Computing $P(\limsup S_n=\infty)$ and $P(\liminf S_n=-\infty)$ for a random walk

Question

Let $X_i$ be i.i.d. non-degenerate random variables with $EX_i=0$, and consider the random walk $S_n = \sum_{i=1}^n X_i$.

If $X_i$ was a simple random walk (i.e., $P(X_i = -1) = 1/2 = P(X_i = 1)$), I am able to show that $$P(\limsup S_n = \infty) = 1 = P(\liminf S_n = -\infty).$$

In my proof of the above statement, I do so by first showing that infinitely often the random walk must take $M$ consecutive positive steps infinitely often, for all $M\in\mathbb{N}$, with probability $1$. Similarly, one can show that the random walk must take $M$ consecutive negative steps infinitely often with probability $1$. (I am also aware that there are other questions on this site that address the above statement.)

However, my question is can this be generalized beyond this simple case of a simple random walk for $X_i$ following an arbitrary, non degenerate distribution with mean $0$? And if so, how does one go about a proof? In the case of the simple random walk we had a nice clean characterization of the distribution of the $X_i$, but in a generalized version we do not. Thanks!

Answer 1

The probabilities are both equal to $1$ for any non-trivial random walk $S_n = \sum_{i=1}^n X_i$ satisfying $\mathbb{E}(X_i)=0$; this is a direct consequence of a statement by Chung-Fuchs:

Let $(X_i)_{i \in \mathbb{N}}$ be a sequence of non-trivial iid $\mathbb{R}$-valued random variables, and denote by $S_n := \sum_{i=1}^n X_i$ the associated random walk. Then the following statements hold true.

• $\mathbb{E}X_1>0 \iff \lim_{n \to \infty} S_n = \infty$ almost surely
• $\mathbb{E}X_1<0 \iff \lim_{n \to \infty} S_n = -\infty$ almost surely
• $\mathbb{E}X_1 = 0 \iff - \infty = \liminf_{n \to \infty} S_n < \limsup_{n \to \infty} S_n = \infty$ almost surely.

The proof is relatively easy if the steps $X_i$ are bounded (i.e. $\mathbb{P}(X_1 \leq K)=1$ for some $K>0$); for general random variables $X_i$ the proof is more involved.