For a sequence $ a_{n}\in\left\{ 0,1,2\right\} $ we define $ [a]_{3}=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}} $
(Thats just Base 3 expansion).
Now let $ A=A_n $ be a sequence of i.i.d random variables, each uniformly distributed over {0,2}
Define $ X=[A]_{3}=\sum_{n=1}^{\infty}\frac{A_{n}}{3^{n}} $
Such $ X $ is said to have to Cantor distribution.
That is all the given details, the way I see it, $ X $ is a random variable on some $ \varOmega $ such that for any $ \omega \in \varOmega$ we have:
$ X\left(\omega\right)=\sum_{n=1}^{\infty}\frac{A_{n}\left(\omega\right)}{3^{n}} $
And the value of $ A_n(\omega) $ can be $ 0 $ or $ 2 $, So that $ X(\omega) $ is in $ [0,1] $ for any $ \omega $.
Now based on the way I understood the details, I want to prove that X is continuous RV. Thus, its sufficies to show that
$ \mathbb{P}\left(X=r\right)=0 $
For any $ r \in \mathbb{R} $.
fix some $ r \in \mathbb{R} $. If there is no sequence $ (a_n)_n $ such that $ r=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}} $ and $ a_{n}\in\left\{ 0,2\right\} $ for any $ n $,
Then $ \mathbb{P}\left(X=r\right)=0 $ because $ X^{-1}\left\{ r\right\} =\emptyset $
So let $ R $ be the set such that for any $ r \in R $ we have a sequence $ \left(a_{n}^{r}\right)_{n} $ such that
$ \sum_{n=1}^{\infty}\frac{a_{n}^{r}}{3^{n}}=r $
and $ a_{n}^{r}\in\left\{ 0,2\right\} $ for any $ n $.
CLAIM 1: I calim that since we can only use the digits $ 0 $ and $ 2 $, there is only one unique seuquence $ a_n^r $ for any $ r \in R $ such that all the demands holds, and further more for $ r_1,r_2 $ we have $ r_{1}\neq r_{2}\iff\left(a_{n}^{r_{1}}\right)_{n}\neq\left(a_{n}^{r_{2}}\right)_{n} $.
If this claim is ture, then:
Note that for any $ \omega \in \varOmega $ we have:
$ X\left(\omega\right)=r\iff\sum_{n=1}^{\infty}\frac{A_{n}\left(\omega\right)}{3^{n}}=r\iff\forall n:A_{n}\left(\omega\right)=a_{n}^{r} $
And this is true because of claim 1, the uniquness of the sequence that defines a number $ r \in R $ , there is no other way for the sum $ \sum_{n=1}^{\infty}\frac{A_{n}\left(\omega\right)}{3^{n}} $ to converge to $ r $ unless $ A_{n}\left(\omega\right)=a_{n}^{r} $ because if we'll have just one term such that $ A_{j}\left(\omega\right)\neq a_{j}^{r} $, we'll get other number than $ r $. (as long as claim 1 is correct).
Now if we agree up to this point, then for any $ r \in R $ :
$ \mathbb{P}\left(X=r\right)=\mathbb{P}\left(\omega\in\varOmega:A_{n}\left(\omega\right)=a_{n}^{r},\forall n\in\mathbb{N}\right)=\mathbb{P}\left(A_{1}=a_{1}^{r}\cap A_{2}=a_{2}^{r}...\right) $
$ =\mathbb{P}\left(\bigcap_{j=1}^{\infty}A_{j}=a_{j}^{r}\right) $
Since $ (A_n)_n $ is a sequence of independent RVs, we have that all those events in the intersecntion are mutually independent and so :
$ \mathbb{P}\left(\bigcap_{j=1}^{\infty}A_{j}=a_{j}^{r}\right)=\prod_{j=1}^{\infty}\mathbb{P}\left(A_{j}=a_{j}^{r}\right) $
And since for any $ j $ we have that $ A_j $ is uniformly distributed on $ \left\{ 0,2\right\} $ and $ a_j^r \in$ {0,2} $ then
$ \mathbb{P}\left(A_{j}=a_{j}^{r}\right)=\frac{1}{2} $
So we have:
$ \mathbb{P}\left(X=r\right)=\prod_{j=1}^{\infty}\mathbb{P}\left(A_{j}=a_{j}^{r}\right)=\prod_{j=1}^{\infty}\frac{1}{2}=\lim_{k\to\infty}\prod_{j=1}^{k}\frac{1}{2}=\lim_{k\to\infty}\frac{1}{2^{k}}=0 $
And thus for any $ r \in \mathbb{R} $, $ \mathbb{P}(X=r) =0 $
And so $ X $ is continious.
So questions:
- Is claim 1 correct?
2.Is this proof holds?
There are few things which I would like to say here. The first claim is true, and the idea seems to be true as well. I want to give out some facts I hope will let you feel better about that:
We need to pick the expansion of the interval [0,1], where for every number the expansion is unique.
It turns out that there is , at least for the "most" of points, unique expansion. The problem of uniqueness arise where we try to expand numbers of the form $\frac{p}{3^k} $. In these cases, there are two optional expansions- One s.t $ \forall j>k $ ,$a_j =0$. The other is of the form $a_j=2 $ ($\forall j>k $). If we pick the second expansion, then it is unique on the whole unit interval.
For example, $ \frac{1}{3}=0.10000...$
And the second expansion, which we prefer, is given by $ \frac{1}{3}=0.02222...$\
I hope it makes you feel more confident about that proof.