I have a question of the following:
Assume $Y_{1}, Y_{2} \ldots$ is a sequence of iid rvs so that $\mathcal{L}\left(Y_{j}\right)=\frac{1}{4} \delta_{0}+\frac{3}{4} \delta_{1},$ for $j \in \mathbb{N}$ Show that the distribution $W$ are singular where $ W:=2 \sum_{j=1}^{\infty} Y_{j} 3^{-j} $
What I have tried is
suppose $Y_{1}, Y_{2}, \ldots$ are i.i.d. taking the value 1 with probability $\frac{3}{4},$ and the value 0 with probability $\frac{1}{4} . \operatorname{Set} Y=\sum_{n=1}^{\infty} Z_{n} 3^{-n}$ (i.e. the base- $\left.3 \text { expansion of } W \text { is } 0 . Y_{1} Y_{2} \ldots\right) .$
Further, define $S \subseteq \mathbf{R}$ by $ S=\left\{x \in[0,1] ; \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} d_{i}(x)=\frac{3}{4}\right\} $ where $d_{i}(x)$ is the $i^{\text {th }}$ digit in the (non-terminating) base- 2 expansion of $x$ Then by the strong law of large numbers, we have $\mathbf{P}(W \in S)=1$ while $\lambda(S)=0 .$ Hence, the law of $Y$ cannot be absolutely continuous. But clearly $\mathcal{L}(W)$ has no discrete component. We conclude that $\mathcal{L}(W)$ cannot be written as a mixture of a discrete and an absolutely continuous distribution. In fact, $\mathcal{L}(Y)$ is singular with respect to $(\text { written } \mathcal{L}(W) \perp \lambda).$
However, I realise that $W$ is not really the 3-expansion. Is there any other way to prove that.
Thanks
$W$ takes values in the Cantor set. Since $Y_i \in \{0,1\}$ almost surely almost all values of $W$ are of the form $\sum \frac {a_n} {3^{n}}$ where $a_n \in \{0,2\}$. These sums all belong to the Cantor set $C$. Since $C$ has measure $0$ it follows that $W$ has singular distribution. Note that independence of $Y_i$ is not required for this.