Does "one-sided" random walk converge to anything?

Question

It's well known that if you scale the simple random walk, $X_n = \sum \limits_{i=0}^n \xi_i$ with $\xi_i$ independent and either $-1$ or $1$ with equal probabilities, and let $n \to \infty$, you get Brownian motion.

What if $\xi_i$ were independent and either $0$ or $1$ with equal probabilities? Could $X_n$ be made to converge towards anything in that case?

Answer 1

Let

$$P(\xi_i=-1)=P(\xi_i=1)=\frac{1}{2}$$ $$P(\xi'_i=0)=P(\xi'_i=1)=\frac{1}{2}$$ $$X_n=\sum\limits_n\xi_i$$ $$X'_n=\sum\limits_n\xi'_i$$

Then

$$2\xi'_i-1 = \xi_i$$

And so

$$2X'_n-n=X_n$$

Putting it on words, $\xi'_i$ is just a scaled and translated version of the original $\xi_i$, thus it can be scaled and translated back.

Thus, the answer is yes.

Answer 2

Since $\{(\xi_i=1)\;\text{i.o}\}\subset\{X_n=+\infty\}$, but by Borel-Cantelli Lemma $\mathsf{P}\{(\xi_i=1)\;\text{i.o}\}=1$, therefore $X_n\to+\infty$ a.s..