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Suppose we define for $A\in \mathcal{B}(\mathbb{R^n})$ the first hitting time

$$ T_A:= \inf\{t\ge 0;X_t(\omega)\in A\}$$

where $X=(X_t)$ is a stochastic process, adapted to a Filtration and with right-continuous paths. Now suppose that $A$ is open then I want to show:

$$\{T_A<t\}\in \mathcal{F}_t$$

for all $t\ge0$ and where $\mathcal{F}_t$ is a element of the filtration. In the book there's a hint, we should show $\{T_A<t\}=\cup_{q\in \mathbb{Q},0\le q<t}\{X_q\in A\}$. One inclusion, i.e $\supset$,is obvious. Unfortunately I got stuck at the other. Since the paths are right continuous, my first thought was to construct a sequence of rational decreasing to $T_A$, however I do not see why such a sequence should exist and I do not see how I should use that $A$ is open. Some help would be appreciated!

hulik

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Let $\omega$ in the LHS. You have for a $r<t$ that $X_r(\omega)\in A$. Let $\{r_n\}$ a dicreasing sequence of rational numbers which converges to $r$. Since the path are right continuous, $X_{r_n}(\omega)\to X_r(\omega)\in A$, so for $n$ large enough $X_{r_n}(\omega)\in A$. Now take $n$ such that $r_n<t$ and you are done. –  Davide Giraudo Feb 29 '12 at 19:15
    
@ Davide Giraudo: This was exactly what I did, but why does such a sequence exist? Of course there is a sequence of rationals converging to $r$. But why can I choose a decreasing one? Sorry, this should be a basic Analysis question but I don't see right now.And I don't see where "we" use that $A$ is open in this argument. –  user20869 Feb 29 '12 at 19:28
    
Fix an integer $m$. Then $r$ is in a interval of the form $[j2^{-m},(j+1)2^{-m})$ for some integer $j$. So take $r_j=(j+12^{-m})$. –  Davide Giraudo Feb 29 '12 at 19:35
    
Of course I means $(j+1)2^{-m}$ in the last comment. Openness is need to ensure us that $X_{r_n}(\omega)\in A$ for $n$ large enough. –  Davide Giraudo Feb 29 '12 at 19:42

1 Answer 1

up vote 2 down vote accepted

Hint: Consider the following: $$\{T_A<t\}=\cup_{0\le s<t}\{X_s\in A\}=\cup_{q\in \mathbb{Q},0\le q<t}\{X_q\in A\}.$$ The first equation is easy, but the second equation is troubling you. You should try to understand why if $X_s\in A$ for some real value $s\in[0,t)$, then also $X_q\in A$ for some rational $q\in[0,t)$. This is where you use the fact that $A$ is open and that the sample path is right-continuous.

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@hulik Do you agree that every interval of the form $[s,s+\varepsilon)$ for $\varepsilon>0$ contains a rational? –  Byron Schmuland Feb 29 '12 at 19:38
    
@ Byron Schmuland: Shame on me!I did a very bad error in reasoning! –  user20869 Feb 29 '12 at 19:44
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@hulik Yes, it is crucial! The sample path $s\mapsto |s-\sqrt{2}|$ hits the set $A=\{0\}$, but not at any rational time. –  Byron Schmuland Feb 29 '12 at 19:47

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