If the value of Mertens function follows normal distribution, does this imply Riemann Hypothesis ?

I thought the answer shall be NO, because normal distribution still has "long tail".

  • $\begingroup$ Which Mertens function ? $M(x)=\sum_{n\le x}\mu(x)$ ? $\endgroup$ – Dietrich Burde Jun 1 '16 at 18:35
  • $\begingroup$ What do you mean by your second statement? $\endgroup$ – anomaly Jun 1 '16 at 18:38
  • $\begingroup$ What I mean is that, if the value of Merterns function follows normal distribution, so most of its value will fall into "square root accurate" region, but there still can be some values which will be in the long tail part of distribution, which will NOT "fall into square root region". $\endgroup$ – david Jun 1 '16 at 18:42

If the value of $X_n$ follows a random i.i.d. distribution with zero mean $\mathbb{E}[X_n] = 0$, zero covariance $\mathbb{E}[X_nX_m] = 0, n \ne m$ and bounded variance $\mathbb{E}[X_n^2] < C$) then $$\mathbb{E}[(\sum_{n < x} X_n)^2)] = \mathbb{E}[\sum_{n < x} \sum_{m < x} X_nX_m] = \sum_{n < x} \sum_{m < x} \mathbb{E}[X_nX_m] = \sum_{n < X} \mathbb{E}[X_n^2] < x C$$

Hence the random Dirichlet series $$F(s) = \sum_{n=1}^\infty X_n n^{-s} = s \int_1^\infty (\sum_{n < x} X_n) x^{-s-1}dx$$

is almost surely holomorphic for $Re(s) > 1/2$.

It follows that if the probabilistic model for $\mu(n)$ was true (it is not, since the primes are perfectly deterministic) : namely that $\mu(n)$ follows an i.i.d. distribution with $$P(\mu(n)= 1) = P(\mu(n)= -1) = \frac{3}{\pi^2}$$ then $\displaystyle\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}$ would be holomorphic on $Re(s) > 1/2$ and the Riemann hypothesis would be true.

(see arguments_for_and_against_the_Riemann_hypothesis )


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