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Feb 13, 2012 at 11:33 comment added Did @covstat: You are welcome. Why did you choose this site to ask the question, rather than math.stackexchange?
Feb 11, 2012 at 2:26 comment added Douglas Zare In dimension $1$, you can calculate the probability that a Brownian motion will have hit both boundaries by time $t$.
Feb 10, 2012 at 19:40 comment added covstat @Didier Piau: I think you've answered it. Thank you.
Feb 10, 2012 at 16:06 comment added Did The random path up to a given finite time is not dense in the cube. The full random path is (almost surely) dense in the cube, in every dimension. This is basically a consequence of Kolmogorov zero-one law: for every tiny part $S$ of the box, the path of time-length $1$ visits $S$ with probability at least $u>0$, uniformly over its starting point. Iterating this, one sees that $S$ is never visited before time $n$ with probability at most $(1-u)^n\to0$.
Feb 10, 2012 at 15:53 comment added covstat I edited the question.
Feb 10, 2012 at 15:51 history edited covstat CC BY-SA 3.0
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Feb 10, 2012 at 2:49 comment added JRN @covstat, are you sure you want a random walk and not a billiard path?
Feb 10, 2012 at 1:56 comment added Yemon Choi If the question is asked on MSE then it should give more details about the tacit assumptions, as per Nate Eldredge's comments. Without specifying the law of your stochastic process the question is ill-posed
Feb 10, 2012 at 0:58 comment added Nate Eldredge This question is probably better suited for math.stackexchange.com as it is not really research level. Also, "random walk" usually refers to a process moving in discrete space; perhaps you are thinking of Brownian motion?
Feb 10, 2012 at 0:17 history asked covstat CC BY-SA 3.0