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Suppose, $\{X_n\}_{n = 1}^{\infty}$ is a sequence of i.i.d random variables, such that $P(X_i = 1) = P(X_i = -1) = \frac{1}{2}$. Now, suppose $\{S_n\}_{n = 1}^\infty$ is a sequence of random variables defined in the following way: $S_n = \Sigma_{i = 1}^n X_i$. Now suppose $T_n$ is the number of integers $m$, such that $\exists k < n, S_k = m$ and $\forall k > n S_k \neq m$ ($\{T_n\}_{n = 1}^\infty$ is also a sequence of random variables.

Is it true, that $P(\lim_{n \to \infty} \frac{T_n}{n} = P(\forall k \in \mathbb{N}, S_k \neq 0))=1$?

I see, that $P(\forall k \in \mathbb{N}, S_k \neq 0) = 2P(\forall k \in \mathbb{N}, S_k > 0)$, but that does not help much.

Also, I used to think, that $P(\forall k \in \mathbb{N}, S_k > 0) = 0$, but now I am not so sure about it…

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