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I want to solve the following task:

Let $p \in (0,1)$ and $X_n$ be a sequence of i.i.d. random variables with $$\mathbb{P}(X_n = 1)=1-\mathbb{P}(X_n=-1)=p$$ for all $n$. Let $$S_0:=0 \text{ and } S_n:=\sum_\limits{i=1}^n X_i$$ and $A=\limsup \{S_n = 0\}$.

Show $$ p \ne \frac{1}{2} \Rightarrow \mathbb{P}(A) = 0.$$

I'm sure that I need the first Borel Cantelli Lemma here. So I have to show $\sum_\limits{n \in \mathbb{N}} \mathbb{P}(S_n=0) < \infty$ and probably this does only hold if $p \ne \frac{1}{2}$ but how do I calculate $\mathbb{P}(S_n=0)$? I know that $S_n$ is only equal to $0$ if I have as many ones as I have minus ones but this approach didn't help me.

Any ideas?

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    Is it possible that $S_1=0$? When $S_2=0$? How many $\pm 1$ should be for $S_{2n}=0$? – 2017-02-14
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    $S_n$ has (almost) a binomial distribution, so there should be no problem computing this probability. Alternatively, you can apply the strong law of large numbers. – 2017-02-14
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    Yes I already had $P(S_n = 0) = {{n}\choose{n/2}} p^{n/2} (1-p)^{n/2}$ (if n is even, otherwise $S_n$ is never $0$). But what do I do now?? – 2017-02-14

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When $n=2k+1$ is odd, $$\{S_{2k+1}=0\}=\emptyset$$ because the sum will be odd.

When $n=2k$ is even, $$\mathbb{P}(S_{2k}=0)=\frac{1}{2^{2k}}\binom{2k}{k}p^k(1-p)^k$$ because exactly $k$ of $X_i$ have to be $1$, and the other $k$ have to be $-1$.

Using the Stirling's approximation, we get

$$\mathbb{P}(S_{2k}=0)\approx \frac{1}{2^{2k}}\frac{4^{2k}} {\sqrt{2\pi k}}(p(1-p))^k = \frac{2^{2k}}{\sqrt{2\pi k}}(p(1-p))^k = \frac{(4p(1-p))^k}{\sqrt{2\pi k}} $$

and when $p\not = \frac{1}{2}$, we have $$4p(1-p)<1$$

so the series converges

$$\sum \mathbb{P}(S_{2k}=0) < \infty$$

by the Root test.

By Borel-Cantelli lemma,

$$ \mathbb{P}(A) = \mathbb{P}(\lim\sup\{S_n=0\}) = 0$$

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    The Stirling approximation is an overkill for this problem. It is enough that ${2k \choose k}< 4^k$. – 2017-02-14
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    Thank you! I also found the way to show it with the strong law of large numbers – 2017-02-14