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Consider the SDE (stochastic differential equation) as follows:

$$dX_t=X_t\big(b(X_t)dt+a(X_t)dW_t\big)$$

where $b,a:\mathbb R\to\mathbb R$ are Lipschitz and bounded and $W$ is a real-valued Brownian motion. Under which conditions on $r>0, \theta>0$ and $b,a$, one has

$$\mathbb P\big[\exists T>0 \mbox{ such that } |X_t|<r \mbox{ for all } t\in [T,T+\theta]\big]=1?$$

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  • $\begingroup$ as soon a process has a drift, then we lose recurrence eg. see here math.stackexchange.com/questions/4570678/… for the probability of hitting a value. $\endgroup$ Commented Jul 11, 2023 at 16:29
  • $\begingroup$ @ThomasKojar I don't think recurrence is related to my desired result... Even $X_\infty$ is infinite, while it is possible to stay in some open ball for a while when $t$ is not that large... $\endgroup$
    – Fawen90
    Commented Jul 11, 2023 at 17:36
  • $\begingroup$ is $\bar{x}$ the initial condition or a generic point different from initial condition? I took it to mean a generic point. If so, then the probability of visiting a ball around it is strictly less than one if the process have a drift. $\endgroup$ Commented Jul 11, 2023 at 17:43
  • $\begingroup$ $\bar x$ is the point that we seek to satisfy this desired equality $\endgroup$
    – Fawen90
    Commented Jul 11, 2023 at 18:05
  • $\begingroup$ even in the case of $\bar{x}$ being the initial condition this doesn't seem possible. For example, suppose we set $T=\tau_{r}-\theta$, for $\tau_{r}$ the exit time from ball $B_{\bar{x}}(r)$. Then the moment we ask $T>0 \Rightarrow \tau_{r}>\theta$ for deterministic $\theta>0$, we get a reduced probability $P[\tau_{r}>\theta]<1$. $\endgroup$ Commented Jul 11, 2023 at 20:20

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