TITLE: Is there a connection of prime numbers and extreme value theory?
QUESTION [9 upvotes]: As most others are, so am I fascinated by primes. 
By the theorem of Euclid and the sieve of Eratosthenes the $ k \ge 2$ - th prime is given by:
$$ p_k = \min_{x>1,\gcd(x,p_1 \cdots p_{k-1})=1} x $$
which might be seen as one definition of the $k$-th prime.
Thank God this process is deterministic, so that each time we apply it, with $p_1=2$ we get the same prime numbers.
Suppose now that we add some randomness to this process:
$$y = \min_{x_i > 1, \gcd(n, x_i)=1} (x_1, \ldots, x_k)$$
where we draw each $x_i$ with replacement and independent of each other from $1,\cdots,n$.
I read that the extreme value theory is concerned with $\min, \max$ of iid random variables as $k$ goes to infinity.
Another hint might be the Gumbel distribution which shows values $\zeta(2), \zeta(3)$ in variance and mean where $\zeta$ is the Riemann zeta function.
Yet another hint comes from empirical connection of primes to extreme value theory:
https://arxiv.org/abs/1301.2242
My question is, if one can make this heuristic more precise, if this is not asked too much. (So maybe there is a way to think of a random process where one can apply the extreme value theory? )
Thanks for your help!
Reference:
https://arxiv.org/abs/1301.2242
https://en.wikipedia.org/wiki/Gumbel_distribution
https://stats.stackexchange.com/questions/220/how-is-the-minimum-of-a-set-of-random-variables-distributed
https://en.wikipedia.org/wiki/Extreme_value_theory
Edit:
I did some computations based on the following model:
Given $N$, choose with replacement $y_1,\cdots,y_m$ from the set $\{ x | \gcd(x,N)=1, N \ge x>1\}$ which has $\phi(N)-1$ elements.
The probability that $Y$ is the least prime $p$ not dividing $N$ is:
$$P(Y=p) = \frac{1}{\phi(N)-1}$$
The probability $P(Y_{\min}=y)$ is:
$$P(Y_{\min}=y)=(1-F(y-1))^m - (1-F(y))^m$$
where 
$$F(y) = \sum_{a \le y} P(Y=a) = \frac{1}{\phi(N)-1} \cdot \chi(N,y)$$
and
$$\chi(N,y) = |\{a | a \le y, a > 1, \gcd(a,N)=1 \}|$$.
Hence the expected value of $Y^{(N,m)}_{\min}$ is given by:
$$E(Y^{(N,m)}_{\min}) = \sum_{k=1,k>1,\gcd(k,N)=1}^N k \cdot P(Y_{\min}=k)$$
For $m \rightarrow \infty$ we "should" have:
$$\lim_{m \rightarrow \infty} E(Y^{(P_k,m)}_{\min})=p_{k+1}$$
where $P_k$ is the $k$-th primorial.
It would be nice, if someone with more experience in probability theory and number theory can have a look at this computation.
I also did some computations with SAGEMATH to this theoretical consideration:
def PK(k):
    return prod(primes(nth_prime(k)))

def FF(N,m,k):
    return 1/(euler_phi(N)-1)*len([ a for a in range(1,k+1) if gcd(a,N)==1 and a > 1])

def EE(N,m):
    EN = euler_phi(N)
    return sum([k*((1-FF(N,m,k-1))**m-(1-FF(N,m,k))**m) for k in range(1,N+1) if gcd(k,N)==1 and k>1])


for n in range(3,6+1):
    print nth_prime(n),EE(PK(n),PK(n)).n()

5 5.00000000000000
7 7.03931627655029
11 11.0222933198474
13 13.0321521439774

Thanks for your help.

REPLY [2 votes]: This answer is maybe two years too late, but better too late then not answered:
Let $N \ge 3$ be a natural number. Consider the set $\Omega_N:= \{y \in \mathbb{N}| 1 < y \le N, \gcd(y,N) =1 \}$.
For a subset $A \in \Omega_N$ we define the probability as:
$$P(A) = \frac{|A|}{|\Omega_N|} = \frac{|A|}{\phi(N)-1}$$,
where $\phi$ is the Euler totient function, defined as $\phi(N) = \{ k | 1 \le k \le N , \gcd(k,N)=1 \}$. From the definition it follows that for $y \in \Omega_N$ we have:
$$P(y) = P(\{y\}) = \frac{1}{\phi(N)-1}$$
and we also have:
$$P(Y \le y ) = \frac{|\{k \in \Omega_N| k \le y\}|}{\phi(N)-1} = \frac{\chi(N,y)}{\phi(N)-1}$$
where $\chi(N,y) = |\{a| 1 < a \le y, \gcd(a,N)=1\}|$
We draw with replacement and independent of each other with equal probability $\frac{1}{\phi(N)-1}$ some $m$ numbers $y_1,\cdots,y_m$ from $\Omega_N$ and we let now
$$Y_{\min} :=  \min_{y_i \in \Omega_N} \{ y_1,\cdots,y_m\}$$
It follows then, that:
$$P(Y_{\min} \le y) = 1-(1-P(Y \le y))^m$$
and so:
$$P(Y_{\min} = y )= P(Y_{\min} \le y) - P(Y_{\min} \le y-1) = (1-P(Y \le y-1))^m-(1-P(Y \le y))^m$$
Let $r := \min( \Omega_N ) $. We notice that $r$ is the smallest prime, which does not divide $N$.
We also notice that $\chi(N,r) = 1$ since the set $\{a | 1 < a \le r, \gcd(a,N)=1 \}$ is equal to the set $\{r\}$, by definition of $r$.
From this las observation, we also observe that $\chi(N,r-1) = 0$.
Hence we get:
$$P(Y_{\min}=r) = (1-P(Y \le y-1))^m-(1-P(Y \le y))^m = (1-\frac{\chi(N,r-1)}{\phi(N)-1})^m-(1-\frac{\chi(N,r)}{\phi(N)-1})^m $$
$$= (1-0)^m-(1-\frac{1}{\phi(N)-1})^m = 1-(1-\frac{1}{\phi(N)-1})^m$$
We further then observe that:
$$\lim_{m \rightarrow \infty} P(Y_{\min} = r) = \lim_{m \rightarrow \infty} 1-(1-\frac{1}{\phi(N)-1})^m = 1$$
The expected value $E(Y_{\min})$ is given by:
$$E(Y_{\min}) = \sum_{ y \in \Omega_N} y \cdot P(Y_{\min} = y)$$
It follows that:
$$\lim_{m \rightarrow \infty} E(Y_{\min}) = \sum_{ y \in \Omega_N} y \cdot \lim_{m \rightarrow \infty} P(Y_{\min} = y)$$
$$ = r \cdot 1 + 0 \cdot \sum_{ y \in \Omega_N, y \neq r} y  = r = \min(\Omega_N)$$
For $N = P_k =$ the $k$-th primorial, it follows that (with $k \ge 2$):
$$\lim_{m \rightarrow \infty} E(Y_{\min}) = \min(\Omega_{P_k}) = p_{k+1},$$
since the prime $p_{k+1}$ is the smallest prime $q$ which does not divide the primorial $P_k$ and for which $1 < q \le P_k$ holds.