Strong Stationary Duality for Diffusion Processes

Abstract

We develop the theory of strong stationary duality for diffusion processes on compact intervals. We analytically derive the generator and boundary behavior of the dual process and recover a central tenet of the classical Markov chain theory in the diffusion setting by linking the separation distance in the primal diffusion to the absorption time in the dual diffusion. We also exhibit our strong stationary dual as the natural limiting process of the strong stationary dual sequence of a well-chosen sequence of approximating birth-and-death Markov chains, allowing for simultaneous numerical simulations of our primal and dual diffusion processes. Lastly, we show how our new definition of diffusion duality allows the spectral theory of cutoff phenomena to extend naturally from birth-and-death Markov chains to the present diffusion context.

Appendix: Details of Theorem 4.2

In proving Theorem 4.1, we made use of Theorem 4.2 adapted from [8], Theorems 4.8.2 and 1.6.5 and Corollary 4.8.9]. We restate the theorem here:

Theorem 4.2 Let $A$ be the generator (as in Sect. 2 ) of a regular diffusion process $Y$ with state space $\mathcal {Y}$. Assume $h_n > 0$ converges to $0$ as $n \rightarrow \infty$. Let $X^{n}\sim (\pi _0^{n},P^{n})$ be a Markov chain on metric state space $\mathcal {Y}^{n} \subset \mathcal {Y}$ and define $Y^{n}_t := X^{n}_{\lfloor t/h_{n} \rfloor }$. Further assume $Y^{n}_0\Rightarrow Y_0$. Letting $\mathcal {B}(\mathcal {Y}^n)$ be the space of real-valued bounded measurable functions on $\mathcal {Y}^n$, define $T^{n}:\mathcal {B}(\mathcal {Y}^{n})\rightarrow \mathcal {B}(\mathcal {Y}^n)$ via

$$\begin{aligned} T^{n} f(x)={\mathbb {E}}_x f(X^{n}_1). \end{aligned}$$

Let $\rho _{n}:C(\mathcal {Y})\rightarrow \mathcal {B}(\mathcal {Y}^{n})$ be defined via $\rho _{n}f(\cdot )=f |_{\mathcal {Y}^{n}}(\cdot )$. If ${\mathcal {D}}_{A}$ is an algebra that strongly separates points, and

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{y\in \mathcal {Y}^{n} }\left| (A^{n}\rho _{n} f)(y) - (A f)(y) \right| =0 \end{aligned}$$

(7.1)

for all $f\in {\mathcal {D}}_{A}$, then $Y^n \Rightarrow Y$.

The purpose of this appendix is to carefully spell out the proof of the above theorem, as the notation in [8] differs considerably from the notation we have adopted. The following chart gives the notational equivalences between the present work and [8]; in connection with $\mu _n(x, \cdot )$, see Corollary 4.8.5 in [8].

Notation in present work:	Notation in [8]:
$\mathcal {Y}$ with the Euclidean metric	$(E, r)$
$\mathcal {Y}^{n},\ A^{n},\ T^{n}$	$E_n,\ A_n,\ T_n$
$P^{n}(x,\cdot )$	$\mu _n(x,\cdot )$
$\{(f,Af)\,|\,f\in {\mathcal {D}}_{A}\}$	$A$
${\mathcal {D}}_{A}$	${\mathcal {D}}_{A}$
$C({\mathcal {S}})$	$L$
$1 / h_n$	$\alpha _n$
$\text {id}$	$\eta _n$
$\rho _{n}$	$\pi _n$

Here $\rho _{n}:C(\mathcal {Y})\rightarrow \mathcal {B}(\mathcal {Y}^{n})$ is defined via $\rho _{n}f(\cdot )=f |_{\mathcal {Y}^{n}}(\cdot )$, and $\text {id}:\mathcal {Y}^{n}\rightarrow \mathcal {Y}$ is the inclusion function embedding $\mathcal {Y}^{n}$ into $\mathcal {Y}$.

Proof of Theorem 4.2

Clearly $C(\mathcal {Y})$ is convergence determining, and by considering suitably smooth uniform approximations in ${\mathcal {D}}_{A}$ to the indicator function of $\{x\}$ for each $x\in \mathcal {Y}$, it follows that ${\mathcal {D}}_{A}\subset C(\mathcal {Y})$ is an algebra that strongly separates points. In the notation of [8], Corollary 4.8.9], we have $G_n=E_n=\mathcal {Y}^n$ , and so to prove $Y^n\Rightarrow Y$, it suffices to prove that for each $T > 0$ and $f \in C(\mathcal {Y})$ we have

$$\begin{aligned} \lim _{n \rightarrow \infty }\sup _{y\in \mathcal {Y}^{n}}\left| (T^{n})^{\left\lfloor t / h \right\rfloor } \rho _{n} f(y)-\rho _{n} T_t f(y)\right| , \qquad 0\le t\le T. \end{aligned}$$

(7.2)

From [8], Theorem 1.6.5], to prove (7.2) it suffices to establish that for all $f\in {\mathcal {D}}_{A}$ we have that $\rho _{n} f\in \mathcal {B}(\mathcal {Y}^{n}) (=L_n$ in the notation of [8], Theorem 1.6.5]) satisfies

$$\begin{aligned} \lim _{n \rightarrow \infty } \sup _{y\in \mathcal {Y}^{n} }\left| \rho _{n} f(y) - f(y) \right| = 0 \end{aligned}$$

(7.3)

and

$$\begin{aligned} \lim _{n \rightarrow \infty } \sup _{y\in \mathcal {Y}^{n} }\left| (A^{n}\rho _{n} f)(y) - (A f)(y) \right| =0. \end{aligned}$$

(7.4)

But (7.3) is clearly true, and (7.4) is assumed (for all $f\in {\mathcal {D}}_{A}$) in the statement of Theorem 4.2. $\square$