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As an illustrative example, consider a modified O-U process $dX_t = -X_tdt + \exp(-t)dW_t$. It is not too hard to understand that after a while the behaviour is dominated by the deterministic component, which will drive the process to $0$ almost surely (for instance by showing that its asymptotic mean and variance converge to 0).

My question is the following: are there any results on proving such a convergence from the generator directly? Formally, I have a (as well-behaved as you wish) generator $\mathcal{A}_t f(x) = \lim_{s \to 0+}\frac{\mathbb{E}[f(X_{t+s}) \mid X_t=x] - f(x)}{s}$, and a known value $x^*$, for which I want to verify that $X_t \to x^*$ a.s.. Can this be done directly by checking some properties of $\mathcal{A}_t$.

My first idea was to simply find conditions under which $\mathcal{A}_t$ converges to a $\mathcal{A}_{\infty}$ for which the Dirac $\delta_{x^*}$ is invariant. This is of course insufficient: it may happen that the decay in the process is too fast to eventually reach the desired asymptotic (imagine adding a $\exp(-t)$ in the drift of the OU above).

For context, I am trying to understand the convergence of continuous-time approximations of stochastic gradient descent (see, e.g., https://jmlr.org/papers/v20/17-526.html) in terms of their generator.

Thanks!

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  • $\begingroup$ In your first paragraph you claim that in order to show convergence to zero almost surely it is enough to show the asymptotic mean and variance tend to zero, how do you go about showing this? $\endgroup$ Commented Jan 13, 2024 at 17:27
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    $\begingroup$ The distribution at time t is Gaussian, so it's fully characterised by its mean and covariance $\endgroup$ Commented Jan 14, 2024 at 18:19
  • $\begingroup$ This characterizes the distribution yes but why does this necessarily imply the paths of the process will tend to zero almost surely, in general this need not be the case ? $\endgroup$ Commented Jan 14, 2024 at 18:48
  • $\begingroup$ It would be of great interest if there was a way to characterize convergence to zero almost surely by looking only at the asymptotic moments but I wouldn’t be sure if this can be done in general (by in general I mean even if you have an OU process similar to the above but with a general time dependent function in the diffusion rather than a decaying exponential) , this was why I was wondering if you had any insight after your comment in the first paragraph. $\endgroup$ Commented Jan 14, 2024 at 18:51
  • $\begingroup$ In this particular case the standard strong law of large numbers for Brownian motion gives convergence to zero almost surely as the solution is just a Brownian motion times a decaying exponential, but if one could obtain convergence to zero almost surely by ONLY using the asymptotic moments that would be very interesting indeed! Is this just a general fact about continuous Gaussian processes perhaps ? $\endgroup$ Commented Jan 14, 2024 at 18:53

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Set initial condition $X_{0}=x_{0}=x^{*}=0$ for simplicity. We can identify its long-term behaviour by studying its density $p(t,x)$ and thus the corresponding Fokker-Plank equation (in terms of the adjoint of the generator $A^{*}$)

$$\partial_{t}p(t,x)=A^{*}p(t,x)=\partial_{x}(-xp(t,x))-\frac{1}{2}\partial_{xx}(e^{-2t}p(t,x))$$ with initial data $p(x,0)=\delta_{x}$.

I plan to return to try to solve this or at least use parabolic estimates to get limiting behaviour, but I am not sure how doable it is especially because of the $e^{-2t}$ factor (here they discuss the similar issue of time-dependent coefficients). Did you also want an exact solution?

If you also want an SDE perspective on more general drifts see "An SDE perspective on stochastic convex optimization" where they study invariant measure limits.

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  • $\begingroup$ Thanks I'll check the ref! $\endgroup$ Commented Oct 4, 2023 at 6:04

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