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Suppose $a_n > 0$ is a sequence of real numbers in $l^2 \setminus l^1$. i.e. $\sum a_n^2 < \infty$ but $\sum a_n = \infty$.

If $B_n$ are an infinite sequence of independent Bernoulli random variables with parameter $\frac{1}{2}$ then the infinite sum $\sum a_n (-1)^{B_n}$ exists almost surely and converges in $L^2$ and almost surely to some random variable, say $A$ (standard result from Martingale theory, although you can prove it directly).

$A$ has the property for any $u < v$, $P(u < A < v) > 0$, so it is "almost continuous", but it's not obvious to me that the distribution can't have atoms. i.e. values $u$ such that $P(A = u) > 0$.

So my question is:

  1. Does a sequence $a_n$ exist with $P(A = 0) > 0$? (if a sequence with any atom exists, one with $0$ being an atom exists because you can just shift $a_1$).
  2. For the specific sequence $a_n = \frac{1}{n}$ does $A$ have any atoms?

I conjecture the answer to both is no, but I'm struggling with the details of proving it.

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  • $\begingroup$ By Bernoulli do you mean Bernoulli with parameter $1/2$? Otherwise the sum diverges almost surely. $\endgroup$ – Robert Israel Jan 5 '17 at 19:20
  • $\begingroup$ Yes, sorry, I do. $\endgroup$ – David R. MacIver Jan 5 '17 at 19:56
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The answer is no.

For any $T \subseteq \mathbb N$, let
$$ A(T) = \sum_{n \in T} a_n (-1)^{B_n}$$ Now $A = A(T) + A(\mathbb N \backslash T)$ where $A(T)$ and $A(\mathbb N \backslash T)$ are independent, and so if $A$ had atoms both $A(T)$ and $A(\mathbb N \backslash T)$ would have atoms. Thus it suffices to find $T$ such that $A(T)$ has no atoms.

In particular, take an increasing sequence $T = \{t_1, t_2, \ldots\}$ such that $0 < a_{t_{j+1}} < a_{t_j}/2$. Then the function $f_T(b_1, b_2, \ldots) = \sum_{j} a_{t_j} b_j$ on $\{-1,1\}^\mathbb N$ is one-to-one. For any real $c$ there is at most one sequence of values of $B_t$, $t \in T$ that makes $A(T) = c$, and thus $P(A(T)=c) = 0$.

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