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Let $\sigma:\mathbb R_+\times\mathbb R\to [1,2]$ be measurable. Consider the SDE $dX_t = \sigma(t,X_t)dW_t$, where $X_0>0$ is independent of Brownian motion $(W_t)_{t\ge 0}$. For every $T>0$ and $R>0$, can we always show

$$\mathbb P[\inf_{0\le t\le T}X_t>0]>0 \mbox{ and } \mathbb P[X_T>R]>0?$$

Here we assume the existence of the solution to the above SDE.

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  • $\begingroup$ The tag ergodic-theory looks inappropriate. There is no measure-preserving map related to your question. $\endgroup$ Jul 7, 2022 at 19:56
  • $\begingroup$ @ChristopheLeuridan Thanks for the comment. Sorry for my poor English... $\endgroup$
    – GJC20
    Jul 7, 2022 at 20:26

2 Answers 2

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The second inequality is also true. Note that $\mathbb P(X_T>R)>0$ iff $\mathbb E\big((X_T-R)^+\big)>0$. Following D-S theorem (see e.g. https://almostsuremath.com/2010/04/20/time-changed-brownian-motion/), there exists a Brownian motion $B$ with respect to some filtration ${\{\mathcal{G}_t\}_{t\ge 0}}$ s.t. for each ${t\ge 0}$, $\omega\mapsto\langle X\rangle_t(\omega)$ is a ${\mathcal{G}_\cdot}-$stopping time and ${X_t-X_0=B_{\langle X\rangle_t}}$. By assumption, one has $t\le\langle X\rangle_t \le 2t$ for all $t>0$, which implies by Jensen's inequality and Brownian motion's properties

$$\mathbb E\big((X_T-R)^+\big) = \mathbb E\big((X_0+B_{\langle X\rangle_t}-R)^+\big) \ge \mathbb E\big((X_0+B_T-R)^+\big)>0,\quad \forall T>0,$$

as the function $(\cdot-R)^+$ is convex. This allows to conclude.

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  • $\begingroup$ Thanks a lot for the answer $\endgroup$
    – GJC20
    Jul 12, 2022 at 12:10
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This is a partial answer. Conditionning on $\{X_0=x\}$, one has

$$\mathbb P[\inf_{0\le t\le T}X_t> 0]= \int_{(0,\infty)}\mathbb P[\inf_{0\le t\le T}(X_t-X_0)>-x|X_0=x]\mathbb P[X_0\in dx].$$

As any continuous martigale starting from zero is a time changed Brownian motion, there exists some Brownian motion, denoted by $B$, s.t. $X_t-X_0=B_{\langle X\rangle_t}$, where $\langle X\rangle_t:=\int_0^t\sigma(s,X_s)^2ds \in [t, 2t]$. Thus \begin{eqnarray} \mathbb P[\inf_{0\le t\le T}(X_t-X_0)> -x|X_0=x]&=&\mathbb P[\inf_{0\le t\le T}B_{\langle X\rangle_t}> -x|X_0=x]\\ &\ge& \mathbb P[\inf_{0\le t\le 2T}B_t>-x|X_0=x]\\ &=& \mathbb P[|B_1|<x/\sqrt{2T}], \end{eqnarray} which yields $$\mathbb P[\inf_{0\le t\le T}X_t> 0]\ge \mathbb P[|W_1|<X_0/\sqrt{2T}]>0.$$

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  • $\begingroup$ I agree with that, provided $B$ independent of $X_0$. You use that twice, and you should mention it. That $B$ is independent of $X_0$ is not so obvious, although it looks true. $\endgroup$ Jul 7, 2022 at 19:49
  • $\begingroup$ @ChristopheLeuridan That's a good point. Indeed, the D-S theorem is applied to $X_t-X_0$ which is independent to $X_0$. Therefore the independence between $B$ and $X_0$ still holds $\endgroup$
    – GJC20
    Jul 7, 2022 at 20:27
  • $\begingroup$ @ChristopheLeuridan Otherwise, we may use D-S theorem to X_t-X_0 conditioning on the event $\{X_0=x\}$ $\endgroup$
    – GJC20
    Jul 7, 2022 at 20:28
  • $\begingroup$ Thank you. I believe more on your second argument. $X-X_0$ is not independent of $X_0$ in general since $d\langle X \rangle_t/dt$ equals $\sigma(0,X_0)^2$ at time $0$. A more complete argument could be that the quadratic variation computed under $P$ is still the quadratic variation computed or under $P[\cdot|A]$ when $A \in \sigma(X_0)$ has positive probability. Calling $\tau_\cdot$ the inverse of the quadratic variation, we derive that $B := X_{\tau_\cdot}$ is a Brownian motion under $P$ and also under $P[\cdot|A]$ when $A \in \sigma(X_0)$ has positive probability. $\endgroup$ Jul 7, 2022 at 21:14

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