$\require{mathtools}
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% .. over and under
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Adapted from the video “Random walks in 2D and 3D are fundamentally different (Markov chains approach)” by Mathemaniac.
In a Markov chain, a state $s$ is called recurrent if when starting from $s$, you know with probability $1$ that you will return to $s$ eventually. If $s$ is not recurrent it is called transient.
Using this notion, we will show the truth in this fantastic quote by Shizuo Kakutani:
A drunk man will find his way home, but a drunk bird may get lost forever.
That is, random walks on the 2D grid will keep coming back to the origin (wherever you start from), while random walks in 3D might never come back.
A characterization of recurrent states
It turns out that a state is recurrent iff the expected number of times you return is infinite.
Let random variable $X$ represent the number of times you return. We can compute its expectation as
\[\E[X] = \sum_{t=1}^\infty \Pr[X \geq t],\]
where $\Pr[X \geq t]$ is the chance that you return at least $t$ times. But if we denote the probability of returning at least once as $r$, then $\Pr[X \geq t] = r^t$ (since the walk is memoryless), so we get
\[\E[X] = r + r^2 + r^3 + \cdots = \frac{r}{1-r},\]
which is indeed infinite iff $r=1$.
Using the characterization for random walks
Of course, this is not the only way to compute the expectation $\E[X]$. We also have
\[\E[X] = \sum_{t=0}^\infty\Pr[\text{at the origin at time $t$}].\]
If we can figure out whether this series is infinite, this will tell us whether the origin is recurrent!
2D case
How do we estimate for the probability of being back at the origin at time $t$?
Say that we’ve taken $h$ horizontal steps (left, right) and $v$ vertical steps (with $h+v=t$). First, we clearly need both $h$ and $v$ to be even. Assuming $t$ is even, this will happen with constant probability.
In addition to that, we need to have taken as many “up” as “down” steps, and as many “left” as “right” steps. For $t$ big enough (say $t \geq 200$), $h$ will be about $t/2$ whp, so the probability of taking as many “up” as “down” steps is comparable to the probability that a random string in $\zo^{t/2}$ has Hamming weight exactly $t/4$, which is $\Theta(1/\sqrt{t})$. Using the same reasoning in the horizontal direction, this gives:
\[\Pr[\text{at the origin at time $t$}] = \underbrace{\Theta(1)}_{\Pr[\text{$h,v$ even}]} \cdot \underbrace{\Theta\p{1/\sqrt{t}}}_{\Pr[\text{up = down}]} \cdot \underbrace{\Theta\p{1/\sqrt{t}}}_{\Pr[\text{left = right}]} = \Theta(1/t)\]
for even $t$ (and $0$ for odd $t$).
Therefore, we get
\[\E[X] = \sum_{t=0}^\infty \Pr[\text{at the origin at time $2t$}] = O(1) + \sum_{t=100}^\infty \Theta(1/t) = \infty,\]
which means the origin is recurrent!
3D case
In the 3D case, the same reasoning holds, just with one more direction to deal with, and we get
\[\Pr[\text{at the origin at time $t$}] = \Theta(1) \cdot \Theta\p{1/\sqrt{t}} \cdot \Theta\p{1/\sqrt{t}} \cdot \Theta\p{1/\sqrt{t}} = \Theta(1/t^{3/2}),\]
so the series
\[\E[X] = \sum_{t=0}^\infty \Pr[\text{at the origin at time $3t$}] = O(1) + \sum_{t=100}^\infty \Theta(1/t^{3/2}) = O(1)\]
converges, which means the origin is transient! Poor bird. :(