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A one-dimensional symmetric simple random walk starts at $S_0 = 1$. How to show with probability one it passes $x = 0$ (or I guess equivalently, the stopping time of hitting $x = 0$ at the first time is finite almost surely)?

It's just a new-learner's doubt, not a homework problem, I would appreciate any advise. Thanks in advance!

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We have \begin{align} \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\}\right) &= \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\mid X_1=-1\}\right)\mathbb P(X_1=-1)\\ &\quad+ \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\mid X_1=1\}\right)\mathbb P(X_1=1)\\ &= \mathbb P(X_1=-1) + \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant -1\}\right)\mathbb P(X_1=1)\\ &= \frac12 + \frac12\mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\}\right)^2, \end{align} from which it follows that $$ \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\}\right)=1.$$

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  • $\begingroup$ @Stupid_Guy Does this answer your question? $\endgroup$ – Math1000 May 28 '16 at 6:58

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