This is my third question on this site regarding Montgomery's conjecture -- and I apologize
if this is too much -- but I am still not understanding well why this conjecture is believed to be true. 


The conjecture I am talking about is as follows (I am giving the slightly corrected version of Freidlander-Granville). 
et $q>1$ be an integer, $a$ an integer coprime to $q$, $\psi(q,a,x) = \sum\_{p^\alpha < x, p^\alpha \equiv a \pmod{q}} \Lambda(n)$. Then:


**Conjecture** for $x>q$, one has $\psi(x,q) = \frac{x}{\phi(q)} + O(x^{1/2+\epsilon} q^{-1/2})$, with an implied constant depending only on $\epsilon$. 


In his answer to my preceding [question](https://mathoverflow.net/questions/115438), Matt Young gives the following heuristic:
For $\chi$ a non-principal Dirichlet character of $(\mathbb Z/q\mathbb Z)^\ast$, one has under GRH
$\psi(\chi,x) = O(x^{1/2+\epsilon})$. Now $\psi(q,a,x)$ is the arithmetic average of the (approximately $q$) terms $\chi^{-1}(a) \psi(\chi,x)$, and if those (weighted) terms are in random position one should expect their sum to have a norm roughly $q^{1/2}$ the norm of the individual term (by Einstein's Brownian motion theorem if you like), giving the conjecture.


This heuristic helped me a lot then but now I'd like to go further. 



> 
> Is there a good reason to believe that the $\chi^{-1}(a) \psi(\chi,x)$ are kind of randomly distributed, for every $a$ relatively prime to $q$? Could this expectation have some link,
>  through the explicit formulas, with the conjectures on the position of zeros on the critical line of Dirichlet L-functions (assuming GRH)? 
> 
> 
> 



Heuristic for Montgomery's conjecture
This is not exactly what you asked, but I believe that there is a simpler heuristic for Montgomery's conjecture which does not involve characters.


Let $A$ be the arithmetic progression $a\pmod q$, where $a$ and $q$ are coprime to each other. We consider a sequence of Bernoulli random variables $(X\_n)\_{n\in A}$ such that


$$\textbf{Prob}(X\_n=1)=\frac{q}{\phi(q)\log n} \quad\text{and}\quad \textbf{Prob}(X\_n=0)=1-\frac{q}{\phi(q)\log n}.$$


The event $X\_n=1$ models the event that $n$ is prime. (There are about $\le x/q$ elements in $A$ up to $x$ and about $x/(\phi(q)\log x)$ of them are primes.) Let $\Psi(x)=\sum\_{n\le x} X\_n\log n$. This is a random model for $\psi(x)$ (or rather $\theta(x)=\sum\_{p\le x}\log p$, but this is merely a technicality). Note that


$$
\mathbb{E}[\Psi(x)] = \frac{q}{\phi(q)}\sum\_{n\le x,\,n\equiv a\pmod q}1 
= \frac{q}{\phi(q)}(x/q+O(1))=\frac{x}{\phi(q)} + O(q/\phi(q))
$$
and
$$
\text{Var}[\Psi(x)] = \frac{q}{\phi(q)} \sum\_{n\le x,\, n\equiv a\pmod q}\log n 
\sim \frac{x\log x}{\phi(q)}.
$$
The Central Limit Theorem implies that the random variable $(\Psi(x) - x/\phi(q))/\sqrt{x\log x/\phi(q)}$ tends to the standard normal distribution. In particular, almost surely as $x\to\infty$, we have that
$$
\left|\Psi(x) - \frac{x}{\phi(q)} \right| \ll x^\epsilon \sqrt{\frac{x\log x}{\phi(q)}}.
$$
(In fact, one should expect this to become accurate as soon as $x/q$ becomes a bit large, that is to say there are enough summands in $\Psi(x)$.) So we deduced Montgomery's conjecture for $\Psi(x)$, the random model of $\psi(x)$.


