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I am reading about recurrence random walks and I see the following expression:

$$\mathbb{P}(\exists n\geq 1,X_n = 0) = 1$$

So, I wonder: what is this precisely? I know, for instance,

$$\mathbb{P}(X_n = 0) = \mathbb{P}(\{\omega:X_n(\omega) = 0\})$$

rigorously. But I couldn't make sense of the first expression in the same language I gave in the second expression. In some sources, they say that this implies that $X_n$ visiting the zero position infinitely often (implies or same thing?). This is probably an issue of notation but if someone explains, it would be truly helpful. Thanks!

(see e.g. Proposition 3 from this pdf)

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3 Answers 3

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Let $R$ denote a property each $\omega$ in $\Omega$ may have or not, and $A=\{\omega\in\Omega\mid R(\omega)\}$, in your case $R(\omega)$ is: $\exists n\geqslant1,X_n(\omega)=0$.

Then, $P(R)$ is simply $P(A)$, that is, $$ P(\exists n\geqslant1,X_n=0)=P(\{\omega\in\Omega\mid \exists n\geqslant1,X_n(\omega)=0\}). $$ By the way, this is exactly for the same reason that $$ P(X_n=0)=P(\{\omega\in\Omega\mid X_n(\omega)=0\}). $$

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  • $\begingroup$ Thanks Did! I am more or less aware of this notation but what I couldn't make sense is for instance what is the meaning of $P(\{\omega\in\Omega\mid \exists n\geqslant1,X_n(\omega)=0\})$ precisely? There exists $n\geq 1$ s.t. $X_n = 0$ with probability one? How this implies that the walk will come to the position $X_n = 0$ infinitely often? $\endgroup$
    – user48547
    Mar 6, 2014 at 16:48
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    $\begingroup$ Yes this is the meaning. Why infinitely often? This is a result, not a prioru obvious (but true for Markov chains). $\endgroup$
    – Did
    Mar 6, 2014 at 16:51
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Formulas involving the existential and universal quantifiers $\exists$ and $\forall$ can be rephrased in terms of unions and intersections. In this case, you can write: $$P\left[\exists n\geq 1: X_{n}=0\right] = P\left[\bigcup_{n\geq 1}\{\omega: X_{n}(\omega)=0\}\right]$$

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It is saying that there is some index $n$ such that $X_n$ will be at the origin with probability 1. This statement makes no claims as to what range of values that $n$ might take; rather, that with probability 1 there is some finite value $n$ such that $X_n = 0$.

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