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Let $n\in\mathbb N$, $B$ a $n$-dimensional Brownian motion, $\sigma_t$ a positive (deterministic) caglad (LCRL) function in $\mathcal L^2([0,\infty[)$ and $f\in\mathcal C^2(\mathbb R^n\to\mathbb R)$ a nonnegative function taking its global minimum at least at one point. Further $f$ has a Lipschitz continuous gradient and $\lim_{||x||\to\infty}f(x)=\infty$. We define the stochastic process $X$ by the SDE $$ dX_t=-\nabla f(X_t)dt+\sigma_tdB_t,\qquad t\geq0\\ X_0=\xi. $$ I already proofed that a unique solution exist and the process $\nabla f(X_t)$ converges a.s. to zero and $f(X_t)$ is convergent a.s. to a finite value. By looking on the SDE it is absolutely obvious that this finite value cannot be the point of a maximum of $f$, since the process is nowhere constant (which can be easily proofen) and then the process gets pushed away of a maximum point by the first term. But I can't create a contradiction when I assume that it is a maximum point. Can anyone give a hint how to proof it?


I already found an example in which the process $X$ got stuck in a saddle point with probability unequal zero. So it is not possible to poof directly that it converges to a minimum point.

If $\sigma$ were zero it would be easy to show, since we then have the deterministic case in which $f(X_t)$ is differentiable. I also tryed to approximate the process $X$ by the process $$ dY^{t_0}_t=-\nabla f(Y^{t_0}_t)dt,\qquad t\geq t_0\\ Y^{t_0}_{t_0}=X_{t_0}, $$ of which I know that it cannot converge to a maximum, since it is monotonically decreasing. But it wasn't succesfull. A fruther idea is this (for the one dimensional case): As we know, that $f(X_t)$ converges a.s., let denote $Y$ its limit. Further we know $\nabla f(X_t)\to0$ a.s. Assume we have a set of meassure greater than zero on which $Y$ is a maximum point. We now consider such a path. With Itô we get $$ (f(X_t)-Y)^2=(f(X_0)-Y)^2+\int_0^t (f(X_s)-Y)(\sigma_s^2 f''(X_s)-2f'(X_s)^2)+\sigma_s^2 f'(X_s)^2\mathrm ds\\ +2\int_0^t \sigma_s f'(X_s)(f(X_s)-Y)\mathrm dB_s $$ Now for $t$ larger than a $t_0$ we can assume, that the integrand in the first integral is positve, since the process is a.s. not constant. As the LHS has to decrease I want to create a contradiction. But I can't control the Itô integral. I know that it is a $\mathcal L^2$ Martingale and thus convergent, but I don't see that this is enough. Anyway a more general way than using the martingale property would be desirable (but not necessary).

If there are further assumptions necessary to show the desired, we can assume it. Any help would be appreciated.

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  • $\begingroup$ I don't understand. In one sentence you claim that $f(X_t)$ converges a.s. to a finite value. In the next you say that $X_t$ "is nowhere constant". Aren't these two claims in contraddiction? Are you sure that $f(X_t)$ is converging almost surely? $\endgroup$
    – Kore-N
    Aug 4, 2021 at 12:10
  • $\begingroup$ Also, are you mixing up maxima and minima of $f$? I guess you would like to find the minimum of $f$, given the way you wrote the problem. $\endgroup$
    – Kore-N
    Aug 4, 2021 at 12:12
  • $\begingroup$ The convergence of $f(X_t)$ doesn't contradict that $X_t$ is nowhere constant. Just think about $1/t\to0(t\to\infty)$ which is also not constant. You could also think about that $f$ has level sets and $X_t$ might move on a level set, thus $f(X_t)$ is constant. And yes, I'm pretty sure about that $f(X_t)$ is converging a.s. The proof is just to long to write it here. $\endgroup$
    – mag
    Sep 6, 2021 at 9:38
  • $\begingroup$ To your second comment: Yes, I would like to find a minimum $f$. But as it might be possible that the process could be captured by a saddle point, it would be nice to know a proof that $X_t$ is at least not in the neighbourhood of a maximum point. Maybe you can telll me at which point I'm mixing up maxima and minima. $\endgroup$
    – mag
    Sep 6, 2021 at 9:47

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