Brownian motion stopping time ( Durrett, Probability : Theory and Examples )

115 Views Asked by At

This is excercise 7.3.3 from durrett's book.

Let $T_n$ be a sequence of stopping times.
Show that $\sup_n T_n$is stopping time

The solution says, let $R_n = max(R_{n-1},T_n), R_1 = T_1$. Then $R_n$ is stopping time, and as $n\rightarrow \infty$, $R_n \uparrow \sup\limits_n T_n$.

There are two things I can't understand.

First, I know that maximum of two stopping time is stopping time, but how can we conclude from that $R_n$ is stopping time? What if $R_{n-1}$ is not a stopping time?

Secondly, how does the equality $\lim\limits_{n\rightarrow\infty} R_n = \sup\limits_n T_n$ hold?

1

There are 1 best solutions below

3
On BEST ANSWER

Use induction to show that $R_n$ is a stopping time.

For any increasing sequence $a_n$ we have $\lim a_n=\sup_n a_n$.

Finally note that $(\sup_n R_n >a)=\bigcup_n (R_n >a) \in \mathcal F_a$ for all $a$ so $\sup_n R_n$ is stopping time.