# Convergence of sum of a sequence of random variables

Consider $(X_{1},\ldots,X_{n})$ a sequence of random variable i.i.d such as $P(X_j=1)=P(X_j=-1)=\frac 12$ for all $j \geq 1$. Consider now the sequence $Y_{n} = \sum_{j=1}^{n} 2^{-j} X_{j}$ for all $n \geq 1$. Proof that $Y_{n}$ converges in distribution to $\operatorname{Unif}(-1,1)$.

It's quite easy to proof that $P(Y_n \leq -1) = P(Y_n\geq 1) = 0$. But how can i get $P\left(\sum_{j=1}^{n} 2^{-j} X_j \leq z\right)$ for $z \in [-1,1]$?

Thanks in advance for any help.

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Divide the inteval $[-1,1]$ into two intervals f equal length. What is the probability of the sequence convergin to a value in each of the two intervals? What if you divide it into four subintervals? Eight? If you can answer that for all subdivisions into $2^n$ subintervals for arbitrary $n$, you can do the general case by using them as approximations. –  Michael Greinecker Dec 31 '11 at 17:39
Thanks, your hint solved my problem :) –  Hernium Dec 31 '11 at 19:02

One can compute the probability of each dyadic event $[k\leqslant 2^nY_n\lt k+1]$, as Michael suggested in comments, or one can use characteristic functions.
For every real number $t$, let $\varphi_n(t)=\mathrm E(\mathrm e^{\mathrm itY_n})$. Note that $\mathrm E(\mathrm e^{\mathrm itX_1})=\cos(t)$. By independence, $$\varphi_n(t)=\prod\limits_{k=1}^{n}\mathrm E(\mathrm e^{\mathrm itX_1/2^k})=\prod\limits_{k=1}^{n}\cos(t/2^k).$$ Here comes a trick: multiply the product on the RHS by $\sin(t/2^n)$ and use recursively from $k=n$ to $k=1$ the relation $2\sin(t/2^k)\cos(t/2^k)=\sin(t/2^{k-1})$. For every $t$ which is not a multiple of $2^n\pi$, this yields $$\varphi_n(t)=\frac{\sin(t)}{2^n\sin(t/2^n)}.$$ When $n\to\infty$, $2^n\sin(t/2^n)\to t$ hence, for every $t\ne0$, $\varphi_n(t)\to\varphi(t)=\sin(t)/t$. Since $\varphi(t)=\mathrm E(\mathrm e^{\mathrm itU})$ with $U$ uniform on $(-1,1)$, this proves that $Y_n\to U$ in distribution.
Note: A stronger result holds: $Y_n\to Y$ almost surely, where $Y=\sum\limits_{n=1}^{+\infty}\frac{X_n}{2^n}$ is uniform on $(-1,1)$.