This question comes from a step in a proof in Mao's book on SDEs (page 120), where the author states that the below is true but he doesn't justify it.

Question: Consider the SDE on $t \geq 0$ $$dX_t = f(x_t,t)dt + g(X_t,t)dB_t, \quad X_0 = x_0 \in \mathbb R-{0}. \tag{1} \label{1}$$

Assume $f$ and $g$ satisfy the local Lipschitz and linear growth conditions so there exists a unique solution $X$ for SDE \eqref{1}. Define $$\tau := \inf \{t \geq 0 :X_t = 0\}$$ and assume $$P(\{\tau < \infty\})>0. \tag{2}$$

Show that we can find a pair of constants $T>0$ and $\theta > 1$ sufficiently large for $P(B)>0,$ where $$B = \{\tau < T \text{ and } |X_t| \leq \theta -1 \text{ for all } 0 \leq t \leq \tau\}.$$

My answer: I think that $P(\tau < \infty) >0$ implies that we can find a large enough $T$ such that $P(\tau < T)>0.$ So the original problem is equivalent to show that there exists a $\theta > 1$ such that $$P(\sup_{0 \leq t \leq \tau}|X_t|< \theta-1)>0. \tag{3} \label3$$ By contradiction, assume that \eqref{3} is false. Then, for all $$\theta >1,P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1)=1. \tag{4} \label4$$ But by Markov inequality $$P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1) \leq \frac{E(\sup_{0 \leq t \leq \tau}|X_t|)}{\theta -1} \leq \frac{E(\sup_{0 \leq t \leq T}|X_t|)}{\theta -1}.$$ Since $f$ and $g$ satisfied the linear growth condition, it can be shown (e.g. page 69 in the same book) that $E(\sup_{0 \leq t \leq T}|X_t|) < C$ for some constant $C,$ which doesn't depend on $\theta.$ Therefore, by taking $\theta$ such that $\theta - 1 > C,$ We have $P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1)<1,$ which contradicts \eqref{4}, so \eqref{3} is true.

Is my approach correct? Does $P(\tau < \infty) >0$ implies that we can find a large enough $T$ such that $P(\tau < T)>0$ (as I said at the beginning of my answer)?


1 Answer 1


Yes, and lets just do the punchline for the intersection:

$$P(\tau < T,\sup_{0 \leq t \leq \tau}|X_t|\leq \theta -1)=P(\tau < T)-P(\tau < T,\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1).$$

Let $c_0:=P(\tau < T)$, then we further take theta so large so that

$$P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1)\leq \frac{E(\sup_{0 \leq t \leq T}|X_t|)}{\theta -1}<\frac{c_{0}}{2},$$

and so

$$P(\tau < T,\sup_{0 \leq t \leq \tau}|X_t|\leq \theta -1)\geq \frac{c_{0}}{2}.$$


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