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Let $X$ be the solution to some stochastic differential equation

$$dX_t =b(X_t) \, dt+a(X_t) \, dW_t,\quad \forall t>0.$$

Here $b,a: \mathbb R^d \to\mathbb R^d$ are bounded and Lipschitz and $W$ denotes a real Brownian motion. Fix some $\theta>0$ (small enough). For every $\varepsilon>0$, does there exist $\delta\equiv \delta(\varepsilon)>0$ (under suitable conditions on $b,a$) such that for $x\in \mathbb R^d$ it holds

$$\mathbb P[\sup_{t-\theta\le s\le t}|X_s-x|\le \varepsilon \mid X_t=x]\ge \delta\text{?}$$

The same question is asked for more general Itô (multidimensional) process $X$, i.e. $dX_t=\alpha_t \, dt + \beta_t \, dW_t$ (with suitable $\alpha,\beta$).

PS: It follows that

$$\mathbb P[\sup_{t-\theta\le s\le t}|X_s-x|\le \varepsilon \mid X_t=x]\ge \mathbb P[\sup_{t-\theta\le s\le t}|X_s-X_{t-\theta}|\le \varepsilon/2 \mid X_t=x]$$

and

$$\mathbb E\Big[\mathbb P[\sup_{t-\theta\le s\le t}|X_s-X_{t-\theta}|\le \varepsilon/2 \mid X_t]\Big]=\mathbb P[\sup_{t-\theta\le s\le t}|X_s-X_{t-\theta}|\le \varepsilon/2] >0. $$

Can we say something about the desired inequality using these the above two inequalities?

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  • $\begingroup$ @ThomasKojar Thanks a lot for the reference. What is not clear to me is how to relate my process $X$ to a gaussian process. Could you please explain why $X$ is in this framework? $\endgroup$
    – Fawen90
    Commented Jul 7, 2023 at 15:26
  • $\begingroup$ sorry, actually I was thinking of something else. Another suggestion is to take the density route (assuming it has one) as done here math.stackexchange.com/questions/2948629/… $\endgroup$ Commented Jul 7, 2023 at 15:54
  • $\begingroup$ There they have a nice bound $$\mathbb{P}^x \left( \sup_{s \leq t} |X_s-x|\leq r \right) \geq 1-ct \sup_{|y-x| \leq r} \sup_{|\xi| \leq r^{-1}} |q(y,\xi)| \quad \text{for all $t \geq 0$}. $$ This is bad for large t. But I think the spirit of the approach of using the kernel to study the small-deviation might be useful for your problem. $\endgroup$ Commented Jul 7, 2023 at 15:55
  • $\begingroup$ @ThomasKojar Thanks for further explanation. Do you mind specifying $\mathbb P^x$? It seems to be the conditional expectation knowing $X_0=x$ for me, while my case is different. My probability is conditioning on the present, but the past $\endgroup$
    – Fawen90
    Commented Jul 7, 2023 at 19:19
  • $\begingroup$ I didn't work through finding the particular pde-kernel/symbol for your case. It might be possible though as with Brownian bridge to come back to unconditional i.e. $X_{cond,t}=X_t-\frac{t}{T}X_{T}+x$. $\endgroup$ Commented Jul 7, 2023 at 19:56

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