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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.

  • 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.2011-12-31
  • 0
    Thanks, your hint solved my problem :)2011-12-31

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