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Considering the (infinite) set of all positive integers that are a product of $2$ primes only, represented in binary $100...01$.

Question: is the distribution of the proportion of $0,1$ digits "uniform" (meaning $Pr(0) = Pr(1)$) ? Or should we expect an asymmetric distribution (eg. the relative frequency of "$1$"s to exceed that of the "$0$"s) ?

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    Which probability distribution are you using for picking numbers to sample bits from? It can't be uniform, because there's no such thing possible for a countably infinite set.2012-12-30
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    @ Henning Makholm Well i am working in the space of "all set of all positive integers that are a product of 2 primes". There is no sampling scheme here. I am looking for a statement regarding the entire population of moduli.2012-12-30
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    My personal hunch is that there should be an asymmetry slighly favoring the "$1$"s. But I am having hard time to find a way to justify that in a rigorous way.2012-12-30
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    Yes I see you're working on that set. But you cannot define _averages_ over such a set without first selecting a distribution to average over.2012-12-30
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    @ Henning Makholm The "distribution" is here the "popolation distribution". There is no sampling scheme here. (Even the original population has a distribution with its own mean, variance, etc.) [The set where you "average over" is the entire set of the moduli.]2012-12-30
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    What does "population distribution" mean? Again, you cannot take averages without specifying what the weights you're going to use for each member of your population are. (There's no such thing as an unweighted average of countably many elements).2012-12-30
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    @Henning Makholm Consider the set of all moduli. Over this set the number of 1 and 0 in the binary representation has a certain (unknown) Distribution. I am interested in the mean of such distribution.2012-12-30
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    >(the number of 1 and 0 in the binary representation) More precisely: the proportion of 0/1 in the binary representation.2012-12-30
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    You're still missing the point: **You can't take a mean of a countably infinite set of values without first choosing a weighting to use for each of those values.**2012-12-30
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    @ Henning Makholm Actually i don't understand your objection. Take the Poisson. It is over a countable infinite set and has mean and variance. Maybe i don't master English enough to correcly describe the problem, and some native English speaker can help here?2012-12-30
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    I will retry a simple explanation of the problem. Take the modulus 2*3, then take 2*5, ... then take 3*5, 3*7... Now imagine the whole set of such moduli in their binary representation. Over this set you will have all these moduli 100....1. For each binary representation, there will be N0 digits = 0 and N1 digits equal to 1. For each binary representation denote by F0 = N0/N and F1 = N1/N the relative frequency of 0/1 (N is the number of digits in the given representation). Basically, the question is: if we consider F0 over the whole population of moduli is it true that F0 = F1 = 0.5 ?2012-12-30
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    Your problem is still unchanged. You have defined a countably infinite sequence of $F(0)$s, but you cannot take an average of those values without first specifying a weighting among them. (When you say "take the Poisson", are you referring to a Poisson distribution? That is exactly a _distribution_, i.e., what you're refusing to specify for your $F(0)$s here).2012-12-30
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    Well if i could specify F1, F0 (so, the "distribution"), i would have answered myself the question. The question is just about the population distribution. Evidently, i am missing to effectively communicate what the problem is. Sorry.2012-12-30
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    I'm afraid I'm out of ways to explain why your statement of what you want is not complete.2012-12-30
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    As Henning says, your question isn’t well-defined. Let $P$ be the set of products of two primes, and let $P_n=\{k\in P:k\le n\}$. Let $a_n$ be the number of $0$’s in the binary representations of elements of $P_n$, and let $b_n$ be the number of $1$’s. It would be meaningful to ask whether $\lim_n\frac{a_n}{a_n+b_n}$ exists and, if so, whether it’s $\frac12$. Is that what you’re after?2012-12-30
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    @Brian M. Scott Yes, essentially that is it. (But maybe that is a bit too strong: it would be enough if the limit holds with probability 1 or with probability converging to 1. So an asymptotically almost sure (a.a.s) or a probability convergence or any other form of weaker convergence would also be interesting). Thanks a lot for a clearer formulation.2012-12-30

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