I think my argument can be made rigorous rather straightforwardly, although perhaps it assumes that there is an answer. It is rather trivial to see that if you start with a [0,1] uniformly distributed random variable and express it as a binary fraction, then each bit must have equal probability of being 0 or 1 (just look at the part of the distribution with all the other bits fixed). Taking the bits as a sequence a[n] of 0s and 1s, then 2*a[n]-1 is your sequence, and it's clear that the new sum is twice the old minus 1, transforming uniform in [0,1] to uniform in [-1,1]. No? --ms On 2014-02-14 00:08, Dan Asimov wrote:
As Fred says, fascinating! Thanks, Victor.
I had no idea that slightly changing the question leads to such subtleties. Though it shouldn’t be surprising that the question has been asked before. (Long before Erdős discussed it? Or not?)
I got the same answer Mike Speciner did: uniform on [-1,1], though maybe a bit more rigorously, using a Fourier transform to get essentially the product for n >= 1 of cos(w/2^n). Which I couldn’t evaluate explicitly but found that Viéte (Vieta) evaluated in 1593 as sinc(w) := sin(w)/w. Then looking up the inverse Fourier transform to get U[-1,1].
It seems that in some sense the density ought to have a spike at any x = 1/2^n in (-1,1), since there are countably many sign patterns with that sum. For all other x in [-1,1], there is a unique sign pattern {e_j} that gives x = Sum e_j/2^j. Don’t know if there is any rigorous way to make sense of this.
—Dan
On Feb 13, 2014, at 2:03 PM, Victor Miller <victorsmiller@gmail.com> wrote:
Your problem is known as the Bernoulli Convolution. Look here for a nice survey.
On 2/13/2014 11:18 AM, Dan Asimov wrote:
Let e_j, j = 1,2,3,... be independent random variables each taking the values +-1 with probability 1/2.
Let the random variable X be defined as
X := Sum_{n=1...oo} e_j/2^j.
PUZZLE: What is the distribution of X ???
--Dan
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