what does it mean that the function $ F(t) $

$$ F(t)= \frac{\arg\zeta (\frac{1}{2}+it)}{\sqrt{\log\log(t)}} $$

is distributed as a 'Gaussian Random variable ?? in the limit $ t \to \infty $

a) $$ \arg\zeta (\frac{1}{2}+it)=(1+o(1))\sqrt{\log\log(t)}$$

b) the Argument of the Zeta function on the critical line is almost $ \sqrt{\log\log(t)} $

  • $\begingroup$ A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t \rightarrow \infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically. $\endgroup$ – Ron Gordon Dec 31 '12 at 17:47

Consider a measurable real-valued function $G$ defined on $(0,+\infty)$. For every $T\gt0$ and real number $x$, define $\ell_T(x)$ as the Lebesgue measure of the set $\{t\leqslant T\mid G(t)\leqslant x\}$.

One says that the function $G$ is asymptotically distributed as a standard normal random variable if, for every real number $x$, $$ \lim\limits_{T\to\infty}T^{-1}\ell_T(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^x\mathrm e^{-z^2/2}\mathrm dz, $$ that is, $$ \lim\limits_{T\to\infty}T^{-1}\int_0^T\mathbf 1_{G(t)\leqslant x}\,\mathrm dt=\mathbb P(Z\leqslant x), $$ where $Z$ is a standard normal random variable.

There exists some variants of this definition but the idea remains the same.


What Selberg proved is that for $E\subset \mathbb R$, we have that the limit as $T\to\infty$ of $$ \frac{1}{T}\mu(T\le t\le 2T\,|\,\arg(\zeta(1/2+i t)/\sqrt{1/2\log\log t}\in E) $$ where $\mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$: $$ \frac{1}{\sqrt{2\pi}}\int_E \exp(-x^2/2)\, dx. $$

Edit: Here's an example that may help clarify, using the harmonic conjugate $\log|\zeta(1/2+i t)|$ (which is implemented in Mathematica). The analog of Selberg's theorem is true for this function as well. The plot is of $\log|\zeta(1/2+i t)|/\sqrt{1/2\log\log(t))}$, for $50\le t\le 100$. Note it looks nothing like a Gaussian.

plot of log(abs(zeta))

Here's a histogram of $50000$ equally spaced values values taken by this function:

histogram of log(abs(zeta))

Extreme negative values (near the Riemann zeros) are extremely scarce, as are large positive values. The fit to the bell curve is not yet good, but the Riemann zeta function approaches its asymptotic behavior very slowly.

  • $\begingroup$ then stopple what you mean is that $$ \frac{1}{T}\int_{T}^{2T}dt \frac{arg \xi(1/2+it)}{\sqrt{(1/2)loglog(t)}}= \frac{1}{\sqrt{2\pi}}\int_{E}exp(-x^{2}/2) $$ is this true $\endgroup$ – Jose Garcia Jan 2 '13 at 11:31
  • $\begingroup$ @Jose: Well, no. Your right side depends on $E$; your left side does not. $\endgroup$ – stopple Jan 2 '13 at 15:29
  • $\begingroup$ @Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $\arg(\zeta(1/2+it))/\sqrt{1/2\log\log(t)}$, it will look like the Gaussian, i.e. a bell curve. $\endgroup$ – stopple Jan 2 '13 at 17:01
  • $\begingroup$ sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ arg\zeta(1/2+it)/\sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience $\endgroup$ – Jose Garcia Jan 2 '13 at 20:05
  • $\begingroup$ @Jose: see edit above. $\endgroup$ – stopple Jan 2 '13 at 21:11

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