a modification on an infinite Bernoulli convolution 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.
I am curious about the following modification:

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?

Has this been investigated? If so, I would be grateful for any reference.
 A: As can be inferred from Nikita's answer, the situation for the family $\mu_\lambda$ is very similar to that of the classical Bernoulli convolutions $\nu_\lambda$, and all the known techniques for the latter situation work almost verbatim for the former. 
I don't have anything to add to Nikita's points 1-3, so let me add some details of what is known about absolute continuity:


*

*For any $k$ there is $\varepsilon_k$ such that $\mu_\lambda$ has a $C^k$ density for almost all $\lambda\in (1-\varepsilon_k,1)$. Erdös proved this for Bernoulli convolutions in 1940, and the same proof works for the measures $\mu_\lambda$ (the argument is based in the fact that the Fourier transform of $\mu_\lambda$ has power decay at infinity).

*$\mu_\lambda$ is absolutely continuous with an $L^2$ density for a.e. $\lambda\in (1/3,1)$. This can be proved with the transversality technique used by Solomyak in 1995 to prove that $\nu_\lambda$ is absolutely continuous for a.e. $\lambda\in (1/2,1)$. See: [Simon, Károly; Tóth, Hajnal R. The absolute continuity of the distribution of random sums with digits $\{0,1,\dots,m-1\}$. Real Anal. Exchange 30 (2004/05), no. 1, 397--409]; they also consider the similar problem for any number of equally spaced translations. The modification is not trivial because the interval on which transversality holds depends on the number of translations.

*It is a special case of Theorem 1.2 in  my paper [Shmerkin, Pablo. On the exceptional set for absolute continuity of Bernoulli convolutions. Geom. Funct. Anal. 24 (2014), no. 3, 946--958. ] that $\mu_\lambda$ is absolutely continuous for all $\lambda\in (1/3,1)$ outside of a set of zero Hausdorff dimension (this is strongly based on Mike Hochman's results for dimensions of self-similar measures).

*The last two results are now superseded by my recent preprint [On Furstenberg's intersection conjecture, self-similar measures, and the Lq norms of convolutions, arXiv:1609.07802]. It follows from the results there that $\mu_\lambda$ has an $L^q$ density for all finite $q$, for all $\lambda\in (1/3,1)$, again outside of a (possible) exceptional set of zero Hausdorff dimension. This is not explicitly stated in my paper, but see Theorem 9.1 and the discussion afterwards.

*What about absolute continuity for specific values of $\lambda$? A recent breakthrough of P.Varju [Absolute continuity of Bernoulli convolutions for algebraic parameters, arXiv:1602.00261] provides many explicit (in terms of a small constant which is not specified) examples of algebraic numbers for which $\nu_\lambda$ is absolutely continuous. I am quite sure the arguments extend to $\mu_\lambda$ but I am less familiar with this. Breuillard and Varju also have some impressive recent results about full dimension of Bernoulli convolutions for many explicit parameters (not only algebraic) that should also extend to more general self-similar measures. 


All of these results hold in fact (for the appropriate range of $\lambda$) for the distribution of the random sums
$$
\sum_{n=1}^\infty \varepsilon_n \lambda^n,
$$
where $\varepsilon_n$ are iid random variables taking finitely many integer values (and for most of the results the values don't even have to be integers).
A: *

*If $\lambda\in(0,1/3)$, then $\mu_\lambda$ is supported by a Cantor set of zero measure and therefore, is singular. 

*If $\lambda=1/3$, then it is the Lebesgue measure on its support.

*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.)

*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. 
