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

Question

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".

Answer 1

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 )