Show sequence of hitting times is sequence of stopping times.

341 Views Asked by At

Let $(M_t)_{t\geq0}$ be a continuous process, $\epsilon >0$ and define a sequence $(S_n)_{n\geq 0}$ by $S_{i+1}=\inf\{t > S_{i} : M_t - M_{S_i} > \epsilon \}$ and $S_0=0$.

Clearly $S_0$ is a stopping time and I would like to use induction to show the entire sequence is. I'm considering just saying that if $S_i$ is a stopping time $S_{i+1}$ is a hitting time of a open set for a continuous process - but the open set is then stochastic and I'm not sure how this affects things.

Is the argument correct? If not what can one do instead?

1

There are 1 best solutions below

1
On BEST ANSWER

I'd say it was easier to do the inductive step directly.

Assume $S_i$ is a stopping time, then for every $t\in\mathbb R$ $M_{S_i\wedge t}$ is $\mathscr F_t$-measurable.

Therefore the event $$[S_{i+1}<t] = \bigcup_{0<s_1<s_2<t\in\mathbb Q} [S_i< s_1 ]\cap \left[\left|M_{S_i\wedge s_1}-M_{s_2}\right|>\varepsilon\right]\in\mathscr F_t$$