2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for the answer $\endgroup$
    – GJC20
    Jul 12, 2022 at 12:10
1
$\begingroup$

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 $$\mathbb P[\inf_{0\le t\le T}X_t> 0]\ge \mathbb P[|W_1|<X_0/\sqrt{2T}]>0.$$

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.