# The necessary sufficient condition for recurrence of a Markovian random walk

Suppose $\sigma_{1},\sigma_{2},...$are i.i.d random variables.$S_{0}=0$. Define $S_{n}=S_{0}+\sum_{i=1}^{n}\sigma_{i}$, then ${S_{n}}$ is a Markovian random walk.

I want to figure out the necessary sufficient condition for $S_{n}$ to be recurrent.

It seems natural that if $\mu = E(\sigma_{1})\neq0$, $S_{n}$ must be trasient (By central-limit theorem, $S_{n} \rightarrow N(n\mu,n^{2}\sigma^{2})$ and the chance for $S_{n}$ to return $0$ decays exponentially, so $E_{0}(V_{0}) = \sum_{n=1}^{\infty}P_{0}({S_{n}=0}) < \infty$).

And that makes $\mu=0$ a necessary requirement, which seems natural, too.

But the question is, is it sufficient? How to prove it or find a counter example? Are there any other properties equivalent to the recurrence of $S_{n}$?

Only the sketch of the proof is necessary should it be too long or too complicated (detailed proof can be provided by book references or external links). Some intuitive remarks are most appreciated.

************************** Update on 4/8/2016 ***************************

It seems that I forgot to mention the type of $\sigma_{n}$, some specifications are expected in subsequent answers(one of the two following):

1. $\sigma_{n}$ is discrete, i.e. $\sigma_{n} \in \mathbb{Z}$
2. $\sigma_{n}$ is continuous, i.e. $\sigma_{n} \in \mathbb{R}$

No, it's not sufficient, you need assumptions on the tails of $\sigma$'s. See Chapter 5 of http://www.ime.unicamp.br/~popov/book_lyapunov.pdf
• Thank you for your prompt answer. However I was more concerned in the case where ${\sigma_{n}}$ are discrete. What's the corresponding result under this assumption? – Lotayou Apr 8 '16 at 5:59
• Essentially, Chapter 5 covers both discrete and continuous cases. For example, Theorem 5.3.1 is formulated in the continuous case, but (as noted there in the text) the results in the discrete case are the same. Up to Section 5.2, the results are general (the chain lives on a set $\Sigma$, which can be $\mathbb{R}$ or $\mathbb{Z}$). – Serguei Popov Apr 8 '16 at 13:23