1
$\begingroup$

let $f : [0, +\infty[\to\mathbb R$ be bounded and continuous, and $(X_t)_{t\geq 0}$ be an adapted process such that X$_0 = 0$ and $X_t = \int_0^t f(s, X_s){\rm d}\!s$ for all $t\geq0$, also define $\pmb t = \inf\{t\ge 0 \;;\; X_t\gt1\}$.

I have to prove that $(X_t)_{t\geq0}$ is almost surely continuous and that $\pmb t$ is a stopping time.

The first thing pops up in my brain is that $f$ is bounded meaning ${\rm d}\!X_t$ is bounded, so even $\frac{{\rm d}\!X}{{\rm d}\!t}$ might be discontinuous, we can still see that integrating a small amount of time will have an upper and lower bound, this implies $t\mapsto X_t$ is continuous as its change for a given change in t can be bounded.

But I am not sure if this technic is applicable for stochastic process or adapted process or not. also I am very confused by the term "almost surely continuous", I know its a property in Brownian motion, but have no idea to prove it in this question. I searched a bit, got the idea that almost sure continuity could be understood as sample continuity.. https://en.wikipedia.org/wiki/Sample-continuous_process

I also tried to take advantages of an theorem about almost sure continuity: https://en.wikipedia.org/wiki/Kolmogorov_continuity_theorem But still didn't figure this out.

Sincerely hope any expert could give me some hints! Many thanks in advance!

$\endgroup$
  • $\begingroup$ Something is missing in $ X_t = \int_0^t f(s,X_s)\,d s$. $\endgroup$ – Sayantan Sep 15 at 6:02
  • $\begingroup$ @Sayantan Hi Sayantan! Thanks for your comment! I am sure I didn't miss any information here, the context is $X_0$=$0$ and $X_t$ is defined as $\int_0^t$f(s, X$_s$)ds for all t$\geq$$0$. $\endgroup$ – Wendy Zhang Sep 15 at 6:10
  • $\begingroup$ In that case $X$ solves the ODE $$ \frac{\, d X_t }{\,d t} = f(t,X_t), X_0 =0.$$ Where does the randomness come from? $\endgroup$ – Sayantan Sep 15 at 7:12
  • $\begingroup$ @Sayantan Thank you so much for your explanation! I presumably think that would be the solution, but I am still got stuck on how to use bounded derivative to prove the almost sure continuity. $\endgroup$ – Wendy Zhang Sep 15 at 7:18
0
$\begingroup$

The process $X$ is differentiable, thus continuous.

Now, $]-\infty,1[$ is an open set. But the first instant of leaving an open set for a continuous process is a stopping time.

Therefore, $\mathbf t$ is a stopping time.

(If the set is closed, it suffices that $X$ is right-continuous.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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