# Recurrent Set and i.i.d. random sequence

Consider an i.i.d. discrete random sequence $\{X_i\}$, suppose $EX_1 \neq 0$ and define $R:=\{x: \text{$x$is recurrent value for$S_n$}\}$. I was trying to show the set $R = \emptyset$ where $S_n := \sum\limits_{i=1}^nX_i$.

The definition of recurrent value $x$ for $S_n$ is: for $\varepsilon>0$, $P( |S_n - x| < \varepsilon\; \text{infinitely often})=1$

I am stuck at the very first step. Here is my try: Proceeding by proof of contradiction: assume $R \neq \emptyset$; hence, there exists $x$ be a recurrent value; i.e., for $\varepsilon>0$, $P( |S_n - x| < \varepsilon\; \text{infinitely often})=1$

and I also notice that $EX_1 \neq 0$ hence there exists $c$ such that $EX_1 = c$ and hence, $ES_n = nc$ by the i.i.d. property.

But I can't conclude/infer further from these two results.

• Sorry, what does i.o. stand for? Commented Nov 10, 2014 at 9:09
• @Stefanos: Infinitely often. Commented Nov 10, 2014 at 11:21
• @gmath ok, thank you... I should have seen it Commented Nov 10, 2014 at 11:22

Without loss of generality, assume that $a= E(X_1) >0$. By the strong law of large numbers, $$P(\lim_{n \to \infty}S_n/n =a)=1$$ Also, $\lim_{n \to \infty} S_n/n =a$ implies that $S_n\to+\infty$. Hence $$P(S_n\to+\infty) =1$$ Hence for any $x \in \mathbb R$ , $$P(S_n -x \to+\infty)=1.$$ Thus $R$ is empty.