# Does a simple symmetric random walk in an orthant always exits the orthant with probability 1?

Let $$S = \{(x_1,x_2,\ldots,x_d) \in \mathbb{Z}^d : x_i > 0, i=1,\ldots,d \}$$ and consider a simple and symmetric random walk starting in a point $$x_0 \in S$$.

I wish to know more properties about the probability $$P(x_0)$$ that this random walk always stays in $$S$$. From Polya's recurrence theorem, I know that $$P(x_0) = 0$$ for $$d=1$$ and for $$d=2$$.

What can be said for $$d \ge 3$$?

1. In particular, is there an explicit expression for $$P(x_0)$$?
2. If not, it is true that $$P(x_0) > 0$$ for all $$x_0 \in S$$ ?

There is at least one direction in which the walk takes infinitely many steps. (In fact it almost surely takes infinitely many steps in all $$d$$ directions, but we don’t need that.) The movement in this direction is a simple symmetric random walk in one dimension. The probability that it stays on the same side of the origin forever is $$0$$. Thus the probability that the $$d$$-dimensional walk stays in the same orthant forever is also $$0$$.