Distribution of sum of iid binary random variables I have a sequence $X_i$ of random variables which can with probability $1/2$ each, take values $+1$ and $-1$. How do I find $\lim_{n \to \infty} P(\Sigma X_i \le x)$? Pretty obviously the sum has equal masses at $+\infty$ and $- \infty$, but I am not able to write a proof down. Should I use all moments from the characteristic function and use it to describe the cdf?
 A: We assume that $x$ is fixed for the duration of this solution.  The first solution is deliberately pretty silly.  For no good reason except familiarity, we let $Y_i=(X_i+1)/2$.  Then the $Y_i$ are independent Bernoulli Random variables, and $\sum_{1}^n X_i=2\sum_1^n Y_i -n$.
We are interested in $P(\sum_1^n x_i \le x)$. This is 
$$P\left(\sum_1^n Y_i -\frac{n}{2} \le \frac{x}{2}\right).$$
The random variable $\sum_1^n Y_i$ has Binomial distribution. 
Divide both sides by $n$. We want
$$P\left(\overline{Y} -\frac{1}{2} \le \frac{x}{2n}\right).$$
The random variable $\overline{Y}$ has mean $\frac{1}{2}$ and variance $\frac{1}{4n}$. By the Central Limit Theorem, the probability that $\overline{Y}-\frac{1}{2}$ is $\le y$ is approximately the probability that $Z \le y/\sqrt{1/4n}$, where $Z$ is standard normal. 
Putting things together, we find that for $n$ large, the probability that $\overline{Y}-\frac{1}{2}$ is $\le \frac{x}{2n}$ is approximately the probability that $Z\le \frac{2x\sqrt{n}}{2n}$, that is, the probability that $Z\le \frac{x}{\sqrt{n}}$.   By approximately we mean that the difference between the two probabilities approaches $0$ as $n\to\infty$.
As $n\to \infty$, the number $\frac{x}{\sqrt{n}}$ approaches $0$, so our probability approaches $\frac{1}{2}$.
Remark: The approach we used was very wasteful. In particular, the $Y_i$ were completely unnecessary! The $X_i$ have mean $0$ and variance $1$, what could be nicer than that?  The random variable $\sum_1^n X_i$ has mean $0$ and variance $n$, and therefore standard deviation $\sqrt{n}$. So the probability that $\sum_1^n X_i <x$ is approximately the probability that $Z\le \frac{x}{\sqrt{n}}$. This approaches $\frac{1}{2}$ as $n\to\infty$.
We have deliberately used the Central Limit Theorem imprecisely. That can be replaced by the precise limit version.  
A: You can also compute directly. Fix $x<0$ and $k$ as the biggest integer not over $x$. 
$$P(S_n \leq x)=\{1-P(S_n=0)\}\frac{1}{2}-P(S_n=-1)-...-P(S_n=k+1)$$ and note that those probabilities converge to $0$.
