Random walk on $\mathbb{Z}^3$. Expected number of visits and probability of return

Question

I am working with the simple symmetric random walk on $\mathbb{Z}^3$. Using the Fourier identity I have been able to prove: $$ P(S_n = 0) = \frac{1}{(2\pi)^3} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left(\frac{\cos x + \cos y + \cos z}{3}\right)^n \, dx \, dy \, dz $$

and that the expected number of visits to zero $N$ (counting $t=0$) is: $$E[N] = \frac{1}{(2\pi)^3} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left(\frac{3}{3 - \cos x - \cos y - \cos z}\right)\, dx \, dy \, dz$$

Now i am asked to show that the probability that the random walk returns to the origin at least once is given by: $$1 - \frac{1}{E[N]}$$

I have no clue on how to approach this.

Answer 1

This is a a general fact about transient Markov chains and the random walk on $\mathbb{Z}^3$ is transient. Denote by $T_0$ the moment of first return to the origin. (The moment $t=0$ is not considered a moment of return.) This is a random quantity with possible values $2,4,6,\dotsc, \infty$.

The transience of the random walk on $\mathbb{Z}^3$ signifies that $p:=P(T_0<\infty)<1$, i.e., the probability of return to $0$ is $<1$. This was proved by George Polya almost a century ago.

If we denote by $E(N)$ the expectation of $N$, then

$$ E(N)=\sum_{n=1}^\infty P(N\geq n)= \sum_{n\geq 1} p^n, $$

where the last equality follows from the equality $p=P(N\geq 1)$ and the strong Markov principle.

Hence

$$E(N)=\frac{p}{1-p}.$$

This is what you wanted to prove. Indeed, if $N^*=N+1$ (so $t=0$ is included), then $E(N^*)=E(N)+1=\frac{1}{1-p}$ and $$ p=1-\frac{1}{E(N^*)}. $$ For more details and proofs the facts claimed above see section 4.2.3 of these notes.