# Convergence of Bernoulli distributed random variables with parameter $1/2$?

In my personal study of convergence of random variables I get stuck on this:

I have random variables $X_i$ that are independent and identically distributed with $P(X_i=0)=P(X_i=1)=1/2$.

We define $S_n= \sum_{i=1}^n X_i2^{-i}$ and $R_n=\sum_{i=1}^n 2X_i3^{-i}$. I wanna find out about the distribution of $S_n$ (in this case, I thought I had computed it but it ends up being no distribution I know, so I must be wrong) and show that both converge almost surely (for $S_n$ determining the distribution of the limit and for $R_n$ showing it's not a continuous random variable). I'm not sure I started this properly and my main problem is about proofs with a.s. convergence. I still thought of showing they were Cauchy Sequences (to get rid of the long summations), can I use that?

Can you help me solve this? I think it'll be interesting because it doesn't look like any classic in the matter!

• If you only need to show convergence, then since $X_i \le 1$, we have $S_n = \sum_{i=1}^n X_i 2^{-i} \le \sum_{i=1}^n 2^{-i}$ -- the latter of which converges. (Ditto for the other case). Commented Oct 21, 2014 at 17:37
• How can I know to which random variable it converges? Commented Oct 21, 2014 at 17:58
• Oh, are you trying to find another random variable to which this sum converges into? (e.g., without the $2^{-i}$ factor, it would converge to poisson)? I think I misunderstood your question. Commented Oct 21, 2014 at 18:05
• Yes, for each sequence I'd like to find as much information as possible about the limit. Commented Oct 21, 2014 at 18:49

$\sum_i X_i 2^i$ should converge to Uniform measure on the interval $(0,1)$ and the second one should converge to the ''uniform measure'' on the Cantor set, otherwise known as the cantor distribution (see http://en.wikipedia.org/wiki/Cantor_distribution).
This is because, every number in the unit interval has a binomial expansion given by the a sequence in $\{0,1\}^{\mathbb N}$ and your random sequence gives a way to choose an ''uniform'' element from $\{0,1\}^{\mathbb N}$. Of course, choosing uniformly from this set does not make sense, but you can try to show that every dyadic interval of the form $(j2^{-n}, (j+1)2^{-n})$ has probability $2^{-n}$ for every $S_k$, $k \ge n$ and use Kolmogorov extension theorem.
For the second sequence, the argument is similar if you note that every element in the Cantor set can be written as $\sum_{i}a_i/2^i$ where $a_i \in \{0,2\}$. You can try to show every intervals in the support of the $t$th level (see the link above to see what I mean) of the Cantor set has probability $2^{-t}$, and again use the Kolmogorov extension. To show that the limit is continuous, but not absolutely continuous (it does not have a density function), this is because the limit is supported on the Cantor set and the Cantor set has Lebesgue measure 0.