# Joint $q$–moments and shift invariance for the multi–species $q$–TAZRP on
the infinite line

Jeffrey Kuan

###### Abstract

This paper presents a novel method for computing certain particle locations in
the multi–species $q$–TAZRP (totally asymmetric zero range process). The
method is based on a decomposition of the process into its discrete–time
embedded Markov chain, which is described more generally as a monotone process
on a graded partially ordered set; and an independent family of exponential
random variables. A further ingredient is explicit contour integral formulas
for the transition probabilities of the $q$–TAZRP. The main result of this
method is a shift invariance for the multi–species $q$–TAZRP on the infinite
line.

By a previously known Markov duality result, these particle locations are the
same as joint $q$–moments. One particular special case is that for step
initial conditions, ordered multi–point joint $q$–moments of the $n$–species
$q$–TAZRP match the $n$–point joint $q$–moments of the single–species
$q$–TAZRP. Thus, we conjecture that the Airy2 process describes the joint
multi–point fluctuations of multi–species $q$–TAZRP.

As a probabilistic application of this result, we find explicit contour
integral formulas for the joint $q$–moments of the multi–species $q$–TAZRP in
the diffusive scaling regime.

##  1 Introduction

In the mathematics and physics community, it has been known since at least the
1990s that the KPZ equation [KPZ86] describes a large class of processes. For
just one example, the relationship between the ASEP (totally asymmetric simple
exclusion process) and KPZ was noted as early as [DM97]. In [BDJ99], it was
shown that the asymptotic fluctuations of the longest increasing subsequence
of random permutations are the same as the asymptotic fluctuations of the
largest eigenvalue of the GUE (Gaussian Unitary Ensemble) [TW94]. This
asymptotic distribution describing the fluctuations, the Tracy–Widom $F_{2}$
distribution, is universal in the KPZ universality class. More recently, it
was shown that WASEP (weakly asymmetric simple exclusion process) converges to
the Cole–Hopf solution of the KPZ equation [ACQ11], as well as to energy
solutions of the KPZ equation (see e.g. the survey [GO], which also discusses
the multi–species case).

The multi–point fluctuations in the KPZ universality class are widely believed
to be the Airy2 process, although a rigorous proof has not yet been
discovered. However, more recently, a few papers have been able to take multi-
point asymptotics for partially asymmetric, non–determinantal models in the
KPZ class ([QS] for the ASEP and [Dim] for two–point asymptotics in the
stochastic six vertex model). For totally asymmetric models with a
determinantal structure, such results go back to [Joh00], and many others.

In the context of multi–species (also called colored, or multi–class or
mult–type) models introduced in [Lig76], significantly less is known, or even
conjectured, about the asymptotics fluctuations. In fact, it is not even
entirely clear which statistic to analyze. It has only been in recent years
that non–rigorous heuristics or conjectures have appeared in the literature,
such as [Spo14]. Even more recently, [CdGH+] rigorously proves that the
crossing probability of a two–species model called the AHR model (introduced
in [AHR98]), converges to a $F_{2}$ times a Gaussian. Also see [dGMW] for
another paper on multi–species ASEP. In this present paper, a different
statistic is studied, perhaps indicating that multi–species models have
multiple interesting statistics in the asymptotic limit.

The model studied here is the multi–species $q$–deformed totally asymmetric
zero range process (TAZRP), introduced in [Tak], generalizing the
single–species version in [SW98]. It was recently shown in [Hay] that in the
$q\rightarrow 1$ limit, the single–species $q$–TAZRP converges to the usual
KPZ equation, and additionally, the fluctuations of the $q$–TAZRP are also
governed by Tracy–Widom $F_{2}$ [FV15, Bar15]. In words, the main result of
this paper (Corollary 2.2) states that (at finite–time) the distributions of
the height functions of the multi–species $q$–TAZRP can be matched to the
multi–point distributions of the height functions of the single–species
$q$–TAZRP. Since those asymptotic fluctuations are conjectured to be the Airy2
process, it is reasonable to conjecture that the asymptotic fluctuations of
the multi–species $q$–TAZRP are also the Airy2 process. Actually, an even more
general statement can be proved (Theorem 2.1), which is somewhat easier to
prove than proving Corollary 2.2 directly – Chronologically, the author
discovered the proof of Corollary 2.2 first, using the methods in section 6,
before discovering the more aesthetically pleasing proof of Theorem 2.1. The
theorem states that the height functions are invariant under certain shift
transformations.

One probabilistic application of the main result is to write explicit formulas
for the joint $q$–moments in the diffusive scaling regime (i.e. when time is
scaled as $L$ and space is scaled as $L^{1/2}$). These joint $q$–moments can
be expressed as contour integral formulas. From these formulas, it is possible
to numerically calculate the fluctuations of the multi–species $q$–TAZRP from
its hydrodynamic limit.

Note that similar–looking statements to Corollary 2.2 have occurred in [BGW]
and [Gal]. The key difference here is that the initial conditions in this
paper occur on the infinite line, rather than with boundary conditions. On the
infinite line, asymptotic results are more tractable. As far as the author is
aware, the main result in this paper do not directly follow from the results
in the aforementioned papers. In any case, the method presented here is
significantly different.

The outline of this paper is as follows: section 2 defines the multi–species
$q$–TAZRP and states the main theorem of this paper, as well as a corollary.
Section 3 briefly covers necessary background and notation. Section 4 proves
the main theorem, using the methods described in the abstract. Section 5
provides the asymptotic limit in the diffusive scaling limit. Section 6
provides an alternative proof of the corollary, and section 7 contains some
numerical and computer data demonstrating some examples of the results of the
paper. A brief discussion on the applicability of these methods to
multi–species ASEP occurs at the end of section 4.

Acknowledgments. The author would like to thank Alexei Borodin and Ivan Corwin
for helpful discussions. Additionally, the author would like to thank an
attendee of the University of Bristol Probability Seminar, whose name the
author did not catch because the seminar was held in a hybrid format, for
helpful research suggestions. The numerical simulations were computed with the
help of the Texas A&M High Performance Research Computing. Finally, the author
would like to thank the organizers of the workshop “Asymptotic Algebraic
Combinatorics”, held at the Institute for Pure and Applied Mathematics in Los
Angeles in February 2020, where helpful ideas were sparked.

##  2 Statement of Main Result

###  2.1 Definition of the multi–species $q$–TAZRP

Recall the definition of the $n$–species $q$–TAZRP, also known as $q$–Boson.
The model was introduced in [Tak], with the single–species model going back to
[SW98]. The state space for $n$–species $q$–TAZRP consists of particle
configurations on a one–dimensional lattice. Here, we take that lattice to be
$\mathbb{Z}$. At each lattice site, there may be arbitrarily many particles,
with $n$ different species of particles. The state space is therefore
$\left(\mathbb{Z}_{\geq 0}^{n}\right)^{\mathbb{Z}}$. Each
$\eta\in\left(\mathbb{Z}_{\geq 0}^{n}\right)^{\mathbb{Z}}$ can be written as
$\eta=(\eta_{i}^{x})$, for $x\in\mathbb{Z}$ and $1\leq i\leq n$, where
$\eta_{i}^{x}$ denotes the number of particles of species $i$ located at
lattice site $x$.

With the state space defined, the jump rates can be stated. Informally, for a
particle configuration $\xi=(\xi_{i}^{x})$, the jump rates for an $i$th
species particle at lattice site $x$ to jump one step to the right is

$q^{\xi_{1}^{x}+\ldots+\xi_{i-1}^{x}}\frac{1-q^{\xi_{i}^{x}}}{1-q},$

Only one particle may jump at a time. More specifically, the generator can be
written explicitly. If $\eta=(\eta_{i}^{x})$ and $\xi=(\xi_{i}^{x})$ are
related by

$\eta_{j}^{y}=\xi_{j}^{y}-1,\quad\eta_{j}^{y+1}=\xi_{j}^{y+1}+1$

for some $j,y$, with all $\eta_{i}^{x}=\xi_{i}^{x}$ for all other values of
$i,x$, then write $\eta=\xi(j,y)$. Then the generator is defined by

$L_{\text{mqTAZRP}}(\xi,\xi(j,y))=q^{\xi_{1}^{y}+\ldots+\xi_{j-1}^{y}}\frac{1-q^{\xi_{j}^{y}}}{1-q}.$

If the particle configuration $\xi(j,y)$ does not exist, then formally set

$L_{\text{mqTAZRP}}(\xi,\xi(j,y))=0.$

In this notation, particles of species $i$ have priority over particles of
species $j$ for $i<j$. An example of the jump rates is shown below in Figure
1.

Figure 1: Example of jump rates for the homogeneous $q$–TAZRP. The left image
shows single–species and the right image shows multi–species. Note that
$q^{4}[1]_{q}+q^{2}[2]_{q}+[2]_{q}=[5]_{q}$.

###  2.2 Height Functions and Shift Invariance

Given any $1\leq j\leq n$ and a particle configuration $\eta=(\eta_{x}^{i})$,
set

$N^{(n+1-j)}_{y}(\eta)=\sum_{x=y}^{\infty}\sum_{i=1}^{n+1-j}\eta^{x}_{i}.$

In words, $N_{y}^{(n+1-j)}(\eta)$ counts the number of species
$1,\ldots,n+1-j$ particles that are weakly to the right of $y$. The reason for
using the notation $n+1-j$ in the superscript is related to the space–reversal
in the duality statement of Proposition 3.1 below; These are sometimes called
the height functions.

Given a pair of particle configurations $(X,Y)$, define the intersection
numbers $I_{ij}(X,Y)$ by

$I_{ij}(X,Y)=\left|[x_{i},y_{i}]\cap[x_{j},y_{j}]\right|$

for $1\leq i,j\leq n$. These are similar to the so–called “intersection
matrices” is made in [Gal].

With these definitions in place, the main theorem can now be stated:

######  Theorem 2.1.

Suppose that $\eta(t)$ evolves as the multi–species $q$–TAZRP and begins in
initial conditions with infinitely many $i$th species particles at $y_{i}$ for
some integers $y_{1},\ldots,y_{n}$. Fix lattice sites $x_{1},\ldots,x_{n}$
(not necessarily distinct). Let $\xi(t)$ also be a multi–species $q$–TAZRP
that begins in initial conditions with infinitely many $i$th species particles
at $y^{\prime}_{i}$ for some integers $y^{\prime}_{1},\ldots,y_{n}^{\prime}$.
Fix lattice sites $x^{\prime}_{1},\ldots,x^{\prime}_{n}$ (not necessarily
distinct).

Assume that there exists an integer $j_{0}\in[1,n]$ such that

$(x_{1}^{\prime},\ldots,x_{n}^{\prime})-(x_{1},\ldots,x_{n})=(y_{1}^{\prime},\ldots,y_{n}^{\prime})-(y_{1},\ldots,y_{n})=\epsilon_{j_{0}}$

where $\epsilon_{j_{0}}$ has a $1$ in the $j_{0}$ component and zeroes
elsewhere. Furthermore assume that there is a sequence $\\{(X(l),Y(l)):1\leq
l\leq L\\}$ of pairs of particle configurations such that the intersection
numbers $I_{ij}(X(l),Y(l))$ are invariant with respect to $l$ and that

$X(1)=(x_{1}^{\prime},\ldots,x_{n}^{\prime}),\quad
Y(1)=(y_{1}^{\prime},\ldots,y_{n}^{\prime}),\quad X(L)_{1}=\ldots X(L)_{n}$

and

$X(l+1)-X(l)=Y(l+1)-Y(l)=\epsilon_{j_{l}}$

for some $j_{1},\ldots,j_{K}$.

Then for any non–negative integers $k_{1}\ldots,k_{n}$

$\mathbb{E}_{y}\left[\prod_{i=1}^{n}q^{k_{j}N_{x_{j}}^{(n+1-j)}(\eta(t))}\right]=\mathbb{E}_{y^{\prime}}\left[\prod_{i=1}^{n}q^{k_{j}N_{x^{\prime}_{j}}^{(n+1-j)}(\xi(t))}\right].$

Furthermore, this quantity has an explicit contour integral formula. Set
$N=k_{1}+\ldots+k_{n}$ and define $M_{1},\ldots,M_{N}$ by

${M}_{k_{1}+\ldots+k_{m}+1}={M}_{k_{1}+\ldots+k_{m}+2}=\cdots={M}_{k_{1}+\ldots+k_{m+1}}=[y_{m},x_{m}].$

Then

$\mathbb{E}\left[\prod_{j=1}^{n}q^{k_{j}N_{x_{j}}^{(n+1-j)}(\eta(t))}\right]=\frac{(-1)^{N}q^{N(N-1)/2}}{(2\pi
i)^{N}}\int\frac{dw_{1}}{w_{1}}\cdots\int\frac{dw_{N}}{w_{N}}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}(1-w_{j})^{-{M}_{j}}e^{-w_{j}t}.$

where the $w_{r}$–contour contains $qw_{r+1},\ldots,qw_{n}$ and $1$, but not
$0$.

######  Remark 1.

The theorem can be applied to inductively to any sequence
$x,x^{\prime},x^{\prime\prime},\ldots,$ and
$y,y^{\prime},y^{\prime\prime},\ldots$ as long as the consecutive intersection
numbers remain unchanged.

######  Remark 2.

It is worth noting that if infinitely many $i$th species and infinitely many
$j$th species particles occupy a lattice site, where $i<j$, then the $j$th
species particles have jump rate $q^{\infty}=0$, and therefore never jump.

By applying the theorem inductively, the initial conditions can be shifted so
that all particles start at the same lattice site:

######  Corollary 2.2.

Suppose $y_{1}<\ldots<y_{n}$ and $x_{1}\geq x_{2}\geq\ldots\geq x_{n}$. Then

$\mathbb{E}\left[\prod_{j=1}^{n}q^{k_{j}N_{x_{j}}^{(n+1-j)}(\eta(t))}\right]=\mathbb{E}\left[\prod_{j=1}^{n}q^{k_{j}N_{M_{j}}^{(n+1-j)}(\xi(t))}\right],$

where $\xi(t)$ is a single–species $q$–TAZRP with step initial conditions,
i.e. all particles starting at $0$.

######  Remark 3.

The reader may naturally be curious if this method can be applied to other
models, such as the multi–species ASEP. This is discussed in section 4.4.

##  3 Background results and notation

###  3.1 $q$–notation

Fix $0<q<1$. For any $k\geq 0$, let

$[k]_{q}=\frac{1-q^{k}}{1-q}=1+q+q^{2}+\ldots+q^{k-1}$

be the $q$–deformed integer. Let $[k]_{q}^{!}=[1]_{q}\cdots[k]_{q}$ be the
$q$–deformed factorial. The $q$–Pochhammer symbol is

$(\alpha;q)_{k}=(1-\alpha)(1-q\alpha)\cdots(1-q^{k-1}\alpha),\quad 0\leq
k\leq\infty.$

Observe that $(1-q)^{k}[k]_{q}^{!}=(q;q)_{k}$.

For each integer $r\geq 1$ and each finite sequence of non–negative integers
$\mathbf{m}=(m_{1},m_{2},\ldots,m_{r})$ whose sum is $N=m_{1}+\ldots+m_{r}$,
define the $q$–multinomial

$\left[\begin{array}[]{c}N\\\
\mathbf{m}\end{array}\right]_{q}:=\left[\begin{array}[]{c}m_{1}+\ldots+m_{r}\\\
m_{1},m_{2},\ldots,m_{r}\end{array}\right]_{q}=\frac{[N]^{!}_{q}}{[m_{1}]^{!}_{q}\cdots[m_{r}]^{!}_{q}}.$

###  3.2 Duality formula for $q$–moments

Let $\vec{k}=(k_{1},\ldots,k_{n})$ and $\vec{M}=(M_{1},\ldots,M_{n})$. Let
$\mathcal{S}(\vec{k},\vec{M})$ denote the set of all particle configurations
$\xi$ with exactly $k_{j}$ particles of species $j$, with all species $j$
particles located strictly to the left of $M_{n+1-j}$. More precisely,

$\xi\in\mathcal{S}(\vec{k},\vec{M})\Leftrightarrow\sum_{x\in\mathbb{Z}}\xi_{j}^{x}=k_{j}\text{
and }\xi^{z}_{j}=0\text{ for all }1\leq j\leq n,z\geq M_{n+1-j}.$

######  Proposition 3.1.

Suppose that $\eta(t)$ begins in the initial conditions with infinitely many
$i$th class particles at $-M_{i}$ for some integers $0<M_{n}<\ldots<M_{1}$.
Fix non–negative integers $k_{1},\ldots,k_{n}$. Then

$\mathbb{E}\left[\prod_{j=1}^{n}q^{k_{j}N_{0}^{(n+1-j)}(\eta)}\right]=\sum_{\xi\in\mathcal{S}(\vec{k},\vec{M})}\mathrm{Prob}_{0}(\xi\text{
at time }t).$

The initial conditions on the right–hand–side consist of $k_{j}$ particles of
species $j$ located at $0$ for $1\leq j\leq n$.

###### Proof.

This follows from the duality result of Theorem 2.5(b) [Kua17]; see also
Corollary 5.3 of [Kua18] for the same duality result with a different proof.
The only additional observation required is that for infinitely many particles
of species $i$ at a single site, the duality function only takes the values
$1$ or $0=q^{\infty};$ i.e. the duality function becomes an indicator
function. ∎

###  3.3 Transition Probabilities for the single–species $q$–TAZRP

Set

$S(w_{a},w_{b})=-\frac{qw_{b}-w_{a}}{qw_{a}-w_{b}}$

and for $\sigma\in S_{n}$ (the permutation group on $n$ letters),

$A_{\sigma}=\prod_{\begin{subarray}{c}\sigma(i)>\sigma(j)\\\ 1\leq i<j\leq
n\end{subarray}}S(w_{\sigma(j)},w_{\sigma(i)}).$

Let

$B\left(w_{1},\ldots,w_{n}\right)=\prod_{1\leq i<j\leq
n}\frac{w_{i}-w_{j}}{w_{i}-qw_{j}}$

By equation (5.1) of [WW09],

$\sum_{\sigma\in
S_{n}}A_{\sigma}\left(w_{\sigma^{-1}(1)},\ldots,w_{\sigma^{-1}(n)}\right)=[n]_{q}!B\left(w_{1},\ldots,w_{n}\right)$
(1)

######  Proposition 3.2.

Let $\mathbf{N}=(N_{1},\ldots,N_{n})$ and $N=N_{1}+\ldots N_{n}$. Fix a
particle configuration $x$ where $x=(x_{1},\ldots,x_{N})$. Given initial
conditions where all $N$ particles begin at $0$,

$\mathrm{Prob}_{0}(X(t)=x)\text{ at time }t)=\\\
\times\frac{(-1)^{N}}{W(x)}\left(\frac{1}{2\pi i}\right)^{N}\sum_{\omega\in
S(N)}\int_{\mathcal{C}_{R}}\cdots\int_{\mathcal{C}_{R}}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\left[\prod_{k=y_{\omega(j)}}^{x_{j}}\left(\frac{1}{1-w_{\omega(j)}}\right)e^{-w_{j}t}\right]dw_{1}\cdots
dw_{N},$

where the $\mathcal{C}_{R}$ are large contours, centered at the origin, that
enclose the poles $b_{k}$ and $w_{j}=qw_{i}$. Here,

$W(x)=\prod_{z\in\mathbb{Z}}\frac{1}{[k_{z}]_{q}^{!}},\quad
k_{z}=\left|\\{k:x_{k}=z\\}\right|$

The paper [KL14] also gives the formula in terms of small contours, which will
be described later.

######  Remark 4.

The results actually hold for more general initial conditions, but these will
not be needed, at least until section 6.

###  3.4 Markov projection property

The multi–species $q$–TAZRP has a property called the Markov projection
property, which generalizes a property originally known as lumpability [KS60].
This is also called the “color–blind” projection property. This states that if
one views all particles as identical, the system is reduced to a
single–species model. A specific case of this is used in the proof of the
corollary.

##  4 Proof of the Main Theorem

By the duality result, the $N$th $q$–moments can be written in terms of
transition probabilities of the $N$–particle system. For the remainder of the
proof, only the transition probabilities of the finite–particle system will be
studied. Note that applying the duality reverses the direction of the jumps,
so that the roles of $x$ and $y$ are now reversed.

###  4.1 Markov processes and embedded Markov chains on posets

The first step is to decompose the continuous–time multi–species $q$–Boson
into its discrete–time embedded Markov chain and its local times. It will
actually be easier to state this step in terms of a general monotone Markov
process on a partially ordered set. The next subsection contains essentially a
stand–alone result that can be read independently of the rest of the paper. As
far as the author knows, these results are new, as are their applications in
integrable probability.

The following definitions are standard in probability theory:

######  Definition 4.1.

Consider a continuous–time Markov process $X(t)$ on a discrete state space,
with generator $L(x,y)$. The discrete–time embedded Markov chain is defined on
the same state space, with transition probabilities

$P(x,y)=\begin{cases}-\dfrac{L(x,y)}{L(x,x)},&x\neq y\\\ 0,&x=y.\end{cases}$

######  Definition 4.2.

Consider a Markov process or Markov chain $X(t)$ on a state space which is
also a partially ordered set with ordering denoted by $\leq$. It is monotone
if for any times $s<t$, the event $\\{X(t)\geq X(s)\\}$ occurs almost surely.

The next proposition has possibly occurred somewhere in the literature before,
but the author is unaware of any appropriate references.

######  Proposition 4.3.

Let $X(t)$ be a time–homogeneous monotone Markov process on a poset
$\mathcal{S}$ with generator $L(x,y)$, and let $Y(n)$ denote its embedded
Markov chain. Additionally let $\\{\mathcal{E}_{z}:z\in\mathcal{S}\\}$ be
independent exponential random variables with respective parameters
$\\{-L(z,z):z\in\mathcal{S}\\}$.

Let $\Pi_{xy}$ denote the set of all sequences $\\{(x=x_{0}\leq
z_{1}<\cdots<z_{n}=y)\\}$ with the property that every
$\mathbb{P}(Y(1)=z_{n}|Y(0)=z_{n-1})$ is nonzero. In other words, there is a
positive probability that $Y(k)$, starting from $x$, is located at $z_{k}$
(for $1\leq k\leq n$) up until it hits $y$. For any $\pi\in\Pi_{xy}$, let

$\mathcal{E}_{\pi}=\sum_{z\in\pi}\mathcal{E}_{z}.$

Then

$\mathbb{P}_{x}(X(t)\leq y)=\sum_{n\geq
0}\sum_{\begin{subarray}{c}\pi\in\Pi_{xy}\\\
l(\pi)=n\end{subarray}}\mathbb{P}_{x}(Y(1)=z_{1},\ldots,Y(n-1)=z_{n-1},Y(n)=y)\mathbb{P}\left(\mathcal{E}_{\pi}\geq
t\right),$

where $\pi=(x<z_{1}<\ldots<z_{n}=y)$ and $l(\pi)$ is the length of $\pi$.

###### Proof.

Let $\mathcal{S}^{\infty}$ denote the set of all infinite paths in
$\mathcal{S}$, irrespective of the jump rates or the partial ordering on
$\mathcal{S}$. As before, let $\pi=(z_{0},z_{1},z_{2},\ldots)$, where
$z_{n}\in\mathcal{S}$ for all $n\geq 0$. Note that the discrete embedded
Markov chain $Y(t)$ and the Markov process $X(t)$ can be coupled together
naturally. Thus

$\mathbb{P}_{x}(X(t)\leq
y)=\sum_{\pi\in\mathcal{S}^{\infty}}\mathbb{P}(Y(n)=z_{n}\text{ for all }n\geq
0)\cdot\mathbb{P}_{x}(X(t)\leq y|Y(n)=z_{n}\text{ for all }n\geq 0)$

by the usual total probability rule. By monotonicity, the first probability
$\mathbb{P}(Y(n)=z_{n}\text{ for all }n\geq 0)$ is only nonzero for paths that
begin with $\pi\in\Pi_{xy}$ where $l(\pi):=n\geq 0$, so the summation can be
replaced with the summation in the right–hand–side of the statement of the
proposition.

For any state $y$, let $\mathbb{H}(z)$ denote the holding time of $X(t)$ at
$z$. Then

$\mathbb{P}_{x}(X(t)\leq y|Y(n)=z_{n}\text{ for all }n\leq
l(\pi))=\mathbb{P}\left(\sum_{k=0}^{\l(\pi)}\mathbb{H}(z_{k})\geq t\right)$

Since $\\{\mathbb{H}(z):z\in\mathcal{S}\\}\stackrel{{\scriptstyle
d}}{{=}}\\{\mathcal{E}_{z}:z\in\mathcal{S}\\}$, this implies that
$\mathcal{E}_{\pi}\stackrel{{\scriptstyle
d}}{{=}}\sum_{k=0}^{\l(\pi)}\mathbb{H}(z_{k})$, this completes the proof.

∎

Now suppose that $\mathcal{S}$ is a graded poset. In other words, there is a
rank function $\rho:\mathcal{S}\rightarrow\mathbb{N}$ such that

  * •

$\rho(x)<\rho(y)$ if $x<y$.

  * •

Whenever $x\lessdot y$, the rank function satisfies $\rho(x)+1=\rho(y)$.

Recall that $x\lessdot y$ means that $x<y$ and that there is no element $z$
such that $x<z<y$. In words, $x$ is covered by $y$.

######  Definition 4.4.

For a discrete–time Markov chain on a discrete state space define

$\mathbb{P}_{\mathcal{H},x}(y)=\mathbb{P}\left(Y(n)=y\text{ for some
}n|Y(0)=x\right).$

In words, $\mathbb{P}_{\mathcal{H},x}(y)$ is the probability that the Markov
process $X(t)$ or the chain $Y(t)$ ever hits $y$ when started from $x$. (The
$\mathcal{H}$ stands for hitting). For a continuous–time Markov chain $X(t)$,
similarly define

$\mathbb{P}_{\mathcal{H},x}(y)=\mathbb{P}\left(X(t)=y\text{ for some
}n|X(0)=x\right).$

Note that if $X(t)$ has discrete–time embedded Markov chain $Y(n)$, then
$\mathbb{P}_{\mathcal{H},x}(y)$ is the same measure for both processes.

The next definition has likely occurred before, but the author was unable to
find a citation.

######  Definition 4.5.

A discrete–time Markov chain $Y(n)$ is compatible with the rank function
$\rho$ if the transition probability $T(x,y)$ is nonzero if and only if
$x\lessdot y$. A continuous–time Markov process $X(t)$ is compatible with the
rank function $\rho$ if its discrete–time embedded Markov chain is compatible
with $\rho$.

Before continuing, here are a few properties of compatible Markov chains and
processes that will be useful later.

######  Lemma 4.6.

Suppose that the discrete–time Markov chain $Y(n)$ on a graded poset is
compatible with the rank function $\rho$. Then:

(a) Almost surely $\rho(Y(n))=\rho(Y(0))+n\text{ for all }n\geq 0$.

(b) $\mathbb{P}_{x,\mathcal{H}}(\cdot)$ is a probability measure when
restricted to $\\{y\in\mathcal{S}:\rho(y)=\rho(x)+n\\}$ for each fixed $n$.

###### Proof.

These both follow immediately from the definitions. ∎

With this additional assumption that $\mathcal{S}$ is a graded poset and the
compatibility assumption, the next proposition can be now stated. Roughly
speaking, one can prove certain distributional identities by induction on the
value of the rank function. In the context of this paper, which examines the
continuous–time multi–species $q$–TAZRP, part (b) below will be most useful.
Part (a), which holds for both discrete and continuous–time processes, is an
intermediate step to prove part (b), for the purposes of this paper. However,
it is certainly possible that future applications in different models will be
able to use part (a) without using part (b).

######  Proposition 4.7.

Suppose that $X(t)$ and $Y(n)$ are compatible with the graded poset
$\mathcal{S}$ with partial ordering $\leq$.

(a) Define an equivalence relation $\equiv$ on $\mathcal{S}\times\mathcal{S}$
by

$(x,y)\equiv(x^{\prime},y^{\prime})\text{ if
}\mathbb{P}_{x,\mathcal{H}}(y)=\mathbb{P}_{x^{\prime},\mathcal{H}}(y^{\prime})\text{
and }\rho(x)-\rho(y)=\rho(x^{\prime})-\rho(y^{\prime}).$

Let $E^{\equiv}$ denote the subset of $(\mathcal{S}\times\mathcal{S})^{2}$
which represents $\equiv$, in other words

$((x,y),(x^{\prime},y^{\prime}))\in E^{\equiv}\text{ if and only if
}(x,y)\equiv(x^{\prime},y^{\prime})$

For any $y\in\mathcal{S}$, let $\mathcal{C}(y)\subset\mathcal{S}$ be

$\mathcal{C}(y)=\\{z:z\lessdot y\\}.$

In other words, $\mathcal{C}(y)$ is the set of all states that can possibly
occur the time–step before the Markov chain hits $y$.

Then $E^{\equiv}$ satisfies the following “inductive property”:

  * •

If $((x,y),(x^{\prime},y^{\prime}))\in(\mathcal{S}\times\mathcal{S})^{2}$
satisfies the “inductive step” that there exists a bijection
$\iota:\mathcal{C}(y)\rightarrow\mathcal{C}^{\prime}(y)$ such that

$T(z,y)=T(\iota(z),y^{\prime})\text{ for all }z\in\mathcal{C}(y),\\\ $

and

  * •

if $E^{\equiv}$ satisfies as well the “inductive hypothesis” that

$((x,z),(x^{\prime},\iota(z))\in E^{\equiv}\text{ for all
}z\in\mathcal{C}(y),$

then $((x,y),(x^{\prime},y^{\prime}))$ is also in $E^{\equiv}$.

(b) Assume that all the conditions in part (a) still hold. Define another
equivalence relation $\sim$ on $\mathcal{S}\times\mathcal{S}$ by

$\displaystyle(x,y)\sim(x^{\prime},y^{\prime})\text{ if
}\mathbb{P}_{x}(X(t)\leq y)$ $\displaystyle=\mathbb{P}_{x^{\prime}}(X(t)\leq
y^{\prime})\text{ for all }t\geq 0$ $\displaystyle\text{ and
}\mathbb{P}_{x}(X(t)=y)$
$\displaystyle=\mathbb{P}_{x^{\prime}}(X(t)=y^{\prime})\text{ for all }t\geq
0$ $\displaystyle\text{ and }\rho(x)-\rho(y)$
$\displaystyle=\rho(x^{\prime})-\rho(y^{\prime})$

Let $E^{\sim}$ denote the subset of $(\mathcal{S}\times\mathcal{S})^{2}$ which
represents $\sim$, in other words

$((x,y),(x^{\prime},y^{\prime}))\in E^{\sim}\text{ if
}(x,y)\sim(x^{\prime},y^{\prime}).$

Then $E^{\sim}$ satisfies the following “inductive property”:

  * •

If the bijection from part (a) also satisfies

$L(z,z)=L(\iota(z),\iota(z)),\text{ for all }z\in\mathcal{C}(y),$

and furthermore, the value of $L(y,y)=L(y^{\prime},y^{\prime})$, and:

  * •

if $E^{\sim}$ satisfies as well the “inductive hypothesis” that

$((x,z),(x^{\prime},\iota(z))\in E^{\sim}\text{ for all }z\in\mathcal{C}(y);$

then $((x,y),(x^{\prime},y^{\prime}))$ is also in $E^{\sim}$ as long as
$\rho(x)-\rho(y)=\rho(x^{\prime})-\rho(y^{\prime})$.

###### Proof.

(a) Proving that $((x,y),(x^{\prime},y^{\prime}))\in E^{\equiv}$ entails
proving that $\rho(x)-\rho(y)=\rho(x^{\prime})-\rho(y^{\prime})$ and also that
$\mathbb{P}_{x,\mathcal{H}}(y)=\mathbb{P}_{x^{\prime},\mathcal{H}}(y^{\prime})$.
The former is easier to prove. To start, let $n=\rho(y)-\rho(x)$. But then
$n=\rho(y^{\prime})-\rho(x^{\prime})$ is true as well because of Lemma 4.6.
Thus it remains to prove the latter statement.

Because of the compatibility criterion (Lemma 4.6),

$\mathbb{P}_{x,\mathcal{H}}(y)=\mathbb{P}_{x}(Y(n)=y),\quad\mathbb{P}_{x^{\prime},\mathcal{H}}(y^{\prime})=\mathbb{P}_{x^{\prime}}(Y(n)=y^{\prime}).$

Thus

$\displaystyle\mathbb{P}_{x,\mathcal{H}}(y)$
$\displaystyle=\sum_{z\in\mathcal{C}(y)}\mathbb{P}_{x}(Y(n-1)=z)\mathbb{P}_{x}(Y(n)=y|Y(n-1)=z)$
$\displaystyle=\sum_{z\in\mathcal{C}(y)}\mathbb{P}_{x}(Y(n-1)=z)T(z,y).$

Likewise,

$\mathbb{P}_{x^{\prime},\mathcal{H}}(y^{\prime})=\sum_{z^{\prime}\in\mathcal{C}(y)}\mathbb{P}_{x^{\prime}}(Y(n-1)=z^{\prime})T(z^{\prime},y^{\prime}).$

To show that these two quantities are equal, we will use the inductive step to
match the $T$ terms and the inductive hypothesis to match the $\mathbb{P}$
terms.

Using the bijection $\iota$, the last equation can be rewritten

$\mathbb{P}_{x^{\prime},\mathcal{H}}(y^{\prime})=\sum_{z\in\mathcal{C}(y)}\mathbb{P}_{x^{\prime}}(Y(n-1)=\iota(z))T(\iota(z),y^{\prime}).$

By the inductive step $T(\iota(z),y^{\prime})=T(z,x^{\prime})$; and by the
inductive hypothesis
$\mathbb{P}_{x^{\prime}}(Y(n-1)=\iota(z))=\mathbb{P}_{x}(Y(n-1)=z)$. So thus

$\mathbb{P}_{x^{\prime},\mathcal{H}}(y^{\prime})=\sum_{z^{\prime}\in\mathcal{C}(y)}\mathbb{P}_{x^{\prime}}(Y(n-1)=z)T(z,y)=\mathbb{P}_{x,\mathcal{H}}(y).$

This proves (a).

(b) For the continuous–time case, the argument is slightly more involved
because the time steps no longer exactly line up with the rank function
$\rho$. The equality $\rho(y)-\rho(x)=\rho(y^{\prime})-\rho(x^{\prime})$
follows again from the Lemma 4.6, as in part (a). Let
$d=\rho(y)-\rho(x)=\rho(y^{\prime})-\rho(x^{\prime})$.

There is the identity of events

$\\{X(t)\leq y\\}=\\{X(t)=y\\}\cup\coprod_{z\in\mathcal{C}(y)}\\{X(t)\leq
z\\},$

so by countable additivity

$\mathbb{P}_{x}(X(t)\leq
y)=\mathbb{P}_{x}(X(t)=y)+\sum_{z\in\mathcal{C}(y)}\mathbb{P}_{x}(X(t)\leq
z).$

Similarly,

$\mathbb{P}_{x^{\prime}}(X(t)\leq
y^{\prime})=\mathbb{P}_{x^{\prime}}(X(t)=y^{\prime})+\sum_{z^{\prime}\in\mathcal{C}(y^{\prime})}\mathbb{P}_{x^{\prime}}(X(t)\leq
z^{\prime}).$

By the inductive hypothesis, the $\Sigma$ terms are equal, so it suffices to
show

$\mathbb{P}_{x}(X(t)=y)=\mathbb{P}_{x^{\prime}}(X(t)=y^{\prime})$

Now, the left–hand–side can be written as (Proposition 4.3)

$\mathbb{P}_{x}(X(t)=y)=\sum_{\pi\in\Pi_{xy}}\mathbb{P}_{x}(Y(1)=z_{1},\ldots,Y_{d-1}=z_{d-1},Y_{d}=y)\mathbb{P}(\mathcal{E}_{x}+\mathcal{E}_{z_{1}}+\ldots\mathcal{E}_{z_{d-1}}<t,\mathcal{E}_{\pi}\geq
t),$

since the holding times $\mathcal{E}_{\cdot}$ are independent of the embedded
Markov chain $Y(n)$. First, let us rewrite the term depending on
$\mathcal{E}$’s. To save space, let
$\vec{\mathcal{E}}=\mathcal{E}_{x}+\ldots+\mathcal{E}_{z_{d-1}}$. Since
$\vec{\mathcal{E}}$ and $\mathcal{E}_{y}$ are independent, and
$\mathcal{E}_{\pi}=\vec{\mathcal{E}}+\mathcal{E}_{y}$, this term can be
rewritten as

$\displaystyle\mathbb{P}(\vec{\mathcal{E}}<t,\vec{\mathcal{E}}+\mathcal{E}_{y}>t)$
$\displaystyle=\int_{0}^{t}\mathbb{P}(\vec{\mathcal{E}}<t,\vec{\mathcal{E}}+{\mathcal{E}}_{y}>t|\vec{\mathcal{E}}=s)f_{\vec{\mathcal{E}}}(s)ds$
$\displaystyle=\int_{0}^{t}\mathbb{P}(\mathcal{E}_{y}>t-s)f_{\vec{\mathcal{E}}}(s)ds$
$\displaystyle=\int_{0}^{t}e^{-\lambda_{y}(t-s)}f_{\vec{\mathcal{E}}}(s)ds$

where the notation $f_{A}(s)$ denotes the probability distribution function of
a random variable $A$. Now by integration by parts,

$\displaystyle\mathbb{P}(\vec{\mathcal{E}}<t,\vec{\mathcal{E}}+\mathcal{E}_{y}>t)$
$\displaystyle=e^{-\lambda_{y}t}\left(e^{\lambda_{y}s}\mathbb{P}(\vec{\mathcal{E}}\leq
s)\Big{|}_{s=0}^{t}-\int_{0}^{t}\lambda_{y}e^{\lambda_{y}s}\mathbb{P}(\vec{\mathcal{E}}\leq
s)ds\right)$ $\displaystyle=\mathbb{P}(\vec{\mathcal{E}}\leq
t)-e^{-\lambda_{y}t}\int_{0}^{t}\lambda_{y}e^{\lambda_{y}s}\mathbb{P}(\vec{\mathcal{E}}\leq
s)ds.$

Therefore, plugging this back in to the equation above,

$\mathbb{P}_{x}(X(t)=y)=\sum_{\pi\in\Pi_{xy}}\mathbb{P}_{x}(Y(1)=z_{1},\ldots,Y_{d-1}=z_{d-1},Y_{d}=y)\\\
\times\left(\mathbb{P}(\mathcal{E}_{\pi}\leq
t)-e^{-\lambda_{y}t}\int_{0}^{t}\lambda_{y}e^{\lambda_{y}s}\mathbb{P}({\mathcal{E}}_{\pi}\leq
s)ds\right).$

Now rewriting the term that depends on $Y^{\prime}s$,

$\mathbb{P}_{x}(X(t)=y)=\sum_{z\in\mathcal{C}(y)}\sum_{\pi\in\Pi_{xz}}\mathbb{P}_{x}(Y(1)=z_{1},\ldots,Y(d-2)=z_{d-2},Y_{d-1}=z)T(z,y)\\\
\times\left(\mathbb{P}(\mathcal{E}_{\pi}\leq
t)-e^{-\lambda_{y}t}\int_{0}^{t}\lambda_{y}e^{\lambda_{y}s}\mathbb{P}({\mathcal{E}}_{\pi}\leq
s)ds\right).$

This then becomes

$\mathbb{P}_{x}(X(t)=y)=\sum_{z\in\mathcal{C}(y)}T(z,y)\sum_{\pi\in\Pi_{xz}}\mathbb{P}_{\mathcal{H},x}(z)\left(\mathbb{P}(\mathcal{E}_{\pi}\leq
t)-e^{-\lambda_{y}t}\int_{0}^{t}\lambda_{y}e^{\lambda_{y}s}\mathbb{P}({\mathcal{E}}_{\pi}\leq
s)ds\right).$

Now, note that

$\sum_{\pi\in\Pi_{xz}}\mathbb{P}_{\mathcal{H},x}(z)\mathbb{P}(\mathcal{E}_{\pi}\leq
t)=\mathbb{P}_{x}(\tau_{z}+\mathcal{E}_{z}<t)$

where $\tau_{z}$ is the first hitting time at the state $z$, since both sides
equal the probability that the process hits $z$ and then leaves $z$ before
time $t$. Therefore

$\mathbb{P}_{x}(X(t)=y)=\sum_{z\in\mathcal{C}(y)}T(z,y)\left(\mathbb{P}_{x}(\tau_{z}+\mathcal{E}_{z}\leq
t)-e^{-\lambda_{y}t}\int_{0}^{t}\lambda_{y}e^{\lambda_{y}s}\mathbb{P}_{x}(\tau_{z}+\mathcal{E}_{z}\leq
s)ds\right).$

By an identical calculation,

$\mathbb{P}_{x^{\prime}}(X(t)=y^{\prime})=\sum_{z^{\prime}\in\mathcal{C}(y^{\prime})}T(z^{\prime},y^{\prime})\left(\mathbb{P}_{x^{\prime}}(\tau_{z^{\prime}}+\mathcal{E}_{z^{\prime}}\leq
t)-e^{-\lambda_{y^{\prime}}t}\int_{0}^{t}\lambda_{y^{\prime}}e^{\lambda_{y^{\prime}}s}\mathbb{P}_{x^{\prime}}(\tau_{z^{\prime}}+\mathcal{E}_{z^{\prime}}\leq
s)ds\right).$

By the additional assumptions on the bijection $\iota$, we can insert
$\lambda_{y}=\lambda_{y^{\prime}}$, $\mathcal{E}_{z}\stackrel{{\scriptstyle
d}}{{=}}\mathcal{E}_{\iota(z)}$ and $T(\iota(z),y^{\prime})=T(z,y)$, so that
the proof is completed once it is shown that

$\mathbb{P}_{x}(\tau_{z}<s)=\mathbb{P}_{x^{\prime}}(\tau_{z^{\prime}}<s)$

for all $s\geq 0$. Here $z^{\prime}=\iota(z)$ to save space in notation.

This final identity can be seen to follow from the inductive hypothesis. We
know that

$\mathbb{P}_{x}(X(t)\leq z)=\mathbb{P}_{x^{\prime}}(X(t)\leq z^{\prime})$

To use this, note that for all $z\in\mathcal{C}(y)$,

$\displaystyle\mathbb{P}_{x}(\tau_{z}<s)$
$\displaystyle=\lim_{m\rightarrow\infty}\sum_{a=0}^{m-1}\mathbb{P}_{x}\left(\tau_{z}\in[as/m,(a+1)s/m)\right)$
$\displaystyle=\lim_{m\rightarrow\infty}\sum_{a=0}^{m-1}\mathbb{P}_{x}\left(X\left(\frac{a}{m}s\right)=z,X\left(\frac{a+1}{m}s\right)\neq
z\right)$
$\displaystyle=\lim_{m\rightarrow\infty}\sum_{a=0}^{m-1}\mathbb{P}_{x}\left(X\left(\frac{a}{m}s\right)=z\right)(1-e^{-\lambda_{z}/m}+O(m^{-2})),$

and a similar expression holds for
$\mathbb{P}_{x^{\prime}}(\tau_{z^{\prime}}<s)$. Since, by the induction
hypothesis,
$\mathbb{P}_{x}\left(X\left(\frac{a}{m}s\right)=z\right)=\mathbb{P}_{x^{\prime}}\left(X\left(\frac{a}{m}s\right)=z^{\prime}\right)$
for all $z\in\mathcal{C}(y)$, we conclude that
$\mathbb{P}_{x}(\tau_{z}<s)=\mathbb{P}_{x^{\prime}}(\tau_{z}^{\prime}<s)$.
This finishes the proof. ∎

###  4.2 Proof of Theorem 2.1

By Proposition 3.1, the $n$th $q$–moments of the multi–species $q$–TAZRP with
infinitely many particles can be expressed in terms of the distributions of
the $n$–particle system. The theorem in the previous section will be applied
to these distributions, as described below.

The multi–species $q$–TAZRP fits into the general set-up of a stochastic
Markov process on a graded poset in the last section. Let $X(t)$ denote the
multi–species $q$–TAZRP and $Y(n)$ its discrete–time embedded Markov chain.
Define a partial order $<$ on the state space so that
$\mathbb{P}(Y(1)=y|Y(0)=x)$ is nonzero if and only if $x\lessdot y$. Let
$\rho(\cdot)$ be the rank function that is compatible with the Markov process,
and normalized so that $\rho((0,\ldots,0))=0$. In general,
$\rho((x_{1},\ldots,x_{n}))=x_{1}+\ldots+x_{n}$.

Now to proceed with the proof of the main theorem. After applying Proposition
3.1, the suffices to prove that, assuimng the intersection numbers match,

$\mathbb{P}_{x}(X(t)\leq y)=\mathbb{P}_{x^{\prime}}(X(t)\leq y^{\prime}),$

where $X(t)$ is the multi–species $q$–TAZRP and
$x=(x_{1},\ldots,x_{n}),y=(y_{1},\ldots,y_{n}),x^{\prime}=(x_{1}^{\prime},\ldots,x_{n}^{\prime}),y^{\prime}=(y_{1}^{\prime},\ldots,y_{n}^{\prime})$.
A proof of this identity will be split up into two cases. The reason for
requiring two cases is related to the state space of the multi–species
$q$–TAZRP, in which multiple particles can occupy a site. The first case is
when the $j_{0}$ in Theorem 2.1 represents a particle which does not share a
site with another particle, and the second case is when it does share a site
with another particle. Theorem 2.1(b) will be applied, but actually only to
the first case; the second case requires explicit contour integral formulas.

####  4.2.1 The first case: applying Proposition 4.7

Suppose that the set $\\{j\in[1,n]:y_{j}=y_{\epsilon_{j_{0}}}\\}$ only has one
element. In words, this means that to obtain $y^{\prime}$ from $y$, the
$j_{0}$th–species particle is moved, and it does not share a lattice site with
any other particle. In this case, the necessary bijection
$\iota:\mathcal{C}(y)\rightarrow\mathcal{C}^{\prime}(y)$ such that

$T(z,y)=T(\iota(z),y^{\prime})\text{ and }L(z,z)=L(\iota(z),\iota(z))\text{
for all }z\in\mathcal{C}(y)$

can be constructed explicitly. More specifically, if
$z=(z_{1},\ldots,z_{n})\in\mathcal{C}(y)$ then $y-z=\epsilon_{j}$ for some
$1\leq j\leq n$, noting that $j$ depends on $z$. Define $\iota(z)$ by
$\iota(z)=y^{\prime}-\epsilon_{j}$ for all $z\in\mathcal{C}(y)$. By the
definition of the jump rates of the multi–species $q$–TAZRP, the above
identity holds, completing the proof in this case.

######  Remark 5.

In the general case, computer and numerical simulations show that
$\mathbb{P}_{x,\mathcal{H}}(y)$ does not only depend on the intersection
numbers $I(x,y)$.

####  4.2.2 The second case: Contour integral formulas for the transition
probabilities

At this point, the reader may have noticed that neither the Markov projection
property nor the contour integral formulas have been used yet. It is at this
step that both of these important ingredients are used in the proof.

Although this section does not explicitly Proposition 4.7, the proof will be
set up in the general framework of a monotone Markov process on a poset and an
induction argument on the rank function. In particular, the induction will be
on the value of $d=\rho(y)-\rho(x)$. When $d=0$, the theorem holds because the
intersection numbers are equal. Now assume that the theorem holds for all
$x,y,x^{\prime},y^{\prime}$ such that
$\rho(y)-\rho(x)=\rho(y^{\prime})-\rho(x^{\prime})<d$. By Proposition 3.1,
this assumption implies that

$\mathbb{P}_{x}(X(t)\leq y)=\mathbb{P}_{x^{\prime}}(X(t)\leq y^{\prime})$

for all $t\geq 0$ whenever $I(x,y)=I(x^{\prime},y^{\prime})$ and
$\rho(y)-\rho(x)=\rho(y^{\prime})-\rho(x^{\prime})<d$. The next step is to
manipulation the equality on the last line to obtain contour integral formulas
(Proposition 3.2).

We have, for general monotone Markov processes on posets, that the
left–hand–side equals

$\displaystyle\mathbb{P}_{x}(X(t)\leq y)$
$\displaystyle=\mathbb{P}_{x}(X(t)\leq y|X(t)\neq x)\mathbb{P}(X(t)\neq
x)+\mathbb{P}_{x}(X(t)\leq y|X(t)=x)\mathbb{P}(X(t)=x)$
$\displaystyle=\sum_{\\{z:x\lessdot
z\\}}\mathbb{P}_{x}(Y(1)=z)\int_{0}^{t}\mathbb{P}_{z}(X(t-s)\leq
y)\lambda_{x}e^{-\lambda_{x}s}ds+\int_{t}^{\infty}\lambda_{x}e^{-\lambda_{x}s}ds$

Suppose that the value of the lattice site $x_{j_{0}}$ occurs a total or $r$
times in $(x_{1},\ldots,x_{n})$. In other words, there are $r$ particles
located at $x_{j_{0}}$. Split the sum over $\\{z:x\lessdot z\\}$ into two
sums: the first where $z$ differs from $x$ at one of those $r$ particles, and
the second is the remaining $z$. The first sum can be collapsed into a single
term by the Markov projection property. The $z$ in the first sum (call this
set $S_{1}$) can be indexed as $z(1),\ldots,z(r)$ such that

$\mathbb{P}_{x}(Y(1)=z(r))=q^{r}\mathbb{P}_{x}(Y(1)=z(1));$

this follows from the definition of the jump rates. Here, we are also using
the Markov projection property, which tells us that it suffices to consider
the case when there is only one particle of each species. Then, we can
consider $\mathbb{P}_{\text{q-exc}}(X(t-s)\leq y)$ where q–exc indicates
random initial conditions supported on $z(1),\ldots,z(r)$ which are
$q$–exchangeable (i.e. the probability of starting in $z(k)$ is equal to
$q^{k-1}/[r]_{q}$. By [Kua19], the dynamics of multi–species $q$–TAZRP
preserve $q$–exchangeability (this is proved in the proof of Lemma 3.5 of
[Kua19]). This means that the sum over $z(1),\ldots,z(r)$ can be replaced with
a single term and the induction hypothesis can be applied. In other words,
letting $S$ denote the

$S=\\{x:x\lessdot z\\}-\\{z(2),\ldots,z(r)\\},$

we have

$\displaystyle\mathbb{P}_{x}(X(t)\leq y)$
$\displaystyle=\mathbb{P}_{x}(X(t)\leq y|X(t)\neq x)\mathbb{P}(X(t)\neq
x)+\mathbb{P}_{x}(X(t)\leq y|X(t)=x)\mathbb{P}(X(t)=x)$
$\displaystyle=\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)\int_{0}^{t}\mathbb{P}_{z}(X(t-s)\leq
y)\lambda_{x}e^{-\lambda_{x}s}ds+\int_{t}^{\infty}\lambda_{x}e^{-\lambda_{x}s}ds$
(2)

For now, focus on the sum over $z$. This sum can be expressed, using the
induction hypothesis, as

$\sum_{z\in S}\mathbb{P}_{x}(Y(1)=z)\int_{0}^{t}\mathbb{P}_{z}(X(t-s)\leq
y)\lambda_{x}e^{-\lambda_{x}s}ds\\\ =\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)\int_{0}^{t}q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\lambda_{x}e^{-\lambda_{x}s}ds\\\ \times\int\cdots\int
dw_{1}\cdots
dw_{n}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}e^{-(w_{1}+\cdots+w_{N})(t-s)}.$

Here, the integers $M_{j}$ depend on $z$, so we write $M_{j}(z)$ to emphasize
this dependence, and also to point out that no other term in the integrand
depends on $z$. The $M_{j}(z)$ also depend on $y$ but that notation will not
be included. The integral over $t$ simplifies, so that the right–hand–side
above equals

$\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots
dw_{n}e^{-(w_{1}+\ldots+w_{N})t}\frac{e^{t(w_{1}+\ldots+w_{N}-\lambda_{x})}-1}{w_{1}+\ldots+w_{N}-\lambda_{x}}\\\
\times
B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}.$

This expression is analytic in $t=0$, so to compute it, it suffices to
calculate its $n$th derivative $\mathcal{D}_{n}$ with respect to $t$ at $t=0$.
The $n$th derivative will be related to the $(n-1)$th derivative, and an
induction argument on $n$ will be used. The $n$th derivative of the previous
equation equals

$\mathcal{D}_{n}(x,y):=\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots dw_{N}\\\
\times\frac{(-\lambda_{x})^{n}-(-1)^{n}(w_{1}+\ldots+w_{N})^{n}}{w_{1}+\ldots+w_{N}-\lambda_{x}}\frac{1}{n!}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}.$
(3)

To write $\mathcal{D}_{n}$ in terms of $\mathcal{D}_{n-1}$, use the identity

$\frac{a^{n}-b^{n}}{a-b}=a^{n-1}+a^{n-2}b+\ldots
ab^{n-2}+b^{n-1}=a^{n-1}+b\frac{a^{n-1}-b^{n-1}}{a-b}$

with $a=\lambda_{x}$ and $b=w_{1}+\ldots+w_{N}$. So set
$\mathcal{D}_{n}=\mathcal{D}_{n}^{(1)}+\mathcal{D}_{n}^{(2)}$ where

$\mathcal{D}_{n}^{(1)}(x,y):=-\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N-1}q^{N-2}\cdots q^{K}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots
dw_{N}\lambda_{x}^{n-1}\frac{(-1)^{n}}{n!}\\\ \times
B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}.$

Then $\mathcal{D}_{n}^{(1)}$ simplifies as

$\mathcal{D}_{n}^{(1)}(x,y):=-\lambda_{x}^{n}\frac{(-1)^{n}}{n!}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)\mathbb{P}_{x}(X(0)\leq
y)=-\lambda_{x}^{n}\frac{(-1)^{n}}{n!}$

But

$\frac{d^{n}}{dt^{n}}\Big{|}_{t=0}\int_{t}^{\infty}\lambda_{x}e^{-\lambda_{x}s}ds=\frac{d^{n}}{dt^{n}}\Big{|}_{t=0}e^{-\lambda_{x}t}=(-1)^{n}\frac{\lambda_{x}^{n}}{n!}$

so that $\mathcal{D}_{n}^{(1)}$ cancels with the $n$th derivative of the
second term in (2). In other words,

$\frac{d^{n}}{dt^{n}}\Big{|}_{t=0}\mathbb{P}_{x}(X(t)\leq
y)=\mathcal{D}_{n}^{(2)}(x,y).$

Using that $\mathcal{D}_{n}-\mathcal{D}_{n}^{(1)}=\mathcal{D}_{n}^{(2)}$,

$\mathcal{D}_{n}^{(2)}(x,y):=-\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots dw_{N}(w_{1}+\ldots+w_{N})\\\
\times\frac{\lambda_{x}^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{\lambda_{x}-(w_{1}+\ldots+w_{N})}\frac{(-1)^{n}}{n!}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}.$
(4)

As this point, the contours themselves are relevant. The claim is that for
each $1\leq j\leq k\leq N$ such that $M_{j}(z)=M_{k}$,

$\int\cdots\int dw_{1}\cdots
dw_{N}w_{k}\frac{\lambda_{x}^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{\lambda_{x}-(w_{1}+\ldots+w_{N})}\frac{(-1)^{n}}{n!}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}\\\
=q^{j-k}\int\cdots\int dw_{1}\cdots
dw_{N}w_{j}\frac{\lambda_{x}^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{\lambda_{x}-(w_{1}+\ldots+w_{N})}\frac{(-1)^{n}}{n!}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}$
(5)

To see this, it suffices to consider when $j=k-1$, as the general case follows
by induction. Then the goal is to prove that

$\int\cdots\int dw_{1}\cdots
dw_{N}(w_{k}-q^{-1}w_{k-1})\frac{\lambda_{x}^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{\lambda_{x}-(w_{1}+\ldots+w_{N})}\frac{(-1)^{n}}{n!}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}=0.$

But this is true because

$B\left(w_{1},\ldots,w_{N}\right)\left(\frac{q}{w_{k}}-\frac{1}{w_{k+1}}\right)=\left[\prod_{\begin{subarray}{c}1\leq
i<j\leq N\\\
(i,j)\neq(k,k+1)\end{subarray}}\frac{w_{i}-w_{j}}{w_{i}-qw_{j}}\right]\frac{w_{k+1}-w_{k}}{w_{k}w_{k+1}},$

making the integrand anti–symmetric in $w_{k}$ and $w_{k+1}$, so the integral
evaluates to $0$, as needed. Thus (5) is true.

Therefore plugging (5) into (4), we get

$\mathcal{D}_{n}^{(2)}(x,y)=-\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots dw_{N}\cdot c_{q,k}w_{k}\\\
\times\frac{\lambda_{x}^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{\lambda_{x}-(w_{1}+\ldots+w_{N})}\frac{(-1)^{n}}{n!}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}},$

where $c_{q,k}$ is some constant depending only on $q,N$ and $k$, and the $k$
can be chosen arbitrarily. (In fact, $c_{q,k}=q^{k-N}[N]_{q}$ but this is not
needed). This can then be rewritten as

$\mathcal{D}_{n}^{(2)}(x,y)=-\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots dw_{N}\cdot c_{q,k}\\\
\times\frac{\lambda_{x}^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{\lambda_{x}-(w_{1}+\ldots+w_{N})}\frac{(-1)^{n}}{n!}B(w_{1},\ldots,w_{N})\left(\prod_{\begin{subarray}{c}j=1\\\
j\neq
k\end{subarray}}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}\right)(1-w_{k})^{-M_{k}(z)-1},$
(6)

Now suppose that $\hat{x},\hat{y},\hat{x}^{\prime},\hat{y}^{\prime}$ satisfy
$\rho(\hat{y})-\rho(\hat{x})=d=\rho(\hat{y}^{\prime})-\rho(\hat{x}^{\prime})$
and $I(\hat{x},\hat{y})=I(\hat{x}^{\prime},\hat{y}^{\prime})$. The goal then
is to prove that

$\mathbb{P}_{\hat{x}}(X(t)\leq\hat{y})=\mathbb{P}_{\hat{x}^{\prime}}(X(t)\leq\hat{y}^{\prime})$

for all $t\geq 0$. By the same arguments as before, namely that the expression
is analytic in $t$, it suffices to show that the $n$th derivative with respect
to $t$ at $t=0$ are equal; i.e.

$\mathcal{D}_{n}^{(2)}(\hat{x},\hat{y})=\mathcal{D}_{n}^{(2)}(\hat{x}^{\prime},\hat{y}^{\prime})$

for all $n\geq 0$. To proceed, use induction on $n$. The base case, when $n=0$
is true because both sides just equal $1$. So suppose that the expression is
true for all values less than $n$.

Returning to $\eqref{DefnD}$, with $n-1$ in place of $n$, plug in the series

$1+(1-w_{k})+(1-w_{k})^{2}+(1-w_{k})^{3}+\ldots=\frac{1}{w_{k}},$

to obtain

$\mathcal{D}_{n-1}(x,y)=\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots dw_{N}\\\
\times\frac{(-\lambda_{x})^{n-1}-(-1)^{n-1}(w_{1}+\ldots+w_{N})^{n-1}}{w_{1}+\ldots+w_{N}-\lambda_{x}}\frac{1}{(n-1)!}B(w_{1},\ldots,w_{N})\prod_{\begin{subarray}{c}j=1\\\
j\neq k\end{subarray}}^{N}\frac{(1-w_{j})^{-M_{j}(z)-1}}{w_{j}}\\\
\times\sum_{s=0}^{\infty}(1-w_{k})^{-M_{k}(z)-1+s}.$

At this point, the value of $k$ needs to be chosen. It will depend only on
$x,y,x^{\prime},y^{\prime}$. Because the intersection numbers $I(x,y)$ and
$I(x^{\prime},y^{\prime})$ are all equal, the interval $[x_{j_{0}},y_{j_{0}}]$
must be either contained in or contain every other interval $[x_{j},y_{j}]$
for $1\leq j\leq N$. Set $k$ to be the value which minimizes the length of the
interval $[x_{j},y_{j}]$; we must have
$[x_{k},y_{k}]\subseteq[x_{j_{0}},y_{j_{0}}].$ Then $M_{k}=y_{k}-z_{k}$ and
the sum over $s$ need only range of $0$ to $y_{k}-z_{k}$ (otherwise the
integral evaluates to $0$). For each such $s$, let $y^{(s)}=y+-s\epsilon_{k}$.
Then rearranging the summations and using this new notation, we can conclude
from the induction assumption, that for all $n\geq 0$,

$\mathcal{D}_{n-1}(x,y)=\lambda_{x}\sum_{z\in
S}\mathbb{P}_{x}(Y(1)=z)q^{N(N-1)/2}\left(\frac{-1}{2\pi
i}\right)^{N}\int\cdots\int dw_{1}\cdots dw_{N}\\\
\times\frac{(\lambda_{x})^{n-1}-(w_{1}+\ldots+w_{N})^{n-1}}{w_{1}+\ldots+w_{N}-\lambda_{x}}\frac{(-1)^{n-1}}{(n-1)!}B(w_{1},\ldots,w_{N})\prod_{\begin{subarray}{c}j=1\\\
j\neq k\end{subarray}}^{N}\frac{(1-w_{j})^{-M^{(s)}_{j}(z)-1}}{w_{j}}\\\
\times\sum_{s=0}^{y_{k}-z_{k}}(1-w_{k})^{-M^{(s)}_{k}(z)-1},$

where we recall that the integers $M_{j}(z)$ depend both on $y$ and $z$, so
that the superscript indicates the dependence is on $y^{(s)}$ rather than $y$.
Comparing this to (6), this means that

$\mathcal{D}_{n-1}(x,y)=\frac{n}{c_{q,k}}\sum_{s=0}^{y_{k}-x_{k}}\mathcal{D}_{n}^{(2)}(x,y^{(s)}),$

which rearranges to

$\mathcal{D}_{n}^{(2)}(x,y)=c_{q,k}n^{-1}\mathcal{D}_{n-1}(x,y)-\sum_{s=1}^{y_{k}-x_{k}}\mathcal{D}_{n}^{(2)}(x,y^{(s)})$

By the induction assumption on $n$, the first term matches, meaning that

$c_{q,k}n^{-1}\mathcal{D}_{n-1}(\hat{x},\hat{y})=c_{q,k}n^{-1}\mathcal{D}_{n-1}(\hat{x}^{\prime},\hat{y}^{\prime}).$

By the induction assumption on $d$, and the choice of $k$ (so that the
intersection numbers match), the sum matches too. This completes the proof.

###  4.3 Remarks on multi–species ASEP

A natural question is if this method applies to the multi–species ASEP. Based
on numerical simulations, the shift invariance of the finite–particle
multi–species ASEP on the line will still hold. A proof of this statement
could possibly follow from the general set up of Markov processes and embedded
Markov chains on graded posets, although in the ASEP case the process is no
longer monotone. The author hopes to pursue this step in later work.

##  5 Asymptotic Limits

As an application of Theorem 2.1, we consider the asymptotic limits of the
joint $q$–moments of the multi–species $q$–TAZRP.

######  Theorem 5.1.

Use the same notations as in Theorem 2.1. Let $t,M_{1},\ldots,M_{N}$ depend on
a parameter $L$ in such a way that

$t/L\rightarrow 1,\quad L^{-1/2}(M_{j}-L)\rightarrow\sigma_{j}.$

Then as $L\rightarrow\infty$,

$\mathbb{E}_{y}\left[\prod_{j=1}^{n}q^{k_{j}N_{x_{j}}^{(n+1-j)}(\eta(t))}\right]\\\
\rightarrow\int_{-\infty}^{\sigma_{1}}dv_{1}\cdots\int_{-\infty}^{\sigma_{N}}dv_{N}\frac{q^{N(N-1)/2}}{(2\pi
L^{1/2})^{N}}\int_{-\infty}^{\infty}du_{1}\cdots\int_{-\infty}^{\infty}du_{N}B(u_{1},\ldots,u_{N})\prod_{j=1}^{N}\exp(-iv_{j}u_{j}-u_{j}^{2}/2).$

###### Proof.

Recall that the contour integral formula is:

$\frac{(-1)^{N}q^{N(N-1)/2}}{(2\pi
i)^{N}}\int\frac{dw_{1}}{w_{1}}\cdots\int\frac{dw_{N}}{w_{N}}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}(1-w_{j})^{-{M}_{j}}e^{-w_{j}t}$

Let $t$ and $M_{j}$ depend on a parameter $L$ as $t=L$ and $M_{j}=\lfloor
L+\sigma_{j}L^{1/2}\rfloor$. By the usual steepest descent method, the
integrand is asymptotically

$\frac{B(w_{1},\ldots,w_{N})}{w_{1}\cdots
w_{N}}\exp(S(w_{1}))\cdots\exp(S(w_{N}))$

where $S(w)$ is the function

$S(w)=-w-\log(1-w).$

The contours need to be deformed to regions where the real part of $S(w)$ is
negative; this is shown in Figure 2. Explicitly, the contours can be circles
of radius $1$ centered at $1$. However, then the contours pass through the
simple poles at $0$.

Figure 2: The shaded region is where $\Re S(w)>0$. The contours must go
through the unshaded region.

To circumvent the issue, we will take the partial derivatives with respect to
$\sigma_{j}$. By elementary analysis arguments, the partial derivatives can be
interchanged with the integral and with the limit. Then the contour integral
becomes, in the asymptotic limit,

$\frac{q^{N(N-1)/2}}{(2\pi
i)^{N}}\int\frac{\log(1-w_{1})}{w_{1}}dw_{1}\cdots\int\frac{\log(1-w_{N})}{w_{N}}dw_{N}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\exp(LS(w_{j})-L^{1/2}\sigma_{j}\log(1-w_{j}))$

which no longer has poles at $0$. Near $w=0$, the function $S(w)$ and
$\log(1-w)$ has the Taylor series

$S(w)=\frac{w^{2}}{2}+\frac{w^{3}}{3}+O(w^{4}),\quad\log(1-w)=-w-\frac{w^{2}}{2}+O(w^{3}).$

Therefore we make the substitution $w_{j}=-iL^{-1/2}u_{j}$. The
$u_{j}$–contours then becomes the real line, so the limit becomes

$\frac{q^{N(N-1)/2}}{(2\pi
L^{1/2})^{N}}\int_{-\infty}^{\infty}du_{1}\cdots\int_{-\infty}^{\infty}du_{N}B(u_{1},\ldots,u_{N})\prod_{j=1}^{N}\exp(-i\sigma_{j}u_{j}-u_{j}^{2}/2).$

Here we used that $B(cu_{1},\ldots,cu_{N})=B(u_{1},\ldots,u_{N})$ for any
constant $c$. Finally, we need to integrate over
$\sigma_{1},\ldots,\sigma_{N}$ since the partial derivatives with respect to
$\sigma_{j}$ were taken before.

∎

Note that the integral is real, despite the presence of the imaginary number
in the integrand. In fact, for $N=1$ recall that

$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\sigma
u-u^{2}/2}du=e^{-\sigma^{2}/2}.$

Also note that the $N=1$ case does not depend on $q$. This can be understood
in the following way: as $q$ decreases towards $0$, the height function
$N_{x}(\eta(t))$ also decreases, because the jump rates slow down. Both of
these decreases cancel out in the value of the expectation
$\mathbb{E}[q^{N_{x}(\eta(t))}]$.

In principle, this theorem allows us to find (computationally) the joint
distributions of the height functions $N_{x}^{j}$. The author hopes to provide
an explicit, non–numerical formula in later work.

##  6 Alternative Proof Of the Corollary 2.2

This section provides an alternative proof of the corollary. The argument uses
the Markov duality and the transition probabilities for the multi–species
$q$–TAZRP with $q$–exchangeable initial conditions. Because the corollary has
already been proved with a more general method, and because this alternative
proof is specific to the $q$–TAZRP and not to other models like ASEP, and
because this proof uses some more algebraic notation, some of the exposition
here is deliberately kept terse.

###  6.1 More notation

In general, a particle configuration can have infinitely many particles. When
restricting to states with finitely many particles, there is another
convenient way of writing particle configurations. Assume there are $N_{k}$
particles of species $k$ ($1\leq k\leq n$). Set $N=N_{1}+\ldots+N_{n}$ to be
the total number of particles. Let $\mathbf{N}$ denote $(N_{1},\ldots,N_{n})$.
For $i\leq j$ let $N_{[i,j]}$ denote $N_{i}+\ldots+N_{j}$. A particle
configuration can be expressed as a pair $(\mathbf{x},\sigma)$, where

$\mathbf{x}=(x_{1}\geq x_{2}\geq\ldots\geq x_{N})$

indicates the location of the particles. Let $\sigma\in S(N)$ denote the
ordering of the species, in the sense that if $\sigma$ is written in two–line
notation as

$\left(\begin{array}[]{cccc}\sigma_{1}&\sigma_{2}&\cdots&\sigma_{N}\\\
1&2&\cdots&N\end{array}\right),$

so that $\sigma_{j}=\sigma^{-1}(j)$, then the $N_{k}$ particles of species $k$
are located at the lattice sites

$x_{\sigma_{N_{[1,k-1]}+1}},\ldots,x_{\sigma_{N_{[1,k]}}}.$ (7)

An equivalent description of the particle configuration $(\mathbf{x},\sigma)$
is as follows. For

$\mathbf{x}\in\mathcal{W}_{N}:=\\{(x_{1},\ldots,x_{N}):x_{1}\geq\ldots\geq
x_{N}\\}\subset\mathbb{Z}^{N},$

define $\mathbf{m}(\mathbf{x})=(m_{1},\ldots,m_{r})$ so that

$x_{1}=\cdots=x_{m_{1}}>x_{m_{1}+1}=\cdots=x_{m_{1}+m_{2}}>x_{m_{1}+m_{2}+1}=\cdots=\cdots>x_{m_{1}+\ldots+m_{r-1}+1}=\cdots=x_{N},$

where $m_{r}$ is defined by $m_{1}+\ldots+m_{r}=N$. Also define
$k_{1},\ldots,k_{N}$ by
$k_{1}=\cdots=k_{N_{1}}=1,k_{N_{1}+1}=\cdots=k_{N_{1}+N_{2}}=2,\ldots$. Then
the particles located at the lattice site
$x_{m_{1}+\ldots+m_{s}+1}=\cdots=x_{m_{1}+\ldots+m_{s+1}}$ have species

$k_{\sigma(m_{1}+\ldots+m_{s}+1)},\ldots,k_{\sigma(m_{1}+\ldots+m_{s+1})}.$
(8)

Note that $\sigma(j)$ is not the same as $\sigma_{j}$ in this notation.

Because more than one particle can occupy a site, the map
$\mathcal{W}_{N}\times S(N)\rightarrow(\mathbb{Z}^{n}_{\geq 0})^{\mathbb{Z}}$
is not injective.

Example 1. Consider the particle configuration shown in the left side of
Figure 3. There is more than one $\sigma\in S(N)$ which defines this particle
configuration, and it is not hard to see that $\sigma=21467358$ has the fewest
inversions. The permutations $21476358,21567438,$ and $35178426$ also describe
the same particle configuration as $\sigma$, but have more inversions.

Figure 3: The particle configuration referenced in Example 1.

From now on, a particle configuration of the $n$–species $q$–TAZRP will be
uniquely defined by a pair $(\mathbf{x},\sigma)$, where
$\mathbf{x}\in\mathcal{W}_{N}$ and $\sigma\in S(N)$ has the fewest inversions
among the permutations that define the particle configuration. Uniqueness is
proven in Proposition 2.7 of [Kua19].

Let $\mathrm{inv}(\sigma)$ denote the number of inversions of $\sigma$.

Given a particle configuration $(\mathbf{x},\sigma)$, where
$\mathbf{m}(\mathbf{x})=(m_{1},\ldots,m_{r})$ and the species numbers are
given by $(N_{1},\ldots,N_{n})$, let $L_{ij}$ denote the number of species $j$
particles located at the lattice site $x_{m_{1}+\ldots+m_{i}}$. Here the
ranges of $i$ and $j$ are given by $1\leq i\leq r$ and $1\leq j\leq n$.

Let us define the dynamics of the multi–species $q$–TAZRP. A visual example is
shown below in Figure 1.

Consider a particle configuration $\xi$ of the $N$–particle $q$–TAZRP with
exactly one particles of each species. For any
$\mathbf{N}=(N_{1},\ldots,N_{n})$ such that $N_{1}+\ldots+N_{n}=N$, there is a
projection map $\Pi_{\mathbf{N}}$ defined by

$(\Pi_{\mathbf{N}}(\xi))^{x}_{i}=\xi^{x}_{N_{1}+\ldots+N_{i}+1}+\ldots+\xi^{x}_{N_{1}+\ldots+N_{i+1}}$

In words, $\Pi_{\mathbf{N}}$ forgets the distinction between the species
$N_{1}+\ldots+N_{i}+1,\ldots,N_{1}+\ldots+N_{i+1}$, where $1\leq i\leq n$. It
is not hard to see from the jump rates that

$\mathrm{Prob}_{\Pi_{\mathbf{N}}\eta}(\Pi_{\mathbf{N}}\xi;t)=\mathrm{Prob}_{\eta}(\xi;t).$

In other words, $\Pi_{\mathbf{N}}$ is a Markov projection.

We also remark on the connection to $q$–TASEP and the Tracy–Widom
fluctuations. Let $x_{t}(t)>x_{2}(t)>\ldots$ denote the positions of the
$q$–TASEP. Then $\eta^{k}(t)=x_{k}(t)-x_{k+1}(t)-1$, where $x_{0}(t)$ is set
to be $\infty$. Thus

$N_{k}(\eta(t))=\eta^{k}(t)+\eta^{k+1}(t)+\ldots=\lim_{L\rightarrow\infty}x_{k}(t)-x_{k+L}(t)-L=x_{k}(t)+k$

for step initial condition $x_{k}(0)=-k$.

###  6.2 Transition Probabilities in the multi–species case

The following describes some transition probabilities in the multi–species
case.

$\sum_{\sigma\in
S_{n}}A_{\sigma}\left(w_{\sigma^{-1}(1)},\ldots,w_{\sigma^{-1}(n)}\right)=[n]_{q}!B\left(w_{1},\ldots,w_{n}\right)$
(9)

######  Proposition 6.1.

Let $\mathbf{N}=(N_{1},\ldots,N_{n})$ and $N=N_{1}+\ldots N_{n}$. Fix a
particle configuration $(\mathbf{x},\sigma)$ where
$\mathbf{x}=(x_{1},\ldots,x_{N})$ and $\sigma\in S(N)$. Given initial
conditions where all $N$ particles begin at $0$,

$\mathrm{Prob}_{0}((\mathbf{x},\sigma)\text{ at time
}t)=q^{\mathrm{inv}(\sigma)}\left(\frac{\prod_{j=1}^{n}\left[N_{j}\right]_{q}^{!}}{\prod_{j=1}^{n}\prod_{i=1}^{r}\left[L_{ij}\right]_{q}^{!}}\right)\left(\prod_{k=1}^{N}\frac{-1}{b_{x_{k}}}\right)\\\
\times\left(\frac{1}{2\pi i}\right)^{N}\sum_{\omega\in
S(N)}\int_{\mathcal{C}_{R}}\cdots\int_{\mathcal{C}_{R}}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}\left[\prod_{k=y_{\omega(j)}}^{x_{j}}\left(\frac{b_{k}}{b_{k}-w_{\omega(j)}}\right)e^{-w_{j}t}\right]dw_{1}\cdots
dw_{N},$

where the $\mathcal{C}_{R}$ are large contours, centered at the origin, that
enclose the poles $b_{k}$ and $w_{j}=qw_{i}$.

###### Proof.

This is a special case of Theorem 3.2 of [Kua19], and follows from Corollary
1.3 of [WW16]; see also [KL14] for the homogeneous case when all $b_{x}\equiv
1$. ∎

###  6.3 Completing the Alternative Proof

Set

$c(q,N,K,m)=\sum_{K\leq i_{1}<i_{2}<\cdots<i_{m}\leq
N-1}q^{i_{1}}q^{i_{2}}\cdots q^{i_{m}}$

For example,

$c(q,N,K,0)=1,\quad\quad c(q,N,K,1)=q^{N-1}+q^{N-2}+\ldots+q^{K},\quad\quad
c(q,N,K,N-K)=q^{N-1}q^{N-2}\cdots q^{K}.$

Note that (splitting the sum into $i_{1}=K-1$ and $i_{1}\neq K-1$)

$c(q,N,K-1,a)=c(q,N,K,a)+q^{K-1}c(q,N,K,a-1),$ (10)

For $\sigma\in S(K)$ and $K\leq L\leq N$, define the “large contour” integral

$\mathcal{I}(q,N,L,K,\vec{M},\sigma,\vec{x})=c(q,N,K,L-K)q^{\mathrm{inv}(\sigma)}(-1)^{L}\left(\frac{1}{2\pi
i}\right)^{L}\int_{\mathcal{C}_{R}}dw_{1}\cdots\int_{\mathcal{C}_{R}}dw_{L}B(w_{1},\ldots,w_{L})\\\
\times\prod_{j=1}^{L-K}\left(\frac{(1-w_{j})^{-M_{j}-1}}{w_{j}}\right)\prod_{j=L-K+1}^{L}\left((1-w_{j})^{-x_{\sigma(j+K-L)}-1}\right)e^{-(w_{1}+\cdots
w_{L})t},$

where by convention an empty product $\prod_{j=a+1}^{a}$ equals $1$.
Additionally define the “small contour” integral by

$\tilde{\mathcal{I}}(q,N,K,\vec{M},\sigma,\vec{x})=q^{N-1}q^{N-2}\cdots
q^{K}q^{\mathrm{inv}(\sigma)}(-1)^{N}\left(\frac{1}{2\pi
i}\right)^{N}\int_{\tilde{\mathcal{C}}^{1}}dw_{1}\cdots\int_{\tilde{\mathcal{C}}^{N}}dw_{N}B(w_{1},\ldots,w_{N})\\\
\times\prod_{j=1}^{N-K}\left(\frac{(1-w_{j})^{-M_{j}-1}}{w_{j}}\right)\prod_{j=N-K+1}^{N}\left((1-w_{j})^{-x_{\sigma(j+K-N)}-1}\right)e^{-(w_{1}+\cdots
w_{N})t},$

where the contour $\tilde{\mathcal{C}}^{r}$ contains
$q\tilde{\mathcal{C}}^{r+1},\ldots,q\tilde{\mathcal{C}}^{N}$ and $1$, but not
$0$.

The next proposition gives both a “large contour” and a “small contour”
integral formula for the probability mentioned in the abstract. Note that the
two formulas reveal the structure of the two Markov projections.

######  Proposition 6.2.

Fix all $b_{x}\equiv 1$. Consider a $N$–particle $q$–TAZRP with exactly one
particle of each species. Let the initial condition consist of all particles
at $0$. For $1\leq i\leq N$, let $z_{i}(t)$ denote the location of the
$i$th–species particle at time $t$. For any fixed $K<N$ and any
$x_{1},\ldots,x_{K}\geq 0$, let $\sigma\in S(K)$ denote the permutation
satisfying $x_{\sigma(1)}\geq\cdots\geq x_{\sigma(K)}$ with the fewest number
of inversions. Then

(a)

$\mathrm{Prob}(z_{1}(t)=x_{1},\ldots,z_{K}(t)=x_{K},z_{K+1}(t)\leq
M_{N-K},\ldots,z_{N}(t)\leq
M_{1})=\sum_{L=K}^{N}\mathcal{I}(q,N,L,K,\vec{M},\sigma,\vec{x}).$

(b)

$\mathrm{Prob}(z_{1}(t)=x_{1},\ldots,z_{K}(t)=x_{K},z_{K+1}(t)\leq
M_{N-K},\ldots,z_{N}(t)\leq
M_{1})=\tilde{\mathcal{I}}(q,N,K,\vec{M},\sigma,\vec{x}).$

###### Proof.

(a) Proceed by induction on the value of $N-K$. The base case is when $K=N-1$.
We then need to show that the probability on the left–hand–side equals (where
$\iota_{L}=(-2\pi i)^{L}$)

$\displaystyle
q^{\mathrm{inv}(\sigma)}\iota_{N-1}\int_{\mathcal{C}_{R}}dw_{1}\cdots\int_{\mathcal{C}_{R}}dw_{N-1}B(w_{1},\ldots,w_{N-1})\prod_{j=1}^{N-1}\left((1-w_{j})^{-x_{\sigma(j)}-1}e^{-w_{j}t}\right)$
$\displaystyle+q^{N-1}q^{\mathrm{inv}(\sigma)}\iota_{N}\int_{\mathcal{C}_{R}}dw_{1}\cdots\int_{\mathcal{C}_{R}}dw_{N}B(w_{1},\ldots,w_{N})\prod_{j=2}^{N}\left((1-w_{j})^{-x_{\sigma(j-1)}-1}e^{-w_{j}t}\right)\left(\frac{(1-w_{1})^{-M_{1}-1}}{w_{1}}e^{-w_{1}t}\right).$

Define $\bar{x}=(\bar{x}_{1}\geq\ldots\geq\bar{x}_{N-1})$ so that
$x_{\sigma(j)}=\bar{x}_{j}$. By Proposition 6.1, and the additivity property
of inversions (namely, that
$\mathrm{\mathrm{inv}}(ab)=\mathrm{inv}(a)+\mathrm{inv}(b)$),

$\displaystyle\mathrm{Prob}(z_{1}(t)=x_{1},\ldots,z_{K}(t)=x_{N-1},z_{N}(t)\leq
M_{1})$
$\displaystyle=q^{\mathrm{inv}(\sigma)}\sum_{k=0}^{N-1}q^{N-1-k}\sum_{y=\bar{x}_{k+1}+1}^{\bar{x}_{k}}(-1)^{N}\left(\frac{1}{2\pi
i}\right)^{N}\int_{\mathcal{C}_{R}}\cdots\int_{\mathcal{C}_{R}}B(w_{1},\ldots,w_{N})$
$\displaystyle\quad\quad\times\left[\prod_{j=1}^{k}(1-w_{j})^{-\bar{x}_{j}-1}\right](1-w_{k+1})^{-y-1}\left[\prod_{j=k+2}^{N}(1-w_{j})^{-\tilde{x}_{j-1}-1}\right]e^{-(w_{1}+\ldots+w_{N})t}dw_{1},\ldots
dw_{N},$

where by convention $\bar{x}_{0}=M_{1}$ and $\bar{x}_{N}+1=0$, and a sum of
the form $\sum_{a+1}^{a}$ equals $0$, and a product of the form
$\prod_{a+1}^{a}$ equals $1$. Each summation is a geometric series, so

$\displaystyle
q^{N-1-k}\sum_{y=\bar{x}_{k+1}+1}^{\bar{x}_{k}}(1-w_{k+1})^{-y-1}$
$\displaystyle=q^{N-1-k}\frac{1-(1-w_{k+1})^{\bar{x}_{k+1}-\bar{x}_{k}}}{1-(1-w_{k+1})^{-1}}(1-w_{k+1})^{-\bar{x}_{k+1}-2}$
$\displaystyle=q^{N-1-k}\frac{(1-w_{k+1})^{-\bar{x}_{k+1}-1}-(1-w_{k+1})^{-\bar{x}_{k}-1}}{-w_{k+1}}.$

Note that the identity still holds when $\bar{x}_{j}=\bar{x}_{j+1}$. Therefore

$\displaystyle\mathrm{Prob}(z_{1}(t)=x_{1},\ldots,z_{K}(t)=x_{N-1},z_{N}(t)\leq
M_{1})$
$\displaystyle=q^{\mathrm{inv}(\sigma)}\sum_{k=0}^{N-1}(-1)^{N-1}\left(\frac{1}{2\pi
i}\right)^{N}\int_{\mathcal{C}_{R}}\cdots\int_{\mathcal{C}_{R}}B(w_{1},\ldots,w_{N})dw_{1}\cdots
dw_{N}$
$\displaystyle\quad\quad\times\left[\prod_{j=1}^{k}(1-w_{j})^{-\bar{x}_{j}-1}\right]q^{N-1-k}\frac{(1-w_{k+1})^{-\bar{x}_{k+1}-1}-(1-w_{k+1})^{-\bar{x}_{k}-1}}{w_{k+1}}\left[\prod_{j=k+2}^{N}(1-w_{j})^{-\tilde{x}_{j-1}-1}\right]e^{-(w_{1}+\ldots+w_{N})t}.$

This sum turns out to be a telescoping sum. To see this, we need to show that
for each $1\leq k\leq N-1$,

$\int_{\mathcal{C}_{R}}\cdots\int_{\mathcal{C}_{R}}B(w_{1},\ldots,w_{N})\left[\prod_{j=1}^{k-1}(1-w_{j})^{-\bar{x}_{j}-1}\right](1-w_{k})^{-\bar{x}_{k}-1}(1-w_{k+1})^{-\bar{x}_{k}-1}\left[\prod_{j=k+2}^{N}(1-w_{j})^{-\tilde{x}_{j-1}-1}\right]\\\
\times\left(\frac{q^{-(k+1)}}{w_{k+1}}-\frac{q^{-k}}{w_{k}}\right)e^{-(w_{1}+\ldots+w_{N})t}dw_{1}\cdots
dw_{N}=0.$

This integral equals $0$ because

$B\left(w_{1},\ldots,w_{N}\right)\left(\frac{q}{w_{k}}-\frac{1}{w_{k+1}}\right)=\left[\prod_{\begin{subarray}{c}1\leq
i<j\leq N\\\
(i,j)\neq(k,k+1)\end{subarray}}\frac{w_{i}-w_{j}}{w_{i}-qw_{j}}\right]\frac{w_{k+1}-w_{k}}{w_{k}w_{k+1}},$
(11)

making the integrand anti–symmetric in $w_{k}$ and $w_{k+1}$. Thus its
integral is zero, showing that the sum over $k$ is a telescoping sum. After
all cancelations, the remaining terms equal

$q^{\mathrm{inv}(\sigma)}(-1)^{N-1}\left(\frac{1}{2\pi
i}\right)^{N}\int_{\mathcal{C}_{R}}dw_{1}\cdots\int_{\mathcal{C}_{R}}dw_{N}B(w_{1},\ldots,w_{N})\left[\prod_{j=1}^{N-1}(1-w_{j})^{-\bar{x}_{j}-1}\right]\frac{(1-w_{N})^{-\bar{x}_{N}-1}}{w_{N}}e^{-(w_{1}+\ldots+w_{N})t}\\\
+q^{N-1}q^{\mathrm{inv}(\sigma)}(-1)^{N}\left(\frac{1}{2\pi
i}\right)^{N}\int_{\mathcal{C}_{R}}dw_{1}\cdots\int_{\mathcal{C}_{R}}dw_{N}B(w_{1},\ldots,w_{N})\frac{(1-w_{1})^{-\bar{x}_{0}-1}}{w_{1}}\left[\prod_{j=2}^{N}(1-w_{j})^{-\tilde{x}_{j-1}-1}\right]e^{-(w_{1}+\ldots+w_{N})t}.$

Recalling that $\bar{x}_{N}+1=0$, the $w_{N}$–contour of first term has only a
simple pole at $w_{N}=0$. Using that
$B(w_{1},\ldots,w_{N-1},0)=B(w_{1},\ldots,w_{N-1})$, the first term reduces to
the first term in the statement of the lemma. It is immediate that the second
terms match as well. This completes the base case.

The inductive step follows from (10) and the same argument.

(b) Start with the sum in the right–hand–side of part (a). The idea is that
deforming the large contours to small contours results in many cancelations,
leaving only the $N$–fold contour integral. Define a “mixed contour” integral
by

$J(q,K,L,P,\vec{M},\sigma,\vec{x})=q^{\mathrm{inv}(\sigma)}(-1)^{L}\left(\frac{1}{2\pi
i}\right)^{L}\int_{\mathcal{C}_{R}}dw_{1}\cdots\int_{\mathcal{C}_{R}}dw_{P}\int_{\tilde{C}^{P+1}}dw_{P+1}\cdots\int_{\tilde{\mathcal{C}}^{L}}dw_{L}\\\
\times
B(w_{1},\ldots,w_{L})\prod_{j=1}^{L-K}\left(\frac{(1-w_{j})^{-M_{j}-1}}{w_{j}}\right)\prod_{j=L-K+1}^{L}\left((1-w_{j})^{-x_{\sigma(j+K-L)}-1}\right)e^{-(w_{1}+\cdots
w_{L})t}.$

Note that the deformation from the large contour to the small contour only
passes through the residue at $0$ (if it exists), and none of the residues of
the form $qw_{i}=w_{j}$ in $B(w_{1},\ldots,w_{L})$. Therefore

$J(q,K,L,L-K,\vec{M},\sigma,\vec{x})c(q,N,K,L-K)=\mathcal{I}(q,N,L,K,\vec{M},\sigma,\vec{x})$
(12)

and

$J(q,K,L,P-1,\vec{M},\sigma,\vec{x})=J(q,K,L,P,\vec{M},\sigma,\vec{x})+q^{P-L}J(q,K,L-1,P-1,\vec{M},\sigma,\vec{x}),$
(13)

with the latter identity using that

$B(w_{1},\ldots,w_{P-1},0,w_{P+1},\ldots,w_{L})=(q^{-1})^{L-P}B(w_{1},\ldots,w_{P-1},w_{P+1},\ldots,w_{L})$

Now, note that the expression in (a) equals (by (12))

$\sum_{L=K}^{N}J(q,K,L,L-K,\vec{M},\sigma,\vec{x})c(q,N,K,L-K)$

while the expression in (b) equals

$J(q,K,N,0,\vec{M},\sigma,\vec{x})c(q,N,K,N-K)=\sum_{L=N}^{N}J(q,K,L,L-N,\vec{M},\sigma,\vec{x})c(q,N,K,L-K).$

Comparing these two expressions, it is seen that it suffices to show that
there exist constants $\tilde{c}(q,N,K,L-K,a)$ such that

$\sum_{L=K+a}^{N}J(q,K,L,L-K-a,\vec{M},\sigma,\vec{x})\tilde{c}(q,N,K,L-K,a)\\\
=\sum_{L=K+a+1}^{N}J(q,K,L,L-K-a-1,\vec{M},\sigma,\vec{x})\tilde{c}(q,N,K,L-K,a+1)$
(14)

where the constants $\tilde{c}(q,N,K,m,a)$ satisfy

$\tilde{c}(q,N,K,L-K,0)=c(q,N,K,L-K),\quad\tilde{c}(q,N,K,N-K,N-K)=c(q,N,K,N-K).$

We will show that (14) holds

$\tilde{c}(q,N,K,m,a)=q^{K}q^{K+1}\cdots q^{K+a-1}c(q,N,K+a,m-a).$

Start by writing (dropping the $\vec{M},\sigma,\vec{x}$ dependence from the
notation)

$\displaystyle\sum_{L=K+a}^{N}J(q,K,L,L-K-a)\tilde{c}(q,N,K,L-K,a)$
$\displaystyle\stackrel{{\scriptstyle\eqref{cCur}}}{{=}}$
$\displaystyle\sum_{L=K+a}^{N}J(q,K,L,L-K-a)q^{K}q^{K+1}\cdots q^{K+a-1}$
$\displaystyle\quad\quad\quad\times(c(q,N,K+a+1,L-K-a)+q^{K+a}c(q,N,K+a+1,L-K-a-1))$
$\displaystyle=$
$\displaystyle\sum_{L=K+a}^{N}J(q,K,L,L-K-a)q^{K}q^{K+1}\cdots
q^{K+a-1}c(q,N,K+a+1,L-K-a)$ $\displaystyle+$
$\displaystyle\sum_{L=K+a+1}^{N}J(q,K,L,L-K-a)q^{K}q^{K+1}\cdots
q^{K+a}c(q,N,K+a+1,L-K-a-1).$

Since $c(q,N,K+a+1,N-K-a)=0$, the two summations can be re–indexed into a
single summation as

$\sum_{L=K+a}^{N}c(q,N,K+a+1,L-K-a)q^{K}q^{K+1}\cdots q^{K+a}\\\
\times\left(J(q,K,L+1,L+1-K-a)+q^{-(K+a)}J(q,K,L,L-K-a)\right).$

By (13), this equals

$\sum_{L=K+a}^{N}c(q,N,K+a+1,L-K-a)q^{K}q^{K+1}\cdots
q^{K+a}J(q,K,L+1,L-K-a),$

which also equals

$\sum_{L=K+a+1}^{N}c(q,N,K+a+1,L-K-a-1)\tilde{c}(q,N,K,L-K,a+1).$

Thus (14) is true, finishing the proof. ∎

Recall that $N^{(j)}(\eta)$ counts the number of species $1,\ldots,n+1-j$
particles that are weakly to the right of $0$.

######  Corollary 6.3.

Suppose that $\eta(t)$ evolves as the multi–species $q$–TAZRP, and begins in
the initial conditions with infinitely many $i$th class particles at
$-\mathcal{M}_{i}$ for some integers
$0<\mathcal{M}_{n}<\ldots<\mathcal{M}_{1}$. Fix non–negative integers
$k_{1},\ldots,k_{n}$. Set $N=k_{1}+\ldots+k_{n}$ and define ${M}_{j}$ ($1\leq
j\leq N$) by

${M}_{k_{1}+\ldots+k_{m}+1}={M}_{k_{1}+\ldots+k_{m}+2}=\cdots={M}_{k_{1}+\ldots+k_{m+1}}=\mathcal{M}_{m+1}$

Then

$\mathbb{E}\left[\prod_{j=1}^{n}q^{k_{j}N_{0}^{(n+1-j)}(\eta(t))}\right]=\frac{(-1)^{N}q^{N(N-1)/2}}{(2\pi
i)^{N}}\int\frac{dw_{1}}{w_{1}}\cdots\int\frac{dw_{N}}{w_{N}}B(w_{1},\ldots,w_{N})\prod_{j=1}^{N}(1-w_{j})^{-{M}_{j}+0}e^{-w_{j}t}.$

where the $w_{r}$–contour contains $qw_{r+1},\ldots,qw_{n}$ and $1$, but not
$0$.

###### Proof.

Use Proposition 3.1 and part (b) of Proposition 6.2. Take $K=1$ in the
small–contour integral $\tilde{I}$. The term $\mathrm{inv}(\sigma)$ is simply
$0$ since the only $\sigma$ is the identity. The resulting sum over $x_{1}$ is
a geometric series, resulting in $z_{1}^{-1}(1-(1-z_{1})^{-M_{1}})$. Because
the integrand does not contain the pole $z_{1}=0$, the result follows. ∎

######  Remark 6.

By comparing the formula to Theorem 2.11 of [BCS14], we see the equality

$\mathbb{E}\left[\prod_{j=1}^{n}q^{k_{j}N_{0}^{(n+1-j)}(\eta(t))}\right]=\mathbb{E}_{\text{step}}\left[\prod_{j=1}^{n}q^{x_{M_{j}}(t)+M_{j}}\right]$

where $x_{M}(t)$ denotes the position of the $M$–th particle in the
single–species $q$–TASEP (which is equivalent to the single–species
$q$–TAZRP). This proves the corollary.

##  7 Computer Simulations

###  7.1 The embedded Markov chain

######  Example 1.

Let $X=(x_{1},x_{2},x_{3})=(0,-1,-2)$ and $Y=(y_{1},y_{2},y_{3})=(1,3,2)$.
Then

$\mathbb{P}_{X,\mathcal{H}}(Y)=\frac{q^{2}\left(17q^{6}+349q^{5}+2500q^{4}+8610q^{3}+15932q^{2}+16454q+7168\right)}{729\left(q^{8}+13q^{7}+73q^{6}+232q^{5}+460q^{4}+592q^{3}+496q^{2}+256q+64\right)}$

For $X=(0,-2,-3)$ and $Y=(1,2,1)$, the value of
$\mathbb{P}_{X,\mathcal{H}}(Y)$ is the same. For $X=(0,-2,-2)$ and
$Y=(1,2,2)$, the values of $\mathbb{P}_{X,\mathcal{H}}(Y)$ is

$\frac{q^{2}\left(89q^{5}+903q^{4}+3325q^{3}+5905q^{2}+5091q+1697\right)}{729\left(q^{7}+11q^{6}+51q^{5}+130q^{4}+200q^{3}+192q^{2}+112q+32\right)}$

Same for $X=(0,-1,-1)$ and $Y=(1,3,3)$ it is the same.

######  Example 2.

Here is a non–example. Suppose $X=(0,-1,-2)$ and $Y=(1,3,4)$. Then

$\mathbb{P}_{\mathcal{H},X}(Y)=\frac{q^{3}\left(808q^{6}+9507q^{5}+45927q^{4}+119125q^{3}+179061q^{2}+151389q+55513\right)}{6561\left(q^{9}+15q^{8}+99q^{7}+378q^{6}+924q^{5}+1512q^{4}+1680q^{3}+1248q^{2}+576q+128\right)}.$

However, for $X=(0,-1,-3)$ and $Y=(1,3,3)$

$\mathbb{P}_{\mathcal{H},X}(Y)=\frac{q^{2}\left(808q^{6}+9507q^{5}+45927q^{4}+119125q^{3}+179061q^{2}+151389q+55513\right)}{19683\left(q^{8}+13q^{7}+73q^{6}+232q^{5}+460q^{4}+592q^{3}+496q^{2}+256q+64\right)}$

even though the intersection numbers are the same. Similarly, for
$X=(0,-2,-2)$ and $Y=(1,2,4)$,

$\mathbb{P}_{\mathcal{H}}(X,Y)=\frac{q^{3}\left(162q^{5}+1600q^{4}+6011q^{3}+11259q^{2}+10793q+4195\right)}{729\left(q^{8}+13q^{7}+73q^{6}+232q^{5}+460q^{4}+592q^{3}+496q^{2}+256q+64\right)}$

###  7.2 The continuous–time multi–species $q$–TAZRP

######  Example 3.

A few examples of the shift–invariance are shown below. Set $q=0.6$ and $t=2$.
In the table below, the coordinates on the vertical axis show the initial
conditions $x$, while the coordinates on the horizontal axis show $y-x$, where
the probability is on the event that the bounds on $X(2)$ being less than or
equal to $y$. The entries show the probability of $\mathbb{P}_{x}(X(2)\leq
y)$, estimated from 10000000 samples. Some of the entries are blank because
the author used up the allotted time on the Texas A&M High Performance
Research Computing Center.

| (0,1,0) | (0,0,1) | (0,0,2) | (0,1,4) | (0,1,3) | (0,1,2) | (0,1,1)
---|---|---|---|---|---|---|---
(0,0,0) | 0.0332632 | 0.0279420 | 0.0343266 | 0.0727376 | 0.0695906 | 0.0626983 | 0.0513806
$(0,0,-1)$ | 0.0100777 | 0.0278481 | 0.0343091 | 0.0727394 | 0.0695814 | 0.0629251 | 0.0482535
$(0,-1,0)$ | 0.0278073 | 0.0100938 |  | 0.0727483 | 0.0673652 | 0.0582948 |
$(0,-1-1)$ | 0.0165454 | 0.0121191 | 0.0152055 | 0.0726808 | 0.0695527 | 0.0628361 | 0.0515214
$(0,0,-2)$ |  |  |  | 0.0726787 | 0.0695210 |  |

######  Example 4.

The contour integral can also be evaluated explicit using residues. For
example, under the fifth column, when $y-x=(0,1,3)$, evaluating the residue
explicitly yields,

$\frac{1}{2(-1+q)^{4}(1+q)^{3}}e^{-\left(\left(6+2q+q^{2}\right)t\right)}\left(2e^{(5+q)t}+2e^{\left(3+2q+q^{2}\right)t}q^{3}(1+q)^{3}(-3+q-t+qt)-\right.$

+$\left.e^{\left(4+q+q^{2}\right)t}q\left(2+4q^{2}(-3+t)-2q^{4}(3+2t)+q^{5}\left(2+t^{2}\right)-2q^{3}\left(11+3t+t^{2}\right)+q\left(6+6t+t^{2}\right)\right)\right)$,
which at $t=2$ and $q=0.6$ numerically evaluates to 0.0695753…. This is
approximately the values in the fifth column.

Similarly, for the fourth column, the residues yield

$\begin{array}[]{l}\frac{1}{6(-1+q)^{5}(1+q)^{4}}\\\
e^{-\left(\left(6+2q+q^{2}\right)t\right)}\left(-6e^{(5+q)t}+3e^{\left(3+2q+q^{2}\right)t}q^{3}(1+q)^{4}\left(12+6t+t^{2}-2q(2+t)^{2}+q^{2}\left(2+2t+t^{2}\right)\right)-\right.\\\
e^{\left(4+q+q^{2}\right)t}q\left(-6+q^{2}(60-18t)+6q^{6}t+6q^{4}(15+2t)-3q^{5}t\left(2+4t+t^{2}\right)+\right.\\\
\left.\left.q^{7}\left(6+6t+3t^{2}+t^{3}\right)+3q^{3}\left(52+8t+5t^{2}+t^{3}\right)-q\left(24+24t+6t^{2}+t^{3}\right)\right)\right)\end{array}$
(15)

which evaluates to 0.0727076….

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