Consider a random walk on $\mathbb Z^2 \subset \mathbb C$ starting from the orgin $(0,0)$. At each step, it randomly go one of the $4$ directions with length $1$ (i.e. $\pm 1, \pm \sqrt{-1}$). It stops if it hits one of the $4$ points $(\pm 5 ,\pm 5)$ in the first time. The question is, what is the expectation of the stopping time?
I can write down explicitly the number of paths of hitting any given point at the $k$-th step. Then the expectation can be written down as a summation of infinite sequences. However, I wonder if there is a simple expression, and I would like to know if there is a more elegant approach to this. Thanks in advance!
Best Answer
This expectation is infinite.
There is clearly a non-zero probability for the walk to reach $x=6$. Thus, if we can show that the expected time for it to get back to $x=5$ is infinite, it will follow that the desired expected first hitting time is also infinite. This, in turn, will follow if the same is true for a one-dimensional random walk, since the fact that we sometimes take steps in the $y$ direction in between only increases the expected time.
So consider a simple symmetric random walk in the $x$ direction. To obtain a contradiction, let $t$ be the finite expected time required for the walk to get back to $x=5$ from $x=6$. Then by translation invariance the expected time to get back to $x=6$ from $x=7$ is also $t$, and thus the expected time to get back to $x=5$ from $x=7$ is $2t$. Since the walk goes to $x=5$ and $x=7$ from $x=6$ with equal probability, this yields $t=1+\frac12(0+2t)=1+t$, a contradiction.