\left[(f\circ\iota)(X^{W_s}_t)\mid X_s\right]\\&=\operatorname E\left[(f\circ\iota)(X^{X_s}_t)\mid X_s\right]\\&=\operatorname E\left[f(W_{s+t})\mid X_s\right].\end{split}\tag h\end{equation}

Finally, let $A$ and $\tilde A$ denote the generator (with respect to the space of bounded Borel measurable functions on $\mathbb R^d$ and $[0,1)^d$, respectively, equiiped with the supremum norm) of $X$ (i.e. of $(\kappa_t)_{t\ge0}$) and $W$ (i.e. of $(\tilde\kappa_t)_{t\ge0}$), respectively. Noting that every $\tilde f\in C_c^2([0,1)^d)$ can be extended by 0 to a function in $f\in C_c^2(\mathbb R^d)$, we easily see that $$\tilde A\tilde f=\left.(Af)\right|_{[0,\;1)^d}\tag i.$$