See Point Processes for the general set up of the cylinder $\sigma$-algebra, but it suffices to say here that we are interested in computing probabilities of events of the form $\{N_{B_1} = n_1, \ldots, N_{B_M} = n_M\}$. These events are called cylinder sets, and we call $\mathbf n = (n_1, \ldots, n_M)$ an occupation vector, and we abbreviate the cylinder set $\{N_{\mathbf B} = \mathbf n\}$.
The simplest cylinder set is $\{N_{\mathbb Z_p} = N\}$, and from the definition of the grand canonical partition function we may write its probability in terms of the canonical and grand canonical partition functions.
Proposition: $$\mathbb P \{N_{\mathbb Z_p} = N\} = \frac{t^N}{N!} \frac{Z(N, \beta, \lambda)}{Z(t, \beta, \lambda)}.$$
To exploit the self-similarity of $\mathbb Z_p$, we name our balls according to which coset of $p \mathbb Z_p$ they are subsets of. Starting with $\mathbf B = (B_m)$, we denote the vector of balls in the $j$th coset $j + p \mathbb Z_p$ by $\mathbf B^j$. It is possible that $\mathbf B^j$ is empty if there were no balls of $\mathbf B$ in $j + p \mathbb Z_p$. By re-organizing and relabeling the population vector $\mathbf n$ we may write $\mathbf n^j$ for the population vector for those balls in $j + p \mathbb Z_p$. Thus we may write $$\{ N_{\mathbf B} = \mathbf n \} = \{N_{\mathbf B^1} = \mathbf n^1, \ldots, N_{\mathbf B^{p-1}} = \mathbf n^{p-1} \}.$$
The proof of the $p$-power theorem can be extended to probabilities for cylinder sets.
Theorem: We get a recursion for $$\mathbb P\{N_{\mathbf B^1} = \mathbf n^1, \ldots, N_{\mathbf B^{p-1}} = \mathbf n^{p-1} \} = \prod_{j=0}^{p-1} \mathbb P\{N_{\mathbf B^j} = \mathbf n^j\}.$$
This recursion terminates in one of the following condition.
- ${\displaystyle \mathbb P \{N_{\mathbb Z_p} = N\} = \frac{t^N}{N!} \frac{Z(N, \beta, \lambda)}{Z(t, \beta, \lambda)}}$.
- If $\mathbf B$ and $\mathbf n$ are empty, then $\{N_{\emptyset} = \emptyset\} = \Omega$ and $ \mathbb P \{N_{\emptyset} = \emptyset\} = 1.$
This follows because the particles in the different cosets can’t detect each other. The cylinder sets in $j + p \mathbb Z_p$ and $j’ + p \mathbb Z_p$ are independent of each other (when $j \neq j’$). This lemma is useful because it provides the inductive step for recursion for probabilities of cylinder sets.
Corollary: $$\mathbb P\{N_{\mathbf B^1} = \mathbf n^1, \ldots, N_{\mathbf B^{p-1}} = \mathbf n^{p-1} \} = \prod_{j=0}^{p-1} T_1 \mathbb P\{\pi (N_{\mathbf B^j} = \mathbf n^j)\}.$$
We are almost to the end of our journey. We have expresses the probability of a cylinder set as the product of tranformed probabilities of cylinder sets formed from the inflated cylinder sets in each coset. Because our original cylinder set only involved specifying the population of a finite number of balls, the number of balls in each inflated cylinder set is less than the original number of balls this allows for a recursion.
One of the terminating conditions is the first proposition in this section. The other terminating condition is that when $\{\pi (N_{\mathbf B^j} = \mathbf n^j)\}$ is empty. However in the case where the $\mathbf B^j$ and $\mathbf n^j$ are empty, we are placing no condition on the particles and thus the probability is simply 1.
Calculations
Consider the given cylinder set in $\mathbb Z_5.$ The theorem tells us that we may look at probabilities of cylinder sets in the cosets.
The first and last of these coset probabilities are terminating condition. Specifically, the last term is simply 1, and the first term is $$T_1 \frac{t^2}{2!} \frac{Z(2, \beta, \lambda)}{Z(t, \beta, \lambda)} = \frac{(t/p)^2}{2} p^{-\beta} \frac{Z(2, \beta, \lambda)}{Z(t, \beta, \lambda)}.$$ Thus,
Similarly,
And,
And,
Following through the definitions, this is equal to $$\begin{eqnarray} &&T_1 \left[\left((t/p) \frac{Z(1,\beta, \lambda)}{Z(t,\beta, \lambda)}\right)^2 \cdot \bigg(\frac{Z(0, \beta, \lambda)}{Z(t, \beta, \lambda)} \bigg) \right] \bigg(\frac{(t/p^2)^5}{5!} p^{-20 \beta} \frac{Z(5,\beta, \lambda)}{Z(t, \beta, \lambda)}\bigg) \\ && \quad = \left((t/p)^2 p^{-\beta} \frac{Z(1,\beta, \lambda)}{Z(t,\beta, \lambda)}\right)^2 \cdot \bigg(\frac{Z(0, \beta, \lambda)}{Z(t, \beta, \lambda)} \bigg) \bigg(\frac{(t/p^2)^5}{5!} p^{-20 \beta} \frac{Z(5,\beta, \lambda)}{Z(t, \beta, \lambda)}\bigg).\end{eqnarray}$$
That is,
Which can be simplified further to yield $$\mathbb P\{N_{\mathbf B} = \mathbf n\} = \frac{t^{33}}{5^{64} 2! 2! 4! 5! 5! 5! 6!} \frac{Z(1, \beta, \lambda)^2 Z(2, \beta, \lambda)^2 Z(4, \beta, \lambda) Z(5, \beta, \lambda)^3 Z(6, \beta, \lambda)}{Z(t, \beta, \lambda)^{11}}.$$