Let $\zeta=\sum_{n=1}^{\infty}2^{-n}\,\omega_{n}$ , where $\omega_{n}$ are independent, $\mathbb{P}(\omega_{n}=1)=p\neq\frac{1}{2}$ . Also $\mathbb{P}(\omega_{n}=0)=1-p$ for all $n\geq1$ . I want to show that the distribution of $\zeta$ does not have a density, using Radon-Nikodym and Borel-Cantelli Lemma.
- Radon-Nikodym establishes a connection between probability distribution and probability density function. A probability distribution exists if and only if the probability measure is absolutely continuous with respect to the Lebesgue measure. To show that the probability measure $Pr$ is absolutely continuous with respect to the Lebesgue measure $\mu$, we need to first realize that there is a one-to-one correspondence between probability distribution functions and probability measure on the real line. Therefore, for some subset $A$ on the real line it suffices for us to show that $Pr(A)$$\geq$ $\mu(A)$, so that absolute continuity does not hold and the density does not exist.
So, writing the probability distribution of partial sum $\zeta=\sum_{n=1}^{N}2^{-n}\,\omega_{n}$ $=$ $\sum_{m=1}^{N}(1-p)^{m}p^{N-1-m}\left(\sum_{\beta\in\{_{N-1}C_{m}\}}\Pr\left(2\omega_{1}\leq(j-\beta)\right)\right)$, where $\beta$ is just a different combination in the indicies of the sum $\sum 2^{-n}$.
I suspect Borel-Cantelli should come handy here to control the infinite sum perhaps? I am a bit lost, but I would appreciate any advice. Thanks!