distribution of infinite sum of $\sum (2x_n -1)/2^n$ $\{X_n\}\sim\mathrm{Bernoulli}(\frac {1}{2})$ $$Y=\sum_{n=0} ^{\infty} \frac {2X_n -1}{2^n}$$
Find the distribution of $Y$
$X_n$ are independent 
 A: It's a uniform distribution on $[-2,2]$.  Suppose $W\sim\mathrm{Uniform}[-2,2]$.  Then $(W+1)/4\sim\mathrm{Uniform}[0,1]$.  Then show that the usual binary digits of $W$ are distributed exactly as you have said $\{X_i\}$ are distributed.
PS: By popular demand expressed in comments below, here are some details of the last step above.  What is the distribution of the $n$th binary digit of a random number uniformly distributed in $[0,1]$?  As a variable $x$ runs from $0$ to $1$, the $n$th binary digit alternates between $0$ and $1$, being $0$ on the leftmost interval of length $1/2^n$, then $1$ in the next interval of that length.  So the probability that it is $1$ is $1/2$.  Next, what is the probability that the $n_1\text{th}, n_2\text{th},\ldots,n_k\text{th}$, digits are equal to some specified sequence of $0$s and $1$?  Suppose these are in increasing order so $n_k$ is the biggest.  The $n_k$th digit of $x$ alternates between $0$ and $1$ as $x$ moves through any of the intervals where the $n_1\text{th}, n_2\text{th},\ldots,n_{k-1}\text{th}$ digits have their specified values, constantly having each value while going through an interval of length $1/2^{n_k}$.  Thus the probability is $1/2$ the probability that the $n_1\text{th}, n_2\text{th},\ldots,n_{k-1}\text{th}$ digits have their specified values.  Procede by mathematical induction to show that it's $1/2^k$, and there you have independence of digits.
A: I am going to use characteristic function. You are going to need to translate this into moment generating somehow (with the appropriate identities)
For $t\neq 0$
$$\mathbb{E}e^{i\lambda Y} = \prod_{i=1}^\infty \mathbb{E}\exp \left(\frac{2i\lambda X_n-i\lambda}{2^n}\right)=\prod_{n=0}^{\infty}(0.5e^{-i\lambda/2^n}+0.5e^{i\lambda/2^n})=\prod_{n=0}^{\infty}\cos\left(\frac{\lambda}{2^n}\right)=\frac{\sin(\lambda)\cos(\lambda)}{\lambda}=\frac{\sin(2\lambda)}{2\lambda}$$
Note this is the characterstic function for a uniform distribution on $[-2,2]$.
I assume doing it using MGF is the same except you have to drop the $i$ and find an identity for $\prod_{n=0}^{\infty}\cosh(\frac{\lambda}{2^n})$
A: Note that $X\sim \mathrm{Bin}(1,\frac{1}{2})$  is just a $X\sim \mathrm{Bernoulli}(\frac{1}{2})$ Thus distribution of $2X_{n}-1$
$$
f(x) =
\begin{cases}
1, & p=\frac{1}{2} \\
-1, & p=\frac{1}{2}
\end{cases}
$$
well than letting $\dfrac{2X_{n}-1}{2^n}$ has distribution 
$$
f(x) =
\begin{cases}
\dfrac{1}{2^n}, & p=\frac{1}{2} \\
-\dfrac{1}{2^n}, & p=\frac{1}{2}
\end{cases}
$$
Thus in order to find distribution of $\sum^\infty W_n$ We can see what moment generating function converges to and see match the distribution that has that as its moment generating function. And since these $W_{n}$ are independent it will be product of each $X_{n}$ MGFs. Thus we have where MGF for $W_{n}$ will be 
$$M(t)_{W_n}=E(e^{W_n t})=\frac{1}{2}\left(e^{\frac{1}{2^n}t}+e^{\frac{-1}{2^n}t}\right)=\cosh\left(\frac{1}{2^n}t\right)$$
So we have that 
$$M(t)_{\sum^{\infty}W_{n}}=\prod_{n=1}^{\infty}M(t)_{W_{n}}=\prod_{n=1}^{\infty}\cosh(\frac{t}{2^{n}})$$
where there as this site says http://functions.wolfram.com/ElementaryFunctions/Cosh/introductions/Cosh/05/ShowAll.html
$$M(t)_{\sum^\infty W_n}=\frac{\sinh(t)}{t}$$
Now I couldn't find what distribution matched this but if you can then there is your answer
