Intuitive reason why a simple symmetric random walk is recurrent on $\Bbb Z^2$ and transient on $\Bbb Z^3$.

Solution 1:

A heuristic argument: The random walk in any dimension tends to stay in a ball of radius $\sqrt{N}$ for the first $N$ steps. In dimension $d$, there are roughly $c N^{d/2}$ points inside the ball of radius $\sqrt{N}$ (for some constant $c$). So, if $d \leq 2$, then it is likely that the $N$ steps of the walk will cover the whole ball and, if $d > 2$, it is unlikely.

Regarding why the walk stays in a ball of radius $N$: Let the individual steps of the walk be $s_1$, $s_2$, ..., $s_N$, so the final position of the walk is $s_1 + s_2 + \cdots + s_N$. Write $E$ for expected value.

We assume that the individual steps are vectors independently drawn from some distribution with $E(s_j) = \vec{0}$ (there is no drift) and $E(|s_j|)$ a constant. Then $$E\left( (s_1 + s_2 + \cdots + s_N)^2 \right) = \sum_{i,j} E(s_i \cdot s_j) = \sum_i E(|s_i|^2) + \sum_{i \neq j} E(s_i) \cdot E(s_j).$$ The first step is linearity of expectation and the second is the statement that $s_i$ and $s_j$ are independent. But we assumed $E(s_i)=\vec{0}$, and the $s_i$ all have the same distribution, so this is just $N E(|s_1|^2)$.

We have shown that $E\left( (s_1 + s_2 + \cdots + s_N)^2 \right)$ grows like $c_1 N$, so we should expect $\left| s_1 + s_2 + \cdots + s_N \right|$ to grow like $c_2 \sqrt{N}$. Making this argument precise will require filling in the exact kind of random walk involved, but hopefully this makes it clear why we expect $\sqrt{N}$.