8
$\begingroup$

Let $(X_t)_{t \geq 0}$ be a diffusion process with dynamics governed by the stochastic differential equation \begin{equation} dX_t = b(X_t)dt + \sigma(X_t)dW_t, ~~ X_0 = x_0, \end{equation} where $b,\sigma$ are Lipschitz and $(W_t)_{t \geq 0}$ is a standard $d$-dimensional Wiener process. Assume $(X_t)_{t \geq 0}$ possesses a unique invariant probability measure $\mu(\cdot)$.

The Euler-Maruayama approximation of this SDE with step size $h$ is a Markov chain defined recursively as $Y_0 = x_0$, and for $n \in \mathbb{Z}^+$ \begin{equation} Y_{(n+1)h}|Y_{nh} = Y_{nh} + b(Y_{nh})h + \sqrt{h}\sigma(Y_{nh})\varepsilon, ~~ \varepsilon \sim \mathcal{N}(0,I_{d \times d}). \end{equation}

I can see that under certain conditions $\{Y_i\}_{i \in \mathbb{Z}^+}$ will also possess a unique invariant probability measure, $\mu^h(\cdot)$. I've seen certain results discussing how the ergodicity properties of $(X_t)_{t \geq 0}$ and $\{Y_i\}_{i \in \mathbb{Z}^+}$ can differ (e.g. scenarios in which the first can be exponentially ergodic but the second transient, see: http://projecteuclid.org/euclid.bj/1178291835).

What I've been unable to find are results discussing how $\mu^h(\cdot)$ and $\mu(\cdot)$ differ. Can we make some statement about how 'close' the two measures are, in total variation \begin{equation} \| \mu^h(\cdot) - \mu(\cdot) \|_{TV} \end{equation} or some other appropriate distance. And can we say anything about the distance \begin{equation} \|\mathbb{E}_{\mu^h}f - \mathbb{E}_{\mu}f\| \end{equation} for some suitable class of functions $f$, assuming both processes are defined on the same space which has norm $\|\cdot\|$? I've searched around online a bit, the closest I've come is here http://www.newton.ac.uk/preprints/NI03065.pdf, where the result is alluded to, which makes me think it might be fairly well known. So if anyone could point me in the right direction or explain why it's obvious then that would be great, thanks!

Apologies if this isn't an appropriate question for overflow, also asked here on math.stackexchange https://math.stackexchange.com/questions/751991/invariant-measure-of-euler-maruyama-discretisation-of-an-ito-diffusion

$\endgroup$

4 Answers 4

3
$\begingroup$

Here I use the notation given in the question above, and the statements made assume the domain of the SDE problem is unbounded, which seems to be the context of the question.

Assume the following conditions hold.

  1. (Regularity) Every derivative of $b(x)$ and $\sigma(x)$ exists and is bounded, and $\sigma(x)$ itself is bounded.
  2. (Ellipticity) The matrix $\sigma(x) \sigma(x)^T$ is positive definite for all $x \in \mathbb{R}^d$.
  3. (Weak Dissipativity) There exist $\beta >0$ and $\alpha \ge 0$ such that $$ b(x)^T x \le - \beta |x|^2 + \alpha $$ for all $x \in \mathbb{R}^d$.

Let $h$ be the time step size. Under these assumptions Talay and Tubaro 1990 prove that there exists $C>0$ such that $$ | \mathbb{E}_{\mu} f - \mathbb{E}_{\mu_h} f| \le C h $$ for all smooth functions $f$ with at most polynomial growth at infinity and for $h$ sufficiently small.

In other words, the weak accuracy of the Euler-Maruyama scheme at finite-time can be extended to infinite-time horizons. The proof uses an expansion of the global error of the Euler-Maruyama scheme that is commonly referred to as a Talay-Tubaro expansion. (Caveat: when the derivative of the drift is not bounded, the moments of the Euler-Maruyama scheme may diverge on finite-time intervals.)

I am not aware of a similar estimate for the TV distance: $$ \| \mu - \mu_h \|_{\text{TV}} = \sup\left\{ \left| \int_{\mathbb{R}^d} f \; d \mu - \int_{\mathbb{R}^d} f \; d \mu_h \right| ~~\text{s.t.}~~ \begin{array}{c} f: \mathbb{R}^d \to \mathbb{R} \;, \\ |f(x)| \le 1 ~\text{for all $x \in \mathbb{R}^d$} \end{array} \right\} \;. $$ The issue with making the desired statement in the TV distance is that the supremum in this definition is taken over functions which are not necessarily smooth, and so, the derivatives of $\mathbb{E}_x f(X_t)$ (where $\mathbb{E}_x$ denotes expectation conditional on $X_0=x$) may become unbounded if $t$ is small enough. A finite-time, TV error estimate for Euler-Maruyama is available in Lemma 4.2 of Bou-Rabee and Hairer 2013.

$\endgroup$
5
  • $\begingroup$ Thanks, this is just what I was looking for. Very much appreciate it! $\endgroup$ Mar 9, 2015 at 15:33
  • $\begingroup$ Old question, I know, but instead of the relation between $\mu$ and $\mu^h$, I am more interested in a formula for $\mu^h$. We can give such a formula for $\mu$ by considering the adjoint of the generator of the diffusion applied to the density of $\mu$. But I don‘t know how to deal with $\mu^h$. $\endgroup$
    – 0xbadf00d
    Jul 1, 2023 at 14:08
  • 1
    $\begingroup$ @0xbadf00d in general, there is no explicit formula for $\mu^h$; see, e.g., arxiv.org/abs/2108.00682 for a well-written and authoritative account of the state-of-the-art $\endgroup$ Jul 3, 2023 at 15:35
  • $\begingroup$ @NawafBou-Rabee Thank you for the reference; I will take a look. My idea was: If $\mu^h$ has an invariant measure, shouldn't it be a Gaussian distribution as well? In that case, it would suffice to determine mean and variance. I've asked for that and explained it further here: math.stackexchange.com/q/4728844/47771. Maybe you can take a look. $\endgroup$
    – 0xbadf00d
    Jul 3, 2023 at 19:24
  • $\begingroup$ Obviously not: it is only Gaussian in the additive Ornstein-Uhlenbeck case. $\endgroup$ Jul 3, 2023 at 20:59
1
$\begingroup$

The paper Rational Construction of Stochastic Numerical Methods for Molecular Sampling does this for a family of numerical methods. Besides, the next to last paragraph of Section 1 points to relevant references.

$\endgroup$
0
1
$\begingroup$

This paper might be relevant: http://arxiv.org/abs/0908.4450

$\endgroup$
1
  • $\begingroup$ Certainly looks to be, I'll take a look at that one too, thanks. $\endgroup$ Apr 15, 2014 at 19:47
0
$\begingroup$

If instead of TV, you work in $W_2$, then isnt Proposition $3.3$ here https://arxiv.org/pdf/1702.03849.pdf giving such a statement about convergence of distributions for their SDE (as given in their equation $1.4$) and its discretization (as given in their equation $2.2$) ?

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.