Consider a deterministic dynamical system on $mathbb{R}^n$ defined by the recurrence $x_{t+1} = f(x_t)$.

Suppose the dynamical system is stable in the following sense: there exists a $Q : mathbb{R}^n rightarrow mathbb{R}_{geq 0}$ and $lambda in (0, 1)$ such that for all $x$ we have $Q(f(x)) leq lambda Q(x)$. Furthermore, for simplicity of this post, let us also assume that (a) $Q(x) geq psi | x|_2^2$ for all $x$ and (b) $Q$ is twice differentiable and its Hessian $nabla^2 Q(x)$ is uniformly upper bounded: $| nabla^2 Q(x) |_{mathrm{op}} leq L$ for all $x$.

Now consider the Markov chain ${X_t}_{t geq 0}$ given by the transition $X_{t+1} = f(X_t) + varepsilon_t$, where each $varepsilon_t$ is sampled independently across time from a distribution that is absolutely continuous wrt Lebesgue measure on $mathbb{R}^n$. Let $P$ denote the transition kernel of this Markov chain.

From standard results in Markov chain theory we know that there exists an invariant measure $pi$ and constants $R > 0$ and $r in (0, 1)$ such that for all measures $mu$:

$$

| mu P^t – pi |_{mathrm{TV}} leq R r^{t} (1 + mu(Q)), :: mu(Q) = int Q(x) mu(dx), :: t=0,1,2,….

$$

The standard way to check this is to verify a Lyapunov drift condition.

For simplicity let us assume that $varepsilon_t$ is an isotropic Gaussian in $mathbb{R}^n$. Then we use the Lyapunov stability property of $Q$, in addition to the assumptions (a) and (b) above, to argue:

$$

begin{align*}

mathbb{E}( Q(X_{t+1}) | X_t ) &leq mathbb{E}( Q(f(X_t)) + frac{L}{2} | varepsilon_t |_2^2 |X_t ) \

&leq lambda Q(X_t) + frac{L n}{2} leq frac{1+lambda}{2} Q(X_t) – frac{(1-lambda)psi}{2} |X_t|_2^2 + frac{Ln}{2}.

end{align*}

$$

Therefore, we have the drift condition:

$$

begin{align*}

&mathbb{E}( Q(X_{t+1}) | X_t ) leq frac{1+lambda}{2} Q(X_t) + frac{Ln}{2} mathbf{1}{X_t in C}, \

&C := left{ x in mathbb{R}^n : |x|_2 leq sqrt{frac{Ln}{psi(1-lambda)}} right}.

end{align*}

$$

It remains to check that $C$ is small set. In particular, we need to find a probability measure $nu$ and positive constant $eta > 0$ such that:

$$

inf_{x in C} P(x, A) geq eta nu(A) :: forall mathcal{B}(A).

$$

The standard way of doing this (see e.g. Section 5.3.5 of Meyn and Tweedie) is to set $nu(A) = frac{mu(A cap C)}{mu(C)}$, with $mu$ as the $n$-dimensional Lebesgue measure.

Writing $C = { x : |x|_2 leq R }$ and letting $q$ denote the density of an $n$-dimensional isotropic Gaussian, we have:

$$

begin{align*}

inf_{x in C} P(x, A) &= int_{A} q(f(x) – y) : dy \

&geq int_{A cap C} q(f(x) – y) : dy \

&geq inf_{x in C, y in C} q(f(x) – y) mu(A cap C) \

&= inf_{x in C, y in C} q(f(x) – y) cdot mu(C) cdot nu(A).

end{align*}

$$

This means we can take $eta = inf_{x in C, y in C} q(f(x) – y) cdot mu(C)$,

and since $q(cdot)$ is continuous and $C$ compact, we have that $eta > 0$, demonstrating $C$ is a small set.

This puts us in a position to recover the geometric ergodicty claim.

The problem here though is that the $eta$ constant will scale like $exp{ -R^2 }$, and since $R$ scales as $sqrt{n}$, this gives us that $eta asymp exp{-n}$.

This in turn tells us that the ergodicity rate $r$ will scale like $1 – e^{-n}$ (see e.g. Theorem 1.3 of Hairer and Mattingly).

**Question:** Is this exponential dependence in the ergodicity rate a limitation of the analysis, or is it actually unavoidable in general?