Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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 $(-1,1)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.