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Defintion of hitting time in my textbook is written as so:

Let $(X_n)_{n\geq 0}$ be a Markov $(\lambda, P)$. The hitting time of a subset $A \subseteq I$ is a stopping time denoted by $H^A$ such that:

$$H^A(\omega)= inf\{n\geq 0 : X_n(\omega) \in A \}$$

This is basically "the first time you hit A".

Now the notation then that the book provides me with is:

$$h^A_i = \Bbb P_i(H^A \lt \infty)$$ and it mean that this is the probability that of hitting $A$ given I started at state $i$. The only thing I don't understand here is the "$H^A \lt \infty$" part. What does this intuitively mean?

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The set $\{ H^A < \infty \} $ is $\{ \omega, H^A(\omega) < \infty \} = \{\omega, \inf(n, X_n(w) \in A) < \infty\}$ where it is assumed that $\inf ( \emptyset) = \infty$.

Thus $\{H^A < \infty \}$ is the set of all the possible outcomes (the $\omega$) for which there is at least of $n \in \mathbb N$ such that $X_n(\omega) \in A$

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  • $\begingroup$ could you maybe explain the intuition behind this condition of it being less than $\infty$ $\endgroup$
    – user255658
    Commented Jan 21, 2020 at 11:29
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    $\begingroup$ actually i understood it. Basically if there are no $n$ such that $\{X_n(\omega) \in A\}$ then $\{X_n(\omega) \in A\}=\{\emptyset\}$ and hence $H^A$ will never happen and hence the hitting time is $\infty$ $\Rightarrow$ the hitting time $h^A_i$ is the probabolility that the set $H^A$ is some existing number $\endgroup$
    – user255658
    Commented Jan 21, 2020 at 11:35
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    $\begingroup$ Since $\inf(\emptyset)= \infty$ then being (stricly) less than $\infty$ means the set of all $n$ such that $X_n(\omega) \in A$ is not empty, hence there is at least such one $n$ for which $X_n(\omega) \in A$ $\endgroup$
    – periwinkle
    Commented Jan 21, 2020 at 11:36

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