Random walk on $\mathbb{Z}^3$ crosses $XY$ plane infinitely often

Question

I need to show that the simple symmetric random walk on $\mathbb{Z}^3$ crosses the $XY$ plane infinitely often. The hint given is to consider the movement relative to the $XY$ plane.

I know that $\mathbb{P}[X_1=(1,0,0)]=...=\mathbb{P}[X_1=(0,0,-1)]=1/6$. I also know that the plane would be denoted by $\{(x,y,0): x,y \in \mathbb{R}\}$. In order to show it crosses it infinitely often, I can try proving that $G(i,j)=\mathbb{E}_i(N(j))=\sum_{n=1}^{\infty} \mathbb{E}_i (\mathbb{1}_{\{j\}}(X_n))$, which is the expected number of times that the Markov Chain visits state $j$ starting from state $i$, goes to infinity.

However, I do not seem to understand how to proceed further. Any help/source would be appreciated!

Answer 1

Hint: consider what happens to the $z$-coordinate during the random walk. It either:

• increases by $1$ (probability $1/6$)

• stays the same (probability $2/3$)

• decreases by $1$ (probability $1/6$)

Showing that the simple symmetric random walk on $\Bbb{Z}^3$ crosses the $xy$-plane infinitely often, then, is equivalent to showing that the random walk on $\Bbb{Z}$ that takes steps of $\pm 1$ with probability $1/6$ each, and otherwise stays in place with probability $2/3$, crosses the point $0$ infinitely often. Can you finish from here?

Answer 2

You only need to consider the $z$-axis.

Imagine the random walk only crosses the plane a finite number of times. Then all remaining walks would have either more ups than downs, or vice versa.

This gives Dyck paths, and the probability of a Dyck path over all paths tends to zero.