We give probabilistic proofs to well-known combinatorial identities; the Durfee rectangles identity, Euler's identity and the $q$-Binomial Theorem. We use the asymmetric simple exclusion process on $\mathbb{Z}$ under its natural product blocking measure to achieve this. The results we derive also allow us to determine the stationary distribution of second class particles in the blocking scenario.

§ INTRODUCTION

The Asymmetric Simple Exclusion Process (ASEP) is a classic nearest neighbour interacting particle system on $\mathbb{Z}$, introduced in 1970 by Spitzer [18]. In 1976, Liggett [16] showed that ASEP has reversible stationary measures known as blocking measures. These blocking measures have a special algebraic structure which allowed Balázs and Bowen [5] to derive a probabilistic proof of the Jacobi triple product identity. In that paper the authors considered ASEP and its equivalent description in terms of the Asymmetric Zero-Range Process. Comparing the blocking measures for each process led to their result. In 2022, this method was generalised by Balázs, Fretwell and Jay [6], where new identities of combinatorial significance were proved by considering blocking measures of more general particle systems.
Here we return to ASEP only, this time we extend the previous work of [5] by determining the distribution of the number of particles in specific regions of the integer lattice. This way we give purely probabilistic proofs to other well-known and non-trivial combinatorial identities. The ones we cover are the Durfee rectangles identity, Euler's identity and the $q$-Binomial Theorem, each fundamental to the theory of integer partitions. Probabilistic methods are certainly not the common way of proving such identities, which gives a unique perspective to this paper.
Attractive interacting particle systems, like ASEP, give rise to coupling methods from which the notion of second class particles arises. The coupled particle systems are referred to as a multi-species particle system. Stationary measures of such systems have drawn lots of interest, for examples in the translation invariant case see Angel [4], Ferrari and Martin [13] and also, in terms of last passage percolation, Fan and Seppäläinen [11].
Informally, a two-species ASEP can be described in the following way. Both species of particles attempt to evolve as the single species ASEP would, i.e. obeying the exclusion rule within their own species. The two species can interact as follows, a first class particle may jump into the space of a second and hence swap position with the second class particle. However if a second class particle attempts to jump into the space of a first class particle the jump is blocked.
The quantities we define and the combinatorial identities we prove allow us to give an explicit formula for the distribution of positions for a given number of second class particles. To the best of our knowledge, the distribution of second class particles under the blocking measure has not been determined before. From private communications, the authors understand that similar results will be extracted from arguments involving the Mallows measure by A. Bufetov and K. Chen in a forthcoming paper.  With the use of a second class particle we also give an alternative proof to the distribution of the number of particles on the half infinite line.
The second class particle arguments rely on an interpretation of the two species ASEP as a single species ASEP with a second ASEP on the particle labels using the first as a dynamic background. It turns out that both these ASEPs are in their stationary blocking distributions, which allows us to perform calculations. Both for the distribution of the number of particles and positions of second class particles, we use probabilistic arguments that lead to various recursions. Finding the solutions of these gives rise to our proofs. For the Durfee rectangles identity we also exploit the well-known symmetry between particles and holes in ASEP.

§.§ Notation

Throughout the paper we will use the $q$-Pochhammer symbol as well as the $q$-Binomial coefficient. For some $a\in\mathbb{R}$ and $n \in \mathbb{Z}_{>0}$, the $q$-Pochhammer symbol is given by
$(a;q)_n\defeq \prod\limits_{i=0}^{n-1}(1-aq^i)$.
Also we set, $(a;q)_0\defeq 1$ and $(a;q)_\infty\defeq\prod\limits_{i=0}^{\infty}(1-aq^i)$.
The $q$-Binomial coefficient is then,

\end{bmatrix}_q\defeq \frac{(q;q)_n}{(q;q)_m(q;q)_{n-m}}=\prod\limits_{i=0}^{m-1}\frac{(1-q^{n-i})}{(1-q^{i+1})}.$$

§.§ Results

§.§.§ Combinatorial Identities

In Section <ref>, we look at the symmetry between particles and holes in ASEP under its blocking measure. Using this symmetry we prove a well-known combinatorial identity, the Durfee rectangles identity.
thmsymidenDurfee Rectangles Identity (for example see equation (4) in Gessel [14])

For $q\in(0,1)$ and any fixed $n\in\mathbb{Z}$,

Combinatorially this identity identity says that for a fixed integer $n$, any integer partition has a maximal rectangle of side lengths $n+k$ and $k$ (for some $k\geq \max\{-n,0\}$) in its Ferrers diagram known as its Durfee rectangle. When $n=0$, this identity is the well-known Durfee square identity (for example see Andrews and Eriksson [3]).
thmeulerEuler's Identity (for example see equation E1 in Andrews [1])
For $q\in(0,1)$ and $z\in\mathbb{R}$,

$$\sum\limits_{k=0}^\infty \frac{q^{\frac{k(k-1)}{2}}z^k}{(1-q)(1-q^2)...(1-q^k)}=(-z;q)_\infty.$$

In combinatorics, this identity gives equivalent ways of writing the generating function for integer partitions into distinct parts. In a general term of the expansion, the power of $z$ gives the number of parts in the partition.

thmqbin$q$-Binomial Theorem (Heine [15], also Andrews [2])
For $q\in(0,1)$, $z\in\mathbb{R}$ and any $m \in \mathbb{Z}_{\geq 0}$,

\end{bmatrix}_{q}=(-z;q)_m.$$

This identity gives equivalent forms of the generating function for integer partitions into distinct parts of size at most $m$. Again here in a general term in the expansion, the power of $z$ is the number of parts in the partition. Euler's identity can be thought of as the limiting identity when we take $m\rightarrow\infty$ in the $q$-Binomial theorem.
The proofs we give here are purely probabilistic using ASEP and give natural probabilistic interpretations to these combinatorial identities.

§.§.§ Distributional results for single and two species ASEP

We denote the family of product blocking measures for ASEP by $\underline{\mu}^c$ for any $c\in\mathbb R$, these measures are written in terms of the asymmetry parameter for the process, $q\in(0,1)$. We also let
$N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})$ denote the number of particles at and to the left of site $m$ in the state $\underline{z}$. We will derive the distribution under this blocking measure of a given number of particles on a half infinite volume.
For any $m \in \mathbb{Z}$, $k\in\mathbb{Z}_{\geq 0}$ and $c \in \mathbb{R}$,

We also consider the distribution under the blocking measure of a given number of particles in a finite volume (see Lemma <ref>).

Using the basic coupling of two ASEPs, $\{\underline{\eta},\underline{\xi}\}$, we find the probability that a second class particles is at a given site. At any time, $\underline{\eta}$ and $\underline{\xi}$ differ in exactly $d>0$ places (at these sites $\xi_i>\eta_i$), these differences define $d$ second class particles.
Assume that the coupled system is stationary and $\underline{\xi}\sim\underline{\mu}^c$. For a given $m\in\mathbb{Z}$, we find a second class particle at site $m$ with probability,

We can also find the stationary distribution of the $d$ second class particle positions process,
Assuming that the coupled system is stationary and $\underline{\xi}\sim \underline{\mu}^c$, the stationary distribution of $\underline{X}$, the position of $d>0$ second class particles, is given by,

$$\mathbb{P}(\underline{X}=\underline{m})=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}}{\prod\limits_{j=1}^d(1+q^{c+d-j-m_j})(1+q^{c+d+1-j-m_j})}, \hspace{5mm} \text{for all } m_1<m_2<...<m_d\in\mathbb{Z}.$$

Of course, when $d=1$ Theorems <ref> and <ref> are the same. Then, in this case, the position of a single second class particle under $\underline{\mu}^c$ is distributed according to $-Y$ where $Y$ is a discrete logistic with parameters $q$ and $-c$ (introduced by Chakraborty and Chakravarty in 2016 [9]).

In order to prove Theorem <ref>, we first condition on the label process for the second class particles. The label of a particle is defined to be the number of particles to its left, which by the blocking nature of the process will be finite. The label process can be viewed as the positions of particles in an asymmetric simple exclusion process on $\mathbb{Z}_{\geq 0}$ with exactly $d$ particles (meaning closed left boundary). Then we split states up into the sites to the left of site $m_1$ and for $j\in\{1,...,d-1\}$ the sites between $m_j$ and $m_{j+1}$. From here we use the distributional results for particles in half infinite and finite volumes along with Euler's identity and the $q$-Binomial Theorem to complete the proof.

§ ACKNOWLEDGEMENTS

The authors would like to thank Dan Fretwell for helpful discussions regarding viewing particle states as integer partitions and the combinatorial identities we have found. Also many thanks to Alexey Bufetov for discussions about the multi-species ASEP.
M. Balázs was partially supported by the EPSRC EP/R021449/1 and EP/W032112/1 Standard Grants of the UK. J. Jay was funded by the Heilbronn Institute for Mathematical Research.

This study did not involve any underlying data.

§ INTERACTING PARTICLE SYSTEMS WITH PRODUCT BLOCKING MEASURE

§.§ Blocking Family

Let us recall the blocking family of particle systems on $\mathbb{Z}$ as introduced by Balázs and Bowen in [5], using the notation as in the paper of Balázs, Fretwell and Jay [6]. We consider nearest neighbour exclusion type particle systems on some state space. The state space we consider is some countable  $\Omega \subseteq I^\Lambda$, where:

* $\Lambda\defeq \{i:\ell-1<i<\mathfrak{r}+1\}\subseteq \mathbb{Z}$, for some (possibly infinite) $-\infty\leq \ell \leq \mathfrak{r}\le \infty$, defines the set of sites

* $I\defeq\{z:\omega^\textrm{min}-1<z<\omega^\textrm{max}+1\}\subseteq \mathbb{Z}$, for some (possibly infinite)  $-\infty\leq \omega^\textrm{min} \leq 0 < \omega^\textrm{max}\leq \infty $, defines the set of possible particle configuration per site.

We consider the following functions $N_{m+\frac{1}{2}}^{\overset{\leftarrow}{p}},N_{m+\frac{1}{2}}^{\overset{\rightarrow}{h}}:I^\Lambda \rightarrow \mathbb{Z}_{\geq 0}\cup \{\infty\}$  for any $m \in \Lambda$ defined by,

$$N_{m+\frac{1}{2}}^{\overset{\leftarrow}{p}}(\underline{\eta})\defeq \sum\limits_{i=\ell}^m (\eta_i-\omega^\textrm{min})\hspace{5mm} \textrm{ and } \hspace{5mm} N_{m+\frac{1}{2}}^{\overset{\rightarrow}{h}}(\underline{\eta})\defeq\sum\limits_{i=m+1}^\mathfrak{r}(\omega^\textrm{max}-\eta_i)$$

(this can be thought of as just the number of `particles' to the left and `holes' to right of $m+\frac{1}{2}$ whenever $\omega^\text{min}$ and $\omega^\text{max}$ are finite). When $m=0$ we simplify notation to $N^{\overset{\leftarrow}{p}}$ and $N^{\overset{\rightarrow}{h}}$.
Now we can define the state space as follows:

* if $\ell\leq0\leq\mathfrak{r}$ then,

$$\Omega\defeq \{\underline{\eta} \in I^\Lambda: (\ell >-\infty \textrm{ or } N^{\overset{\leftarrow}{p}}(\underline{\eta})< \infty)\textrm{ and }(\mathfrak{r}<\infty \textrm{ or } N^{\overset{\rightarrow}{h}}(\underline{\eta})<\infty)\}$$

* if $\mathfrak{r}<0$ then,

$$\Omega\defeq \{\underline{\eta} \in I^\Lambda: \ell >-\infty \textrm{ or } N_{\mathfrak{r}+\frac{1}{2}}^{\overset{\leftarrow}{p}}(\underline{\eta})< \infty\}$$

* if $\mathfrak{\ell}>0$ then,

$$\Omega\defeq \{\underline{\eta} \in I^\Lambda: \mathfrak{r}<\infty \textrm{ or } N_{\ell+\frac{1}{2}}^{\overset{\rightarrow}{h}}(\underline{\eta})<\infty\}$$

Let $\underline{\eta}^{(i,j)}$ denote the state reached from $\underline{\eta}\in \Omega$ by a particle jumping from site $i$ to site $j$ (when this is possible), that is,
\eta_k &\textrm{if } k \neq i,j\\
\eta_i-1 &\textrm{if } k=i \\
\eta_j+1 &\textrm{if }k=j.
\end{cases}$$

The systems we consider evolve according to the following Markov generators. In the bulk (sites $\ell \leq i\leq \mathfrak{r}-1$) the generator is of the form,

for some cylinder function $\varphi:\Omega \rightarrow \mathbb{R}$ and right/left jump rates $p,q:I^2\rightarrow [0,\infty)$. When either $\ell>-\infty$ or $\mathfrak{r}<\infty$ we consider open infinite type boundaries with boundary generators given by,

\begin{align*}
\left(L^\ell \varphi\right)(\underline{\eta})&=p_\ell(\eta_\ell)\left(\varphi(\underline{\eta}^{(\ell-1,\ell)})-\varphi(\underline{\eta})\right)+q_\ell(\eta_\ell)\left(\varphi(\underline{\eta}^{(\ell,\ell-1)})-\varphi(\underline{\eta})\right)\\
\left(L^\mathfrak{r} \varphi\right)(\underline{\eta})&=p_\mathfrak{r}(\eta_\mathfrak{r})\left(\varphi(\underline{\eta}^{(\mathfrak{r},\mathfrak{r}+1)})-\varphi(\underline{\eta})\right)+q_\mathfrak{r}(\eta_\mathfrak{r})\left(\varphi(\underline{\eta}^{(\mathfrak{r}+1,\mathfrak{r})})-\varphi(\underline{\eta})\right).
\end{align*}

Here $\underline{\eta}^{(\ell-1,\ell)}$ and $\underline{\eta}^{(\mathfrak{r}+1,\mathfrak{r})}$ denote the states reached from a particle entering via the boundary with $p_\ell$ and $q_\mathfrak{r}$ being the corresponding jump rates. Similarly $\underline{\eta}^{(\ell,\ell-1)}$ and $\underline{\eta}^{(\mathfrak{r},\mathfrak{r}+1)}$ denote the states reached from a particle leaving the system via the boundary with jump rates $q_\ell$ and $p_\mathfrak{r}$.
Particle systems of this type are members of the blocking family if they satisfy the blocking conditions:

* For $-\infty<\omega^{\textrm{min}}< \omega^{\textrm{max}}<\infty$ we have that

$$p(\omega^{\textrm{min}}, \cdot)=p(\cdot, \omega^{\textrm{max}})=q(\omega^{\textrm{max}}, \cdot)=q(\cdot, \omega^{\textrm{min}})=0.$$

If $\ell>-\infty$,

and if $\mathfrak{r}<\infty$,

* The system is attractive; that is $p(\cdot,\cdot)$ is non-decreasing in the first variable and non-increasing in the second whilst $q(\cdot,\cdot)$ is non-increasing in the first variable and non-decreasing in the second. If $\ell>-\infty$, $p_\ell$ is non-increasing and $q_\ell$ non-decreasing. If $\mathfrak{r}<\infty$, $p_\mathfrak{r}$ is non-decreasing and $q_\mathfrak{r}$ non-increasing.

* There exist $p_{\text{asym}},q_{\text{asym}}\in\mathbb{R}$ satisfying $\frac{1}{2}<p_{\textrm{asym}}=1-q_{\textrm{asym}} \leq 1$, and functions $f: I \rightarrow [0,\infty)$ and $s: I \times I \rightarrow [0,\infty)$ such that,

$$p(y,z)=p_{\textrm{asym}}\cdot s(y,z+1)\cdot f(y) \hspace{5mm}\textrm{ and } \hspace{5mm}q(y,z)=q_{\textrm{asym}}\cdot s(y+1,z)\cdot f(z).$$

If $\omega^\textrm{min}$ is finite then $f(\omega^\textrm{min})=0$, and if $\omega^\textrm{max}$ is finite  we extend the domain of $s$ and require that $s(\omega^\textrm{max}+1,\cdot)=s(\cdot,\omega^\textrm{max}+1)=0$. (Note that attractivity implies that $s$ is non-increasing in both of its variables and $f$ is non-decreasing).

Particle systems which satisfy these conditions will in general have many stationary distributions. In particular, they have reversible stationary product blocking measures.
([5], Theorem $3.1$)

$$c^\text{min} \defeq \begin{cases}
-\infty & \text{ when } \omega^\text{max}<\infty\\
-\frac{\lim\limits_{z\rightarrow \infty}\ln{f(z)}}{(\ln{p}-\ln{q})} &\text{ when } \omega^\text{max}=\infty
\end{cases} \qquad c^\text{max}\defeq \begin{cases}
\infty &\text{ when } \omega^\text{min}>-\infty\\
-\frac{\lim\limits_{z\rightarrow -\infty}\ln{f(z)}}{(\ln{p}-\ln{q})}& \text{ when } \omega^\text{min}=-\infty.
\end{cases}$$

Take $c^\text{min}<c<c^\text{max}$. If $\ell>-\infty$ then assume that $p_\ell$ and $q_\ell$ are such that,

Similarly, if $\mathfrak{r}<\infty$ then assume that $p_\mathfrak{r}$ and $q_\mathfrak{r}$ are such that,

Then there is a product reversible stationary blocking measure $\underline{\mu}^c$ on $\Omega$, given by the marginals

$$\mu_i^c(z)=\frac{1}{Z_i^c}\frac{\left(\frac{p_\textrm{asym}}{q_\textrm{asym}}\right)^{(i-c)z}}{f(z)!}\hspace{5mm} \textrm{ for } i\in \Lambda \text{ and } z \in I,$$

where $Z_i^c\defeq \sum\limits_{z=\omega^\text{min}}^{\omega^\text{max}}\frac{\left(\frac{p_\text{asym}}{q_\text{asym}}\right)^{(i-c)z}}{f(z)!}$ is the normalising factor and $f(z)!\defeq\begin{cases} \prod\limits_{y=1}^zf(y) &\textrm{for } z>0 \\
1 &\textrm{for } z=0 \\
\frac{1}{\prod\limits_{y=z+1}^{0}f(y)} &\textrm{for } z<0.
\end{cases}$

By definition of the state space $\Omega$ and by a Borel-Cantelli argument for $\underline{\mu}^c$ we see that any state of a process in the Blocking family has a left most particle and right most hole a.s.

§.§ Ergodic Decomposition

When either $\ell$ or $\mathfrak{r}$ are finite we see that the state space $\Omega$ is irreducible. This is not true in the doubly infinite case, but we can decompose $\Omega$ into irreducible components in the following way.
Let us concentrate on the case when $\Lambda=\mathbb{Z}$ and $I=\{\omega^\text{min},...,\omega^\text{max}\}$ for some $-\infty<\omega^\text{min}\leq 0<\omega^\text{max}<\infty$. Let $\tilde{q}\defeq \frac{q_\text{asym}}{p_\text{asym}}<1$, then we  can write the one parameter family of stationary blocking measures as,

$$\underline{\mu}^c(\underline{\eta})=\prod\limits_{i=-\infty}^0\frac{\tilde{q}^{-(i-c)(\eta_i-\omega^\text{min})}}{\tilde{q}^{\omega^\text{min}(i-c)}f(\eta_i)!\cdot Z_i^c}\prod\limits_{i=1}^\infty \frac{\tilde{q}^{(\omega^\text{max}-\eta_i)(i-c)}}{\tilde{q}^{\omega^\text{max}(i-c)}f(\eta_i)!\cdot Z_i^c}.$$

It is clear to see that for any $m\in\mathbb{Z}$ the quantity,

$$N_{m+\frac{1}{2}}(\underline{\eta})\defeq N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{\eta})-N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{\eta})=\sum\limits_{i=m+1}^\infty (\omega^\text{max}-\eta_i)-\sum\limits_{i=-\infty}^m (\eta_i-\omega^\text{min})$$

is finite by definition and is conserved by the dynamics of the process. As before when $m=0$ we simplify notation to $N$.
Using this fact, we see that the state space decomposes into irreducible components,
$$\Omega=\bigcup_{n \in \mathbb{Z}}\Omega^n \hspace{2mm} \text{ where } \hspace{2mm}\Omega^n\defeq\{\underline{\eta} \in \Omega: N(\underline{\eta})=n\}.$$

We note that the left shift operator, $\tau$ such that $(\tau\underline{\eta})_i=\eta_{i+1}$, defines a bijection $\Omega^n \xrightarrow[]{\tau} \Omega^{n-(\omega^\text{max}-\omega^\text{min})}$ (i.e if $\underline{\eta} \in \Omega^n$ then, $N(\tau \underline{\eta})=n-(\omega^\text{max}-\omega^\text{min})$).
We can then find the unique stationary distribution on $\Omega^n$,

where $\underline{\mu}^c(\{N=n\})$ is found as the solution to the recurrence relation given in the following lemma.
([5], Lemma $6.1$ and Corollary $6.2$)
The following relation holds,

$$\underline{\mu}^c(\tau \underline{\eta})=\tilde{q}^{c(\omega^\text{max}-\omega^\text{min})-N(\underline{\eta})}\underline{\mu}^c( \underline{\eta}).$$

This gives the recursion,

§.§ Distributional Results

We will now prove some distributional results for our processes in blocking measure. Here we assume that $\Lambda=\mathbb{Z}$ and $I=\{\omega^\textrm{min},...,\omega^\text{max}\}$ for some $-\infty<\omega^\textrm{min}\leq0<\omega^\text{max}<\infty$.  Firstly, we see that if we shift the entire system this corresponds to a shift in the parameter $c$.

For any $c \in \mathbb{R}$ and $\underline{\eta}\in \Omega$

Writing the blocking measure compactly we have,
$$\underline{\mu}^c(\tau\underline{\eta})=\prod\limits_{i=-\infty}^\infty\frac{\tilde{q}^{(c-i)(\tau\underline{\eta})_i}}{f((\tau\underline{\eta})_i)!\cdot Z_i^c}=\prod\limits_{i=-\infty}^\infty\frac{\tilde{q}^{(c-i)\eta_{i+1}}}{f(\eta_{i+1})!\cdot Z_i^c}=\prod\limits_{j=-\infty}^\infty\frac{\tilde{q}^{(c-(j-1))\eta_j}}{f(\eta_j)!\cdot Z_{j-1}^c}=\prod\limits_{j=-\infty}^\infty\frac{\tilde{q}^{((c+1)-j)\eta_j}}{f(\eta_j)!\cdot Z_{j}^{c+1}}=\underline{\mu}^{c+1}(\underline{\eta}).$$

For any $c \in \mathbb{R}$ and $m,k \in \mathbb{Z}$,

* $\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})=\underline{\mu}^{c+1}(\{N^{\overset{\leftarrow}{p}}_{m+1+\frac{1}{2}}(\underline{z})=k\})$

* $\underline{\mu}^c(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k\})=\underline{\mu}^{c+1}(\{N^{\overset{\rightarrow}{h}}_{m+1+\frac{1}{2}}(\underline{z})=k\})$

* $\underline{\mu}^c(\{N_{m+\frac{1}{2}}(\underline{z})=k\})=\underline{\mu}^{c+1}(\{N_{m+1+\frac{1}{2}}(\underline{z})=k\})$

Firstly we note that for any $r \in \mathbb{Z}$,

\begin{align*}
N^{\overset{\rightarrow}{p}}_{r+\frac{1}{2}}(\tau\underline{\eta})&=\sum\limits_{i=-\infty}^r (\eta_{i+1}-\omega^\text{min})=\sum\limits_{i=-\infty}^{r+1}(\eta_i-\omega^\text{min})=N^{\overset{\rightarrow}{p}}_{r+1+\frac{1}{2}}(\underline{\eta})\\
N^{\overset{\rightarrow}{h}}_{r+\frac{1}{2}}(\tau\underline{\eta})&=\sum\limits_{i=r+1}^\infty(\omega^\text{max}-\eta_{i+1})=\sum\limits_{i=r+2}^\infty(\omega^\text{max}-\eta_i)=N^{\overset{\rightarrow}{h}}_{r+1+\frac{1}{2}}(\underline{\eta}) \\
\end{align*}

So we have that,

\begin{align*}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})=\underline{\mu}^c&(\{\underline{z}:N^{\overset{\leftarrow}{p}}_{m+1+\frac{1}{2}}(\tau^{-1}\underline{z})=k\}) \\
\end{align*}

Similar arguments hold for $\underline{\mu}^c(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k\})$ and $\underline{\mu}^c(\{N_{m+\frac{1}{2}}(\underline{z})=k\})$.

We can also give results where we only change the parameter $c$ and not the quantity we wish to understand the probability of.

For any $c\in\mathbb{R}$, $m,k\in\mathbb{Z}$ we have that,

* $\underline{\mu}^{c-1}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})=\frac{\tilde{q}^{-k-\omega^\text{min}(m+1-c)}}{Z_{m+1}^c}\cdot \underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})$

* $\underline{\mu}^{c-1}(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k\})=\tilde{q}^{k+\omega^\text{max}(m+1-c)}Z_{m+1}^c\cdot \underline{\mu}^c(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k\})$

* $\underline{\mu}^{c-1}(\{N_{m+\frac{1}{2}}(\underline{z})=k\})=\tilde{q}^{k+(\omega^\text{max}-\omega^\text{min})(m+1-c)}\cdot \underline{\mu}^c(\{N_{m+\frac{1}{2}}(\underline{z})=k\}).$

* We have that,
\begin{align*}
\underline{\mu}^{c-1}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})&=\sum\limits_{\underline{z}:~N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k}\underline{\mu}^{c-1}(\underline{z})\\
&=\sum\limits_{\underline{z}:~N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k}\prod\limits_{j=-\infty}^m\frac{\tilde{q}^{(c-1-j)z_j}}{f(z_j)!\cdot Z_j^{c-1}}\\
&=\sum\limits_{\underline{z}:~N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k}\prod\limits_{j=-\infty}^m\frac{\tilde{q}^{(c-1-j)(z_j-\omega^\text{min})}}{\tilde{q}^{\omega^{min}(j-c+1)}f(z_j)!\cdot Z_{j+1}^{c}}\\
&=\sum\limits_{\underline{z}:~N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k}\frac{\tilde{q}^{-\sum\limits_{j=-\infty}^m(z_j-\omega^\text{min})}}{\tilde{q}^{\omega^\text{min}(m+1-c)}Z_{m+1}^c}\prod\limits_{j=-\infty}^m\frac{\tilde{q}^{(c-j)(z_j-\omega^\text{min})}}{\tilde{q}^{\omega^\text{min}(j-c)}f(z_j)!\cdot Z_{j}^{c}}\\
\end{align*}

* We have that,
\begin{align*}
\underline{\mu}^{c-1}(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k\})&=\sum\limits_{\underline{z}:~N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k}\underline{\mu}^{c-1}(\underline{z})\\
&=\sum\limits_{\underline{z}:~N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k}\prod\limits_{j=m+1}^\infty\frac{\tilde{q}^{(c-1-j)z_j}}{f(z_j)!\cdot Z_j^{c-1}}\\
&=\sum\limits_{\underline{z}:~N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k}\prod\limits_{j=m+1}^\infty\frac{\tilde{q}^{(\omega^\text{max}-z_j)(j-c+1)}}{\tilde{q}^{\omega^\text{max}(j-c+1)}f(z_j)!\cdot Z_j^{c-1}}\\
&=\sum\limits_{\underline{z}:~N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k}\tilde{q}^{\sum\limits_{j= m+1}^\infty(\omega^\text{max}-z_j)}\cdot \tilde{q}^{\omega^\text{max}(m+1-c)}Z_{m+1}^c\prod\limits_{j= m+1}^\infty\frac{\tilde{q}^{(\omega^\text{max}-z_j)(j-c)}}{\tilde{q}^{\omega^\text{max}(j-c)}f(z_j)!\cdot Z_j^{c}}\\
\end{align*}

* We notice that,

\begin{align*}\underline{\mu}^{c-1}(\{N_{m+\frac{1}{2}}(\underline{z})=k\})&=\sum\limits_{k_1,k_2\in \mathbb{Z}}\underline{\mu}^{c-1}(\{N_{m+\frac{1}{2}}(\underline{z})=k\}\cap\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k_1\}\cap\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k_2\})\\
&=\sum\limits_{\substack{k_1,k_2\in\mathbb{Z}:\\k_2-k_1=k}}\underline{\mu}^{c-1}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k_1\}) \cdot\underline{\mu}^{c-1}(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k_2\}).
\end{align*}

Using (a) and (b) we have that,

\begin{align*}
&\sum\limits_{\substack{k_1,k_2: \\ k_2-k_1=k}} \underline{\mu}^{c-1}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k_1\}) \cdot\underline{\mu}^{c-1}(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k_2\}) \\
&= \sum\limits_{\substack{k_1,k_2: \\ k_2-k_1=k}} \frac{\tilde{q}^{-k_1-\omega^\text{min}(m+1-c)}}{Z_{m+1}^c}\cdot\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k_1\}) \cdot \tilde{q}^{k_2+\omega^\text{max}(m+1-c)}Z_{m+1}^c\cdot\underline{\mu}^c(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k_2\})\\
&= \tilde{q}^{k+(\omega^\text{max}-\omega^\text{min})(m+1-c)}\sum\limits_{\substack{k_1,k_2: \\ k_2-k_1=k}} \underline{\mu}^{c}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k_1\}) \cdot\underline{\mu}^{c}(\{N^{\overset{\rightarrow}{h}}_{m+\frac{1}{2}}(\underline{z})=k_2\}) \\
&=\tilde{q}^{k+(\omega^\text{max}-\omega^\text{min})(m+1-c)}\cdot \underline{\mu}^c(\{N_{m+\frac{1}{2}}(\underline{z})=k\}).
\end{align*}

§ THE ASYMMETRIC SIMPLE EXCLUSION PROCESS (ASEP)

For the rest of the paper we will focus on a particular member of the blocking family, the asymmetric simple exclusion process (ASEP) on the state space,

$$\Omega\defeq\{\underline{z}\in\{0,1\}^\mathbb{Z}:N^{\overset{\leftarrow}{p}}(\underline{\eta}), N^{\overset{\rightarrow}{h}}(\underline{\eta})<\infty\}.$$

That is the interacting particle system on $\Omega$ which evolves according to nearest neighbour particle jumps happening with jump rates,

$$p(z_i,z_{i+1})=\mathbb{I}\{z_i \neq 0\}\mathbb{I}\{z_{i+1}\neq 1\} \hspace{5mm}\text{ and }\hspace{5mm} q(z_i,z_{i+1})=q\mathbb{I}\{z_{i+1} \neq 0\}\mathbb{I}\{z_{i}\neq 1\},$$

for some $q\in(0,1)$. Blocking measures for ASEP were first found by Liggett in 1976 [16] and by the general form of blocking measures from Balázs and Bowen [5], we know that the natural product blocking measure for any $c\in\mathbb{R}$ is given by,

$$\underline{\mu}^c(\underline{\eta})=\prod\limits_{i=-\infty}^\infty\frac{q^{-(i-c)\eta_i}}{1+q^{-(i-c)}}=\prod\limits_{i=-\infty}^0\frac{q^{-(i-c)\eta_i}}{1+q^{-(i-c)}}\prod\limits_{i=1}^\infty \frac{q^{(i-c)(1-\eta_i)}}{1+q^{(i-c)}}.$$

We also have the unique stationary measure on the irreducible component, $\Omega^n$, of the state space for each $n$. The results of [16] gives us that,

\begin{equation}\label{eq: mu of N}
\underline{\mu}^c(\{N=n\})=\frac{q^{\frac{n(n+1)}{2}-nc}}{\sum\limits_{\ell=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}-\ell c}},
\end{equation}

and then,

$$\underline{\nu}^n(\underline{\eta})=\frac{\sum\limits_{\ell=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}-\ell c}}{q^{\frac{n(n+1)}{2}-nc}}\prod\limits_{i=-\infty}^0\frac{q^{-(i-c)\eta_i}}{1+q^{-(i-c)}}\prod\limits_{i=1}^\infty \frac{q^{(i-c)(1-\eta_i)}}{1+q^{i-c}}\cdot \mathbb{I}\{N(\underline{\eta})=n\}.$$

§.§ Distributional Results

We will now give some distributional results for ASEP; in particular we focus on the functions $N_{m+\frac{1}{2}}$ and $N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}$ for any $m\in\mathbb{Z}$
In [5] Balázs and Bowen gave the distribution under $\underline{\mu}^c$ of the conserved quantity about the site $0$, that is $\underline{\mu}^c(\{N=n\})$ (given in equation (<ref>)). By considering shifts of ASEP states we can find the distribution of the conserved quantity about any site $m\in\mathbb{Z}$.

For any $n,m\in\mathbb{Z}$ and $c\in\mathbb{R}$,

$$\underline{\mu}^c(\{N_{m+\frac{1}{2}}=n\})=\frac{q^{\frac{(n+m)(n+m+1)}{2}-(n+m)c}}{\sum\limits_{\ell=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}-\ell c}}$$

As we saw in the proof of Corollary <ref>, $N_{m+\frac{1}{2}}(\underline{z})=N_{m-1+\frac{1}{2}}(\tau\underline{z})$. Thus we have that, $N_{m+\frac{1}{2}}(\underline{z})=N(\tau^m\underline{z})$. And as we have seen before, a left shift decreases the conserved quantity by $(\omega^\text{max}-\omega^\text{min})=1$ and so, $N(\tau^m\underline{z})=N(\underline{z})-m$. Then, by (<ref>) we have that,

$$\underline{\mu}^c(\{N_{m+\frac{1}{2}}=n\})=\underline{\mu}^c(\{N=n+m\})=\frac{q^{\frac{(n+m)(n+m+1)}{2}-(n+m)c}}{\sum\limits_{\ell=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}-\ell c}}.$$

We now consider the probability under the blocking measure that there is some $k\in\mathbb{Z}_{\geq 0}$ particles to left of a site $m\in\mathbb{Z}$ (inclusive).

Conditioning on site $m$ we have that (for $k\geq 1$),

\begin{align*}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})&=\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k|z_m=0\})\mu^c_m(0)+\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k|z_m=1\})\mu_m^c(1) \\
\end{align*}

We want to look at $N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})$ not $N^{\overset{\leftarrow}{p}}_{m-1+\frac{1}{2}}(\underline{z})$ so we use the result of Corollary <ref>:

By Lemma <ref>,

And so we have,

Giving the recurrence relation,

which we solve by,

\begin{align*}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})&=\left(\prod\limits_{i=1}^k \frac{q^{i-1+c-m}}{1-q^i} \right)\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=0\})\\
&=\prod\limits_{i=1}^k \frac{q^{i-1+c-m}}{1-q^i}\cdot \prod\limits_{j=-\infty}^ m\frac{1}{1+q^{c-j}}\\
\end{align*}

An alternative proof of this result by considering two coupled ASEPs can be given. We will see this in Section <ref>.

We also consider the probability under the blocking measure that there is some $k$ particles in between two sites.

For any $m_1+1<m_2 \in \mathbb{Z},c \in \mathbb{R}$ and $k \in \{0,1,...,\hat{m}_2\}$ where $\hat{m}_2:=m_2-m_1-1$,

\hat{m}_2 \\
\end{bmatrix}_{q}.$$

Firstly, let us consider the cases where $k=0$ or $k=\hat{m_2}$. These cases are directly computable since there is only 1 state in each case for which $N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k$. If $k=0$ then the state is such that all the sites in $[m_1+1,m_2-1]$ are empty and so,

\begin{equation}\label{eq: 0 particles}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=0\})=\frac{1}{\prod\limits_{i=m_1+1}^{m_2-1}(1+q^{c-i})}=\frac{1}{\prod\limits_{i=1}^{\hat{m}_2}(1+q^{c-m_2+i})}=\frac{1}{(-q^{c-m_2+1};q)_{\hat{m}_2}}.
\end{equation}

If $k=\hat{m}_2$ then the state is such that all the sites in $[m_1+1,m_2-1]$ are full and so,

\begin{equation}\label{eq: full particles}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=\hat{m}_2\})=\prod\limits_{i=m_1+1}^{m_2-1}\frac{q^{c-i}}{(1+q^{c-i})}=\frac{q^{\hat{m}_2(c+1-m_2)+\frac{\hat{m}_2(\hat{m}_2-1)}{2}}}{(-q^{c-m_2+1};q)_{\hat{m}_2}}.\end{equation}

Now, let us consider $k\in\{1,...,\hat{m}_2-1\}$, we see that

\begin{align}\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-&N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k\})\notag\\
=& \underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k|z_{m_2-1}=0\})\cdot \mu^c_{m_2-1}(0)\notag\\
&+\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-2+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k-1\})\cdot \frac{q^{c-m_2+1}}{1+q^{c-m_2+1}}.\label{eq: finite asep particles}
\end{align}

We now show that $\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k\})=\frac{q^{k(c+1-m_2)+\frac{k(k-1)}{2}}}{(-q^{c-m_2+1};q)_{\hat{m}_2}}\begin{bmatrix}
\hat{m}_2 \\
\end{bmatrix}_{q}$ satisfies (<ref>). This formula is inspired by a combinatorial argument using integer partitions. As additional material this argument is given in Appendix <ref> but those parts are not needed for any of the proofs in this article, since these are self-contained and purely probabilistic.
Start with the RHS of (<ref>),

\begin{align*}
&\hspace{8mm}=\frac{1}{1+q^{c-m_2+1}}\left( \frac{q^{k(c+2-m_2)+\frac{k(k-1)}{2}}}{(-q^{c-m_2+2};q)_{\hat{m}_2-1}}\begin{bmatrix}
\hat{m}_2-1 \\
\end{bmatrix}_{q} + q^{c-m_2+1}\frac{q^{(k-1)(c+2-m_2)+\frac{(k-1)(k-2)}{2}}}{(-q^{c-m_2+2};q)_{\hat{m}_2-1}}\begin{bmatrix}
\hat{m}_2-1 \\
\end{bmatrix}_{q} \right)\\
\hat{m}_2-1 \\
\end{bmatrix}_{q}+\begin{bmatrix}
\hat{m}_2-1 \\
\end{bmatrix}_{q}\right).
\end{align*}

Using the $q$-binomial analogue of Pascal's identity (Lemma <ref>) we have that,

\hat{m}_2-1 \\
\end{bmatrix}_{q}+\begin{bmatrix}
\hat{m}_2-1 \\
\end{bmatrix}_{q}=\begin{bmatrix}
\hat{m}_2 \\
\end{bmatrix}_{q}.$$

Thus (<ref>) holds with $\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k\})=\frac{q^{k(c+1-m_2)+\frac{k(k-1)}{2}}}{(-q^{c-m_2+1};q)_{\hat{m}_2}}\begin{bmatrix}
\hat{m}_2 \\
\end{bmatrix}_{q}$.

To complete the proof we first notice that the boundary conditions given by (<ref>) and (<ref>) conform with the solution above and show that with the recursion (<ref>) the distribution is completely fixed. To find the value of $\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_1+\frac{1}{2}}(\underline{z})=k\})$ at a point $(\hat{m_2},k)$ the recursion (<ref>) requires values at the points $(\hat{m_2}-1,k)$ and $(\hat{m_2}-1,k-1)$. Using the recursion iteratively we arrive to the boundary points $(x,x)$ and $(x,0)$ for any $x\in\{1,\dots,k\}$. The former case is handled by (<ref>) and the latter by (<ref>).

Notice that Theorem <ref> can be seen by taking $m\defeq m_2-1$  and the limit as $m_1 \to -\infty$ in Lemma <ref>. This gives it yet another alternative proof.

§.§ Symmetry of ASEP

In the previous section we considered the function $N_{m+\frac{1}{2}}^{\overset{\leftarrow}{p}}$, the number of particles to the left of a site. Now we consider the distribution under the blocking measure of the function $N_{m+\frac{1}{2}}^{\overset{\rightarrow}{h}}$. To do this we notice that there is a symmetry between particles and holes in ASEP, this symmetry is made precise in the following Lemma.

Let $\underline{\eta}$ be an ASEP with right drift and blocking measure (for any $c\in\mathbb{R}$),

Consider the process given by $\hat{\underline{\eta}}\defeq \underline{1}-\underline{\eta}$. This process is an ASEP with left drift on the state space

$$\hat{\Omega}\defeq\{\underline{z}\in\{0,1\}^\mathbb{Z}: \sum\limits_{i=1}^\infty z_i, \sum\limits_{i=-\infty}^0 (1-z_i)<\infty\}$$

and has stationary measure (for any $c\in\mathbb{R}$),

Then for any $n,m\in\mathbb{Z}$ and $c\in\mathbb{R}$,

where $N^{\overset{\rightarrow}{p}}_{m+\frac{1}{2}}(\underline{z})\defeq \sum\limits_{i=m+1}^\infty z_i$.

By definition of $\hat{\underline{\eta}}$, particles in $\hat{\underline{\eta}}$ move like holes in $\underline{\eta}$. That is, particles jump to the left at rate $1$ and to the right at rate $q$, unless the jump is blocked. Thus, the probability there is a particle at site $i$, is exactly the probability that there isn't in $\underline{\eta}$ and similarly for when there isn't a particle in $\hat{\underline{\eta}}$. So we have that $\hat{\underline{\mu}}^c=\bigotimes\limits_{i\in\mathbb{Z}}\hat{\mu}_i^c$ given by,

\begin{align*}
\hat{\mu}^c_i(1)&=\mu^c_i(0)=\frac{1}{1+q^{-(i-c)}}=\frac{q^{i-c}}{1+q^{i-c}}\\
\hat{\mu}_i^c(0)&=\mu^c_i(1)=\frac{q^{-(i-c)}}{1+q^{-(i-c)}}=\frac{1}{1+q^{i-c}}
\end{align*}

is reversible stationary for this ASEP with left drift.
Hence if we define, $N^{\overset{\rightarrow}{p}}_{m+\frac{1}{2}}(\underline{z})\defeq \sum\limits_{i=m+1}^\infty z_i$, it immediately follows that  for any $n,m\in\mathbb{Z}$ and $c\in\mathbb{R}$,

We will now see that considering particles to the right of say $m+\frac{1}{2}$ in $\hat{\underline{\eta}}$ is similar to considering particles to the left of $m+\frac{1}{2}$ in $\underline{\eta}$.

For any $n,m \in\mathbb{Z}$ and $c\in\mathbb{R}$,

We have that,

\begin{align*}
&\hat{\underline{\mu}}^c(\{N^{\overset{\rightarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=n\})=\sum\limits_{\substack{\underline{z}\in\hat{\Omega}:\\ \sum\limits_{i=m+1}^\infty z_i=n}}\prod\limits_{i=m+1}^\infty\hat{\mu}_i^c(z_i)=\sum\limits_{\substack{\underline{z}\in\hat{\Omega}:\\ \sum\limits_{i=m+1}^\infty z_i=n}}\prod\limits_{i=0}^\infty\hat{\mu}_{m+1+i}^c(z_{m+1+i})\\
\text{and}&\\
&\underline{\mu}^{2m+1-c}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=n\})=\sum\limits_{\substack{\underline{z}\in\Omega:\\ \sum\limits_{i=-\infty}^m z_i=n}}\prod\limits_{i=-\infty}^ m\mu_i^{2m+1-c}(z_i)=\sum\limits_{\substack{\underline{z}\in\Omega:\\ \sum\limits_{i=-\infty}^m z_i=n}}\prod\limits_{i= 0}^\infty\mu_{m-i}^{2m+1-c}(z_{m-i}).
\end{align*}

We see that for $i\geq 0$ the marginals $\mu^{2m+1-c}_{m-i}$ and $\hat{\mu}^c_{m+1+i}$ agree, for $z\in\{0,1\}$

Thus for any $n,m \in\mathbb{Z}$ and $c\in\mathbb{R}$,

By combining Lemmas <ref> and <ref> we have the following result.

For any $n,m\in\mathbb{Z}$ and $c\in\mathbb{R}$,

§.§ Combinatorial Identities coming from ASEP

Certain probabilistic questions for blocking family interacting particle systems, such as stationarity and equivalences between systems, lead to proofs of identities of combinatorial significance. For example, in 2018 Balázs and Bowen [5] gave a probabilistic proof of the Jacobi triple product identity using the Exclusion - Zero-range correspondence. Also, in 2022 Balázs, Fretwell and Jay [6] found new 3-variable combinatorial identities as well as known identities by considering the family of $0$-$1$-$2$ blocking particle systems and also the $k$-Exclusion process. Here we see that other well-known combinatorial identities with interpretations in terms of integer partitions can be explained probabilistically by considering certain quantities for ASEP.
By considering the symmetry between particles and holes for the asymmetric simple exclusion process we can give a probabilistic proof to a well-known identity from the theory of integer partitions, the Durfee rectangles identity.

It is clear by definition of $N(\cdot)$ that,

$$\underline{\mu}^c(\{N(\underline{z})=n\})=\sum\limits_{k=\max\{-n,0\}}^\infty\underline{\mu}^c(\{N^{\overset{\rightarrow}{h}}(\underline{z})=n+k\})\cdot \underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}(\underline{z})=k\}).$$

By (<ref>) as well as Theorem <ref> and Corollary <ref> (for $m=0$) we have that,

\begin{align*}
\frac{q^{\frac{n(n+1)}{2}-nc}}{\sum\limits_{\ell=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}-\ell c}}&=\sum\limits_{k=\max\{-n,0\}}^\infty \frac{q^{-(n+k)c+\frac{(n+k)(n+k+1)}{2}}}{(-q^{1-c};q)_\infty(q;q)_{n+k}}\cdot \frac{q^{k(c-1)+\frac{k(k+1)}{2}}}{(-q^c;q)_\infty(q;q)_k}\\
&=\sum\limits_{k=\max\{-n,0\}}^\infty \frac{q^{\frac{n(n+1)}{2}+k(n+k)-nc}}{(-q^{1-c};q)_\infty(q;q)_{n+k}(-q^c;q)_\infty(q;q)_k}.
\end{align*}

The Jacobi Triple Product identity states that, for $q,z\in\mathbb{R}$  such that $|q|<1$ and $z\neq 0$,

$$\sum\limits_{l=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}}z^{\ell}=(q;q)_\infty(-qz;q)_\infty(-z^{-1};q)_\infty.$$

Then by setting $z\defeq q^{-c}$ we have that,

$$\sum\limits_{l=-\infty}^\infty q^{\frac{\ell(\ell+1)}{2}-\ell c}=(q;q)_\infty(-q^{1-c};q)_\infty(-q^c;q)_\infty.$$

Hence we find that,

Notice that if we instead consider $\underline{\mu}^c(\{N_{m+\frac{1}{2}}(\underline{z})=n\})$ for any $m\in\mathbb{Z}$ we prove the same identity. This is since $m$ is just a shift for the the state and the factors involving $m$ cancel out.

This proof is purely probabilistic; we give an alternative, combinatorial proof of Theorem <ref> in Appendix <ref>, this sheds light on why we should expect combinatorial identities from these probabilistic questions. We will now demonstrate that the identity above has a combinatorial interpretation in terms of integer partitions.
Let us first consider the case when $n=0$, then the identity is,

This is a well known identity, namely the identity for Durfee squares of integer partitions (for example see Chapter 8 in Andrews and Eriksson [3]). For a given integer partition the Durfee square is the largest square (anchored in the upper left-hand corner) in its Ferrers diagram. Equivalently, a given partition with $\ell$ parts, $(\lambda_1\geq\lambda_2\geq...\geq\lambda_\ell)$, has a Durfee square of side length $k\leq \ell$ if $\lambda_k\geq k$ and $\lambda_{k+1}\leq k$. With this definition, any integer partition decomposes into 3 pieces: its unique Durfee square of side length $k$, a partition into up $k$ parts to its right, and a partition with parts at most size $k$ underneath (for example see Figure <ref> below). We denote this decomposition in the following way,

where $\lambda^{(r)}=(\lambda_1-k,...,\lambda_k-k)$ is the partition to the right of the Durfee square and $\lambda^{(d)}=(\lambda_{k+1},...,\lambda_{\ell})$ the partition underneath the Durfee square.
From this decomposition we see where the RHS of the identity comes from. Consider one term in the sum, say $\frac{q^{k^2}}{(q;q)_k^2}$, for some $k$. This is the generating function for integer partitions whose Durfee square is of side length $k$:

* $q^{k^2}$ corresponds to the Durfee square.

* $\frac{1}{(q;q)_k}$ is the generating function for partitions into up to $k$ parts (see Section 6.2 in Andrews and Eriksson [3]) and so corresponds to the partition to the right of the Durfee square.

* By conjugation $\frac{1}{(q;q)_k}$ is also the generating function for partitions into parts of size at most $k$ (see Section 6.2 in Andrews and Eriksson [3]) so this corresponds to partition underneath the Durfee square.

The sum over all $k$ considers Durfee squares of any size and so we recover all integer partitions, for which the generating function is $\frac{1}{(q;q)_\infty}$, hence the identity holds.

[red] (-0.25,3.25)–(1.25,3.25);
[red] (-0.25,1.75)–(1.25,1.75);
[red] (-0.25,1.75)–(-0.25,3.25);
[red] (1.25,1.75)–(1.25,3.25);
[black] (0,3) circle (2pt);
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(a) at (1.5,-1.5) [label=$\lambda$(8873211)];
(a) at (5,1) [label=$\Longleftrightarrow$];
(a) at (9,0.8) [label=$\lambda\equiv \{3^2\comma (5\comma5\comma4)\comma(3\comma2\comma1\comma1)\}$];

An example of the Durfee square for an integer partition, $\lambda$, of $30$.

Now let us consider what the identity means combinatorially for any fixed $n\in\mathbb{Z}$. The identity now states that, for $n$ fixed, every integer partition has a unique Durfee rectangle, a generalisation of the notion of Durfee square (see for example equation (4) in Gessel [14]). For some fixed $n$ we say that a given partition with $\ell$ parts, $(\lambda_1\geq\lambda_2\geq...\geq\lambda_\ell)$ has a Durfee rectangle of side lengths $n+k$ and $k$ if $\lambda_{k}\geq n+k$ and $\lambda_{k+1}\leq n+k$. From this definition we see that for fixed $n$ each integer partition has a unique Durfee rectangle (just as it had a unique Durfee square i.e. the case when $n=0$). Now with this definition, for each fixed $n$, any integer partition decomposes into 3 pieces: its unique Durfee rectangle of side lengths $k$ and $n+k$, a partition into up to $k$ parts to its right, and a partition with parts at most size $n+k$ underneath. We denote this decomposition in the following way,

where $\lambda^{(n,r)}=(\lambda_1-(n+k),...,\lambda_k-(n+k)$ is the partition to the right of the Durfee rectangle and $\lambda^{(n,d)}=(\lambda_{k+1},...,\lambda_\ell)$ the partition underneath the Durfee rectangle. See Figure <ref> below for an example of the Durfee rectangle inside a given integer partition for different values of $n$.
Now we can see where the RHS of the identity comes from. Consider one term in the sum, say $\frac{q^{k(n+k)}}{(q;q)_k(q;q)_{n+k}}$, for some $k$. This is the generating function for integer partitions with Durfee rectangle of side lengths $k$ and $n+k$:

* $q^{k(n+k)}$ corresponds to the Durfee rectangle.

* $\frac{1}{(q;q)_k}$ is the generating function for partitions into up to $k$ parts and so corresponds to the partition to the right of the Durfee rectangle.

* By conjugation $\frac{1}{(q;q)_{n+k}}$ is the generating function for partitions with parts of size at most $n+k$ and so corresponds to the partition underneath the Durfee rectangle.

Summing these generating functions for all $k$ from $\max\{-n,0\}$ to $\infty$ (with $n\in\mathbb{Z}$ fixed) we get all integer partitions, and thus the identity holds.

[red] (-0.25,3.25)–(2.25,3.25);
[red] (-0.25,1.75)–(2.25,1.75);
[red] (-0.25,1.75)–(-0.25,3.25);
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[black] (0,2) circle (2pt);
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[black] (1,2) circle (2pt);
[black] (1.5,2) circle (2pt);
[black] (2,2) circle (2pt);
[black] (2.5,2) circle (2pt);
[black] (3,2) circle (2pt);
[black] (0,1.5) circle (2pt);
[black] (0.5,1.5) circle (2pt);
[black] (1,1.5) circle (2pt);
[black] (0,1) circle (2pt);
[black] (0.5,1) circle (2pt);
[black] (0,0.5) circle (2pt);
[black] (0,0) circle (2pt);
(a) at (1.5,-1.5) [label=$\lambda\equal (8\comma 8\comma 7\comma 3\comma 2\comma 1\comma 1)$];
(a)at(2.7,-2.25)[label=$\equiv \{3*5 \comma (3\comma3\comma2)\comma(3\comma2\comma1\comma1)\}$];
(a) at (1.5, 3.5) [label=If $n\equal2$ then here $k\equal 3$];
[red] (9.75,3.25)–(10.75,3.25);
[red] (9.75,0.75)–(10.75,0.75);
[red] (9.75,0.75)–(9.75,3.25);
[red] (10.75,0.75)–(10.75,3.25);
[black] (10,3) circle (2pt);
[black] (10.5,3) circle (2pt);
[black] (11,3) circle (2pt);
[black] (11.5,3) circle (2pt);
[black] (12,3) circle (2pt);
[black] (12.5,3) circle (2pt);
[black] (13,3) circle (2pt);
[black] (13.5,3) circle (2pt);
[black] (10,2.5) circle (2pt);
[black] (10.5,2.5) circle (2pt);
[black] (11,2.5) circle (2pt);
[black] (11.5,2.5) circle (2pt);
[black] (12,2.5) circle (2pt);
[black] (12.5,2.5) circle (2pt);
[black] (13,2.5) circle (2pt);
[black] (13.5,2.5) circle (2pt);
[black] (10,2) circle (2pt);
[black] (10.5,2) circle (2pt);
[black] (11,2) circle (2pt);
[black] (11.5,2) circle (2pt);
[black] (12,2) circle (2pt);
[black] (12.5,2) circle (2pt);
[black] (13,2) circle (2pt);
[black] (10,1.5) circle (2pt);
[black] (10.5,1.5) circle (2pt);
[black] (11,1.5) circle (2pt);
[black] (10,1) circle (2pt);
[black] (10.5,1) circle (2pt);
[black] (10,0.5) circle (2pt);
[black] (10,0) circle (2pt);
(a) at (11.5,-1.5) [label=$\lambda\equal (8\comma 8\comma 7\comma 3\comma 2\comma 1\comma 1)$];
(a)at(12.5,-2.25)[label=$\equiv \{5*2 \comma (6\comma6\comma5\comma1)\comma(1\comma1)\}$];
(a) at (11.5, 3.5) [label=If $n\equal -3$ then here $k\equal 5$];

An example of the Durfee rectangle for an integer partition, $\lambda$, of $30$ when $n=2$ and $n=-3$.

We now demonstrate that two other well-known combinatorial identities (Euler's identity and the $q$-Binomial Theorem) arise as consequences of Theorem <ref> and Lemma <ref>.

For any $m \in\mathbb{Z}$ and $c \in \mathbb{R}$ it is clear that,
$\sum\limits_{k=0}^\infty \underline{\mu}^c (\{N_m^{\overset{\leftarrow}{p}}(\underline{z})=k\})=1.$ By Theorem <ref>, that is

$$\sum\limits_{k=0}^\infty \frac{q^{k(c-m)}q^{\frac{k(k-1)}{2}}}{(q;q)_{k}(-q^{c-m};q)_\infty}=1.$$

If we let $z:=q^{c-m}$ and rearrange we find Euler's identity,

$$\sum\limits_{k=0}^\infty \frac{q^{\frac{k(k-1)}{2}}z^k}{(1-q)(1-q^2)...(1-q^k)}=\prod\limits_{i=0}^\infty(1+zq^i).$$

Combinatorially this identity gives two ways of writing the 2-variable generating function for integer partitions into distinct parts. A general term is of the form $d_{n,k}q^nz^k$ where $d_{n,k}$ gives the number of partitions of $n$ into exactly $k$ distinct parts. The product on the LHS of the identity clearly counts partitions into distinct parts (with a $z$ in front of $q$ to count the number of non-zero parts). For the sum on the RHS, $k$ gives the number of parts and it is well known that $\frac{q^{\frac{k(k+1)}{2}}}{(q;q)_k}$ is the generating function for partitions into exactly $k$ distinct parts (see for example Section 2 of Cimpoeaş's 2022 paper [10]). In Appendix <ref> we give an alternate proof of Theorem <ref> by considering particles states with $k$ particles to the left of some site $m$ as partitions of some $s$ (left jumps away from a ground state) into up to $k$ parts. Instead we can think of a state that has $k$ particles to the left of $m$ as a partition of some $n$ where each particle denotes a part of size how far this particle is from the boundary site $m$. Since there can only be one particle per site these distances must be distinct and thus we have a partition into exactly $k$ distinct parts. This explains why Euler's identity arises a consequence of Theorem <ref>.
We also see the $q$-Binomial Theorem:

For any $m_1<m_2\in\mathbb{Z}$ and $c \in \mathbb{R}$ it is clear that, $\sum\limits_{k=0}^{m_2-m_1-1}\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_2-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_{1}+\frac{1}{2}}(\underline{z})=k\})=1.$ By Lemma <ref>, that is

$$\sum\limits_{k=0}^{m_2-m_1-1} \frac{q^{k(c+1-m_2)}q^{\frac{k(k-1)}{2}}}{(-q^{c+1-m_2};q)_{m_2-m_1-1}}\begin{bmatrix}
\end{bmatrix}_q=1.$$

Now if we let $z:=q^{c+1-m_2}$, $m:=m_2-m_1-1$ and rearrange we have the $q$-binomial theorem,

\end{bmatrix}_{q}=(-z;q)_m.$$

Combinatorially this identity gives two ways of writing the 2-variable generating function for integer partitions into distinct parts of size at most $m$. It is known that the one-variable generating function for partitions into distinct parts is given by $\prod\limits_{i=1}^\infty(1+q^i)$, clearly if we truncate this product at $i=m$ we have distinct parts of size a most $m$ (see for example Equation 5.4 of Andrews and Eriksson [3]). If we then write $\prod\limits_{i=1}^m(1+zq^i)$ this is the two-variable generating function for these partitions with the power of $z$ counting the exact number of parts used. Now for the sum on the RHS, as before $k$ gives the number of parts. It is known that $\begin{bmatrix}
\end{bmatrix}_{q}$ is the generating function for partitions into up to $k$ part of size up to $m-k$ (for example see Section 7.2 of Andrews and Eriksson [3]). If we add a triangle of side length $k$ to the Young diagram of such a partition we now have a partition with distinct parts of size up to $m$, as desired. Note that adding this triangle is the same as multiplying the generating function by a factor of $q^{\frac{k(k+1)}{2}}$. We note that if we take $m$ to infinity in the $q$-Binomial Theorem we get Euler's Identity. In Appendix <ref> we give an alternative proof of Lemma <ref> by considering particle states with $0\leq k\leq m_2-m_1-1$ particles inside $(m_1,m_2)$ as partitions of some $s$ (left jumps away from a ground state) into up $k$ parts of size up to $m_2-m_1-1-k$. Instead we can think of the states as partitions into exactly $k$ parts with size up to $m\defeq m_2-m_1-1$ by letting each particle represent a part of size its distance away from the right boundary site $m_2$. Again we see that these distances must be distinct due to the exclusion rule and so the parts in the corresponding partition are distinct. This is explains why the $q$-Binomial Theorem arises as a consequence of Lemma <ref>.

§ THE ASYMMETRIC SIMPLE EXCLUSION PROCESS WITH SECOND CLASS PARTICLES

Now we consider the asymmetric simple exclusion process with 2 species of particles, first and second class particles. Both species of particles evolve according to ASEP dynamics obeying the exclusion rule within their own class, with first class particles viewing second class as `holes'. That is if,

* a second class particle tries to jump into a space occupied by a first class particle the jump is blocked and nothing changes

* a first class particle tries to jump into a space occupied by a second class particle it takes that space and the two particles swap positions.

To construct an ASEP with first class particles and some fixed $d>0$ second class particles we use the basic coupling method. This idea of defining second class particles via a basic coupling is similar to the work of Ferrari, Kipnis and Saada [12] and later work of Tracy and Widom [19] as well as Balázs and Seppäläinen [7] (also work of Liggett [16] in 1976 considered coupling for simple exclusion processes).
We will study the behaviour of these second class particles under the blocking measure.

§.§ Constructing a Basic Coupling of Two ASEPs

To construct a basic coupling we start with an ASEP on $\mathbb{Z}$, $\{\underline{\xi}(t)\}$, and also a process on the particle labels. For some given $d>0$, this label process is defined by an ASEP on the half infinite line $\mathbb{Z}_{\geq 0}$ with exactly $d$ particles having left drift, $\{\underline{\gamma}(t)\}$. The dynamics of $\{\underline{\gamma}(t)\}$ is restricted by the state of $\{\underline{\xi}(t)\}$, such that a particle jump over the edge $(i,i+1)$ in $\underline{\gamma}(t)$ can only occur if the particles $i$ and $i+1$ in $\underline{\xi}(t)$ are nearest neighbours. Lastly we construct a second ASEP on $\mathbb{Z}$, $\{\underline{\eta}(t)\}$,  by at each time removing the particles corresponding to the label process from $\underline{\xi}(t)$. We see that this defines a basic coupling of two ASEPs with the second class particles being defined as the particles corresponding to the label process. We make this procedure precise below.

§.§.§ ASEP on $\mathbb{Z}$

We consider an asymmetric simple exclusion process on ,
$\Omega\defeq\{\underline{z}\in\{0,1\}^\mathbb{Z}:N^{\overset{\leftarrow}{p}}(\underline{\eta}), N^{\overset{\rightarrow}{h}}(\underline{\eta})<\infty\}$, under the blocking measure $\underline{\mu}^c$ (as in Section <ref>) At time $t\in\mathbb{R}_{\geq 0}$ we will denote the state of the process by $\underline{\xi}(t)\in\Omega$.
At any time $t \in \mathbb{R}_{\geq 0}$ we attach to the state $\underline{\xi}(t)$ a labelling of the particles. That is, we enumerate the particles in $\underline{\xi}(t)$ from left to right starting from $0$ for the left most particle and increasing as we go to the right. So a particle's label is simply the number of particles to its left. When we consider second class particles we will be interested in the label of only $d>0$ certain particles.

§.§.§ ASEP on $\mathbb{Z}_{\geq 0}$ with exactly $d$ particles (Label process)

Alongside the asymmetric simple exclusion process $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ we also consider an ASEP on the half infinite integer line, with exactly $d>0$ particles moving with a left drift. That is, an interacting particle system on the state space
$\hat{\Omega}\defeq\{\underline{z}\in\{0,1\}^{\mathbb{Z}_{\geq0}}: \sum\limits_{i=0}^\infty z_i=d\}$,
which evolves according to nearest neighbour particle jumps with jump rates, (for $i\geq 0$)

$$\hat{p}(z_i,z_{i+1})=q\mathbb{I}\{z_i \neq 0\}\mathbb{I}\{z_{i+1}\neq 1\} \hspace{5mm}\text{ and }\hspace{5mm} \hat{q}(z_i,z_{i+1})=\mathbb{I}\{z_{i+1} \neq 0\}\mathbb{I}\{z_{i}\neq 1\}.$$

Note that by the definition of the state space $\hat{\Omega}$ any state has exactly $d$ particles and so there is a closed boundary to the left of site $0$. At time $t\in\mathbb{R}_{\geq 0}$ we will denote the state of this process by $\underline{\gamma}(t)$.

Let $\underline{y}(t)=(y_1(t),...,y_d(t))$, with $0\leq y_1(t)<y_2(t)<...<y_d(t)$, be the vector of sites where the particles are in $\underline{\gamma}(t)$, i.e. $\forall j \in \{1,..,d\}$,

$$y_j(t)=i \text{ if } \gamma_i=1 \text{ and } \sum\limits_{k=0}^i\gamma_k=j.$$

In fact, we will consider a restriction of the process $\{\underline{\gamma}(t)\}_{t\in\mathbb{R}_{\geq0}}$ by conditioning on the process $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq0}}$. We say that a particle jump which is possible over the edge $(i,i+1)$ in the half-infinite ASEP can only happen if the particles with labels $i$ and $i+1$ are nearest neighbours at that time in the ASEP on $\Omega$. So, we adjust the jump rates $\hat{p}$ and $\hat{q}$ to,

\begin{align*}
&\hat{p}^*(\gamma_i(t),\gamma_{i+1}(t))=q\mathbb{I}\{\gamma_i(t) \neq 0\}\mathbb{I}\{\gamma_{i+1}(t)\neq 1\}\left(\sum\limits_{j\in\mathbb{Z}}\mathbb{I}\left\{\sum\limits_{k=-\infty}^{j-1}\xi_k(t)=i\right\}\mathbb{I}\{\xi_j(t)=1\}\mathbb{I}\{\xi_{j+1}(t)=1\}\right)\\
\text{ and, }&\hat{q}^*(\gamma_i(t),\gamma_{i+1}(t))=\mathbb{I}\{\gamma_{i+1}(t) \neq 0\}\mathbb{I}\{\gamma_{i}(t)\neq 1\}\left(\sum\limits_{j\in\mathbb{Z}}\mathbb{I}\left\{\sum\limits_{k=-\infty}^{j-1}\xi_k(t)=i\right\}\mathbb{I}\{\xi_j(t)=1\}\mathbb{I}\{\xi_{j+1}(t)=1\}\right).
\end{align*}

For this restricted process, let the vector of sites where particles are in $\underline{\gamma}(t)$ be $\underline{x}(t)$. Once we define a coupling and thus define $d$ second class particles, $\{\underline{x}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ will be process that gives the labels for the second class particles.
So we have an ASEP $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq0}}$ with some label process sitting on top where the $d$ particles of interest can swap only when they are neighbouring another particle. For example see Figure <ref> below (the red particles in $\underline{\xi}(t)$ are those which correspond to the process $\{\underline{x}(t)\}_{t\in\mathbb{R}_{\geq 0}}$). In this example the particle jump at sites 2 and 3 is allowed but not the jump at sites 6 and 7 in $\underline{\gamma}(t)$.

[thick, ->] (-11,0)–(0,0);
in -11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1 [thick, -](cm, -0.03)–(cm, 0.1);
(a) at (-11,-1) [label= 0];
(a) at (-10,-1) [label= 1];
(a) at (-9,-1) [label= 2];
(a) at (-8,-1) [label= 3];
(a) at (-7,-1) [label= 4];
(a) at (-6,-1) [label= 5];
(a) at (-5,-1) [label= 6];
(a) at (-4,-1) [label= 7];
(a) at (-3,-1) [label= 8];
(a) at (-2,-1) [label= 9];
(a) at (-1,-1) [label= 10];
[black] (-11,0.5) circle (4pt);
[black] (-10,0.5) circle (4pt);
[black] (-8,0.5) circle (4pt);
[black] (-6,0.5) circle (4pt);
[black] (-5,0.5) circle (4pt);
(a) at (1, -0.5) [label=$\underline{\gamma}(t)$];
(a) at (7, -0.5) [label=and so $\underline{x}(t)\equal (0\comma 1\comma 3\comma 5\comma 6)$];

[thick, <->] (-11,-3)–(11,-3);
in -10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10 [thick, -](cm, -3.03)–(cm, -2.9)node[anchor=north]$\x$;
[black] (-7,-2.5) circle (4pt);
(a) at (-7.3,-3) [label=2];
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(a) at (-8.3,-3) [label=1];
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(a) at (-9.3,-3) [label=0];
[red] (-6,-2.5) circle (4pt);
(a) at (-6.3,-3) [label=3];
[black] (8,-2.5) circle (4pt);
(a) at (7.7,-3) [label=8];
[black] (9,-2.5) circle (4pt);
(a) at (8.7,-3) [label=9];
[black] (10,-2.5) circle (4pt);
(a) at (9.7,-3) [label=10];
[black] (10.25,-2.75) circle (1pt);
[black] (10.5,-2.75) circle (1pt);
[black] (10.75,-2.75) circle (1pt);
[black] (-4,-2.5) circle (4pt);
(a) at (-4.3,-3) [label=4];
[red] (-1,-2.5) circle (4pt);
(a) at (-1.3,-3) [label=5];
[red] (3,-2.5) circle (4pt);
(a) at (2.7,-3) [label=6];
[black] (5,-2.5) circle (4pt);
(a) at (4.7,-3) [label=7];
[black] (-10.25,-2.75) circle (1pt);
[black] (-10.5,-2.75) circle (1pt);
[black] (-10.75,-2.75) circle (1pt);
(a) at (11.5, -3.5) [label=$\underline{\xi}(t)$];

An example of the pair $(\underline{\xi}(t),\underline{x}(t))$ when $d=5$.

We will be interested in where the second class particles are at time $t$. As discussed the red particles in the above figure for example will eventually be the second class particles. We denote the positions of the particles in $\underline{\xi}(t)$ corresponding to the label process $\underline{x}(t)$ by, $\underline{X}(t)=(X_1(t),...,X_d(t))$, i.e. $X_j(t)$ is the position of the particle with label $x_j(t)$ in $\underline{\xi}(t)$. So for each $j\in\{1,...,d\}$,

$$X_j(t)=i \text{ if } \xi_i(t)=1 \text{ and } \sum\limits_{k=0}^{i-1}\xi_k(t)=x_j(t).$$

It is clear to see by construction that at any time $t\geq 0$ we have that $X_1(t)< X_2(t) < ... <X_d(t)$.

§.§.§ Basic Coupling of Two ASEPs

We now construct another process $\{\underline{\eta}(t)\}_{t\in\mathbb{R}_{\geq 0}}$, by taking the state of the ASEP at time $t$, $\underline{\xi}(t)$, and removing the particles which correspond to the label process $\underline{x}(t)$. That is, at any time $t\geq 0$,

$$\underline{\eta}(t)\defeq\underline{\xi}(t)-\sum_{j=1}^d\underline{\delta}_{X_j(t)} .$$

For example see Figure <ref>

(a) at (-8, 2) [label=Suppose $\underline{\xi}(t)$ and $\underline{x}(t)$ are as follows];
[thick, ->] (-11,0)–(0,0);
in -11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1 [thick, -](cm, -0.03)–(cm, 0.1);
(a) at (-11,-1) [label= 0];
(a) at (-10,-1) [label= 1];
(a) at (-9,-1) [label= 2];
(a) at (-8,-1) [label= 3];
(a) at (-7,-1) [label= 4];
(a) at (-6,-1) [label= 5];
(a) at (-5,-1) [label= 6];
(a) at (-4,-1) [label= 7];
(a) at (-3,-1) [label= 8];
(a) at (-2,-1) [label= 9];
(a) at (-1,-1) [label= 10];
[black] (-11,0.5) circle (4pt);
[black] (-10,0.5) circle (4pt);
[black] (-8,0.5) circle (4pt);
[black] (-6,0.5) circle (4pt);
[black] (-5,0.5) circle (4pt);
(a) at (1, -0.5) [label=$\underline{\gamma}(t)$];
(a) at (7, -0.5) [label=and so $\underline{x}(t)\equal (0\comma 1\comma 3\comma 5\comma 6)$];

[thick, <->] (-11,-3)–(11,-3);
in -10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10 [thick, -](cm, -3.03)–(cm, -2.9)node[anchor=north]$\x$;
[black] (-7,-2.5) circle (4pt);
(a) at (-7.3,-3) [label=2];
[red] (-8,-2.5) circle (4pt);
(a) at (-8.3,-3) [label=1];
[red] (-9,-2.5) circle (4pt);
(a) at (-9.3,-3) [label=0];
[red] (-6,-2.5) circle (4pt);
(a) at (-6.3,-3) [label=3];
[black] (8,-2.5) circle (4pt);
(a) at (7.7,-3) [label=8];
[black] (9,-2.5) circle (4pt);
(a) at (8.7,-3) [label=9];
[black] (10,-2.5) circle (4pt);
(a) at (9.7,-3) [label=10];
[black] (10.25,-2.75) circle (1pt);
[black] (10.5,-2.75) circle (1pt);
[black] (10.75,-2.75) circle (1pt);
[black] (-4,-2.5) circle (4pt);
(a) at (-4.3,-3) [label=4];
[red] (-1,-2.5) circle (4pt);
(a) at (-1.3,-3) [label=5];
[red] (3,-2.5) circle (4pt);
(a) at (2.7,-3) [label=6];
[black] (5,-2.5) circle (4pt);
(a) at (4.7,-3) [label=7];
[black] (-10.25,-2.75) circle (1pt);
[black] (-10.5,-2.75) circle (1pt);
[black] (-10.75,-2.75) circle (1pt);
(a) at (11.5, -3.5) [label=$\underline{\xi}(t)$];
(a) at (-11, -5) [label=then $\underline{\eta}(t)$ is];
[thick, <->] (-11,-7)–(11,-7);
in -10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10 [thick, -](cm, -7.03)–(cm, -6.9)node[anchor=north]$\x$;
[black] (-7,-6.5) circle (4pt);
[black] (8,-6.5) circle (4pt);
[black] (9,-6.5) circle (4pt);
[black] (10,-6.5) circle (4pt);
[black] (10.25,-6.75) circle (1pt);
[black] (10.5,-6.75) circle (1pt);
[black] (10.75,-6.75) circle (1pt);
[black] (-4,-6.5) circle (4pt);
[black] (5,-6.5) circle (4pt);
[black] (-10.25,-6.75) circle (1pt);
[black] (-10.5,-6.75) circle (1pt);
[black] (-10.75,-6.75) circle (1pt);
(a) at (11.5, -7.5) [label=$\underline{\eta}(t)$];

An example of constructing $\underline{\eta}(t)$ from the pair $(\underline{\xi}(t),\underline{x}(t))$ (when $d=5$).

The process $\{\underline{\eta}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ is an ASEP and the pair $\{\underline{\eta}(t), \underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ is a basic coupling of two asymmetric simple exclusion processes which only differ at $d$ many sites at any time.

First we confirm that $\{\underline{\eta}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ is an asymmetric simple exclusion process. Since $\underline{\xi}(t)$ takes values in the state space $\Omega=\{\underline{z}\in\{0,1\}^\mathbb{Z}:N^{\overset{\leftarrow}{p}}(\underline{\eta}), N^{\overset{\rightarrow}{h}}(\underline{\eta})<\infty\}$ and by definition $\underline{\eta}(t)=\underline{\xi}(t)-\sum_{j=1}^d\underline{\delta}_{X_j(t)}$, it is clear that the process $\{\underline{\eta}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ also has state space $\Omega$. Now we will show that the process $\{\eta(t)\}_{t\in\mathbb{R}_{\geq 0}}$ evolves according to ASEP dynamics; that is each particle jumps left with rate $q$ and right with rate $1$ unless the jump is blocked. Let us consider the following possible cases for particle jumps in $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$:

* A particle associated to the label process (red particles in Figure <ref>) makes a particle jump. When this happens the particles in $\underline{\xi}(t)$ that are not associated with the label process $\underline{x}(t)$ are not affected and so there is no change to $\underline{\eta}(t)$.

* A particle not associated to the label process makes a particle jump. When this jump happens in $\underline{\xi}(t)$ it will happen in $\underline{\eta}(t)$ with the same jump rate ($q$ to the left and $1$ to the right).

* Lastly, the label process may evolve resulting in the swapping of neighbouring particles in $\underline{\xi}(t)$ (one associated with the label process and one that is not).

* Suppose a particle at site $i\geq 0$ in $\underline{\gamma}(t)$ jumps right (meaning particle $i$ and $i+1$ are neighbours in $\underline{\xi}(t)$ say at sites $j$ and $j+1$). Then in $\underline{\xi}(t)$ this results in particle $i$ no longer being associated with the label process and particle $i+1$ becomes associated with the label process. That is in $\underline{\eta}(t)$ the particle at site $j+1$ jumps to the left with rate equal to a right jump in $\{\underline{\gamma}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ so with rate $q$.

* Similarly suppose that the particle at site $i>0$ in $\underline{\gamma}(t)$ jumps left (meaning particle $i-1$ and $i$ are neighbours in $\underline{\xi}(t)$ say at sites $j$ and $j+1$). Then in $\underline{\xi}(t)$ this results in particle $i$ no longer being associated with the label process and particle $i-1$ becomes associated with the label process. That is in $\underline{\eta}(t)$ the particle at site $j$ jumps to the right with rate equal to a left jump in $\{\underline{\gamma}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ so with rate $1$.

Thus $\{\underline{\eta}(t)\}_{t \in\mathbb{R}_{\geq 0}}$ evolves according to particle jumps at rates,

\begin{align*}
p(\eta_i(t),\eta_{i+1}(t))=\mathbb{I}\{\xi_i(t) \neq 0\}&\mathbb{I}\{\xi_{i+1}(t)\neq 1\}\\
&+\mathbb{I}\{\gamma_{\sum\limits_{j=-\infty}^{i}\xi(t)}(t)\neq 0\}\mathbb{I}\{\gamma_{\sum\limits_{j=-\infty}^{i}\xi(t)-1}(t)\neq 1 \}\mathbb{I}\{\xi_i(t)=1\}\mathbb{I}\{\xi_{i+1}(t)=1\}
\end{align*}

\begin{align*}
q(\eta_i(t),\eta_{i+1}(t))=q\Big(\mathbb{I}\{\xi_{i+1}(t) \neq 0\}&\mathbb{I}\{\xi_{i}(t)\neq 1\}\\
&+\mathbb{I}\{\gamma_{\sum\limits_{j=-\infty}^{i-1}\xi(t)}(t)\neq 0 \}\mathbb{I}\{\gamma_{\sum\limits_{j=-\infty}^{i-1}\xi(t)+1}(t)\neq 1\}\mathbb{I}\{\xi_i(t)=1\}\mathbb{I}\{\xi_{i+1}(t)=1\}\Big)
\end{align*}

or equivalently written only in terms of $\underline{\eta}(t)$,

$$p(\eta_i(t),\eta_{i+1}(t))=\mathbb{I}\{\eta_i(t) \neq 0\}\mathbb{I}\{\eta_{i+1}(t)\neq 1\} \hspace{3mm}\text{ and }\hspace{3mm} q(\eta_i(t),\eta_{i+1}(t))=q\mathbb{I}\{\eta_{i+1}(t) \neq 0\}\mathbb{I}\{\eta_{i}(t)\neq 1\}.$$

By construction it is clear that the two processes $\{\underline{\eta}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ and $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ evolve together according to ASEP dynamics and thus the pair gives a basic coupling.

This coupling defines the asymmetric exclusion process with $d$ second class particles. Here, $\underline{x}(t)$ is the label process for the second class particles and $\underline{X}(t)$ gives the positions of the second class particles at time $t$.

§.§ Distributional results for the basic coupling

Let us now look at some distributional results for the process with second class particles given we know the distribution of the ASEP with only first class particles.

With the relation $\underline{\eta}=\underline{\xi}-\sum\limits_{j=1}^d\underline{\delta}_{X_j}$, when we reach stationarity for the coupling, we have that,

$$\underline{\xi} \sim \underline{\nu}^n \hspace{1mm}\Longleftrightarrow\hspace{1mm} \underline{\eta}\sim\underline{\nu}^{n+d}.$$

This is clear by definition of $\underline{\nu}^n$ and since,

\begin{align*}
N(\underline{\xi})&=\sum\limits_{i=1}^\infty (1-\xi_i)-\sum\limits_{i=-\infty}^0\xi_i\\
&=\sum\limits_{i=1}^\infty (1-\{\underline{\eta}+\sum\limits_{j=1}^d\underline{\delta}_{X_j}\}_i)-\sum\limits_{i=-\infty}^0\{\underline{\eta}+\sum\limits_{j=1}^d\underline{\delta}_{X_j}\}_i\\
\end{align*}

Since $\underline{\nu}^n$ is the unique stationary distribution for the marginal process conditioned on having conserved quantity $n$ it holds that $\underline{\xi} \sim \underline{\nu}^n \hspace{1mm}\Longleftrightarrow\hspace{1mm} \underline{\eta}\sim\underline{\nu}^{n+d}$.

With the relation $\underline{\eta}=\underline{\xi}-\sum\limits_{j=1}^d\underline{\delta}_{X_j}$, when we reach stationarity for the coupling we have that,

$$\underline{\xi} \sim \underline{\mu}^{c-d} \hspace{1mm}\Longleftrightarrow\hspace{1mm} \underline{\eta}\sim\underline{\mu}^c.$$

First let us show that $\underline{\xi}\sim \underline{\mu}^{c-d}\Rightarrow \underline{\eta}\sim\underline{\mu}^c$. Assume $\underline{\xi}\sim\underline{\mu}^{c-d}$ and consider,

$$\mathbb{P}(\underline{\eta}=z)=\mathbb{P}\left(\underline{\eta}=\underline{z}, N(\underline{\eta})=N(\underline{z})\right)=\underline{\nu}^{N(\underline{z})}(\underline{z}) \cdot \mathbb{P}\left(N(\underline{\eta})=N(\underline{z})\right).$$

Let $\tau$ denote the left shift $(\tau\underline{y})_i=y_{i+1}$, by Lemma <ref> we have that $\underline{\mu}^c(\tau^j\underline{y})=\underline{\mu}^{c+j}(\underline{y})$, and so

\begin{align*}
\mathbb{P}\left(N(\underline{\eta})=N(\underline{z})\right)&=\mathbb{P}\left(N(\underline{\xi})=N(\underline{z})-d\right)\\
&=\sum\limits_{\underline{y}:N(\underline{y})=N(\underline{z})-d} \underline{\mu}^{c-d}(\underline{y})\\
&=\sum\limits_{\underline{u}:N(\underline{u})=N(\underline{z})} \underline{\mu}^{c-d}(\tau^d\underline{u})\\
&=\sum\limits_{\underline{u}:N(\underline{u})=N(\underline{z})} \underline{\mu}^{c}(\underline{u})\\
\end{align*}

Hence, $\mathbb{P}(\underline{\eta}=z)=\underline{\nu}^{N(\underline{z})}(\underline{z}) \cdot \underline{\mu}^c\left(\{N(\underline{\eta})=N(\underline{z})\}\right)= \underline{\mu}^c(\underline{z})$.
By a similar argument we can show that $\underline{\eta}\sim\underline{\mu}^c \Rightarrow \underline{\xi} \sim \underline{\mu}^{c-d}$.

§.§ Stationary Distribution for the Label Process

Let us now look at stationarity for the label process for the second class particles, $\underline{x}(t)$. By construction, $\underline{x}(t)$ gives the positions of the particles in a half infinite ASEP with exactly $d$ particles, $\underline{\gamma}(t)$,  at time $t$. The evolution of this half infinite ASEP depends on the state of the ASEP on $\mathbb{Z}$, $\underline{\xi}(t)$, as previously described.
The stationary distribution for the label process $\{\underline{x}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ is as given below, where we use the notation, $Z^d_{+}\defeq\{\underline{x} \in \mathbb{Z}^d: 0\leq x_1<x_2<...<x_d\}$.
The process $\underline{x}(t)$ has reversible stationary distribution $\underline{\pi}$ where,

$$\underline{\pi}(\underline{x})=\prod\limits_{i=1}^d\left(1-q^i\right)\cdot q^{\sum\limits_{i=1}^d x_i-\frac{d(d-1)}{2}} \hspace{10mm} \text{for any } \underline{x} \in Z_+^d.$$

(We adapt the proofs of Balázs and Seppäläinen [7], Lemma 4.1 and also [8], Proposition 3.1).

First we show that, $\sum\limits_{\underline{x}\in Z_+^d}\underline{\pi}(\underline{x})=1$.

\begin{align*}
\sum\limits_{\underline{x}\in Z_+^d}\underline{\pi}(\underline{x})&=\prod\limits_{i=1}^d(1-q^i)q^{-\frac{d(d-1)}{2}}\cdot\sum\limits_{\underline{x}\in Z_+^d}q^{\sum\limits_{i=1}^d x_i} \\
&= \prod\limits_{i=1}^d(1-q^i)q^{-\frac{d(d-1)}{2}}\cdot\sum\limits_{x_1=0}^\infty\sum\limits_{x_2=x_1+1}^\infty \cdots \sum\limits_{x_d=x_{d-1}+1}^\infty q^{\sum\limits_{i=1}^d x_i} \\
&= \prod\limits_{i=1}^d(1-q^i)q^{-\frac{d(d-1)}{2}}\cdot\sum\limits_{x_1=0}^\infty \cdots \sum\limits_{x_{d-1}=x_{d-2}+1}^\infty\sum\limits_{n=1}^\infty q^{\sum\limits_{i=1}^{d-2} x_i+2x_{d-1}+n}\\
&\vdots \\
&= \prod\limits_{i=1}^d(1-q^i)q^{-\frac{d(d-1)}{2}}\cdot\sum\limits_{x_1=0}^\infty q^{dx_1}\sum\limits_{n_{d-1}=1}^\infty q^{(d-1)n_{d-1}}\cdots \sum\limits_{n_1=1}^\infty q^{n_1}\\
&=\prod\limits_{i=1}^d(1-q^i)q^{-\frac{d(d-1)}{2}}\cdot \frac{1}{1-q^d}\prod\limits_{i=1}^{d-1}\frac{q^i}{1-q^i}\\
&=q^{-\frac{d(d-1)}{2}}\cdot q^{\sum\limits_{i=1}^{d-1}i}=1.
\end{align*}

Let's consider the label process without the restrictions on the evolution given by the underlying asymmetric exclusion process on $\mathbb{Z}$, $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$. That is the process $\underline{y}(t)$, which gives the positions of the particles in an ASEP on $\mathbb{Z}_{\geq 0}$ with exactly $d$ particles and closed left boundary, $\{\underline{\gamma}(t)\}_{t \in \mathbb{R}_{t\geq 0}}$. The jump rates for this process are (for $i\geq 0$),

$$\hat{p}(z_i,z_{i+1})=q\mathbb{I}\{z_i \neq 0\}\mathbb{I}\{z_{i+1}\neq 1\} \hspace{5mm}\text{ and }\hspace{5mm} \hat{q}(z_i,z_{i+1})=\mathbb{I}\{z_{i+1} \neq 0\}\mathbb{I}\{z_{i}\neq 1\}.$$

For $j\in\{1,..,d\}$, let $\underline{e}_j$ be the $d$ long vector such that $(\underline{e}_j)_k=\mathbb{I}\{k=j\}$. When the $j^\text{th}$ particle jumps left  then its position is decreased by 1, $\underline{y}(t)\mapsto \underline{y}(t)-\underline{e}_j$. Similarly, if the $j^\text{th}$ particle jumps right then its position increases by 1, $\underline{y}(t)\mapsto \underline{y}(t)+\underline{e}_j$.
We know show that,

$$\underline{\pi}(\underline{y})=\prod\limits_{i=1}^d\left(1-q^i\right)\cdot q^{\sum\limits_{i=1}^d y_i-\frac{d(d-1)}{2}} \hspace{10mm} \text{for any } \underline{y} \in Z_+^d,$$

is stationary for $\{\underline{y}(t)\}_{t\in\mathbb{R}_{t\geq 0}}$. To do this we check that detailed balance holds; that is for any $\underline{y}\in Z_+^d$ and $j\in\{1,...,d\}$, we check that

$$\underline{\pi}(\underline{y})q\mathbb{I}\{z_i\neq 0\}\mathbb{I}\{z_{i+1}\neq 1\}=\underline{\pi}(\underline{y}+\underline{e}_j)\mathbb{I}\{z^{(i,i+1)}_{i+1}\neq 0\}\mathbb{I}\{z^{(i,i+1)}_{i}\neq 1\},$$

where $\underline{z}\in\hat{\Omega}$ is the current state of $\{\underline{\gamma}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ with the $j^\text{th}$ particle at site $i\geq 0$. This trivially holds when a jump is not possible over the edge $(i,i+1)$. Now consider the case when $z_i=1$ and $z_{i+1}=0$, then

\begin{align*}
\underline{\pi}(\underline{y})q&=\prod\limits_{i=1}^d\left(1-q^i\right)\cdot q^{\sum\limits_{i=1}^d y_i-\frac{d(d-1)}{2}}\cdot q\\
&= \prod\limits_{i=1}^d\left(1-q^i\right)\cdot q^{\sum\limits_{i=1}^d y_i+1-\frac{d(d-1)}{2}}\\
&=\prod\limits_{i=1}^d\left(1-q^i\right)\cdot q^{\sum\limits_{i=1}^d (\underline{y}+\underline{e}_j)_i-\frac{d(d-1)}{2}}\\
&= \underline{\pi}(\underline{y}+\underline{e}_j).
\end{align*}

Now we re-introduce the restriction coming from the process $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$. Consider an approximating process $\underline{x}^m(t)$ for some $m\in\mathbb{Z}_{\geq0}$ which has the same initial value, $\underline{x}^m(0)=\underline{x}(0)$. The process $\{\underline{x}^m(t)\}_{t\in\mathbb{R}_{\geq 0}}$ evolves so that the underlying process $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ only restricts its motion between states in the range $\{\underline{x}\in Z_+^d:x_d<m\}$. We couple the processes together so that $\underline{x}(t)=\underline{x}^m(t)$ until the first time that one of them exits the range $\{\underline{x}\in Z_+^d:x_d<m\}$.
Suppose we fix $m$ for now. Let $0=t_0<t_1<t_2<...$ be a partition of time such that $t_j\nearrow \infty$ and the state of the process $\underline{\xi}(t)$ on the sites $(-\infty,j_m(t_i)]$ (where $j_m(t_i)$ is the site of the $m^\text{th}$ particle at time $t_i$) is constant on each time interval $t\in[t_i,t_{i+1})$. Then on each time interval $[t_i,t_{i+1})$ the process $\underline{x}^m(t)$ is a continuous time Markov chain with jump rates, (for $k\in\{1,...,d\}$) when $\underline{x}+\underline{e}_k\in Z^+_d$,

q\mathbb{I}\{\gamma_{x_k}(t)=1\}\mathbb{I}\{\gamma_{x_k+1}=0\}\left(\sum\limits_{j\in\mathbb{Z}}\mathbb{I}\left\{\sum\limits_{\ell=-\infty}^{j-1}\xi_\ell(t)=k\right\}\mathbb{I}\{\xi_j(t)=1\}\mathbb{I}\{\xi_{j+1}=1\}\right) &\text{ if } x_k\leq m\\
q\mathbb{I}\{\gamma_{x_k}(t)=1\}\mathbb{I}\{\gamma_{x_k+1}=0\} &\text{ if } x_k> m,
\end{cases}$$

with this rate being $0$ if $\underline{x}+\underline{e}_k\notin Z^+_d$; and when $\underline{x}-\underline{e}_k\in Z^+_d$,

\mathbb{I}\{\gamma_{x_k}(t)=1\}\mathbb{I}\{\gamma_{x_k-1}=0\}\left(\sum\limits_{j\in\mathbb{Z}}\mathbb{I}\left\{\sum\limits_{\ell=-\infty}^{j-1}\xi_\ell(t)=k-1\right\}\mathbb{I}\{\xi_j(t)=1\}\mathbb{I}\{\xi_{j+1}=1\}\right) &\text{ if } x_k\leq m+1\\
\mathbb{I}\{\gamma_{x_k}(t)=1\}\mathbb{I}\{\gamma_{x_k-1}=0\} &\text{ if } x_k> m+1,
\end{cases}$$

with this rate being $0$ when $\underline{x}-\underline{e}_k\notin Z^+_d$.
Since in the time intervals $[t_i,t_{i+1})$ the background giving the restriction of $\{\underline{y}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ to $\{\underline{x}^m(t)\}_{t\in\mathbb{R}_{\geq 0}}$ is fixed we are in the setting of Proposition 5.10 of Liggett [II.,[17]] and so on each interval $t\in[t_i,t_{i+1})$ the measure $\underline{\pi}$ is also reversible stationary for $\underline{x}^m(t)$. Thus $\underline{\pi}$ is reversible stationary for $\{\underline{x}^m(t)\}_{t\in\mathbb{R}_{\geq 0}}$.
The coupling between $\{\underline{x}^m(t)\}_{t\in\mathbb{R}_{\geq 0}}$ and $\{\underline{x}(t)\}_{t\in\mathbb{R}_{\geq 0}}$ gives us that $\underline{x}^m(t)\xrightarrow[m\rightarrow\infty]{}\underline{x}(t)$ a.s. , thus the measure $\underline{\pi}$ is also reversible stationary for the process $\{\underline{x}(t)\}_{t\in\mathbb{R}_{\geq 0}}$.

From the proof of the above theorem we see that the distribution of the labels of the second class particles is independent of the distribution of the process $\{\underline{\xi}(t)\}_{t\in\mathbb{R}_{\geq 0}}$.

Using the coupling we give an alternative proof of Theorem <ref>.

We prove this by considering the coupling for the case of a single second class particle, $\underline{\eta}=\underline{\xi}-\underline{\delta}_X$. Using Lemma <ref> we have that,

\begin{equation}\label{eq: Np recurrence}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{\eta})=k\})=\underline{\mu}^{c-1}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{\xi})=k\})\mathbb{P}(x\geq k)+\underline{\mu}^{c-1}(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{\xi})=k+1\})\mathbb{P}(x\leq k).
\end{equation}

Since, either the second class particle is to the right of the site $m$ and so its label is at least $k$ or it is to the left of the site $m+1$ and so its label is at most $k$.
We want the measures above to all have the same parameter so we look to change $c-1$ to $c$; by Lemma <ref> (remembering that $\omega^\text{min}=0$ and $\omega^\text{max}=1$) we have that,

$$ \underline{\mu}^{c-1}((\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})=\frac{q^{-k}}{1+q^{c-1-m}}\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\}).$$

Hence (<ref>) becomes,

\begin{multline*}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})= \frac{q^{-k}}{1+q^{c-1-m}}\underline{\mu}^c((\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})\cdot \mathbb{P}(x\geq k) \\ +\frac{q^{-(k+1)}}{1+q^{c-1-m}}\underline{\mu}^c((\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k+1\})\cdot \mathbb{P}(x\leq k)
\end{multline*}

By Theorem <ref> for $d=1$,

$$\mathbb{P}(x\geq k)=\sum\limits_{i=k}^\infty (1-q)q^i=q^k \hspace{5mm} \text{and} \hspace{5mm} \mathbb{P}(x \leq k)=\sum\limits_{i=0}^k (1-q)q^i=1-q^{k+1}$$

thus we have that,

$$\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})=\frac{\underline{\mu}^c((\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})}{1+q^{c-1-m}}+\frac{q^{-(k+1)}-1}{1+q^{c-1-m}}\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k+1\}). $$

Rearranging gives the following recurrence,

We now solve this recurrence,

\begin{align*}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})&=\frac{q^{c-1-m}}{q^{-k}-1}\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k-1\})
\\
\\
\vdots \\
&= \frac{q^{k(c-1-m)}(-1)^k}{\prod\limits_{j=0}^{k-1}(1-q^{-(k-j)})}\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=0\}).
\end{align*}

We can calculate $\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=0\})$ explicitly to be $\prod\limits_{j\leq m}\frac{1}{1+q^{c-j}}$ and so we have that,

\begin{align*}
\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m+\frac{1}{2}}(\underline{z})=k\})&=   \frac{q^{k(c-1-m)}(-1)^k}{\prod\limits_{j=0}^{k-1}(1-q^{-(k-j)})}\prod\limits_{j\leq m}\frac{1}{1+q^{c-j}}\\
&=\frac{q^{k(c-1-m)}(-1)^k}{(q^{-k};q)_k} \prod\limits_{i=0}^\infty\frac{1}{1+q^{i+c-m}}\\
\end{align*}

Finally we use the Pochhammer relation of Lemma <ref> giving,

§.§ Second Class Particles Positions Distribution

Now we look to describe the distribution of the positions of the second class particles within the system. We will consider two questions:

* For a given site what is the probability (under the blocking measure $\underline{\mu}^c$) that there is a second class particle there?

* For a given $d$ distinct sites what is the probability (under the blocking measure $\underline{\mu}^c$) that the $d$ second class particles lie on these exact sites?

The first question can be easily computed by looking at expectations under the blocking measures for $\underline{\eta}$ and $\underline{\xi}$.

By the coupling we have that, $\eta_m=\xi_m-\sum\limits_{i=1}^d(\underline{\delta}_{X_i})_m$. We see that the event $\{\xi_m>\eta_m\}$ is exactly the event that $\{\sum\limits_{i=1}^d(\underline{\delta}_{X_i})_m>0\}$ or equivalently that $\{\sum\limits_{i=1}^d(\underline{\delta}_{X_i})_m=1\}$ since we are in the simple exclusion setting. Now if we take expectations over the coupling equation we have that,

$$\mathbb{E}[\eta_m]=\mathbb{E}[\xi_m]-\mathbb{E}\left[\sum\limits_{i=1}^d(\underline{\delta}_{X_i})_m\right] \hspace{3mm} \Longleftrightarrow \hspace{3mm} \mathbb{P}(\eta_m=1)=\mathbb{P}(\xi_m=1)-\mathbb{P}\left(\sum\limits_{i=1}^d(\underline{\delta}_{X_i})_m=1\right).$$

Recall Lemma <ref> and so by the assumption that $\underline{\xi}\sim\underline{\mu}^c$, it must be that $\underline{\eta}\sim\underline{\mu}^{c+d}$. Hence,

$$\mathbb{P}(\xi_m>\eta_m)=\mathbb{P}(\xi_m=1)-\mathbb{P}(\eta_m=1)= \frac{q^{c-m}}{1+q^{c-m}}-\frac{q^{c+d-m}}{1+q^{c+d-m}}=\frac{(1-q^d)q^{c-m}}{(1+q^{c-m})(1+q^{c+d-m})}$$

We see that this is equal to $\mathbb{P}(Y=-m)$ for some $Y$ which is distributed according to the discrete logistic distribution with parameters $(q,-c)$ (as defined by Chakraborty and Chakravarty in 2016 [9]).

To answer the second question we will make use of the label process for the second class particles. We then use the product structure of the blocking measure to split the state space up into the piece to the left of where we want the first second class particle and the pieces between the sites for the other second class particles. The following Theorem gives the stationary distribution for the positions of all the $d$ second class particles.

Of course when $d=1$ the two questions we considered are the same and so Theorems <ref> and <ref> are equal:

When $d>1$ we see that this distribution is product in the $m_j$'s other than the exclusion phenomenon represented by the strict ordering of the $m_j$'s.

In order to prove Theorem <ref> we first condition on the labels of the second class particles. By the independence of the label process and $\underline{\xi}$ we have that,

$$\mathbb{P}(\underline{X}=(m_1,..,m_d))=\sum\limits_{\underline{k}\in Z_+^d}\underline{\pi}(\underline{k})\cdot \underline{\mu}^c\left(\left\{\underline{\xi}: \xi_{m_i}=1, \sum\limits_{j=-\infty}^{m_i-1}\xi_j=k_i \text{ for } i \in \{1,..,d\}\right\}\right).$$

Let us focus on the probability under $\underline{\mu}^c$ in the above.

For $m_1<m_2<...<m_d \in \mathbb{Z}$ and $k_1<k_2<...<k_d \in \mathbb{Z}_{\geq0}$,

$$\underline{\mu}^c\left(\left\{\underline{\xi}: \xi_{m_i}=1, \sum\limits_{j=-\infty}^{m_i-1}\xi_j=k_i \text{ for } i \in \{1,..,d\}\right\}\right) = \frac{q^{(k_1+1)(c-m_1)+\frac{k_1(k_1+1)}{2}}}{(q;q)_{k_1}(-q^{c-m_d};q)_\infty}\prod\limits_{j=2}^d\frac{q^{(\hat{k}_j+1)(c-m_j)+\frac{\hat{k}_j(\hat{k}_j+1)}{2}}(q;q)_{\hat{m}_j}}{(q;q)_{\hat{k}_j}(q;q)_{\hat{m}_j-\hat{k}_j}}$$

where $\hat{k}_j:=k_j-k_{j-1}-1$ and $\hat{m}_j:=m_j-m_{j-1}-1$ for $j \in \{2,...,d\}$.

Take some $m_1<m_2<...<m_d \in \mathbb{Z}$ and $k_1<k_2<...<k_d \in \mathbb{Z}_{\geq0}$ and define for $j \in \{2,...,d\}$ $\hat{k}_j:=k_j-k_{j-1}-1$, then

\begin{align*}
\underline{\mu}^c\Bigg(\Bigg\{\underline{\xi}:& \xi_{m_i}=1, \sum\limits_{j=-\infty}^{m_i-1}\xi_j=k_i \text{ for } i \in \{1,..,d\}\Bigg\}\Bigg)  \\
&=\underline{\mu}^c\left(\left\{ \underline{z}: z_{m_i}=1,\sum\limits_{i=-\infty}^{m_1-1}z_i=k_1, \sum\limits_{i=m_{j-1}+1}^{m_j-1}z_i=\hat{k}_j \text{ for } j \in\{2,..,d\} \right\}\right).
\end{align*}

So the event we are interested in (once we have fixed a value $\underline{k}$ for the label process) consists of:

* Particles in $\underline{\xi}$ at sites $m_1,...,m_d$.

* $k_1$ particles to the left of site $m_1$.

* For each $j\in\{2,..d\}$, $\hat{k}_j$ particles on the sites in the range $[m_{j-1}+1,m_{j+1}-1]$.

Moreover, by the product structure of $\underline{\mu}^c$ these are independent and so the above is equal to,

\begin{equation}\label{eq: mu^(...) proof}
\prod\limits_{i=1}^d\mu^c_{m_i}(1)\cdot\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_1-1+\frac{1}{2}}(\underline{z})=k_1\})\cdot\prod\limits_{j=2}^d\underline{\mu}^c(\{N^{\overset{\leftarrow}{p}}_{m_j-1+\frac{1}{2}}(\underline{z})-N^{\overset{\leftarrow}{p}}_{m_{j-1}+\frac{1}{2}}(\underline{z})=\hat{k}_j\}).
\end{equation}

By Theorem <ref> and Lemma <ref> we have,

$$   \text{Equation } (\ref{eq: mu^(...) proof})= \prod\limits_{i=1}^d \frac{q^{c-m_i}}{1+q^{c-m_i}}\cdot \frac{q^{k_1(c-m_1)+\frac{k_1(k_1+1)}{2}}}{(q;q)_{k_1}(-q^{c-m_1+1};q)_\infty}\prod\limits_{j=2}^d\frac{q^{\hat{k}_j(c-m_j)+\frac{\hat{k}_j(\hat{k}_j+1)}{2}}(q;q)_{\hat{m}_j}}{(-q^{c-m_j+1};q)_{\hat{m}_j}(q;q)_{\hat{k}_j}(q;q)_{\hat{m}_j-\hat{k}_j}}.$$

Let us consider,

\begin{align*}
(1+q^{c-m_1})&(-q^{c+m_1+1};q)_\infty\prod\limits_{j=2}^d(1+q^{c-m_j})(-q^{c-m_j+1};q)_{\hat{m}_j} \\
&= (1+q^{c-m_1})\prod\limits_{i=0}^\infty(1+q^{c-m_1+1+i})\cdot \prod\limits_{j=2}^d\left\{(1+q^{c-m_j})\prod\limits_{i=0}^{\hat{m}_j-1}(1+q^{c-m_j+1+i})\right\}\\
&= \prod\limits_{i=0}^\infty(1+q^{c-m_1+i})\cdot\prod\limits_{j=2}^d\left\{\prod\limits_{i=0}^{\hat{m}_j}(1+q^{c-m_j+i})\right\}=\prod\limits_{i=-m_1}^\infty(1+q^{c+i})\cdot\prod\limits_{j=2}^d\left\{\prod\limits_{i=-m_j}^{-m_{j-1}-1}(1+q^{c+i})\right\}\\&=\prod\limits_{i=-m_d}^\infty(1+q^{c+i})=\prod\limits_{i=0}^\infty(1+q^{c-m_d+i})
\end{align*}

$$\underline{\mu}^c\left(\left\{\underline{\xi}: \xi_{m_i}=1, \sum\limits_{j=-\infty}^{m_i-1}\xi_j=k_i \text{ for } i \in \{1,..,d\}\right\}\right) = \frac{q^{(k_1+1)(c-m_1)+\frac{k_1(k_1+1)}{2}}}{(q;q)_{k_1}(-q^{c-m_d};q)_\infty}\prod\limits_{j=2}^d\frac{q^{(\hat{k}_j+1)(c-m_j)+\frac{\hat{k}_j(\hat{k}_j+1)}{2}}(q;q)_{\hat{m}_j}}{(q;q)_{\hat{k}_j}(q;q)_{\hat{m}_j-\hat{k}_j}}.$$

Using Lemma <ref>, Euler's Identity and the $q$-Binomial Theorem we prove Theorem <ref>, and find the stationary distribution of the positions of the second-class particles.

\begin{align*}
&\mathbb{P}(\underline{X}=\underline{m})=\sum\limits_{\underline{k}\in Z_+^d}\underline{\pi}(\underline{k})\cdot \underline{\mu}^c\left(\left\{\underline{\xi}: \xi_{m_i}=1, \sum\limits_{j=-\infty}^{m_i-1}\xi_j=k_i \text{ for } i \in \{1,..,d\}\right\}\right)\\
&=\sum\limits_{k_1=0}^\infty\sum\limits_{k_2=k_1+1}^{k_1+1+\hat{m}_2}\cdots \sum\limits_{k_d=k_{d-1}+1}^{k_{d-1}+1+\hat{m}_d}\prod\limits_{i=1}^d(1-q^i)q^{\sum\limits_{i=1}^d k_i-\frac{d(d-1)}{2}}\underline{\mu}^c\left(\left\{\underline{\xi}: \xi_{m_i}=1, \sum\limits_{j=-\infty}^{m_i-1}\xi_j=k_i \text{ for } i \in \{1,..,d\}\right\}\right)\\
&=\sum\limits_{k_1=0}^\infty\sum\limits_{\hat{k}_2=0}^{\hat{m}_2}\cdots \sum\limits_{\hat{k}_d=0}^{\hat{m}_d}\prod\limits_{i=1}^d(1-q^i)q^{dk_1+\sum\limits_{i=2}^d (d+1-i)\hat{k}_i} \frac{q^{(k_1+1)(c-m_1)+\frac{k_1(k_1+1)}{2}}}{(q;q)_{k_1}(-q^{c-m_d};q)_\infty}\prod\limits_{j=2}^d\frac{q^{(\hat{k}_j+1)(c-m_j)+\frac{\hat{k}_j(\hat{k}_j+1)}{2}}(q;q)_{\hat{m}_j}}{(q;q)_{\hat{k}_j}(q;q)_{\hat{m}_j-\hat{k}_j}}\\
&=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}}{(-q^{c-m_d};q)_\infty}\sum\limits_{k_1=0}^\infty\sum\limits_{\hat{k}_2=0}^{\hat{m}_2}\cdots \sum\limits_{\hat{k}_d=0}^{\hat{m}_d}\frac{q^{k_1(c+d-m_1)+\frac{k_1(k_1+1)}{2}}}{(q;q)_{k_1}}\prod\limits_{j=2}^d\frac{q^{\hat{k}_j(c+d+1-j-m_j)+\frac{\hat{k}_j(\hat{k}_j+1)}{2}}(q;q)_{\hat{m}_j}}{(q;q)_{\hat{k}_j}(q;q)_{\hat{m}_j-\hat{k}_j}}
\end{align*}

Euler's identity gives us that,

$$\sum\limits_{k_1=0}^\infty \frac{q^{k_1(c+d-m_1)+\frac{k_1(k_1+1)}{2}}}{(q;q)_{k_1}}=(-q^{c+d+1-m_1};q)_\infty.$$

For each $j \in \{2,...,d\}$ the $q$-Binomial Theorem gives,

$$\mathbb{P}(\underline{X}=\underline{m})=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}(-q^{c+d+1-m_1};q)_\infty\prod\limits_{j=2}^d(-q^{c+d+2-j-m_j};q)_{\hat{m}_j}}{(-q^{c-m_d};q)_\infty}.$$

By iteratively applying Lemma <ref> we have,

\begin{align*}
\mathbb{P}(\underline{X}=\underline{m})&=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}(-q^{c+d+1-m_1};q)_\infty\prod\limits_{j=2}^d(-q^{c+d+2-j-m_j};q)_{\hat{m}_j}}{(-q^{c-m_d};q)_\infty}\\
&=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}(-q^{c+d+1-m_1};q)_\infty\prod\limits_{j=2}^{d-1}(-q^{c+d+2-j-m_j};q)_{\hat{m}_j}}{(1+q^{c-m_d})(1+q^{c+1-m_d})(-q^{c+1-m_{d-1}};q)_\infty}\\
&=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}(-q^{c+d+1-m_1};q)_\infty\prod\limits_{j=2}^{d-2}(-q^{c+d+2-j-m_j};q)_{\hat{m}_j}}{(1+q^{c-m_d})(1+q^{c+1-m_d})(1+q^{c+1-m_{d-1}})(1+q^{c+2-m_{d-1}})(-q^{c+2-m_{d-2}};q)_\infty}\\
&=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}(-q^{c+d+1-m_1};q)_\infty}{\prod\limits_{j=2}^d(1+q^{c+d-j-m_j})(1+q^{c+d+1-j-m_j})(q^{c+d-1-m_1};q)_\infty}.
\end{align*}

Lastly we see that,

and thus,

$$\mathbb{P}(\underline{X}=\underline{m})=\frac{\prod\limits_{i=1}^d(1-q^i)\cdot q^{dc-\sum\limits_{j=1}^dm_j}}{\prod\limits_{j=1}^d(1+q^{c+d-j-m_j})(1+q^{c+d+1-j-m_j})}.$$

Thus, we have found the distribution under the blocking measure for the positions of any given $d$ second class particles in ASEP.

§ FUTURE DIRECTIONS AND OPEN QUESTIONS

As we saw in Section <ref>, natural questions for the particle system can lead to probabilistic proofs of well-known combinatorial identities. The authors are exploring further directions.
In the paper of Balázs, Fretwell and Jay [6], it is shown that states of blocking particle systems with $0$-$1$-…-$k$ (for $k\geq 2$) particles per site correspond to generalised Frobenius partitions (these are a generalisation of the notion of integer partitions). With this in mind, perhaps by considering the number of particles to the left of some site or within a finite range or symmetry between holes and particles for these more general blocking systems we would see GFP versions of the Euler, $q$-Binomial and Durfee rectangles identities.
In the paper Balázs and Bowen [5], the identity  was derived by comparing the blocking measures for ASEP and zero-range in the infinite volume. A finite or half-infinite volume version will provide further interesting connections to the theory of integer partitions.
Both directions are subjects of upcoming papers.

§ SIMPLE $Q$-POCHHAMMER MANIPULATIONS

The $q$-Binomial analogue of Pascal's identity (see for example equation 7.1 in Andrews and Erikkson [3])
For any $m\in\mathbb Z_{>0}$ and $k\in\{0,\dots,m\}$,

\end{bmatrix}_q=q^k\begin{bmatrix}
\end{bmatrix}_q+\begin{bmatrix}
\end{bmatrix}_q.$$

Let us consider,

\begin{align*}
\end{bmatrix}_q+\begin{bmatrix}
\end{bmatrix}_q&= q^k\frac{(q;q)_{m-1}}{(q;q)_k(q;q)_{m-1-k}}+\frac{(q;q)_{m-1}}{(q;q)_{k-1}(q;q)_{m-k}} \\
\end{bmatrix}_q.
\end{align*}

For any $k\in\mathbb Z_{\ge0}$,

$$(q^{-k};q)_k=(q^{-1};q^{-1})_k=\frac{(q;q)_k}{(-1)^kq^{\frac{k(k+1)}{2}}}. $$

The first equality holds since,

The second equality holds since,

$$(q^{-1};q^{-1})_k=\prod\limits_{i=1}^k(1-q^{-i})=(1-q^{-1})(1-q^{-2})\cdots(1-q^{-k})=q^{-\sum\limits_{i=1}^k i}\prod\limits_{j=1}^k(q^i-1)= \frac{(q;q)_k}{(-1)^kq^{\frac{k(k+1)}{2}}}.$$

For any $d\in\mathbb Z_{>0}$, and any $m_1<m_2<...<m_d \in \mathbb Z$ and for each $j \in \{2,..,d\}$,

where $\hat{m}_j\defeq m_j-m_{j-1}-1$

\begin{align*}
\frac{(-q^{c+d+2-j-m_j};q)_{\hat{m}_j}}{(-q^{c+d-j-m_j};q)_\infty}&=\frac{\prod\limits_{i=1}^{\hat{m}_j}(1+q^{c+d+1-j-m_j+i})}{\prod\limits_{i=0}^\infty(1+q^{c+d-j-m_j+i})}\\
\end{align*}

§ COMBINATORIAL ARGUMENTS

This section is included for completeness. Notice that no other parts of the paper rely on these arguments.

We notice that, any state $\underline{z}$ such that $N_{m+\frac{1}{2}}^{\overset{\leftarrow}{p}}(\underline{z})=k$ is some $s$ number of left jumps away from the state $\tilde{\underline{\omega}}$, which is defined on the sites $(-\infty,m]$ as,

1 &\text{ if } i \in \{m-k+1,...,m\}\\
0 &\text{ o/w. }
\end{cases}$$

We see that a left particle jump changes the probability of a state, under $\underline{\mu}^c$, by a factor of $q$,

$$\underline{\mu}^c(\underline{z}^{(j,j-1)})=\prod\limits_{i\in\mathbb{Z}}\frac{q^{-(i-c)z_i^{(j,j-1)}}}{1+q^{-(i-c)}}=q^{j-c}\cdot q^{-(j-1-c)}\cdot\underline{\mu}^c(\underline{z})=q\cdot\underline{\mu}^c(\underline{z}).$$

Thus we have that,

$$\underline{\mu}^c(\{N_{m+\frac{1}{2}}^{\overset{\leftarrow}{p}}(\underline{z})=k\})=\underline{\mu}^c|_{(-\infty,m]}(\tilde{\underline{\omega}})\sum\limits_{s=0}^\infty q^s \hat{g}(k,s),$$

where $\hat{g}(k,s)$ counts the number of states on $(-\infty,m]$ that are $s$ left jumps away from $\underline{
\tilde{\omega}}$. A state that is $s$ left jumps away from $\underline{\tilde{\omega}}$ can be viewed as a partition of $s$ where the parts are the number of left jumps each particle is away from its position in $\underline{\tilde{\omega}}$. So the partitions that these states correspond to are partitions of $s$ with up to $k$ parts. Then $\sum\limits_{s=0}^\infty q^s \hat{g}(k,s)$ is the generating function for such partitions, by conjugation of partitions we know that $\sum\limits_{s=0}^\infty q^s \hat{g}(k,s)=\frac{1}{(q;q)_k}$ (for example see Section 6.2 of Andrews and Eriksson [3]). Now we find,

$$\underline{\mu}^c|_{(-\infty,m]}(\tilde{\underline{\omega}}) = \prod\limits_{i=-\infty}^m \mu^c_i(\tilde{\omega}_i)=\frac{\prod\limits_{i=m-k+1}^mq^{c-i}}{\prod\limits_{i=-\infty}^m(1+q^{c-i})}=\frac{q^{k(c-m)+\frac{k(k-1)}{2}}}{\prod\limits_{i=0}^\infty(1+q^{c-m+i}}=\frac{q^{k(c-m)+\frac{k(k-1)}{2}}}{(-q^{c-m};q)_{\infty}}.$$

Putting this together we have that,

For some $k \in \{0,...,\hat{m}_2\}$ define a state $\underline{\omega}$ on $[m_1+1,m_2-1]$ such that,

1 &\text{for } i \in \{m_2-k,...,m_2-1\}\\
0 &\text{for } i \in \{m_1+1,...,m_2-k-1\}.
\end{cases}$$

We see that for any state $\underline{z}$ defined on $[m_1+1,m_2-1]$ there is some minimal number of right particle jumps needed to get to $\underline{\omega}$; that is $\underline{z}$ is some $s$ left jumps away from $\underline{\omega}$.
Thus, since a left jump changes the measure by a factor of $q$,

where $g(k,s)$ counts the number of states on $[m_1+1,m_2-1]$ that are $s$ jumps away from $\underline{\omega}$. The probability under $\underline{\mu}^c|_{[m_1+1,m_2-1]}$ of the state $\underline{\omega}$ is,

\underline{\mu}^c|_{[m_1+1,m_2-1]}(\underline{\omega})=\prod\limits_{i=m_1+1}^{m_2-1}\mu^c_i(\omega_i)=\frac{\prod\limits_{i=m_2-k}^{m_2-1}q^{c-i}}{\prod\limits_{i=m_1+1}^{m_2-1}(1+q^{c-i})}=\frac{q^{k(c+1-m_2)+\sum\limits_{i=1}^{k-1} i}}{\prod\limits_{i=1}^{\hat{m}_2}(1+q^{c-m_2+i})}=\frac{q^{k(c+1-m_2)+\frac{k(k-1)}{2}}}{(-q^{c-m_2+1};q)_{\hat{m}_2}}.

A state that is $s$ left jumps away from $\underline{\omega}$ can be viewed as a partition of $s$ where the parts are the number of left jumps each particle is away from its position in $\underline{\omega}$. So the partitions which these states correspond to are partitions of $s$ with up to $k$ parts of size up to $\hat{m}_2-k$. Hence $\sum\limits_{s=0}^{k(\hat{m}_2-k)}q^sg(k,s)$ is the generating function for these such partitions which can be written in the following way (for example see Section 7.2 of Andrews and Eriksson [3]),

\hat{m}_2 \\
\end{bmatrix}_{q} = \frac{(q;q)_{\hat{m}_2}}{(q;q)_k(q;q)_{\hat{m}_2-k}}

Putting all this together we find that,

[1]
G.E. Andrews.
A simple proof of Jacobi's triple product identity.
Proc. Amer. Math. Soc., 16:333–334, 1965.

[2]
G.E. Andrews.
$q$-Series: Their Development and Application in Analysis,
Number Theory, Combinatorics, Physics, and Computer Algebra.
Amer. Math. Soc., 1986.

[3]
G.E. Andrews and K. Eriksson.
Integer Partitions.
Cambridge University Press, 2004.

[4]
O. Angel.
The stationary measure of a 2-type totally asymmetric exclusion
J. Combin. Theory Ser. A, 113(4):625–635, 2006.

[5]
M. Balázs and R. Bowen.
Product blocking measures and a particle system proof of the jacobi
triple product.
Ann. Inst. H. Poincaré Prob. Stat., 54(1):514–528, 02

[6]
M. Balázs, D. Fretwell, and J. Jay.
Interacting particle systems and jacobi style identities.
Research in the Mathematical Sciences, 9:48, 2022.

[7]
M. Balázs and T. Seppäläinen.
Fluctuation bounds for the asymmetric simple exclusion process.
ALEA, 6:1–24, 2009.

[8]
M. Balázs and T. Seppäläinen.
Order of current variance and diffusivity in the asymmetric simple
exclusion process.
Ann. Math., 171(2):1237–1265, 2010.

[9]
S. Chakraborty and D. Chakravarty.
A new discrete probability distribution with integer support on
Communications in Statistics - Theory and Methods,
45(2):492–505, 2016.

[10]
M. Cimpoeaş.
A note on the number of partitions of $n$ into $k$ parts.
U.P.B. Sci. Bull., Series A, 84:131–138, 2022.

[11]
W. Fan and T. Seppäläinen.
Joint distribution of busemann functions in the exactly solvable
corner growth model.
Probability and Mathematical Physics, 1(1):55 – 100, 2020.

[12]
P.A. Ferrari, C. Kipnis, and E. Saada.
Microscopic structure of travelling waves in the asymmetric simple
exclusion process.
Ann. Prob., 19(1):226–244, 1991.

[13]
P.A. Ferrari and J. Martin.
Stationary distributions of multi-type totally asymmetric exclusion
Ann. Probab., 35(3):807–832, 2007.

[14]
I. M. Gessel.
Some generalized durfee square identities.
Discrete Mathematics, 49(1):41–44, 1984.

[15]
E. Heine.
Untersuchungen über die reihe.
J. reine angew. Math., 1847(34):285–328, 1847.

[16]
T.M. Liggett.
Coupling the simple exclusion process.
Ann. Probab., 4:339–356, 1976.

[17]
T.M. Liggett.
Interacting particle systems, volume 276 of Grundlehren
der Mathematischen Wissenschaften [Fundamental Principles of Mathematical
Springer-Verlag, New York, 1985.

[18]
F. Spitzer.
Interaction of Markov processes.
Advances in Math., 5:246–290 (1970), 1970.

[19]
C. A. Tracy and H. Widom.
On the distribution of a second-class particle in the asymmetric
simple exclusion process.
J. Phys. A, 42(42):425002, 6, 2009.