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I have problems with the proof that a first hitting time is a stopping time:

Let $\tau$ be the first hitting time into the set A, for a process $\{ X_n \}$ adapted to a filtration $\mathcal F_n$.

I know that a random time is a stopping time if the set $ \{\tau\le n\}\in\mathcal F_n \, \, \forall n \in[0,\infty)$

Define now the first hitting time as $\tau_A= \mathrm{inf}(n\ge0 : X_n\in A)$

I find everywhere the same proof:

$ \{\omega \in \Omega : \tau_A(\omega) \le n\} =\bigcup_{k=0}^n\{\omega \in \Omega : X_k(\omega) \in A\} \in \mathcal F_n$

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  • $\begingroup$ Go back wayyyy behind, to the first chapter of notational conventions in probability theory: there, they say that $ \{\tau_A \le n\}$ is a shorthand for the set $$ \{\omega\in\Omega\,|\,\tau_A(\omega) \le n\}.$$ If this point is clear to you, you should also be able to correct the definition of $\tau_A$ in your post, which is absurd at the moment since it confuses $ \tau_A$ with $ \{\tau_A\}$. $\endgroup$
    – Did
    Mar 29 '15 at 13:07
  • $\begingroup$ $\tau$ is random variable so it takes as argument $\omega$ and I agree, but in a practical example what could be an $\omega$ in this case? for r.v. which are like "n° of heads in 2 coins tosses" a possible $\omega=\{HT\}$. I imagine that also the notation $\{X_k\in A\}$ means $\{\omega\in\Omega : X_k(\omega)\in A\}$, but again now, which could be an $\omega$ for $X_k$? I have problems in visualizing this in my head since stochastic processes for me are r.v. indexed on the time, but I never asked myself which values they take from $\Omega$ to give back the output real number. $\endgroup$
    – Drush
    Mar 29 '15 at 13:20
  • $\begingroup$ The identity of $\Omega$ is irrelevant, all that counts is that there exists such suitable probability space. Try to solve the exercise and you will see that you never need to know what $\Omega$ is. I might even have explained this in details somewhere on the site. $\endgroup$
    – Did
    Mar 29 '15 at 13:24
  • $\begingroup$ If I had to explain with words the proof, I would say that the omegas for which $\tau_A(\omega)\le n$ are the same as the union of the omegas such that $X_k(\omega)\in A$ but since we are considering a sequence of $X_k$ until n, these omegas are in $\mathcal F_n$, is that correct? $\endgroup$
    – Drush
    Mar 29 '15 at 13:28
  • $\begingroup$ No, that some $\omega$ is or is not in $\mathcal F_n$ is far from being correct. What is the nature of the object $\mathcal F_n$, already? $\endgroup$
    – Did
    Mar 29 '15 at 13:30
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So then I got the answer:

$\{\tau_A\le n\}=\{\bigcup_{k=0}^n X_k\in A\}$ because if the first hitting time of A has occurred before n, the omegas for which this occur are the same omegas for $X_k$ for which "At least a $X_k$ at some point before n has entered A", then these omegas belong to $\mathcal F_n$ because $X_k$ is $\mathcal F_n$-measurable.

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    $\begingroup$ $\{\bigcup_{k=0}^n X_k\in A\}$: incorrect notation. $\endgroup$
    – Did
    Mar 31 '15 at 17:43
  • $\begingroup$ do I have to write $\bigcup _{k=0}^n\{X_k\in A\}$ ? $\endgroup$
    – Drush
    Mar 31 '15 at 17:58
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    $\begingroup$ You have. $ $ $ $ $\endgroup$
    – Did
    Mar 31 '15 at 19:16

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