The distribution $\nu\_{\lambda}$ of the random series $\sum\pm\lambda^n$ is the infinite convolution product of $\frac12(\delta\_{-\lambda^n}+\delta\_{\lambda^n})$. This problem has been [studied extensively](http://www.math.uchicago.edu/~schlag/papers/sixty.pdf).


I am curious about the following modification:



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> **QUESTION.** Let $\mu\_{\lambda}$ be the distribution of $\sum\_{n=0}^{\infty}\epsilon\_n\lambda^n$ where the $\epsilon\_n$ are chosen from $\{-1,0,1\}$ independently with probability $\frac13$. For which $0<\lambda<1$ is $\mu\_{\lambda}$ absolutely continuous or singular?
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Has this been investigated? If so, I would be grateful for any reference.



a modification on an infinite Bernoulli convolution
1. If $\lambda\in(0,1/3)$, then $\mu\_\lambda$ is supported by a Cantor set of zero measure and therefore, is singular.
2. If $\lambda=1/3$, then it is the Lebesgue measure on its support.
3. If $\lambda^{-1}$ is a Pisot number, then $\mu\_\lambda$ is singular, via a standard Erdős type argument. (Its Fourier transform does not vanish at $\infty$, which means that it cannot be absolutely continuous.)
4. For a.e. $\lambda\in(1/3, 1)$ it is absolutely continuous. I can't remember who proved this, but this is a standard result in the theory (via transversality).


Pablo Shmerkin definitely knows this. I suspect the exceptional set has zero Hausdorff dimension, too, which is his result for the Bernoulli convolutions. 


