This is a follow-up question, which is related to the answer of this quesiton: Is there a connection of prime numbers and extreme value theory?
I will duplicate the answer here, so this question is self-contained:
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.
1) Question:For the $k$-th prime $p_k$ consider the random variable $Y^{(k,m)}_{\min} := \min_{ y_i \in \Omega_{P_{k-1}}} \{y_1,\cdots,y_m\}$. Are $Y^{(k,m)}, Y^{(l,m)}$ for $k \neq l$ independent random variables?
2) Question:By linearity of the expected value, we have for the prime gap $p_{k+1}-p_k = \lim_{m \rightarrow \infty} E ( Y_{\min}^{(k,m)}-Y_{\min}^{({k-1},m)})$. Is it possible to use an inequality from probability theory to derive some bounds on the prime gaps?
