In some recent work I found I needed to prove a central limit theorem for the interesting series:

$\sum_{n=1}^\infty \cos (u \log p_n) $

where u is a random variable uniform on the interval $[0,2\pi]$ and $p_n$ is the n-th prime number. If the primes were truly random, this series would essentially be like a random walk and the CLT definitely applies. The primes appear random enough but are not completely random. I checked numerically that the probability distribution for the above series is indeed gaussian (normal), so I believe it does indeed have a CLT.

I'm aware that the CLT is known to apply to some series that are not independently identically distributed random variables if the correlations between them are very weak. The formalism for the latter involves the notion of $\alpha$-mixing which I do not know much about. The above series is similar to the lacunary trigonometric series, but fails the Hadamard gap condition, so those theorems (Salem-Zygmund) do not apply.

Does anyone have any suggestions for how to approach this? Is there a simple way to see if the series would pass the $\alpha$-mixing criterion? I'm not sure I want to invest all the time to learn about the CLT for series with $\alpha$ mixing, so am looking for something simpler.

(the recent work referred to relates this series to the Riemann hypothesis and is at math.NT if you are interested in how the series originated.)

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    $\begingroup$ It is well known that the sequence $\cos(t \log p)$ should behave like independent random variables. See e.g. p.3 of arxiv.org/pdf/1305.4618.pdf or Section 2.2 of arxiv.org/pdf/math/0506218v3.pdf. One can make this rigorous by the Selberg central limit theorem, though for applications this is often not precise enough. $\endgroup$ Nov 6 '14 at 23:13
  • $\begingroup$ Thanks for the comment Peter, but I just looked at those papers and don't see how they directly apply. I could not find the actual series in the paper. Is it implicit? $\endgroup$ Nov 6 '14 at 23:27
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    $\begingroup$ OK Peter, it is clearer in the second paper you mentioned. Thanks. $\endgroup$ Nov 6 '14 at 23:44
  • $\begingroup$ @PeterHumphries: "well known" is a bit of an overstatement, since the second paper you mention explicitly states clearly that they assume they are iid's, so this is not a complete answer. My question is how to actually go about proving it. Nevertheless your comment is useful, since I was not aware of the paper, but only of later papers by some of the same authors. $\endgroup$ Nov 7 '14 at 0:05

If you only look inside $[0,2\pi]$, then it is not true that the distribution is Gaussian.


Set $$ S(x;t) = \sum_{p\le x} p^{it}, $$ so that the sum you are looking at is the real part of $S(x;t)$. Then the Prime Number Theorem implies that for $t\in[0,2\pi]$ we have that $$ S(x;t) = \frac{x^{1+it}}{(1+it)\log x} + O\left(\frac{x}{\log^2x}\right). $$ Therefore $$ \mu:=\frac{1}{2\pi} \int_0^{2\pi} S(x;t)dt \ll \frac{x}{\log^2x}, $$ by integration by parts, and $$ \sigma^2:= \frac{1}{2\pi} \int_0^{2\pi}|S(x;t)|^2 dt \sim \frac{cx^2}{\log^2x}, $$ for some $c>0$. Consequently, $$ \begin{align} M_k&:=\frac{1}{2\pi} \int_0^{2\pi}\left| \frac{S(x;t)-\mu}{\sigma}\right|^{2k} dt \\ & = \frac{1}{2\pi} \int_0^{2\pi}\left| \frac{S(x;t)(1+O(1/\log x))}{\sigma}\right|^{2k} dt \\ &\sim \frac{1}{2\pi c^{k}}\int_0^{2\pi} \frac{dt}{(1+t^2)^k} \in[c_1^k,c_2^k] \end{align} $$ for all fixed $k\in\mathbb{N}$, where $c_1$ and $c_2$ are certain positive constants. If the distribution of $S(x;t)$ were Gaussian, then $M_k$ would grow like $c_0^k k!$ for some $c_0>0$, much faster than its current growth.


You do get a Gaussian distribution, but you have to look at appropriate ranges of $t$ with respect to $x$ (and we have to look at long intervals of $t$, because we don't understand well what happens inside very short intervals). This is part of the context of the Selberg Central Limit Theorem for the Riemann zeta function. In the proof of his theorem, Selberg shows that the statistical behavior of $\log\zeta(1/2+it)$ for $t\in[T,2T]$ can be modeled by the statistical behavior of $\sum_{p\le T^\epsilon} 1/p^{1/2+it}$ for some $\epsilon=\epsilon(T)$ that tends to 0 slowly. Then he proceeds to estimate moments of the latter sum (which is essentially equivalent to studying the distribution of the sum you're interested in).

The rough idea is that $$ \begin{align} \int_T^{2T} S(x;t)^k \overline{S(x;t)}^{\ell} dt &= \sum_{p_1,\dots,p_{k+\ell}\le x} \int_T^{2T} \left(\frac{p_1\cdots p_k}{p_{k+1}\cdots p_{k+\ell}}\right)^{it} dt \\ &= T\sum_{\substack{ p_1,\dots,p_{k+\ell}\le x \\ p_1\cdots p_k=p_{k+1}\cdots p_{k+\ell}}} 1 + O\left( \sum_{\substack{ p_1,\dots,p_{k+\ell}\le x \\ p_1\cdots p_k\neq p_{k+1}\cdots p_{k+\ell}}} \frac{1}{|\log(p_1\cdots p_k/(p_{k+1}\cdots p_{k+\ell}))|}\right). \end{align} $$ The main term above is 0 unless $k=\ell$, in which case it equals asymptotically $T$ times the $2k$-th moment of a Gaussian. The error term can be shown to be small if $x^{\max\{k,\ell\}}\le T^{1/4}$. Indeed, in this case $p_1\cdots p_k,p_{k+1}\cdots p_{k+\ell}\le T^{1/4}$ and thus $|\log(p_1\cdots p_k/(p_{k+1}\cdots p_{k+\ell}))|\ge 1/T^{1/4}$. So the total error term is $T^{1/4} \pi(x)^{k+\ell} \le T^{3/4}$, which is small enough for our purposes. Hence, if $x=T^{o(1)}$, then an increasingly (as $T\to\infty$) large number of moments is shown to match the moments of the Gaussian, and standard probability results then imply that the distribution of $S(x;t)$ for $t\in[T,2T]$ is Gaussian.

  • $\begingroup$ This is not the answer for a simple reason: if you use the leading term in the prime number theorem as you did above, this does not capture the fluctuations in the primes, i.e. their randomness. $Li(x)$ is smooth. Another way of saying this is that if we replace $p_n$ by $n\log n$ in the series, you lose the gaussian for the same reason. $\endgroup$ Nov 10 '14 at 19:57
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    $\begingroup$ The oscillation of $p^{it}$ is very mild when $t$ is bounded and not enough to give you square-root cancellation. The asymptotic $S(x;t)\sim x^{1+it}/\log x$ is a direct consequence of the Prime Number Theorem and partial summation. When $t=0$, this is the original Prime Number Theorem. As a matter of fact, you can replace $p_n$ by $n\log n$ when $t$ is bounded (with some error of course). The Gaussian distribution does not come from the fluctuations in the distribution of the primes, but by their multiplicative independence. $\endgroup$ Nov 10 '14 at 20:26
  • $\begingroup$ Interesting comment. It is true that numerically the gaussian is much better for higher t, where the actual series we are interested in is $\sum_p \cos (u t \log p)$. Can you be more specific, perhaps give a reference, or send me a longer explanation in private? $\endgroup$ Nov 10 '14 at 23:01
  • $\begingroup$ You can look at Selberg's paper "Contributions to the theory of the Riemann zeta-function" (Arch. Math. Naturvid. 48, (1946). no. 5, 89–155) or Tsang's thesis (available here math.sjsu.edu/~goldston/TsangThesis.htm; Section 3 should be relevant). In these references, most of the hard work is devoted to the study of $\zeta$, but you can still find information about $\sum_{p\le x}1/p^{1/2+it}$. Also, Lamzouri has some related results (Theorem 4A here math.yorku.ca/~lamzouri/distribzeta.pdf is a pretty strong manifestation of the multiplicative independence of primes). $\endgroup$ Nov 11 '14 at 0:36
  • $\begingroup$ thanks, I am studying the Selberg CLT now and indeed it is quite close to what I need, so may be very useful. I think your answer is the best one can do at the moment, so I marked it as "answered". $\endgroup$ Nov 11 '14 at 16:07

The previous answer was helpful, thank you. But I think I can see a stronger version. If you consider the series

$\sum_n \cos ( u \lambda_n) $

where $u$ is a random variable, then this series satisfies the CLT if the $\lambda_n$ are linearly independent over the integers. Now, $\lambda_n = \log p_n$ are linearly independent by the unique prime factorization theorem.


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