4
$\begingroup$

I was focusing on Feller test for explosions for a SDE like this $$dX_t=\mu(X_t)\cdot dt + \sigma^2(X_t)\cdot dW_t$$ Particularly, I was focusing on Karatzas, Shreve and attention is on exit time of the process $\{X_t\}$ from an interval on the real line $U=(k,l)$. Starting from $$\tau_{n}=\inf\Big\{t\geq0: \displaystyle{\int}_0^t \sigma^2(X_s)ds\geq n\Big\}; \hspace{0.5cm}n=1,2,...$$ $$T_{u,v}=\inf\{t\geq0: X_t\notin(u,v)\};\hspace{0.5cm}k<u<v<l$$ , after a series of passagges I am not going to mention here, letting both $n$ and $t$ go to $\infty$ I am told that $$\mathbb{E}\{T_{u,v}\}<\infty$$ that is "$X$ exits from every compact subinterval of $(k,l)$ in finite expected time".


I cannot understand why a conclusion on the exit of the process from every compact subinterval of $(k,l)$ can be drawn. What I mean is:
if I let $t$ go to $\infty$, then $T_{u,v}=\inf\{t\geq0: X_t\notin(u,v)\}=\infty$, which - as far as I know - means that there is not a finite stopping time in which the process $\{X_t\}$ at least touches the bounds of the interval $U$.

However, then the two probabilities $$\mathbb{P}[X_{T_{u,v}}=u]$$ $$\mathbb{P}[X_{T_{u,v}}=v]$$ are defined and I am told that they add up to $1$, making me guess that either $X_{T_{u,v}}=u$ or $X_{T_{u,v}}=v$, contradicting my above conclusion that the process $X_t$ does never touch (at least) the bounds $u$ and/or $v$.



What's wrong with my reasoning? Could you possibly clear my mind up and/or suggest some other sources which could help me clear my mind up?

$\endgroup$
8
  • 1
    $\begingroup$ How exactly can you “let $t$ go to infinity” in the expression for $T_{u,v}$? It does not depend on $t$ like a function, $f(t)$, does, rather it takes the infimum over all $t$ such that $X_t\notin (u,v)$, so it is either a finite value or infinite (if the process never leaves that interval). Or am I missing something? (Letting $n\to \infty$ in $\tau_n$ is certainly meaningful though but $\tau_n$ seems not to be mentioned at all after it is introduced). $\endgroup$ Sep 3, 2020 at 15:53
  • $\begingroup$ Yeah, you are right, a silly error of mine. So, at this point, another doubt arises: is Feller test something that has to be done necessarily by considering the process at a very "late" time (that is, as $t\to\infty$)? Which would be the sense of such a thing and, above all, why at the end am I told that $X_{T_{u,v}}$ is necessarily either $u$ or $v$ and CANNOT equal $\infty$? (at least, this is what I can understand from the text) @NapD.Lover $\endgroup$ Sep 3, 2020 at 15:58
  • 1
    $\begingroup$ let me try: for a process with continuous sample paths, at the time of exit from an interval $(u, v)$, by the property of continuous sample paths, we must necessarily have $X_{T_{u, v}}=u$ or $=v$. In other words, in order to exit the open interval $(u, v)$ continuously in time, one must cross $u$ or $v$. For jump processes this is not the case since one could jump discontinuously outside the interval thus requiring study of the distribution of the overshoot. Lastly, here $X$ cannot jump to infinity from inside $(u, v)$ without hitting the endpoints so $X_T$ is finite. $\endgroup$ Sep 3, 2020 at 16:10
  • 1
    $\begingroup$ If $T=\infty$ then $u<X_t<v$ for all $t$ but if $T<\infty$ then it leaves the interval on one side and it must cross and equal a boundary because of continuous sample paths, again. Its analogous to the intermediate value theorem, in a way. If at time $T-\delta$ the process is in $(u, v)$ and at time $T+\delta$ the process is $>v$ then by continuity of $X_t$ in $t$ there is some point $t^* \in (T-\delta, T+\delta)$ such that $X_{t^*}=v$. Necessarily $t^*=T$ because of $T$’s definition as the first exit time of $(u, v)$. $X_T=v+\epsilon$ is only possible with jumps. $\endgroup$ Sep 3, 2020 at 16:29
  • 1
    $\begingroup$ yeah the author must be referring to the case of finite exit time, $T<\infty$, but if this event is almost sure, then $X_T$ is essentially $\{u, v\}$-valued (excuse my omission of subscripts on $T$). When there is no exit, the only possibilities are either $X_T=X_\infty$ settles on some value $x\in (u, v)$ or this limit does not exist as $X_t$ moves about continuously in $(u, v)$ for the rest of time aimlessly, never settling on any one value. I don’t have access to Shreve’s book though to check the relevant passages. $\endgroup$ Sep 3, 2020 at 17:29

0

You must log in to answer this question.