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!

  • $\begingroup$ 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

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.)


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