Random walk on $\mathbb{Z}$ with unusual generating set Polya's recursion theorem says that a simple random walk on $\mathbb{Z}$ is recurrent. In the proofs that I have seen, one considers the steps to be of lengths $+1$ and $-1$. My question is the following: Is this also true changing the generating set of $\mathbb{Z}$? 
For example, the set $\{6,10,15\}$ generates $\mathbb{Z}$ and no subset of it does. Consider the random walk that takes steps of length $6$, $-6$, $10$, $-10$, $15$, $-15$ uniformly. Is this random walk recurrent?
 A: A common approach to the recurrence/transience of random walks on lattices uses discrete Fourier transform. More specifically, one first notes that, for every integer valued random variable $S_n$, $$2\pi P(S_n=0)=\int_{-\pi}^\pi E(e^{itS_n})dt$$ and that, when $(S_n)$ is a random walk starting from $0$ with steps distributed as $X$, $$E(e^{itS_n})=\left(E(e^{itX})\right)^n$$ These two remarks put together yield the key formula

$$2\pi \sum_{n=0}^\infty P(S_n=0)=\sum_{n=0}^\infty\int_{-\pi}^\pi \left(E(e^{itX})\right)^ndt=\int_{-\pi}^\pi\frac{dt}{1-E(e^{itX})}$$

Now, if $X$ is symmetric with $p_x=P(X=x)$ for every $x$, then $$E(e^{itX})=p_0+2\sum_{x\geqslant1}p_x\cos(tx)$$ hence $$1-E(e^{itX})=2\sum_{x\geqslant1}p_x(1-\cos(tx))$$The convergence of the integral $$\int_{-\pi}^\pi\frac{dt}{1-E(e^{itX})}$$ depends on the behaviour of the function near $0$. When, as here, the support of $X$ is bounded, the expansion up to order $x^2$ is always valid, and yields the equivalent $$1-E(e^{itX})=2\sum_{x\geqslant1}p_x\left(\frac12t^2x^2+o(t^2)\right)\sim vt^2$$ when $t\to0$, with $$v=\sum_{x\geqslant1}p_xx^2=\frac12E(X^2)$$ In particular, the integral diverges at $t=0$, thus the series $$\sum_{n=0}^\infty P(S_n=0)$$ diverges, thus the random series $$\sum_{n=0}^\infty\mathbf 1_{S_n=0}$$ is infinite almost surely (can you prove this step?), that is, the random walk $(S_n)$ is recurrent.
A: Yes, it will be recurrent as well (it is not important that the step lengths form a generating set, as long as they are symmetric with respect to the origin). Hint: Loosely speaking, you can split your walk in three classical walks, which just consider steps of one absolute step size. Each walk is recurrent, by Polya's theorem, hence for each there are arbitraryly large times at which they are positive and negative. How can you deduce from that, that there are "times", such that ALL three are positive (and negative) at the same "time" (depending on how you write it down the three walks won't be entirely independent, but in some sense that is enough here). It is hard to write down properly, but the idea should be clear(?).
