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Consider a random variable $\tau$:

$$\tau := \inf \left\{ {k = \sigma , \ldots ,T:S_k \ge u} \right\}\bigwedge T,$$ where $\sigma$ is a stopping time, $S_k$ - stochastic process, $T, u$ are constants and $a\wedge b=\min(a,b)$. I would like to prove that $\tau$ is a stopping time, i.e. I need to show that $\{ \tau\le k\} \in {F_k}$, where ${F_k}$ is natural filtration.

I tried to present $\tau$ as $\{ \tau \le k\} = \mathop \bigcup \limits_{j = \sigma }^k \{ \tau = j\} $ and to show that $\{ \tau = j\}\in F_j, {F_j} \subset {F_k}$, so I can conclude that $\{ \tau\le k\} \in {F_k}$. But can I really use a stopping time $\sigma$ in union operator as starting index? If not, what is the correct way to prove that $\tau$ is a stopping time?

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up vote 2 down vote accepted

Let $F$ denote the filtration $(F_n)_{n\geqslant0}$ and $S$ the process $(S_n)_{n\geqslant0}$. The result follows from the fact that, for every $0\leqslant k\leqslant T-1$, $$ [\tau\leqslant k]=\bigcup_{i=0}^kA_i^k\qquad\text{with}\qquad A_i^k=[\sigma=i]\cup\bigcup_{j=i}^k[S_j\geqslant u]. $$ Fix $0\leqslant k\leqslant T-1$. Then, for every $0\leqslant i\leqslant k$, $[\sigma=i]\in F_i$ because $\sigma$ is an $F$-stopping time, and $F_i\subset F_k$, hence $[\sigma=i]\in F_k$. For every $i\leqslant j\leqslant k$, $[S_j\geqslant u]\in F_j$ because $S$ is $F$-adapted, and $F_j\subset F_k$, hence $[S_j\geqslant u]\in F_k$.

Thus, $A_i^k\in F_k$ for every $0\leqslant i\leqslant k$, which proves that $[\tau\leqslant k]\in F_k$.

Finally, for every $k\geqslant T$, $[\tau\leqslant k]=\Omega\in F_k$, hence $\tau$ is an $F$-stopping time.

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Is it also correct if we replace $S_k$ by a right-continuous stochastic process $X_t$, a right-continuous filtration $\mathcal{F}_t$, an open set $B$ and a stopping time $S$ with values in $[0,\infty[$? Or to be more precise i formulate it as a Theorem:

Theorem: Assume we have a right-continuous stochastic process $X_t$, a right-continuous filtration $\mathcal{F}_t$, an open set $B$ and a stopping time $S$ with values in $[0,\infty[$. Then

$ \tau_B^S:=\inf \left\{t\geq S\ : \ X_t\in B\right\}$

is a stopping time.

I tried to proof it, but failed. I have shown, that for a fixed number $r\in[0,\infty[$ the random variable $\tau_B^r$ is a stopping time. Indeed by $X$ right-continuous and $B$ open we have

$\left\{\tau_B^S<t\right\}=\bigcup\limits_{s\ :\ s\ \in\ [u,t[\cap \mathbb{Q}} \left\{X_s\in B\right\} \in \mathcal{F}_t$

and therefore by the right-continuity of the filtration the claim follows.

Now you can write

$\left\{\tau_B^S<t\right\}=\bigcup\limits_{r\ :\ r \ \in\ [0,t[} \left\{S = r\right\}\cap \left\{\tau_B^r<t\right\}$.

and by the first part the intersection of the two sets is an element of $\mathcal{F}_t$. But now we got to my problem. I don't know how to get a countable union, i.e. something like that

$\left\{\tau_B^S<t\right\} ?=? \bigcup\limits_{r\ :\ r \ \in\ [0,t[\cap \mathbb{Q}} \left\{S = r\right\}\cap \left\{\tau_B^r<t\right\}$.

Is this possible or is there another more simple solution to the problem?

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