Let $B_t$ be a 1-dimensional Brownian motion and $t \in [0,T]$. Suppose we have a diffusion process $X_t$ such that $$ dX_t = u(X_t,t) dt + v(X_t,t) dt \hspace{10pt} \text{ and } \hspace{10pt} X_0= x \in (a,b) \subset \mathbb{R}$$ for some smooth $u$ and $v$. Let $\tau(x,t) = \inf \big\{ s \in [t,T]: X_s\notin (a,b) \big\}$ be the first exit time. Is there any method to obtain some estimate about the expectation of different quotient of $\dfrac{\tau(x+h) - \tau (x)}{h}$? I can not find it some standard textbooks.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ The derivative of $\tau$ with respect to $x$ does not exist even for the Brownian motion. $\endgroup$– Mateusz KwaśnickiCommented Dec 10, 2021 at 23:29
-
$\begingroup$ May I have the reason about it? Actually I would like to some estimate about the expectation of different quotient of $\dfrac{\tau(x+h) - \tau (x)}{h}$. $\endgroup$– mnmn1993Commented Dec 11, 2021 at 4:30
-
1$\begingroup$ Suppose, for simplicity, that $a = 0$ and $b = \infty$ . Let $B_t$ be the Brownian motion started at zero and $X_t^x = x + B_t$. Write $\tau(x) = \tau(x,0)$. Then $\tau$ is the minimum of $T$ and the first passage time below $-x$ for $B_t$, and hence the path of $1/2$-stable subordinator stopped at level $T$. And these are a.s. non-differentiable. $\endgroup$– Mateusz KwaśnickiCommented Dec 11, 2021 at 8:11
Add a comment
|