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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?

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    Defining $Y_n=2X_n-n$ would bring you back exactly to the first situation hence the answer is trivially "yes". – 2017-02-27

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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.

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    Thank you! Can you instantly tell if it is it possible to easily express the limiting process in terms of Brownian motion? – 2017-02-27
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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..