Theorem 6.2 on parameter symmetry. In Section 7 we compute the correlation
kernels from Section 2.2. In Section 8 we prove Theorem 2.28 on conditioned
walks. In Section 9 we prove a precise version of Theorem 2.32, for the edge
particle systems, see Theorems 9.5 and 9.6. In Section 10 we prove Theorem
2.37 on extremal coherent measures. In Section 11 we prove the duality results
from Section 2.6. In Section 12 we explain the connection between the
shuffling algorithm for sampling arbitrary weightings of the Aztec diamond and
certain Bernoulli push-block dynamics and prove Theorem 2.48. In Section 13,
we develop a framework generalising some of the results of [62] and [16] in
order to prove Theorem 2.50. Finally, in Section 14 we prove Theorem 2.20 on
convergence to the discrete Bessel point process.

##  3 One-dimensional dynamics and inhomogeneous Toeplitz-like matrices

###  3.1 Inhomogeneous Toeplitz-like matrices

Given $R>0$ we use the following notation for the half plane
$\mathbb{H}_{-R}=\\{z\in\mathbb{C}:\Re(z)>-R\\}$. We write
$\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ for the set of functions which are
holomorphic in $\mathbb{H}_{-R}$. Recall that a function $f$ is entire if it
is holomorphic in the whole of $\mathbb{C}$.

######  Definition 3.1.

For $f\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ we define the following
quantities, where recall that the counterclockwise contour
$\mathsf{C}_{\mathbf{a}}\subset\mathbb{H}_{-R}$ encircles all the points of
the sequence $\mathbf{a}$,

$\displaystyle\mathsf{T}_{f}(x,y)$
$\displaystyle=-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{y}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{p_{x}(w)}{p_{y+1}(w)}f(w)dw,\
\ x,y\in\mathbb{Z}_{+},$ (30) $\displaystyle\mathsf{T}_{f}(x)$
$\displaystyle=\mathsf{T}_{f}(0,x)=-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{x}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{f(w)}{p_{x+1}(w)}dw,\
\ x\in\mathbb{Z}_{+}.$ (31)

Clearly, $\mathsf{T}_{f}$ depends on $\mathbf{a}$ but we supress it from the
notation since $\mathbf{a}$ is fixed. Observe that, in the homogeneous case
$a_{x}\equiv 1$, $[\mathsf{T}_{f}(x,y)]_{x,y\in\mathbb{Z}_{+}}$ is nothing but
the Toeplitz matrix with symbol $f(1-z)$. For this reason we shall call
$\mathsf{T}_{f}$ the inhomogeneous Toeplitz-like matrix/operator with
symbol161616We prefer this terminology to the somewhat more precise “… with
symbol $f(1-z)$”. $f$.

We note that the only poles for the integrand in the definition of
$\mathsf{T}_{f}$ come from the inhomogeneity $\mathbf{a}$. In particular, we
note that $\mathsf{T}_{f}(x,y)\equiv 0$, for $x>y$, namely the matrix
$[\mathsf{T}_{f}(x,y)]_{x,y\in\mathbb{Z}_{+}}$ is upper-triangular. From a
probabilistic standpoint, for certain special choices of $f$, $\mathsf{T}_{f}$
will be the transition probability, from $x$ to $y$, of a Markov chain on
$\mathbb{Z}_{+}$ which moves only to the right.

More general functions $f$, which are not in
$\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$, can be used to define
$\mathsf{T}_{f}$ and most of the results below have corresponding analogues.
We restrict to $f\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ for simplicity
since they suffice for the probabilistic applications we have in mind.

For functions $g:\mathbb{Z}_{+}\to\mathbb{C}$ (equivalently sequences in
$\mathbb{C}^{\mathbb{Z}_{+}}$), for which the sum converges, we write
$\mathsf{T}_{f}g(x)=\sum_{y=0}^{\infty}\mathsf{T}_{f}(x,y)g(y)$. It is easy to
see that the operator $\mathsf{T}_{f}$ is well-defined on bounded sequences
but it is actually defined on a somewhat larger space, see Proposition 3.8.

In fact, as we show in Proposition 3.10, the matrix
$[\mathsf{T}_{f}(x,y)]_{x,y\in\mathbb{Z}_{+}}$, with general sequence
$\mathbf{a}$, is similar to the standard Toeplitz matrix (namely with
$a_{x}\equiv 1$) with symbol $f(1-z)$ using an explicit change of basis
matrix. However, even in the one-dimensional setting of this section it is not
possible, using this similarity, to simply translate results from the standard
Toeplitz matrix setting to the inhomogeneous one, see Remark 3.11. In the
multidimensional setting of Sections 4, 5 and 6 it is unclear whether this
matrix similarity is of any use at all but it would be interesting if
something can be done using it.

We finally define the adjoint kernel by
$\mathsf{T}_{f}^{*}(x,y)=\mathsf{T}_{f}(y,x)$ and corresponding operator
$\mathsf{T}_{f}^{*}g(x)=\sum_{y=0}^{\infty}\mathsf{T}_{f}(y,x)g(y)$. Clearly,
by its lower-triangular structure, $\mathsf{T}_{f}^{*}$ is well-defined on the
whole of $\mathbb{C}^{\mathbb{Z}_{+}}$. From a probabilistic standpoint
$\mathsf{T}_{f}^{*}$ evolves measures on $\mathbb{Z}_{+}$ according to the
dynamics of the Markov chain governed by $\mathsf{T}_{f}$.

The following constant will appear often throughout the paper.

######  Definition 3.2.

Given an inhomogeneity sequence $\mathbf{a}$, define $R(\mathbf{a})$ by

$R(\mathbf{a})=\sup_{k\in\mathbb{Z}_{+}}a_{k}-\inf_{k\in\mathbb{Z}_{+}}a_{k}.$
(32)

We have the following expansion of functions $f$ in terms of the polynomials
$p_{x}(z)$. When $a_{x}\equiv 1$, $x\in\mathbb{Z}_{+}$, this is simply the
Taylor expansion of the function $f$ around 1.

######  Lemma 3.3.

Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ and $\mathcal{U}$ be a
compact set in $\mathbb{H}_{-R}$ for some $R>0$. Suppose there exists a
(counterclockwise) contour $\mathsf{C}_{\mathbf{a}}\subset\mathbb{H}_{-R}$
containing $\\{a_{x}\\}_{x\in\mathbb{Z}_{+}}$ such that

$\sup_{k\in\mathbb{Z}_{+}}\frac{\sup_{u\in\mathcal{U}}|u-a_{k}|}{\inf_{w\in\mathsf{C}_{\mathbf{a}}}|w-a_{k}|}=r<1.$
(33)

Then, the following expansion converges absolutely and uniformly for
$u\in\mathcal{U}$,

$f(u)=\sum_{x=0}^{\infty}p_{x}(u)\mathsf{T}_{f}(x).$ (34)

If $f$ is entire then for any compact $\mathcal{U}\subset\mathbb{C}$ a contour
$\mathsf{C}_{\mathbf{a}}$ satisfying (33) exists and so (34) converges
uniformly on compact sets in $\mathbb{C}$. Finally, suppose $R>R(\mathbf{a})$.
Then, for $\epsilon>0$ small enough and $\mathcal{U}=\mathcal{U}_{\epsilon}$
given by the rectangle

$\mathcal{U}=\left\\{u\in\mathbb{C}:-\epsilon\leq\Re{u}\leq\sup_{k\in\mathbb{Z}_{+}}a_{k},-\epsilon\leq\Im(u)\leq\epsilon\right\\}$
(35)

a contour $\mathsf{C}_{a}$ satisfying (33) exists.

###### Proof.

Observe that,

$\displaystyle-\frac{1}{a_{y}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathsf{a}}}\frac{p_{x}(w)}{p_{y+1}(w)}dw=\mathbf{1}_{x=y}.$

Thus, we get that with $f(u)=p_{y}(u)$, and more generally by linearity for
$f(u)$ any polynomial, the expansion (34), which is a finite sum, converges
uniformly and absolutely for all $u\in\mathbb{C}$.

We now assume that $\mathsf{C}_{\mathbf{a}}$ is picked as in the statement.
Recall that $f$ is analytic in $\mathbb{H}_{-R}$. Take $f_{N}(u)$ to be the
truncated degree $N$ Taylor polynomial in the Taylor expansion of $f(u)$ about
a point $u_{*}\in\mathbb{R}_{+}$. This expansion converges in the open disk of
radius $R+u_{*}$ centred at $u_{*}$. By taking $u_{*}$ large enough we can get
this disk to contain both $\mathcal{U}$ and $\mathsf{C}_{\mathbf{a}}$ and thus
for the expansion to converge on $\mathcal{U}$ and $\mathsf{C}_{\mathbf{a}}$.
Moreover note that, from the observation in the paragraph above we have,

$\sum_{x=0}^{\infty}p_{x}(u)\mathsf{T}_{f_{N}}(x)=\sum_{x=0}^{\infty}-\frac{1}{a_{x}}\frac{1}{2\pi\textnormal{i}}p_{x}(u)\oint_{\mathsf{C}_{\mathbf{a}}}\frac{f_{N}(w)}{p_{x+1}(w)}dw=f_{N}(u).$
(36)

Let $u\in\mathcal{U}$ be fixed. We then have the following bound, for any
$N\in\mathbb{N}$ and $(x,w)\in\mathbb{Z}_{+}\times\mathsf{C}_{\mathbf{a}}$, by
using (33) and the fact that $f_{N}$ converges uniformly to $f$ in
$\mathsf{C}_{\mathbf{a}}$ and thus it is uniformly bounded,

$\left|\frac{1}{a_{x}}\frac{p_{x}(u)f_{N}(w)}{p_{x+1}(w)}\right|\lesssim\prod_{k=0}^{x}\frac{|u-a_{k}|}{|w-a_{k}|}\lesssim
r^{x}.$

Thus, by the dominated convergence theorem we have, for fixed
$u\in\mathcal{U}$,

$\sum_{x=0}^{\infty}p_{x}(u)\mathsf{T}_{f_{N}}(u)\overset{N\to\infty}{\longrightarrow}\sum_{x=0}^{\infty}p_{x}(u)\mathsf{T}_{f}(u).$

On the other hand, we know that $f_{N}(u)\to f(u)$ uniformly for
$u\in\mathcal{U}$ and this proves the expansion for fixed $u\in\mathcal{U}$.
We now show that the series converges uniformly and absolutely in
$\mathcal{U}$. We can bound for any $x\in\mathbb{Z}_{+}$ and for all
$u\in\mathcal{U}$, using (33) and the fact that $f$ is uniformly bounded in
$\mathcal{U}$,

$\displaystyle\left|p_{x}(u)\mathsf{T}_{f}(x)\right|\lesssim\oint_{\mathsf{C}_{\mathbf{a}}}\prod_{k=0}^{x}\frac{|u-a_{k}|}{|w-a_{k}|}dw\lesssim
r^{x}.$

By the Weirstrass M-test the desired conclusion follows.

When $f$ is entire then for any compact $\mathcal{U}\subset\mathbb{C}$ we can
take $\mathsf{C}_{\mathbf{a}}$ a circle of very large radius centred at the
origin and (33) is seen to hold. Finally, suppose $R>R(\mathbf{a})$ and
$\mathcal{U}=\mathcal{U}_{\epsilon}$ is as in (35). We claim that taking
$\mathsf{C}_{\mathbf{a}}\subset\mathbb{H}_{-R}$ a rectangular contour with
sides parallel to the real and imaginary axes with the left side being part of
the line $\Re(w)=-R+\epsilon$, the right side part of the line $\Re(w)=M$ and
the upper and lower sides parts of the lines $\Im(w)=\pm M$ respectively for
some large $M$ works. See Figure 8 for an illustration of this contour. For
such $\mathsf{C}_{\mathbf{a}}$ we have, by taking $\epsilon$ small enough,

$\sup_{k\in\mathbb{Z}_{+}}\frac{\sup_{u\in\mathcal{U}}|u-a_{k}|}{\inf_{w\in\mathsf{C}_{\mathbf{a}}}|w-a_{k}|}\leq\frac{\sup_{k\in\mathbb{Z}_{+}}a_{k}+\mathcal{O}(\epsilon)}{\inf_{k\in\mathbb{Z}_{+}}a_{k}+R+\mathcal{O}(\epsilon)}<\frac{\sup_{k\in\mathbb{Z}_{+}}a_{k}}{\inf_{k\in\mathbb{Z}_{+}}a_{k}+R(\mathbf{a})}=1.$

This concludes the proof.

$\Re\left(w\right)=-R$$\Re\left(w\right)=-R+\epsilon$$\Im\left(w\right)=M$$\Im\left(w\right)=-M$$\Re(w)=M$$\mathsf{C}_{\mathbf{a}}$$0$$-\epsilon$$\sup_{k}a_{k}$
Figure 8: The contour (not to scale) used in the proof of Lemma 3.3 associated
to the rectangle $\mathcal{U}=\mathcal{U}_{\epsilon}$ from (35).

∎

We have the following composition property for the $\mathsf{T}_{f}$ kernels.
In the probabilistic setting this is simply the Chapman-Kolmogorov equation.

######  Lemma 3.4.

Suppose $f,g\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ with
$R>R(\mathbf{a})$. Then, we have

$\mathsf{T}_{f}\mathsf{T}_{g}=\mathsf{T}_{fg}.$ (37)

###### Proof.

We compute, by deforming the $u$ contour from $\mathsf{C}_{\mathbf{a}}$ to a
contour $\tilde{\mathsf{C}}_{\mathbf{a}}\subset\mathcal{U}$ where
$\mathcal{U}$ is the region defined in (35), which can be done without
crossing any poles, and making use of Lemma 3.3 and then deforming back to
$\mathsf{C}_{\mathbf{a}}$,

$\displaystyle\mathsf{T}_{f}\mathsf{T}_{g}(x,y)$
$\displaystyle=-\frac{1}{a_{y}}\frac{1}{2\pi\textnormal{i}}\oint_{\tilde{\mathsf{C}}_{\mathbf{a}}}\frac{g(u)}{p_{y+1}(u)}\sum_{m=0}^{\infty}p_{m}(u)\oint_{\mathsf{C}_{\mathbf{a}}}\frac{p_{x}(w)f(w)}{p_{m+1}(w)}dwdu$
$\displaystyle=-\frac{1}{a_{y}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{p_{x}(u)}{p_{y+1}(u)}f(u)g(u)du=\mathsf{T}_{fg}(x,y),$

as desired. ∎

We have the following normalisation result for $\mathsf{T}_{f}$.

######  Lemma 3.5.

Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for some
$\epsilon>0$. For any $x\in\mathbb{Z}_{+}$ we have

$\displaystyle\sum_{y=0}^{\infty}\mathsf{T}_{f}(x,y)=f(0).$

###### Proof.

Observe that since $f$ has no poles in $\mathsf{C}_{\mathbf{a}}$,
$\mathsf{T}_{f}(x,y)=0$ for $y\leq x$. Then, we deform
$\mathsf{C}_{\mathbf{a}}$ to a contour
$\tilde{\mathsf{C}}_{\mathbf{a}}\subset{\mathbb{H}_{-\epsilon}}$ such that

$\sup_{w\in\tilde{\mathsf{C}}_{\mathbf{a}}}\sup_{k\in\mathbb{Z}_{+}}\left|\frac{a_{k}}{a_{k}-w}\right|=r<1.$
(38)

Such a contour is always possible to find. We can take a rectangular contour
with sides parallel to the real and imaginary axes with the left side being
part of the line $\Re(w)=-\epsilon^{\prime}$ for some
$0<\epsilon^{\prime}<\epsilon$, the right side part of the line $\Re(w)=M$ and
the upper and lower sides parts of the lines $\Im(w)=\pm M$ respectively for
some very large $M$. See Figure 9 for an illustration. Then, we have uniformly
for $w\in\tilde{\mathsf{C}}_{\mathbf{a}}$,

$\sum_{y>x}\frac{1}{a_{m}p_{y+1}(w)}=-\frac{1}{wp_{x+1}(w)}.$ (39)

This follows, after some relabelling, by taking the $x\to\infty$ limit in the
elementary identity, by virtue of the bound (38) above for
$w\in\tilde{\mathsf{C}}_{\mathbf{a}}$,

$\sum_{k=0}^{x}\frac{1}{a_{k}}\prod_{i=0}^{k}\frac{a_{i}}{a_{i}-w}=-\frac{1}{w}\left(1-\prod_{i=0}^{x}\frac{a_{i}}{a_{i}-w}\right).$
(40)

Thus we obtain, where we deform to the $\mathsf{C}_{\mathbf{a},0}$ contour
without crossing any poles,

$\displaystyle\sum_{y=0}^{\infty}\mathsf{T}_{f}(x,y)=\sum_{y=x}^{\infty}\mathsf{T}_{f}(x,y)$
$\displaystyle=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\left[\frac{1}{wp_{x+1}(w)}-\frac{1}{a_{x}p_{x+1}(w)}\right]p_{x}(w)f(w)dw$
$\displaystyle=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{f(w)}{w}dw.$

Evaluating the integral by picking the residue at $0$ gives the result.

$\Re\left(w\right)=-\epsilon$$\Re\left(w\right)=-\epsilon^{\prime}$$\Im\left(w\right)=M$$\Im\left(w\right)=-M$$\Re(w)=M$$\inf_{k}a_{k}$$0$$\sup_{k}a_{k}$
Figure 9: The contour (not to scale) used in the proof of Lemma 3.5.

∎

For $g:\mathbb{Z}_{+}\to\mathbb{C}$ define the forward and backward discrete
derivatives $\nabla^{+}$ and $\nabla^{-}$ by

$\nabla^{+}g(x)=g(x+1)-g(x),\ \ \nabla^{-}g(x)=g(x-1)-g(x).$

We have the following relation for $\mathsf{T}_{f}$ that we call the duality
relation (the terminology is because this relation is somewhat reminiscent to
dualities for Markov processes [76] and in particular the Siegmund duality
[129]; however it is actually not a case of a Siegmund duality or any Markov
duality for that matter).

######  Lemma 3.6.

Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for some
$\epsilon>0$. We then have

$-\frac{a_{x}}{a_{y}}\nabla_{x}^{+}\mathsf{T}_{f}\mathbf{1}_{\llbracket
0,y\rrbracket}(x)=\mathsf{T}_{f}(x,y).$ (41)

###### Proof.

Observe that, from Lemma 3.5 we have

$\nabla_{x}^{+}\mathsf{T}_{f}\mathbf{1}_{\llbracket
0,y\rrbracket}(x)=-\nabla_{x}^{+}\sum_{m>y}\mathsf{T}_{f}(x,m).$

On the other hand, using identity (40), after deforming first to the contour
$\tilde{\mathsf{C}}_{\mathbf{a}}$ as in the proof of Lemma 3.5 and then to
$\mathsf{C}_{\mathbf{a},0}$, we get that the last display is equal to

$-\nabla_{x}^{+}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{p_{x}(w)f(w)}{wp_{y+1}(w)}dw.$

Bringing the discrete derivative $\nabla^{+}_{x}$ inside the integral, using
that

$a_{x}\nabla_{x}^{+}\frac{p_{x}(w)}{w}=-p_{x}(w)$

and then deforming back to $\mathsf{C}_{\mathbf{a}}$ gives the result. ∎

Finally, we note that, for fixed $\lambda$, the function $x\mapsto
p_{x}(\lambda)$ is an eigenfunction of $\mathsf{T}_{f}$ with explicit
eigenvalue.

######  Lemma 3.7.

Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ with $R>R(\mathbf{a})$ and
$\lambda\in\mathcal{U}$ the set defined in (35). Then, the function
$h_{\lambda}(x)=p_{x}(\lambda)$ is an eigenfunction of $\mathsf{T}_{f}$ with
eigenvalue $f(\lambda)$,

$\mathsf{T}_{f}h_{\lambda}(x)=f(\lambda)h_{\lambda}(x),\ \ \forall
x\in\mathbb{Z}_{+}.$ (42)

If $f$ is entire then the above holds for all $\lambda\in\mathbb{C}$.

###### Proof.

This is an application of Lemma 3.3 with the function
$f(z)h_{z}(x)=f(z)p_{x}(z)$. ∎

In the rest of this subsection we collect some general results and comments
about the operator/matrix $\mathsf{T}_{f}$. These will not be used in any of
our probabilistic applications but they are interesting in their own right so
we present them here. Define the following spaces
$\ell_{\textnormal{exp},M}\left(\mathbb{Z}_{+}\right)$, where
$M\in(1,\infty)$, and $\ell_{\textnormal{exp}}\left(\mathbb{Z}_{+}\right)$ by,

$\displaystyle\ell_{\textnormal{exp},M}\left(\mathbb{Z}_{+}\right)$
$\displaystyle=\left\\{\left(g(k)\right)_{y\in\mathbb{Z}_{+}}\in\mathbb{C}^{\mathbb{Z}_{+}}:\exists\
C_{M}<\infty\textnormal{ such that for all }y\in\mathbb{Z}_{+},|g(y)|\leq
C_{M}M^{y}\right\\},$
$\displaystyle\ell_{\textnormal{exp}}\left(\mathbb{Z}_{+}\right)$
$\displaystyle=\left\\{\left(g(k)\right)_{y\in\mathbb{Z}_{+}}\in\mathbb{C}^{\mathbb{Z}_{+}}:\exists\
M\in(0,\infty)\textnormal{ such that
}\left(g(k)\right)_{y\in\mathbb{Z}_{+}}\in\ell_{\textnormal{exp},M}\left(\mathbb{Z}_{+}\right)\right\\}.$

######  Proposition 3.8.

Suppose $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for some
$\epsilon$. Then, there exists $r<1$ and $C_{f,r}<\infty$ such that for all
$x,y\in\mathbb{Z}_{+}$ we have

$\left|\mathsf{T}_{f}\left(x,y\right)\right|\leq C_{f,r}r^{|y-x|}.$

In particular, $\mathsf{T}_{f}$ is well-defined acting on
$\ell_{\textnormal{exp},M}\left(\mathbb{Z}_{+}\right)$ for $M<r^{-1}$. If $f$
is entire then for any $0<\delta<1$ there exists $C_{f,\delta}<\infty$ such
that for all $x,y\in\mathbb{Z}_{+}$,

$\left|\mathsf{T}_{f}\left(x,y\right)\right|\leq C_{f,\delta}\delta^{|y-x|},$

and so in this case $\mathsf{T}_{f}$ is well-defined acting on
$\ell_{\textnormal{exp}}\left(\mathbb{Z}_{+}\right)$.

###### Proof.

For the first statement we can deform the $\mathsf{C}_{\mathbf{a}}$ contour to
the rectangle from the proof of Lemma 3.5, see also Figure 9. By taking the
absolute value inside the integral we obtain the desired conclusion. When $f$
is entire we can deform $\mathsf{C}_{\mathbf{a}}$ to a very large circle from
which, after bringing the absolute value inside the integral, the result
follows. ∎

######  Remark 3.9.

We give a more abstract interpretation of $\mathsf{T}_{f}^{*}$ viewed as an
operator from a certain sequence space to itself when $f$ is entire. Define
the space $\ell_{\textnormal{hol}}$ by,

$\ell_{\textnormal{hol}}=\left\\{\left(g(k)\right)_{y\in\mathbb{Z}_{+}}\in\mathbb{C}^{\mathbb{Z}_{+}}:\forall\
0<r<1,\ \exists\ C_{r}<\infty\textnormal{ such that for all
}y\in\mathbb{Z}_{+},|g(y)|\leq C_{r}r^{y}\right\\}.$

Observe that, we have the inclusions, with $M,q\geq 1$, where $\ell^{q}$ are
the usual sequence spaces,

$\ell_{\textnormal{hol}}\left(\mathbb{Z}_{+}\right)\subset\ell^{q}\left(\mathbb{Z}_{+}\right)\subset\ell_{\textnormal{exp},M}\left(\mathbb{Z}_{+}\right)\subset\ell_{\textnormal{exp}}\left(\mathbb{Z}_{+}\right).$

Now, we consider the map $\mathcal{V}$ given by, where
$\mathsf{Hol}\left(\mathbb{C}\right)$ denotes the space of entire functions,

$\displaystyle\mathcal{V}:\ell_{\textnormal{hol}}\left(\mathbb{Z}_{+}\right)$
$\displaystyle\to\mathsf{Hol}\left(\mathbb{C}\right),$
$\displaystyle\left(g(y)\right)_{y\in\mathbb{Z}_{+}}$ $\displaystyle\mapsto
h(z)=\sum_{y\in\mathbb{Z}_{+}}g(y)p_{y}(z),$

which by the preceding results of this section it is well-defined and moreover
it is in fact a bijection with inverse given by,

$\displaystyle\mathcal{V}^{-1}:\mathsf{Hol}\left(\mathbb{C}\right)$
$\displaystyle\to\ell_{\textnormal{hol}}\left(\mathbb{Z}_{+}\right),$
$\displaystyle h$
$\displaystyle\mapsto\left(-\frac{1}{a_{y}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{h(z)}{p_{y+1}(z)}dz\right)_{y\in\mathbb{Z}_{+}}=\left(\mathsf{T}_{h}(y)\right)_{y\in\mathbb{Z}_{+}}.$

The fact that this sequence has the correct decay properties again follows by
deforming $\mathsf{C}_{\mathbf{a}}$ to a very large circle. Given
$f\in\mathsf{Hol}\left(\mathbb{C}\right)$ define the multiplication operator
$\mathsf{Mult}_{f}:\mathsf{Hol}\left(\mathbb{C}\right)\to\mathsf{Hol}\left(\mathbb{C}\right)$
by $\mathsf{Mult}_{f}\left(h\right)=fh$. Then, we can see that we have the
following representation of
$\mathsf{T}_{f}^{*}:\ell_{\textnormal{hol}}\left(\mathbb{Z}_{+}\right)\to\ell_{\textnormal{hol}}\left(\mathbb{Z}_{+}\right)$:

$\mathsf{T}_{f}^{*}=\mathcal{V}^{-1}\mathsf{Mult}_{f}\mathcal{V}.$ (43)

We now prove that the matrix $[\mathsf{T}_{f}(x,y)]_{x,y\in\mathbb{Z}_{+}}$,
with general sequence $\mathbf{a}$, is similar to the standard Toeplitz matrix
(namely with $a_{x}\equiv 1$) with symbol $f(1-z)$ using an explicit change of
basis matrix. We restrict to entire $f$ for simplicity. The result below can
also be derived from (43) but we give a direct computational proof which is
instructive.

######  Proposition 3.10.

Let $f$ be entire. Let $\mathsf{T}_{f}(x,y)$ be as in (30) with general and
fixed $\mathbf{a}$ and denote by $\tilde{\mathsf{T}}_{f}(x,y)$ the homogeneous
case (standard Toeplitz case) of (30):

$\tilde{\mathsf{T}}_{f}(x,y)=\tilde{\mathsf{T}}_{f}(y-x)=-\frac{1}{2\pi\textnormal{i}}\oint_{|1-z|=1}\frac{f(z)(1-z)^{x}}{(1-z)^{y+1}}dz=\frac{1}{2\pi\textnormal{i}}\oint_{|z|=1}f(1-z)z^{x-y-1}dz.$

Moreover, consider the matrix $\mathbf{A}(\mathbf{a})$ with entries given by,
with $k,m\in\mathbb{Z}_{+}$,

$\mathbf{A}_{km}(\mathbf{a})=(-1)^{k-m}\left(\prod_{l=0}^{k-1}a_{l}^{-1}\right)e_{k-m}\left(1-a_{0},1-a_{1},\dots,1-a_{k-1}\right)$
(44)

where $e_{l}$ is the $l$-th elementary symmetric polynomial:

$e_{l}(z_{1},z_{2},\dots,z_{N})=\sum_{1\leq j_{1}<j_{2}<\cdots<j_{l}\leq
N}z_{j_{1}}z_{j_{2}}\cdots z_{j_{l}}.$

Then, for any $x,y\in\mathbb{Z}_{+}$, we have

$\mathsf{T}_{f}\left(x,y\right)=\left[\mathbf{A}(\mathbf{a})\tilde{\mathsf{T}}_{f}\mathbf{A}^{-1}(\mathbf{a})\right](x,y).$
(45)

###### Proof.

Observe that, both $\left\\{(1-z)^{x}\right\\}_{x\in\mathbb{Z}_{+}}$ and
$\left\\{p_{x}(z)\right\\}_{x\in\mathbb{Z}_{+}}$ are bases in the ring of
polynomials $\mathbb{C}[z]$. By virtue of Vieta’s formulae and (44) we can see
that $\mathbf{A}(\mathbf{a})$ is actually the change of basis matrix, and in
particular also invertible,

$\displaystyle p_{k}(z)$
$\displaystyle=\sum_{m\in\mathbb{Z}_{+}}\mathbf{A}_{km}(\mathbf{a})(1-z)^{m},$
(46) $\displaystyle(1-z)^{k}$
$\displaystyle=\sum_{m\in\mathbb{Z}_{+}}\mathbf{A}^{-1}_{km}(\mathbf{a})p_{m}(z).$
(47)

Then, using (46), we can write

$\mathsf{T}_{f}(x,y)=-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{y}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{f(z)p_{x}(z)}{p_{y+1}(z)}dz=\sum_{m\in\mathbb{Z}_{+}}\mathbf{A}_{xm}(\mathbf{a})\left(-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{y}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{f(z)(1-z)^{m}}{p_{y+1}(z)}dz\right).$

By virtue of Lemma 3.3, since $f$ is entire, expanding $f(z)(1-z)^{m}$ as a
series in two ways

$\displaystyle f(z)(1-z)^{m}$
$\displaystyle=\sum_{k\in\mathbb{Z}_{+}}\left(-\frac{1}{2\pi\textnormal{i}}\oint_{|1-z|=1}\frac{f(z)(1-z)^{m}}{(1-z)^{k+1}}dz\right)(1-z)^{k}=\sum_{k\in\mathbb{Z}_{+}}\tilde{\mathsf{T}}_{f}(m,k)(1-z)^{k},$
$\displaystyle f(z)(1-z)^{m}$
$\displaystyle=\sum_{k\in\mathbb{Z}_{+}}\left(-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{k}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{f(z)(1-z)^{m}}{p_{k+1}(z)}dz\right)p_{k}(z),$

using (47) and comparing coefficients for $p_{y}(z)$ we get

$-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{y}}\oint_{\mathsf{C}_{\mathbf{a}}}\frac{f(z)(1-z)^{m}}{p_{y+1}(z)}dz=\sum_{l\in\mathbb{Z}_{+}}\tilde{\mathsf{T}}_{f}(m,l)\mathbf{A}_{ly}^{-1}(\mathbf{a})$

and this completes the proof. ∎

######  Remark 3.11.

Using Proposition 3.10 and assuming the corresponding result for the standard
Toeplitz setting, namely $a_{x}\equiv 1$, we can give quick alternative proofs
of Lemmas 3.4 and 3.7 for general $\mathbf{a}$ (and entire $f$). It is also
possible to prove Lemma 3.5 for general $\mathbf{a}$ in the same way if we
observe that $\sum_{m=0}^{\infty}\mathbf{A}_{xm}^{-1}(\mathbf{a})=1$, for all
$x\in\mathbb{Z}_{+}$. However, the important (later on) duality formula from
Lemma 3.6 does not seem to follow along these lines.

###  3.2 One-dimensional dynamics and Markov transition kernels

The following choices for the function $f$ in $\mathsf{T}_{f}$ in Lemmas 3.12,
3.13, 3.14 below are the basic building blocks for the models we study. They
correspond to transition probabilities for an inhomogeneous space Bernoulli
walk, geometric walk and continuous time pure-birth chain.

######  Lemma 3.12.

Let $f(z)=1-\alpha z$. We have, with $x,y\in\mathbb{Z}_{+}$,

$\mathsf{T}_{f}(x,y)=\alpha a_{x}\mathbf{1}_{y=x+1}+(1-\alpha
a_{x})\mathbf{1}_{y=x}.$ (48)

###### Proof.

Immediate evaluation of the contour integral using the residue formula. ∎

Note that, for the above expression to be positive and thus have probabilistic
meaning we need $0\leq\alpha\leq\left(\sup_{k}a_{k}\right)^{-1}$.

######  Lemma 3.13.

Let $f(z)=(1+\beta z)^{-1}$. We have, with $x,y\in\mathbb{Z}_{+}$,

$\mathsf{T}_{f}(x,y)=\frac{1}{1+\beta a_{y}}\prod_{k=x}^{y-1}\frac{\beta
a_{k}}{1+\beta a_{k}}\mathbf{1}_{y\geq x}.$ (49)

###### Proof.

This is again a direct derivation but a little less trivial so we give the
details. For the computation we assume that all the $a_{k}$’s are distinct and
then remove this restriction by continuity. After evaluating the contour
integral in terms of residues (with all the $a_{k}$’s distinct) and some
relabelling, in order to prove (49), we are required to show that

$\sum_{k=0}^{n}\prod_{j\neq k}\frac{1}{a_{j}-a_{k}}\frac{1}{1+\beta
a_{k}}=\beta^{n}\prod_{k=0}^{n}\frac{1}{1+\beta a_{k}}.$

Clearing the denominators we need to show

$\sum_{k=0}^{n}(-1)^{k}\prod_{m>l,l\neq k}(a_{m}-a_{l})\prod_{l\neq k}(1+\beta
a_{l})=\beta^{n}\prod_{m>l}(a_{m}-a_{l}).$

Now, observe that the left hand side can be written as the determinant

$\det\begin{bmatrix}\prod_{l\neq 0}(1+\beta
a_{l})&1&a_{0}&\cdots&a_{0}^{n-2}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
\prod_{l\neq n}(1+\beta a_{l})&1&a_{n}&\cdots&a_{n}^{n-2}\end{bmatrix}.$

Replace $a_{n}$ by a variable $w$ and consider the following polynomial of
degree $n-1$ in $w$

$\displaystyle w\mapsto\det\begin{bmatrix}\prod_{l\neq 0}^{n-1}(1+\beta
a_{l})(1+\beta w)&1&a_{0}&\cdots&a_{0}^{n-2}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\ \prod_{l=0}^{n-1}(1+\beta
a_{l})&1&w&\cdots&w^{n-2}\end{bmatrix}.$

It has roots at $w=a_{0},a_{1},\dots,a_{n-1}$. So it is equal to

$C_{n}^{\beta}\prod_{i=0}^{n-1}(w-a_{i})$

where $C_{n}^{\beta}$ is the coefficient of the $w^{n-1}$ term. This term is
obtained from

$\displaystyle w^{n-2}\det\begin{bmatrix}\prod_{l\neq 0}(1+\beta
a_{l})(1+\beta w)&1&a_{0}&\cdots&a_{0}^{n-2}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\ \prod_{l\neq n-1}(1+\beta a_{l})(1+\beta
w)&1&w&\cdots&w^{n-2}\end{bmatrix}$ $\displaystyle=w^{n-2}(1+\beta
w)\det\begin{bmatrix}\prod_{l\neq 0}(1+\beta
a_{l})&1&a_{0}&\cdots&a_{0}^{n-3}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
\prod_{l\neq n-1}(1+\beta a_{l})&1&a_{n-1}&\cdots&a_{n-1}^{n-3}\end{bmatrix}.$

Hence, we get

$C_{n}^{\beta}=\beta\det\begin{bmatrix}\prod_{l\neq 0}(1+\beta
a_{l})&1&a_{0}&\cdots&a_{0}^{n-3}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
\prod_{l\neq n-1}(1+\beta a_{l})&1&a_{n-1}&\cdots&a_{n-1}^{n-3}\end{bmatrix}$

and the claim follows by induction. ∎

Again, for the above expression for $\mathsf{T}_{f}(x,y)$ to be positive we
need $\beta\geq 0$.

######  Lemma 3.14.

The probability that a pure-birth chain having jump rate $a_{k}$ at location
$k\in\mathbb{Z}_{+}$ goes from $x\in\mathbb{Z}_{+}$ to $y\in\mathbb{Z}_{+}$ in
time $t\in\mathbb{R}_{+}$ is given by $\mathsf{T}_{f}(x,y)$ with
$f(z)=e^{-tz}$.

###### Proof.

This is well-known. One easily shows that $\mathsf{T}_{e^{-tz}}(x,y)$ solves
the corresponding Kolmogorov equation. A derivation can be found, for example,
in [133]. ∎

##  4 An intertwining

###  4.1 Intertwining for non-intersecting paths

In this section, in a sense that will be clearer in the sequel, we prove that
the dynamics on single levels of the interlacing arrays are consistent from
$N$ to $N+1$. We need some definitions and notation.

######  Definition 4.1.

Given $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for some
$\epsilon>0$ define the following kernel $\mathsf{P}^{(N)}_{f}$ on
$\mathbb{W}_{N}$ by

$\mathsf{P}_{f}^{(N)}(\mathbf{x},\mathbf{y})=\det\left(\mathsf{T}_{f}(x_{i},y_{j})\right)_{i,j=1}^{N}.$
(50)

######  Definition 4.2.

We define the following non-negative kernel $\Lambda_{N+1,N}$ from
$\mathbb{W}_{N+1}$ to $\mathbb{W}_{N}$ by

$\Lambda_{N+1,N}(\mathbf{y},\mathbf{x})=\prod_{i=1}^{N}\frac{1}{a_{x_{i}}}\mathbf{1}_{\mathbf{x}\prec\mathbf{y}},\
\ \mathbf{x}\in\mathbb{W}_{N},\ \mathbf{y}\in\mathbb{W}_{N+1}.$ (51)

We have the following alternative description of $\Lambda_{N+1,N}$ using the
well-known fact that $\mathbf{1}_{\mathbf{x}\prec\mathbf{y}}$ can be written
as a determinant with indicator function entries [134, 26]. To do this, we
will use the standard notational device in the setting of extending the set
$\mathbb{Z}_{+}$ by an extra symbol that we denote by $\mathsf{virt}$.

######  Lemma 4.3.

Let $N\geq 1$. Define the function
$\phi:\left(\mathbb{Z}_{+}\cup\\{\mathsf{virt}\\}\right)\times\mathbb{Z}_{+}$
by, with $y\in\mathbb{Z}_{+}$,

$\displaystyle\phi(x,y)=\begin{cases}-a_{x}^{-1}\mathbf{1}_{y>x},\ \
&x\in\mathbb{Z}_{+}\\\ 1,\ \ &x=\mathsf{virt}.\end{cases}$

Then, with $x_{N+1}=\mathsf{virt}$ we have

$\Lambda_{N+1,N}(\mathbf{y},\mathbf{x})=\det\left(\phi(x_{i},y_{j})\right)_{i,j=1}^{N+1}.$
(52)

We observe the composition property.

######  Proposition 4.4.

Let $f,g\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ with $R>R(\mathbf{a})$.
Then, we have

$\mathsf{P}_{f}^{(N)}\mathsf{P}_{g}^{(N)}=\mathsf{P}_{fg}^{(N)}.$ (53)

###### Proof.

This is a direct application of the Cauchy-Binet formula and Lemma 3.4. ∎

For certain choices of functions $f$ we obtain that $\mathsf{P}_{f}^{(N)}$ has
a probabilistic interpretation, for any $N\geq 1$, in terms of non-
intersecting paths. Of course, for $N=1$ this is already a consequence of
Lemmas 3.12, 3.13 and 3.14.

######  Proposition 4.5.

Let $f(z)$ be a (possibly infinite) product of factors of the form
$1-\alpha_{i}z,(1+\beta_{i}z)^{-1},e^{-tz}$, where
$0\leq\alpha_{i}\leq\left(\sup_{k}a_{k}\right)^{-1}$, $0\leq\beta_{i}<\infty$
and $t\geq 0$, such that $f\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$. Then,
$\mathsf{P}^{(N)}_{f}$ is non-negative.

###### Proof.

If $f(z)=e^{-tz}$, then $\mathsf{P}^{(N)}_{f}$, by virtue of Lemma 3.14,
corresponds to the Karlin-McGregor [90] transition kernel of independent pure-
birth chains killed when they intersect. If instead $f(z)$ is a finite product
of factors of the form $1-\alpha_{i}z,(1+\beta_{i}z)^{-1}$ satisfying
$0\leq\alpha_{i}\leq\left(\sup_{k}a_{k}\right)^{-1}$, $0\leq\beta_{i}<\infty$,
then $\mathsf{P}^{(N)}_{f}$, by virtue of Lemma 3.12 and Lemma 3.13 is given
by the Lindstrom-Gessel-Viennot (LGV) formula for non-intersecting walks on a
weighted directed acyclic graph [69], see Figure 10 for an illustration. It is
the transition probability of independent walks, taking either Bernoulli or
geometric steps, in discrete-time, killed when they intersect. Finally, the
general product case follows from Proposition 4.4 and a limit in the number of
factors. ∎

$x$$x+1$$x+2$$x+3$$x+4$$x$$x+1$$x+2$$x+3$$x+4$$1-\alpha a_{x}$$\alpha
a_{x}$$1-\alpha a_{x+1}$$\alpha a_{x+1}$$1-\alpha a_{x+2}$$\alpha
a_{x+2}$$1-\alpha a_{x+3}$$\alpha a_{x+3}$$1-\alpha a_{x+4}$

$x$$x+1$$x+2$$x+3$$x+4$$x$$x+1$$x+2$$x+3$$x+4$$(1+\beta
a_{x})^{-1}$$\frac{\beta a_{x}}{1+\beta a_{x}}$$(1+\beta
a_{x+1})^{-1}$$\frac{\beta a_{x+1}}{1+\beta a_{x+1}}$$(1+\beta
a_{x+2})^{-1}$$\frac{\beta a_{x+2}}{1+\beta a_{x+2}}$$(1+\beta
a_{x+3})^{-1}$$\frac{\beta a_{x+3}}{1+\beta a_{x+3}}$$(1+\beta a_{x+4})^{-1}$

Figure 10: An illustration of part of the LGV graphs corresponding to a single
Bernoulli step with parameter $\alpha$, namely corresponding to the function
$f(z)=1-\alpha z$ and a single geometric step with parameter $\beta$,
corresponding to the function $f(z)=(1+\beta z)^{-1}$. We can join such
elementary graphs one after the other.

The following intertwining relation between the transition kernels (although
note that at this stage positivity is not yet required) $\mathsf{P}_{f}^{(N)}$
and $\mathsf{P}_{f}^{(N+1)}$ is at the heart of many of our results. We give a
direct proof but different arguments for a proof will also be presented in the
sequel.

######  Theorem 4.6.

Let $N\geq 1$. Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for
some $\epsilon>0$, with $f(0)=1$. Then, we have the intertwining

$\mathsf{P}^{(N+1)}_{f}\Lambda_{N+1,N}=\Lambda_{N+1,N}\mathsf{P}^{(N)}_{f}.$
(54)

###### Proof.

We first compute the left hand side using Cauchy-Binet formula to obtain,
where we use Lemma 3.5 and the fact that $f(0)=1$,

$\displaystyle\mathsf{P}^{(N+1)}_{f}\Lambda_{N+1,N}(\mathbf{y},\mathbf{x})$
$\displaystyle=\sum_{\mathbf{z}\in\mathbb{W}_{N+1}}\det\left(\mathsf{T}_{f}(y_{i},z_{j})\right)_{i,j=1}^{N+1}\det\left(\phi(x_{i},z_{j})\right)_{i,j=1}^{N+1}$
$\displaystyle=\det\begin{bmatrix}\mathsf{T}_{f}\phi^{*}(y_{1},x_{1})&\cdots&\mathsf{T}_{f}\phi^{*}(y_{1},x_{N})&1\\\
\vdots&\vdots&\vdots&\vdots\\\
\mathsf{T}_{f}\phi^{*}(y_{N+1},x_{1})&\cdots&\mathsf{T}_{f}\phi^{*}(y_{N+1},x_{N})&1\end{bmatrix}.$

Here, and below, $\phi^{*}(x,y)=\phi(y,x)$ and $\phi^{*}\mathsf{T}_{f}$ and
$\mathsf{T}_{f}\phi^{*}$ denotes convolution (not multiplication). We now
compute the right hand side, by expanding along the last column and using
Cauchy-Binet again

$\displaystyle\Lambda_{N+1,N}\mathsf{P}^{(N)}_{f}(\mathbf{y},\mathbf{x})$
$\displaystyle=\sum_{\mathbf{z}\in\mathbb{W}_{N}}\det\left(\phi(z_{i},y_{j})\right)_{i,j=1}^{N+1}\det\left(\mathsf{T}_{f}(z_{i},x_{j})\right)_{i,j=1}^{N}$
$\displaystyle=\sum_{l=0}^{N+1}(-1)^{N+1-l}\det\left(\phi^{*}\mathsf{T}_{f}(y_{i},x_{j})\right)_{i=1,\dots,N+1,i\neq
l;j=1,\dots,N}$
$\displaystyle=\det\begin{bmatrix}\phi^{*}\mathsf{T}_{f}(y_{1},x_{1})&\cdots&\phi^{*}\mathsf{T}_{f}(y_{1},x_{N})&1\\\
\vdots&\vdots&\vdots&\vdots\\\
\phi^{*}\mathsf{T}_{f}(y_{N+1},x_{1})&\cdots&\phi^{*}\mathsf{T}_{f}(y_{N+1},x_{N})&1\end{bmatrix}.$

We note that from Lemma 3.6 we have

$a_{y}^{-1}\mathsf{T}_{f}(y,x)=a_{x}^{-1}\nabla_{y}^{+}\sum_{z>x}\mathsf{T}_{f}(y,z)$

and thus by summing over $y$ we obtain

$\displaystyle\phi^{*}\mathsf{T}_{f}(y,x)=\mathsf{T}_{f}\phi^{*}(y,x)-\mathsf{T}_{f}\phi^{*}(0,x).$
(55)

Now, using the identity above and column operations we obtain the desired
equality. ∎

We now go on to obtain an analogous intertwining for the normalised versions
of $\mathsf{P}_{f}^{(N)}$ and $\Lambda_{N+1,N}$ to be defined shortly. We need
the following definition.

######  Definition 4.7.

For any $N\geq 1$ we define the following strictly positive function
$\mathfrak{h}_{N}(\mathbf{x})=\mathfrak{h}_{N}(\mathbf{x};\mathbf{a})$ for
$\mathbf{x}\in\mathbb{W}_{N}$ recursively by $\mathfrak{h}_{1}(x)=1$ and

$\mathfrak{h}_{N+1}(\mathbf{x})=\Lambda_{N+1,N}\mathfrak{h}_{N}(\mathbf{x}).$

######  Remark 4.8.

Observe that, by comparing the definitions of the two functions we get that
$\mathfrak{h}_{N}(\mathbf{x})=\textnormal{dim}_{N}(\mathbf{x})$ from Section
2.5 on the inhomogeneous Gelfand-Tsetlin graph.

From Theorem 4.6 we immediately obtain the following.

######  Proposition 4.9.

Let $N\geq 1$. Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for
some $\epsilon>0$, with $f(0)=1$. Then
$\mathsf{P}_{f}^{(N)}\mathfrak{h}_{N}=\mathfrak{h}_{N}$.

Thus, by a Doob $h$-transform, see [61, 125], by $\mathfrak{h}_{N}$ we can
define the following kernels $\mathfrak{L}_{N+1,N}$ and
$\mathfrak{P}_{f}^{(N)}$ from $\mathbb{W}_{N+1}$ to $\mathbb{W}_{N}$ and from
$\mathbb{W}_{N}$ to itself respectively. By construction,
$\mathfrak{L}_{N+1,N}$ is Markov. Similarly, as long as $f$ is so that
$\mathsf{P}_{f}^{(N)}$ is non-negative, then $\mathfrak{P}_{f}^{(N)}$ is also
Markov.

######  Definition 4.10.

Let $N\geq 1$. Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for
some $\epsilon>0$, with $f(0)=1$. Define

$\displaystyle\mathfrak{P}_{f}^{(N)}(\mathbf{x},\mathbf{y})$
$\displaystyle=\frac{\mathfrak{h}_{N}(\mathbf{y})}{\mathfrak{h}_{N}(\mathbf{x})}\mathsf{P}_{f}^{(N)}(\mathbf{x},\mathbf{y}),$
$\displaystyle\mathbf{x},\mathbf{y}\in\mathbb{W}_{N},$ (56)
$\displaystyle\mathfrak{L}_{N+1,N}(\mathbf{y},\mathbf{x})$
$\displaystyle=\frac{\mathfrak{h}_{N}(\mathbf{x})}{\mathfrak{h}_{N+1}(\mathbf{y})}\Lambda_{N+1,N}(\mathbf{y},\mathbf{x}),$
$\displaystyle\mathbf{x}\in\mathbb{W}_{N},\mathbf{y}\in\mathbb{W}_{N+1}.$ (57)

Here, we are slightly abusing notation as we have already defined
$\mathfrak{P}^{(N)}_{f}(\mathbf{x},\mathbf{y})$ for special choices of
$f=f_{s,t}$ in (12). By virtue of Lemma 7.10 the expressions (12) and (56) are
one and the same. Observe that, the following theorem is then a direct
consequence of Theorem 4.6 and the above definitions.

######  Theorem 4.11.

Let $N\geq 1$. Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for
some $\epsilon>0$, with $f(0)=1$. Then,

$\mathfrak{P}^{(N+1)}_{f}\mathfrak{L}_{N+1,N}=\mathfrak{L}_{N+1,N}\mathfrak{P}^{(N)}_{f}.$
(58)

###  4.2 Intertwinings for determinantal kernels

In this subsection we introduce a slightly more general framework for
intertwinings involving determinants which may be of independent interest. As
we explain below, from this setup it would be possible to anticipate Theorem
4.6, although we do not give the details required for a complete independent
proof.

Fix $N\geq 1$. We are given the following data: for $1\leq j\leq N+1$,
parameters $\lambda_{j}\in\mathbb{C}$, $\lambda_{j}\neq\lambda_{k}$ and
corresponding functions $h_{\lambda_{j}}:\mathbb{Z}_{+}\to\mathbb{C}$, for
$i=1,2,3,4$, kernels
$A^{(i)}:\mathbb{Z}_{+}\times(\mathbb{Z}_{+}\cup\\{\mathsf{virt}\\})\to\mathbb{C}$,
such that, with the series assumed to be converging absolutely,

$A^{(i)}h_{\lambda_{j}}(x)=\sum_{y=0}^{\infty}A^{(i)}(x,y)h_{\lambda_{j}}(y)=c_{\lambda_{j}}^{(i)}h_{\lambda_{j}}(x)$
(59)

and moreover $A^{(i)}(x,\mathsf{virt})=h_{\lambda_{N+1}}(x)$. Finally, assume
that for any $\mathbf{x}\in\mathbb{W}_{n}$, with $1\leq n\leq N$,

$\det\left(h_{\lambda_{j}}(x_{i})\right)_{i,j=1}^{n}\neq 0.$

######  Definition 4.12.

Given the data in the above paragraph, define the kernels, for $i=1,2,3,4,$

$\displaystyle A_{n}^{(i)}(\mathbf{x},\mathbf{y})$
$\displaystyle=\prod_{j-1}^{n}\frac{1}{c_{\lambda_{j}}^{(i)}}\frac{\det\left(h_{\lambda_{j}}(y_{i})\right)_{i,j=1}^{n}}{\det\left(h_{\lambda_{j}}(x_{i})\right)_{i,j=1}^{n}}\det\left(A^{(i)}(x_{i},y_{j})\right)_{i,j=1}^{n},\
\ \mathbf{x},\mathbf{y}\in\mathbb{W}_{n},\ \ n=1,\dots,N+1,$ $\displaystyle
A_{N+1,N}^{(i)}\left(\mathbf{y},\mathbf{x}\right)$
$\displaystyle=\prod_{j=1}^{N-1}\frac{1}{c_{\lambda_{j}}^{(i)}}\frac{\det\left(h_{\lambda_{j}}(y_{i})\right)_{i,j=1}^{N}}{\det\left(h_{\lambda_{j}}(x_{i})\right)_{i,j=1}^{N+1}}\det\left(A^{(i)}(y_{i},x_{j})\right)_{i,j=1}^{N+1},\
\ \mathbf{x}\in\mathbb{W}_{N},\mathbf{y}\in\mathbb{W}_{N+1}.$

Then, we have the following proposition.

######  Proposition 4.13.

Assume the above, then we have, for $i=1,2,3,4,$

$\displaystyle\sum_{\mathbf{y}\in\mathbb{W}_{n}}A_{n}^{(i)}(\mathbf{x},\mathbf{y})=1,\
\ \sum_{\mathbf{x}\in\mathbb{W}_{N}}A^{(i)}_{N+1,N}(\mathbf{y},\mathbf{x})=1,$
(60)

where we write $x_{N+1}=\mathsf{virt}$. Moreover, if we assume, for
$x,y\in\mathbb{Z}_{+}$,

$A^{(1)}A^{(2)}(x,y)=A^{(3)}A^{(4)}(x,y)+\tilde{h}(y)h_{\lambda_{N+1}}(x)$
(61)

for some (possibly identically zero) function
$\tilde{h}:\mathbb{Z}_{+}\to\mathbb{C}$ we have the intertwining

$A_{N+1}^{(1)}A^{(2)}_{N+1,N}=A^{(3)}_{N+1,N}A^{(4)}_{N}.$ (62)

###### Proof.

In the case of $A_{n}^{(i)}$, the normalization (60) is a direct consequence
of the Cauchy-Binet formula and the eigenfunction relation (59). In the case
of $A_{N,N-1}^{(i)}$ we expand along the last column, use Cauchy-Binet and the
eigenfunction relation (59), from which (60) follows.

For the intertwining, we compute the left hand side first, with
$x_{N+1}=\mathsf{virt}$ using the Cauchy-Binet formula

$\displaystyle\left[A_{N+1}^{(1)}A_{N+1,N}^{(2)}\right](\mathbf{y},\mathbf{x})=\prod_{j=1}^{N+1}\frac{1}{c_{\lambda_{j}}^{(1)}}\prod_{j=1}^{N}\frac{1}{c_{\lambda_{j}}^{(2)}}\frac{\det\left(h_{\lambda_{j}}(x_{i})\right)_{i,j=1}^{N}}{\det\left(h_{\lambda_{j}}(y_{i})\right)_{i,j=1}^{N+1}}\det\left(A^{(1)}A^{(2)}(y_{i},x_{j})\right)_{i,j=1}^{N+1}.$

Now, we work on the right hand side, we expand along the last column of the
size-$(N+1)$ determinant, and use Cauchy-Binet to obtain

$\displaystyle\left[A_{N+1,N}^{(3)}A_{N}^{(4)}\right](\mathbf{y},\mathbf{x})=\prod_{j=1}^{N+1}\frac{1}{c_{\lambda_{j}}^{(3)}}\prod_{j=1}^{N}\frac{1}{c_{\lambda_{j}}^{(4)}}\frac{\det\left(h_{\lambda_{j}}(x_{i})\right)_{i,j=1}^{N}}{\det\left(h_{\lambda_{j}}(y_{i})\right)_{i,j=1}^{N+1}}\det\left(A^{(3)}A^{(4)}(y_{i},x_{j})\right)_{i,j=1}^{N+1}.$

Now, observe that the last column of the determinants on the left and right
hand sides involving the kernels $A^{(1)}A^{(2)}$ and $A^{(3)}A^{(4)}$ has
entries in the $i$-th row given by
$c_{\lambda_{N+1}}^{(2)}h_{\lambda_{N+1}}(x_{i})$ and
$c_{\lambda_{N+1}}^{(3)}h_{\lambda_{N+1}}(x_{i})$ respectively. Using relation
(61) and column operations we then obtain

$\det\left(A^{(1)}A^{(2)}(y_{i},x_{j})\right)_{i,j=1}^{N+1}=\frac{c_{\lambda_{N+1}}^{(1)}}{c_{\lambda_{N+1}}^{(3)}}\det\left(A^{(3)}A^{(4)}(y_{i},x_{j})\right)_{i,j=1}^{N+1}.$

Thus, we get

$\left[A_{N+1}^{(1)}A_{N+1,N}^{(2)}\right](\mathbf{y},\mathbf{x})=\prod_{j=1}^{N}\frac{c_{\lambda_{j}}^{(3)}c_{\lambda_{j}}^{(4)}}{c_{\lambda_{j}}^{(1)}c_{\lambda_{j}}^{(2)}}\left[A_{N+1,N}^{(3)}A_{N}^{(4)}\right](\mathbf{y},\mathbf{x}).$

Since, as we have proven earlier, both sides sum over
$\mathbf{x}\in\mathbb{W}_{N}$ to $1$ we obtain that the ratio of constants
must be equal to $1$ (for example, when $\tilde{h}\equiv 0$ then this is
trivial to see from the eigenfunction relation) and this gives the desired
intertwining. ∎

Of course, for the above to have any probabilistic meaning one needs to
address the non-trivial question of positivity of the various kernels.

We now explain, without giving the rigorous details, how to get Theorem 4.6
from this framework. Let $f\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$, with
$R>R(\mathbf{a})$ and $f(0)=1$, be arbitrary. Take
$A^{(1)}=A^{(4)}=\mathsf{T}_{f}$. Let
$g_{\lambda_{N+1}}(w)=(w-\lambda_{N+1})^{-1}$. Take

$\displaystyle
A^{(2)}(x,y)=A^{(3)}(x,y)=-\frac{1}{a_{y}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{p_{x}(w)g_{\lambda_{N+1}}(w)}{p_{y+1}(w)}dw.$

Note that,
$g_{\lambda_{N+1}}\in\mathsf{Hol}\left(\mathbb{H}_{\lambda_{N+1}}\right)$ and
observe that $A^{(2)}=A^{(3)}=\mathsf{T}_{g_{\lambda_{N+1}}}$ for
$\lambda_{N+1}$ large and negative (since then the pole at $\lambda_{N+1}$ is
not contained in the contour $\mathsf{C}_{\mathbf{a},0}$). Take, as
$h_{\lambda}(x)=p_{x}(\lambda)$ which, from Lemma 3.7, is an eigenfunction of
the $A^{(i)}$ with eigenvalues
$c^{(1)}_{\lambda}=c^{(4)}_{\lambda}=f(\lambda)$ and
$c^{(2)}_{\lambda}=c_{\lambda}^{(3)}=(\lambda-\lambda_{N+1})^{-1}$. Finally,
from Lemma 3.4 all the $A^{(i)}$ operators commute when
$\lambda_{N+1}<-R(\mathbf{a})$. We now take
$\lambda_{1},\dots,\lambda_{N+1}\to 0$. First, note that
$h_{\lambda_{j}}(x)\to 1$ and as $\lambda_{N+1}\to 0$,
$A^{(2)}(x,y)=A^{(3)}(x,y)\to\phi^{*}(x,y)=\phi(y,x)$ using the representation
of $\phi$ in Lemma 4.3. Moreover, although the operators $A^{(i)}$ no longer
commute if $\lambda_{N+1}$ is small (the reason is that we have an extra
residue coming from the pole of $g_{\lambda_{N+1}}$ at $\lambda_{N+1}$ in
$\mathsf{C}_{\mathbf{a},0}$) we still get a relation of the form (61) which
essentially becomes (55) in the limit. Furthermore, ratios of functions
$\det\left(h_{\lambda_{j}}(x_{i})\right)_{i,j=1}^{n}$ combined with the
eigevalues $c^{(i)}_{\lambda}$ converge to ratios of functions
$\mathfrak{h}_{n}$, by virtue of the representation of the $\mathfrak{h}_{n}$
given in Lemma 7.10. Putting everything together, and after some manipulations
we formally obtain the intertwining in Theorem 4.6.

##  5 Some couplings

In this section we introduce certain couplings between the intertwined
semigroups from Section 4 that give rise to the push-block type dynamics
introduced in Section 2.2. These constructions have their origin, in some
sense, in the most basic coalescing random walk model that we discuss next.
The following space of two-level interlaced configurations will make its
appearance often:

$\mathbb{W}_{N,N+1}=\left\\{(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N}\times\mathbb{W}_{N+1}:\mathbf{x}\prec\mathbf{y}\right\\}.$
(63)

###  5.1 Couplings from coalescing random walks

Suppose we are given a sequence of functions
$\left(f_{t,t+1}(z)\right)_{t=0}^{\infty}$, with either
$f_{t,t+1}(z)=1-\alpha_{t}z$ or $f_{t,t+1}(z)=(1+\beta_{t}z)^{-1}$, where for
all $t\geq 0$, the parameters satisfy
$0\leq\alpha_{t}\leq(\sup_{x\in\mathbb{Z}_{+}}a_{x})^{-1}$ and
$0\leq\beta_{t}<\infty$. In particular, $\mathsf{T}_{f_{t,t+1}}$ corresponds
to either an inhomogeneous Bernoulli or geometric jump. For $s\leq t$, define
the function $f_{s,t}$ (with $f_{t,t}=1$) by

$f_{s,t}(z)=f_{s,s+1}(z)\cdots f_{t-1,t}(z).$

At each space-time point $(x,t)\in\mathbb{Z}_{+}^{2}$ we put an independent
Bernoulli random variable $\mathsf{U}_{x,t}$ which can be one of two kinds
depending on $f_{t,t+1}$:

$\displaystyle\mathsf{U}_{x,t}=\begin{cases}\mathsf{U}_{x,t}^{(\alpha)},\ \
&\textnormal{ if }f_{t,t+1}(z)=1-\alpha_{t}z,\\\ \mathsf{U}_{x,t}^{(\beta)},\
\ &\textnormal{ if }f_{t,t+1}(z)=(1+\beta_{t}z)^{-1},\end{cases}$

where the Bernoulli variables
$\mathsf{U}^{(\alpha)}_{x,t},\mathsf{U}^{(\beta)}_{x,t}$ satisfy

$\displaystyle\mathbb{P}\left(\mathsf{U}_{x,t}^{(\alpha)}=1\right)$
$\displaystyle=1-\mathbb{P}\left(\mathsf{U}_{x,t}^{(\alpha)}=0\right)=\alpha_{t}a_{x},$
$\displaystyle\mathbb{P}\left(\mathsf{U}_{x,t}^{(\beta)}=1\right)$
$\displaystyle=1-\mathbb{P}\left(\mathsf{U}_{x,t}^{(\beta)}=0\right)=\frac{\beta_{t}a_{x}}{1+\beta_{t}a_{x}}.$

Then, we do the following:

  * •

If $\mathsf{U}_{x,t}^{(\alpha)}=1$ we put an arrow going from $(x,t)$ to
$(x+1,t+1)$.

  * •

If $\mathsf{U}_{x,t}^{(\alpha)}=0$ we put an arrow going from $(x,t)$ to
$(x,t+1)$.

  * •

If $\mathsf{U}_{x,t}^{(\beta)}=1$ we put an arrow going from $(x,t)$ to
$(x+1,t)$.

  * •

If $\mathsf{U}_{x,t}^{(\beta)}=0$ we put an arrow going from $(x,t)$ to
$(x,t+1)$.

Note that, every space-time point has exactly one outgoing arrow but possibly
multiple incoming arrows. See Figure 11 for an illustration.

######  Definition 5.1.

For any times $s\leq t$, we define the random map
$\mathcal{Z}_{s,t}:\mathbb{Z}_{+}\to\mathbb{Z}_{+}$ as follows:
$\mathcal{Z}_{s,t}(x)$ is the location at time $t$ of the (deterministic)
motion starting from $x\in\mathbb{Z}_{+}$ at time $s$ which follows the random
arrows in the above construction.

See Figure 11 for an illustration of $\mathcal{Z}_{s,t}$.

$x$$x+1$$x+2$$x+3$$t=0$$t=1$$t=2$$t=3$ Figure 11: An illustration of the
construction of $\mathcal{Z}_{s,t}$. The arrows are random constructed from
the independent random variables $\mathsf{U}_{x,t}$ at each space-time point
$(x,t)$ as explained in the text. The first step from time $0$ to time $1$ is
a Bernoulli step, the second step is a geometric step and the third step is
again a Bernoulli step. The random walks starting at $x$ and $x+1$ move
independently for one step and then they coalesce and move together. For
example, $\mathcal{Z}_{0,1}(x)=\mathcal{Z}_{0,1}(x+1)=x+1$,
$\mathcal{Z}_{0,2}(x)=\mathcal{Z}_{0,2}(x+1)=x+1$,
$\mathcal{Z}_{0,3}(x)=\mathcal{Z}_{0,3}(x+1)=x+2$.

Observe that, by construction, almost surely for all $t_{1}\leq t_{2}\leq
t_{3}$ we have

$\mathcal{Z}_{t_{2},t_{3}}\left(\mathcal{Z}_{t_{1},t_{2}}(x)\right)=\mathcal{Z}_{t_{1},t_{3}}(x),\
\ \forall x\in\mathbb{Z}_{+}.$

We can thus think of $\left(\mathcal{Z}_{s,t}\right)_{s\leq t}$ as a
stochastic flow of maps. Fixing the starting time $s$ and locations
$x_{1},\dots,x_{N}$ the following process is called the $N$-point motion of
the flow

$t\mapsto\left(\mathcal{Z}_{s,t}(x_{1}),\dots,\mathcal{Z}_{s,t}(x_{N})\right),\
t\geq s.$

By its very construction this motion is distributed as $N$ random walks
starting from locations $x_{1},\dots,x_{N}$ at time $s$, moving independently
with Bernoulli or geometric jumps (depending on the corresponding functions
$f_{u,u+1}$ for $u\geq s$) until any two walks meet at a space-time point
$(x,t)$ from which time onwards they move together (as the same random
variables are used to drive both their evolutions).

We can compute the distribution of this process explicitly.

######  Proposition 5.2.

Let $N\geq 1$ and $s\leq t$. Let $x_{1}\leq\cdots\leq x_{N}$ and
$y_{1}\leq\cdots\leq y_{N}$. Then,

$\mathbb{P}\left(\mathcal{Z}_{s,t}(x_{1})\leq
y_{1},\dots,\mathcal{Z}_{s,t}(x_{N})\leq
y_{N}\right)=\det\left(\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}\rrbracket}(x_{i})-\mathbf{1}_{i<j}\right)_{i,j=1}^{N}.$ (64)

###### Proof.

We first assume $\mathbf{x}=(x_{1},\dots,x_{N})\in\mathbb{W}_{N}$ and then
remove this restriction below. Observe that by summing over the LGV-formula we
obtain

$\mathbb{P}\left(\mathcal{Z}_{s,t}(x_{1})\leq
y_{1}<\mathcal{Z}_{s,t}(x_{2})\leq y_{2}<\cdots<\mathcal{Z}_{s,t}(x_{N})\leq
y_{N}\right)=\det\left(\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}\rrbracket}(x_{i})\right)_{i,j=1}^{N}.$

Observe that this formula would have simply been zero and thus not immediately
useful if the $x_{i}$ were not distinct. Then, by writing the indicator

$\mathbf{1}_{\mathcal{Z}_{s,t}(x_{1})\leq
y_{1},\dots,\mathcal{Z}_{s,t}(x_{N})\leq y_{N}}$

in terms of indicators

$\mathbf{1}_{\mathcal{Z}_{s,t}(x_{i_{1}})\leq
y_{j_{1}}<\mathcal{Z}_{s,t}(x_{i_{2}})\leq
y_{i_{2}}<\cdots<\mathcal{Z}_{s,t}(x_{i_{k}})\leq y_{j_{k}}}$

for increasing sequences $i_{1},\dots,i_{k}$ and $j_{1},\dots,j_{k}$ as
explained in detail in Proposition 9 of [134] we obtain the desired formula
for $\mathbf{x}\in\mathbb{W}_{N}$. We now explain how to remove the
restriction that the coordinates of $\mathbf{x}$ are distinct. Suppose we have
$m$ distinct values

$\displaystyle
x_{1}=\cdots=x_{i_{1}-1},x_{i_{1}}=\cdots=x_{i_{2}-1},\dots,x_{i_{m-1}}=\cdots=x_{N}.$

By the coalescence property we have, with $i_{0}=1$,

$\mathbb{P}\left(\mathcal{Z}_{s,t}(x_{i})\leq y_{i},\ \ \textnormal{ for
}1\leq i\leq N\right)=\mathbb{P}\left(\mathcal{Z}_{s,t}(x_{i_{j-1}})\leq
y_{i_{j-1}},\ \ \textnormal{ for }1\leq j\leq m\right).$

Thus, we need to show that the size $N$ determinant on the right hand side of
(64) boils down to the corresponding size $m$ determinant. Using the special
structure of the determinant this is easy to prove. For $j=0,\dots,m-1$
subtract row $i_{j}$ from rows $i_{j}+1,\dots,i_{j+1}-1$. Then, by swapping
consecutive rows bring rows $i_{0},i_{1},\dots,i_{m-1}$ to positions
$1,\dots,m$ and similarly by swapping consecutive columns bring columns
$i_{0},i_{1},\dots,i_{m-1}$ to positions $1,\dots,m$. We then get the
determinant of a block triangular matrix whose top-left corner is the $m\times
m$ matrix of interest and whose bottom-right corner is the identity matrix.
The conclusion follows. ∎

The following definition is central in our argument.

######  Definition 5.3.

Consider the following kernel, defined as the $(2N+1)\times(2N+1)$
determinant, with $s\leq t$ and
$(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{W}_{N,N+1}$,

$\displaystyle\mathcal{Q}_{s,t}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\det\begin{bmatrix}\mathcal{A}_{s,t}(\mathbf{y},\mathbf{y}^{\prime})&\mathcal{B}_{s,t}(\mathbf{y},\mathbf{x}^{\prime})\\\
\mathcal{C}_{s,t}(\mathbf{x},\mathbf{y}^{\prime})&\mathcal{D}_{s,t}(\mathbf{c},\mathbf{x}^{\prime})\end{bmatrix},$

where the entries of the matrices
$\mathcal{A}_{s,t}(\mathbf{y},\mathbf{y}^{\prime})$,
$\mathcal{B}_{s,t}(\mathbf{y},\mathbf{x}^{\prime})$,
$\mathcal{C}_{s,t}(\mathbf{x},\mathbf{y}^{\prime})$ and
$\mathcal{D}_{s,t}(\mathbf{x},\mathbf{x}^{\prime})$ of sizes
$(N+1)\times(N+1)$, $(N+1)\times N$, $N\times(N+1)$ and $N\times N$
respectively are given by

$\displaystyle\mathcal{A}_{s,t}(\mathbf{y},\mathbf{y}^{\prime})_{ij}$
$\displaystyle=-\nabla_{y_{j}^{\prime}}^{-}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}^{\prime}\rrbracket}(y_{i})=\mathsf{T}_{f_{s,t}}(y_{i},y_{j}),$
$\displaystyle\mathcal{B}_{s,t}(\mathbf{y},\mathbf{x}^{\prime})_{ij}$
$\displaystyle=\frac{1}{a_{x_{j}^{\prime}}}\left(\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}^{\prime}\rrbracket}(y_{i})-\mathbf{1}_{j\geq i}\right),$
$\displaystyle\mathcal{C}_{s,t}(\mathbf{x},\mathbf{y}^{\prime})_{ij}$
$\displaystyle=a_{x_{i}}\nabla_{x_{i}}^{+}\nabla_{y_{j}^{\prime}}^{-}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}^{\prime}\rrbracket}(x_{i}),$
$\displaystyle\mathcal{D}_{s,t}(\mathbf{x},\mathbf{x}^{\prime})_{ij}$
$\displaystyle=-\frac{a_{x_{i}}}{a_{x_{j}^{\prime}}}\nabla_{x_{i}}^{+}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(x_{i})=\mathsf{T}_{f_{s,t}}(x_{i},x_{j}^{\prime}).$

######  Remark 5.4.

One could replace the function $f_{s,t}$ in the definition above by a general
$f\in\mathsf{Hol}\left(\mathbb{H}_{-\epsilon}\right)$ for some $\epsilon>0$ to
define a kernel $\mathcal{Q}_{f}^{N,N+1}$. Then, from an analytic standpoint
the results in Propositions 5.7 and 5.9 extend to this setting as well. What
is not clear however is whether they have any probabilistic meaning for more
general $f$.

######  Proposition 5.5.

We have, where
$(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{W}_{N,N+1}$,

$\displaystyle\mathcal{Q}_{s,t}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\prod_{i=1}^{N}\frac{a_{x_{i}}}{a_{x_{i}^{\prime}}}(-\nabla_{x_{i}}^{+})\prod_{i=1}^{N+1}(-\nabla_{y_{i}^{\prime}}^{-})\mathbb{P}\left(\mathcal{Z}_{s,t}(y_{i})\leq
y_{i}^{\prime},\mathcal{Z}_{s,t}(x_{j})\leq x_{j}^{\prime},\textnormal{ for
all }i,j\right).$

###### Proof.

This follows by virtue of Proposition 5.2 and the explicit Definition 5.3 of
$\mathcal{Q}_{s,t}^{N,N+1}$ and careful inspection of the two formulae. ∎

######  Proposition 5.6.

Let $s\leq t$. Viewed as a kernel from $\mathbb{W}_{N,N+1}$ to itself,
$\mathcal{Q}^{N,N+1}_{s,t}$ is sub-Markov:

$\displaystyle\mathcal{Q}_{s,t}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]$
$\displaystyle\geq 0,\ \
\forall(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{W}_{N,N+1},$
$\displaystyle\sum_{(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{w}_{N,N+1}}\mathcal{Q}_{s,t}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]$
$\displaystyle\leq 1,\ \ \forall(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1}.$

###### Proof.

Proving non-negativity directly from the determinant formula is actually
tricky. We use the connection to the flow $\mathcal{Z}_{s,t}$ via Proposition
5.5 instead. Observe that, by the very construction of the coalescing flow
$\mathcal{Z}_{s,t}$ the events

$\displaystyle\left\\{\mathcal{Z}_{s,t}(x_{i})\leq
x_{i}^{\prime},\mathcal{Z}_{s,t}(y_{i})\leq y_{j}^{\prime},\ \textnormal{ for
all }i,j\right\\}$

are increasing as the variables $x_{i}$ decrease and the variables
$y_{j}^{\prime}$ increase. The claim then follows from the representation
given in Proposition 5.5 of $\mathcal{Q}^{N,N+1}_{s,t}$ in terms of the
probability of such events.

We now move to the second item of the statement. We show that

$\sum_{\mathbf{y}^{\prime}:(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{W}_{N,N+1}}\mathcal{Q}_{s,t}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\det\left(\mathcal{D}_{s,t}(x_{i},x_{j})\right)_{i,j=1}^{N}=\mathsf{P}^{(N)}_{f_{s,t}}\left(\mathbf{x},\mathbf{x}^{\prime}\right).$

Note that, this is nothing else but the LGV formula [69] for non-intersecting
paths discussed in Proposition 4.5 so it is sub-Markov and hence the sum over
$\mathbb{W}_{N,N+1}$ of $\mathcal{Q}^{N,N+1}_{s,t}$ is indeed less than $1$.
The above claim can be seen in two ways (which are essentially equivalent).
First, by direct computation. Use multinearity to bring the sum over
$\mathbf{y}^{\prime}$ inside the determinant definition of
$\mathcal{Q}^{N,N+1}_{s,t}$ and then use the relations (we use the convention
$x_{N+1}^{\prime}\equiv\infty$ and $x_{0}^{\prime}\equiv-1$)

$\displaystyle\sum_{y_{j}^{\prime}=x_{j-1}^{\prime}+1}^{x_{j}^{\prime}}\mathcal{A}_{s,t}(\mathbf{y},\mathbf{y}^{\prime})_{ij}$
$\displaystyle=\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(y_{i})-\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j-1}^{\prime}\rrbracket}(y_{i}),$
$\displaystyle\sum_{y_{j}^{\prime}=x_{j-1}^{\prime}+1}^{x_{j}^{\prime}}\mathcal{C}_{s,t}(\mathbf{x},\mathbf{y}^{\prime})_{ij}$
$\displaystyle=-a_{x_{i}}\nabla_{x_{i}}^{+}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(x_{i})+a_{x_{i}}\nabla_{x_{i}}^{+}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j-1}^{\prime}\rrbracket}(x_{i}).$

By some simple row-column operations the claim follows. Second, using the flow
representation in Proposition 5.5 take the sum over $\mathbf{y}^{\prime}$ to
get rid of the discrete derivatives in $y_{j}^{\prime}$, then use the formula
in Proposition 5.2 and conclude by virtue of Lemma 3.6. ∎

A more involved argument, that introduces an inverse $\mathcal{Z}_{s,t}^{-1}$
to the flow $\mathcal{Z}_{s,t}$ can be used to establish the semigroup
property
$\mathcal{Q}^{N,N+1}_{t_{1},t_{2}}\mathcal{Q}^{N,N+1}_{t_{2},t_{3}}=\mathcal{Q}^{N,N+1}_{t_{1},t_{3}}$,
for $t_{1}\leq t_{2}\leq t_{3}$. The actual dynamics of the induced process on
$\mathbb{W}_{N,N+1}$, as far as we can tell, cannot be obtained directly using
this connection and we need to take a different approach in the next
subsections. We believe that $\mathcal{Q}_{s,t}^{N,N+1}$ should correspond to
the transition probabilities of two-level dynamics taking sequential-update
Bernoulli steps or Warren-Windridge geometric steps of certain parameters
depending on whether each factor $f_{u,u+1}(z)$ is given by $(1-\alpha_{u}z)$
or $(1+\beta_{u}z)^{-1}$. We will prove this in the case of Bernoulli jumps in
Proposition 5.10 and take yet another approach to consider geometric jumps in
Section 5.3.

Before discussing such dynamics further we prove a number of intertwining
relations (which in fact hold beyond the probabilistic setting, see Remark
5.4). Towards this end, define the projection
$\Pi_{N}:\mathbb{W}_{N,N+1}\to\mathbb{W}_{N}$ by projecting on the
$\mathbf{x}$-coordinate. In particular, for a function $g$ on $\mathbb{W}_{N}$
we thus define a function $\Pi_{N}g(\mathbf{x},\mathbf{y})=g(\mathbf{x})$ on
$\mathbb{W}_{N,N+1}$.

######  Proposition 5.7.

Let $s\leq t$. We have

$\displaystyle\left[\mathcal{Q}_{s,t}^{N,N+1}\Pi_{N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right)$
$\displaystyle=\left[\Pi_{N}\mathsf{P}_{f_{s,t}}^{(N)}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right),\
$
$\displaystyle(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1},\mathbf{x}^{\prime}\in\mathbb{W}_{N},$
(65)
$\displaystyle\left[\mathsf{P}_{f_{s,t}}^{(N+1)}\Lambda_{N+1,N}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right)$
$\displaystyle=\left[\Lambda_{N+1,N}\mathcal{Q}_{s,t}^{N,N+1}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),\
$
$\displaystyle\mathbf{y}\in\mathbb{W}_{N+1},\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\in\mathbb{W}_{N,N+1},$
(66)

where we view $\Lambda_{N+1,N}$ given in Definition 4.2 as a kernel from
$\mathbb{W}_{N+1}$ to $\mathbb{W}_{N,N+1}$ in the obvious way

$\Lambda_{N+1,N}\left(\mathbf{y},(\mathbf{x},\mathbf{k})\right)=\prod_{i=1}^{N}\frac{1}{a_{x_{i}}}\mathbf{1}_{\mathbf{x}\prec\mathbf{k}}\mathbf{1}_{\mathbf{k}=\mathbf{y}}.$

###### Proof.

The equality (65) is shown using the computation at the end of the proof of
Proposition 5.6. For equality (66), we need to sum over $\mathbf{x}$ such that
$\mathbf{x}\prec\mathbf{y}$. We use multinearity to bring the sum inside the
determinant and the relations

$\displaystyle\sum_{x_{i}={y_{i}}}^{y_{i+1}-1}\frac{1}{a_{x_{i}}}\mathcal{C}_{s,t}(\mathbf{x},\mathbf{y}^{\prime})_{ij}$
$\displaystyle=\nabla_{y_{j}^{\prime}}^{+}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}^{\prime}\rrbracket}(y_{i+1})-\nabla_{y_{j}^{\prime}}^{+}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,y_{j}^{\prime}\rrbracket}(y_{i}),$
$\displaystyle\sum_{x_{i}={y_{i}}}^{y_{i+1}-1}\frac{1}{a_{x_{i}}}\mathcal{D}_{s,t}(\mathbf{x},\mathbf{x}^{\prime})_{ij}$
$\displaystyle=-\frac{1}{a_{x_{j}^{\prime}}}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(y_{i+1})+\frac{1}{a_{x_{j}}^{\prime}}\mathsf{T}_{f_{s,t}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(y_{i}).$

Then, the statement follows from simple row-column operations. ∎

Observe that, by a combination of displays (65) and (66) we obtain yet another
proof of Theorem 4.6.

######  Proposition 5.8.

Let $s\leq t$. The kernel defined by

$\displaystyle\mathsf{Q}^{N,N+1}_{s,t}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\frac{\mathfrak{h}_{N}(\mathbf{x}^{\prime})}{\mathfrak{h}_{N}(\mathbf{x})}\mathcal{Q}^{N,N+1}_{s,t}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]$
(67)

is a Markov kernel from $\mathbb{W}_{N,N+1}$ to itself.

###### Proof.

By combining Propositions 4.9 and 5.7 we obtain

$\displaystyle\left[\mathcal{Q}^{N,N+1}_{s,t}\Pi_{N}\mathfrak{h}_{N}\right](\mathbf{x},\mathbf{y})=\Pi_{N}\mathfrak{h}_{N}(\mathbf{x},\mathbf{y}),$

from which the result follows. ∎

If by abusing notation we view $\mathfrak{L}_{N+1,N}$ as a Markov kernel from
$\mathbb{W}_{N+1}$ to $\mathbb{W}_{N,N+1}$ as we did with $\Lambda_{N+1,N}$,
by combining the results above we readily obtain the intertwinings of Markov
kernels.

######  Proposition 5.9.

Let $s\leq t$. Then, we have

$\displaystyle\left[\mathsf{Q}_{s,t}^{N,N+1}\Pi_{N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right)$
$\displaystyle=\left[\Pi_{N}\mathfrak{P}_{f_{s,t}}^{(N)}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right),\
$
$\displaystyle(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1},\mathbf{x}^{\prime}\in\mathbb{W}_{N},$
(68)
$\displaystyle\left[\mathfrak{P}_{f_{s,t}}^{(N+1)}\mathfrak{L}_{N+1,N}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right)$
$\displaystyle=\left[\mathfrak{L}_{N+1,N}\mathsf{Q}_{s,t}^{N,N+1}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),\
$
$\displaystyle\mathbf{y}\in\mathbb{W}_{N+1},\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\in\mathbb{W}_{N,N+1}.$
(69)

###  5.2 Sequential-update Bernoulli coupling

In this section we prove that the formula for $\mathcal{Q}_{s,t}^{N,N+1}$
given in Definition 5.3 with $f_{u,u+1}(z)=1-\alpha_{u}z$, where
$0\leq\alpha_{i}\leq\left(\sup_{k}a_{k}\right)^{-1}$, actually corresponds to
the sequential-update Bernoulli push-block dynamics. More precisely, we have
the following result. Observe that, in the definition of the stochastic
process $\left(\mathsf{X}(t),\mathsf{Y}(t);t\geq 0\right)$ below the
interaction of the $\mathsf{Y}$-components with the $\mathsf{X}$-components is
exactly given by the interaction between consecutive levels in the sequential-
update Bernoulli dynamics of Definition 2.5.

######  Proposition 5.10.

Consider the following stochastic process
$\left(\mathsf{X}(t),\mathsf{Y}(t);t\geq 0\right)$ in
$\mathbb{W}_{N}\times\mathbb{W}_{N+1}$ in discrete time initialised at
$(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1}$. The component
$\left(\mathsf{X}(t);t\geq 0\right)$ evolves autonomously as $N$ independent
Bernoulli walks, each with transition probabilities from time $s$ to time
$s+1$ given by $\mathsf{T}_{(1-\alpha_{s}z)}$. Then, given the updated
component $\mathsf{X}(s+1)$ we update $\mathsf{Y}(s)$ to $\mathsf{Y}(s+1)$ as
follows:

  * •

If $\mathsf{Y}_{i}(s)=x$ and $\mathsf{X}_{i}(s+1)=x$ then
$\mathsf{Y}_{i}(s+1)=x$ (block).

  * •

If $\mathsf{Y}_{i}(s)=x$ and $\mathsf{X}_{i-1}(s+1)=x$ then
$\mathsf{Y}_{i}(s+1)=x+1$ (push).

  * •

Otherwise, $\mathsf{Y}_{i}$ moves as a Bernoulli random walk with transition
probability $\mathsf{T}_{(1-\alpha_{s}z)}$ independent of the other
coordinates $\mathsf{Y}_{j}$.

The whole process is killed at the first collision time $\tau=\inf\\{t\geq
0:\mathsf{X}(t)\notin\mathbb{W}_{N}\\}$ of the $\mathsf{X}$ coordinates, so
that in particular
$\left(\mathsf{X}(t),\mathsf{Y}(t)\right)\in\mathbb{W}_{N,N+1}$ for all
$t<\tau$. Then, the transition probabilities from time $s$ to time $t$ of
$\left(\mathsf{X}(t),\mathsf{Y}(t);t\geq 0\right)$ killed when it exits
$\mathbb{W}_{N,N+1}$ are given by $\mathcal{Q}_{s,t}^{N,N+1}$ with the
underlying functions given by $f_{i,i+1}(z)=1-\alpha_{i}z$, where
$0\leq\alpha_{i}\leq\left(\sup_{k}a_{k}\right)^{-1}$.

###### Proof.

Take a test function $g:\mathbb{W}_{N,N+1}\to\mathbb{R}_{+}$. Fix an arbitrary
time $T\geq 0$. Define for any $0\leq t\leq T$, the functions (we drop
dependence on $N$)

$\displaystyle F\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)$
$\displaystyle=\sum_{\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\in\mathbb{W}_{N,N+1}}\mathcal{Q}^{N,N+1}_{t,T}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]g(\mathbf{x}^{\prime},\mathbf{y}^{\prime}),$
$\displaystyle G\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)$
$\displaystyle=\mathbb{E}\left[g\left(\mathsf{X}(T),\mathsf{Y}(T)\right)\mathbf{1}_{T<\tau}\big{|}\left(\mathsf{X}(t),\mathsf{Y}(t)\right)=(\mathbf{x},\mathbf{y}),t\leq\tau\right].$

We now show, by backward induction in $t$, that $F$ and $G$ are equal from
which the proposition follows. Observe that,

$F\left(T,T,(\mathbf{x},\mathbf{y})\right)=G\left(T,T,(\mathbf{x},\mathbf{y})\right)=g(\mathbf{x},\mathbf{y}),\
\ (\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1}.$

Then, observe that by the way our dynamics work $G$ satisfies the following
set of equations (the last three equations (71), (72), (73) define the value
of
$G\left(t+1,T,\left(\mathbf{x}+\boldsymbol{\delta},\mathbf{y}+\boldsymbol{\epsilon}\right)\right)$
in (70) below when
$(\mathbf{x}+\boldsymbol{\delta},\mathbf{y}+\boldsymbol{\epsilon})\notin\mathbb{W}_{N,N+1}$),
with the first equation holding for $0\leq t\leq T-1$ and the rest for $0\leq
t\leq T$:

$\displaystyle
G\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)=\sum_{\delta_{i},\epsilon_{i}\in\\{0,1\\}}\prod_{i=1}^{N}\mathsf{T}_{f_{t,t+1}}(x_{i},x_{i}+\delta_{i})\prod_{i=1}^{N+1}\mathsf{T}_{f_{t,t+1}}(y_{i},y_{i}+\epsilon_{i})G\left(t+1,T,\left(\mathbf{x}+\boldsymbol{\delta},\mathbf{y}+\boldsymbol{\epsilon}\right)\right),$
(70)
$\displaystyle\nabla_{y_{i}}^{+}G\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)\big{|}_{y_{i}=x_{i}}=0\
\ (\textnormal{i-th particle is blocked}),$ (71)
$\displaystyle\nabla_{y_{i+1}}^{+}G\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)\big{|}_{y_{i+1}=x_{i}}=0\
\ (\textnormal{(i+1)-th particle is pushed}),$ (72) $\displaystyle
G\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)\big{|}_{x_{i}=x_{i+1}}=0\
\ (\textnormal{process killed at }\tau).$ (73)

These determine $G(t,T,(\mathbf{x},\mathbf{y}))$ starting from
$G\left(T,T,(\mathbf{x},\mathbf{y})\right)=g(\mathbf{x},\mathbf{y})$. We now
show that $F(t,T,(\mathbf{x},\mathbf{y}))$ satisfies the same equations which
completes the proof.

First, it readily follows from the form of the entries
$\mathcal{C}_{s,t}(\mathbf{x},\mathbf{y}^{\prime})_{ij}$ and
$\mathcal{D}_{s,t}(\mathbf{x},\mathbf{x}^{\prime})_{ij}$ that, for $0\leq
t\leq T$:

$F\left(t,T,(\mathbf{x},\mathbf{y})\right)\big{|}_{x_{i}=x_{i+1}}=0.$

The boundary condition

$\nabla_{y_{i}}^{+}F\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)\big{|}_{y_{i}=x_{i}}=0,$

follows from the relations

$\displaystyle\nabla_{y_{i}}^{+}\mathcal{A}_{t,T}\left(\mathbf{y},\mathbf{y}^{\prime}\right)_{ij}\big{|}_{y_{i}=x_{i}}$
$\displaystyle=-a_{x_{i}}^{-1}\mathcal{C}_{t,T}\left(\mathbf{x},\mathbf{y}^{\prime}\right)_{ij},$
$\displaystyle\nabla_{y_{i}}^{+}\mathcal{B}_{t,T}\left(\mathbf{y},\mathbf{x}^{\prime}\right)_{ij}\big{|}_{y_{i}=x_{i}}$
$\displaystyle=-a_{x_{i}}^{-1}\mathcal{D}_{t,T}\left(\mathbf{x},\mathbf{x}^{\prime}\right)_{ij},$

while the boundary condition

$\nabla_{y_{i+1}}^{+}F\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)\big{|}_{y_{i+1}=x_{i}}=0,$

follows from the relations

$\displaystyle\nabla_{y_{i+1}}^{+}\mathcal{A}_{t,T}\left(\mathbf{y},\mathbf{y}^{\prime}\right)_{i+1j}\big{|}_{y_{i+1}=x_{i}}$
$\displaystyle=-a_{x_{i}}^{-1}\mathcal{C}_{t,T}\left(\mathbf{x},\mathbf{y}^{\prime}\right)_{ij},$
$\displaystyle\nabla_{y_{i+1}}^{+}\mathcal{B}_{t,T}\left(\mathbf{y},\mathbf{x}^{\prime}\right)_{i+1j}\big{|}_{y_{i+1}=x_{i}}$
$\displaystyle=-a_{x_{i}}^{-1}\mathcal{D}_{t,T}\left(\mathbf{x},\mathbf{x}^{\prime}\right)_{ij}.$

Finally, the equation

$F\left(t,T,\left(\mathbf{x},\mathbf{y}\right)\right)=\sum_{\delta_{i},\epsilon_{i}\in\\{0,1\\}}\prod_{i=1}^{N}\mathsf{T}_{f_{t,t+1}}(x_{i},x_{i}+\delta_{i})\prod_{i=1}^{N+1}\mathsf{T}_{f_{t,t+1}}(y_{i},y_{i}+\epsilon_{i})F\left(t+1,T,\left(\mathbf{x}+\boldsymbol{\delta},\mathbf{y}+\boldsymbol{\epsilon}\right)\right)$

follows from multinearity of the determinant and the following set of
equations for the individual entries

$\displaystyle\mathcal{A}_{t,T}\left(\mathbf{y},\mathbf{y}^{\prime}\right)_{ij}$
$\displaystyle=\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}\left(y_{i},y_{i}+\epsilon\right)\mathcal{A}_{t+1,T}\left(\mathbf{y}+\epsilon,\mathbf{y}^{\prime}\right)_{ij},$
$\displaystyle\mathcal{D}_{t,T}\left(\mathbf{x},\mathbf{x}^{\prime}\right)_{ij}$
$\displaystyle=\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}\left(x_{i},x_{i}+\epsilon\right)\mathcal{D}_{t+1,T}\left(\mathbf{x}+\epsilon,\mathbf{x}^{\prime}\right)_{ij},$
$\displaystyle\mathcal{B}_{t,T}\left(\mathbf{y},\mathbf{x}^{\prime}\right)_{ij}$
$\displaystyle=a_{x_{j}^{\prime}}^{-1}\left(\mathsf{T}_{f_{t,T}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(y_{i})-\mathbf{1}_{j\geq i}\right)$
$\displaystyle=a_{x_{j}^{\prime}}^{-1}\left(\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}(y_{i},y_{i}+\epsilon)\mathsf{T}_{f_{t+1,T}}\mathbf{1}_{\llbracket
0,x_{j}^{\prime}\rrbracket}(y_{i})-\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}(y_{i},y_{i}+\epsilon)\mathbf{1}_{j\geq
i}\right)$
$\displaystyle=\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}(y_{i},y_{i}+\epsilon)\mathcal{B}_{t+1,T}\left(\mathbf{y}+\epsilon,\mathbf{x}^{\prime}\right)_{ij},$
$\displaystyle\mathcal{C}_{t,T}\left(\mathbf{x},\mathbf{y}^{\prime}\right)_{ij}$
$\displaystyle=a_{x_{i}}\nabla_{x_{i}}^{+}\nabla_{y_{j}}^{-}\mathsf{T}_{f_{t,T}}\mathbf{1}_{\llbracket
0,y_{j}^{\prime}\rrbracket}(x_{i})=-\nabla_{y_{j}^{\prime}}^{-}\left[a_{y_{j}^{\prime}}\mathsf{T}_{f_{t,T}}\left(x_{i},y_{j}^{\prime}\right)\right]$
$\displaystyle=-\nabla_{y_{j}^{\prime}}^{-}\left[a_{y_{j}^{\prime}}\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}\left(x_{i},x_{i}+\epsilon\right)\mathsf{T}_{f_{t+1,T}}\left(x_{i}+\epsilon,y_{j}^{\prime}\right)\right]$
$\displaystyle=\sum_{\epsilon\in\\{0,1\\}}\mathsf{T}_{f_{t,t+1}}\left(x_{i},x_{i}+\epsilon\right)\mathcal{C}_{t+1,T}\left(\mathbf{x}+\epsilon,\mathbf{y}^{\prime}\right)_{ij}.$

This completes the proof. ∎

###  5.3 Warren-Windridge geometric dynamics coupling

In section we take a different approach to study the transition probabilities
of the Warren-Windridge geometric dynamics. We take $f_{0,1}(z)=(1+\beta
z)^{-1}$. To ease notation we write $\mathsf{T}(x,y)$ for
$\mathsf{T}_{f_{0,1}}(x,y)$ and $\mathsf{P}^{(N)}$ and $\mathfrak{P}^{(N)}$
for $\mathsf{P}_{f_{0,1}}^{(N)}$ and $\mathfrak{P}_{f_{0,1}}^{(N)}$
respectively.

######  Definition 5.11.

We write
$\mathcal{G}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]$
for the transition probabilities of the following single-step discrete-time
stochastic process $\left(\mathsf{X}(t),\mathsf{Y}(t);t=0,1\right)$ in
$\mathbb{W}_{N,N+1}$. The $\mathsf{X}$-component moves autonomously, with
$\mathsf{X}(1)$ given $\mathsf{X}(0)=\mathbf{x}$ distributed as
$\mathfrak{P}^{(N)}\left(\mathsf{x},\cdot\right)$. Given the updated component
$\mathsf{X}(1)$, we update component $\mathsf{Y}$ as follows:

  * •

Coordinate $\mathsf{Y}_{i}$ is first moved to the intermediate position
$\max\\{\mathsf{Y}_{i}(0),\mathsf{X}_{i-1}(1)+1\\}$ (push).

  * •

$\mathsf{Y}_{i}$ subsequently attempts to jump to
$z\geq\max\\{\mathsf{Y}_{i}(0),\mathsf{X}_{i-1}(1)+1\\}$ with the
inhomogeneous geometric probability
$\mathsf{T}(\max\\{\mathsf{Y}_{i}(0),\mathsf{X}_{i-1}(1)+1\\},z)$.

  * •

All jumps to $z\geq\mathsf{X}_{i}(0)=x_{i}$ are suppressed and in such case
$\mathsf{Y}_{i}(1)$ is taken to be $x_{i}$ (block).

######  Remark 5.12.

Of course, the process $\left(\mathsf{X}(t),\mathsf{Y}(t);t=0,1\right)$ can be
extended to all times $t\geq 0$. For the $s$-th step we simply replace
$f_{0,1}(z)$ by $f_{s,s+1}(z)=(1+\beta_{s}z)^{-1}$ and all other quantities
are defined analogously using $f_{s,s+1}$. All the results that follow go
through with the obvious notational modification.

Observe that, in the definition of the stochastic process
$\left(\mathsf{X}(t),\mathsf{Y}(t);t=0,1\right)$ above the interaction of the
$\mathsf{Y}$-components with the $\mathsf{X}$-components is exactly given by
the interaction between consecutive levels in the Warren-Windridge geometric
dynamics of Definition 2.7. As mentioned previously we believe171717We have
shown this for $N=1$ by tediously checking all possibilities. that
$\mathsf{Q}_{0,1}^{N,N+1}$, with $f_{0,1}(z)=(1+\beta z)^{-1}$, should
correspond to Warren-Windridge geometric dynamics step with parameter $\beta$.
In particular, we believe that the question mark on the equality below should
be removed

$\mathcal{G}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]\overset{?}{=}\mathsf{Q}_{0,1}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right].$
(74)

However, when geometric jumps are involved, a proof along the lines of the one
given for Proposition 5.10 above turns out to be tricky. So we take a
different route inspired by the work of Warren-Windridge [135] in the
homogeneous case. Observe that, we have

$\mathcal{G}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\mathfrak{P}^{(N)}(\mathbf{x},\mathbf{x}^{\prime})\mathcal{R}^{N,N+1}(\mathbf{y}^{\prime};\mathbf{x}^{\prime},\mathbf{x},\mathbf{y})$

where we write

$\mathcal{R}^{N,N+1}(\mathbf{y}^{\prime};\mathbf{x}^{\prime},\mathbf{x},\mathbf{y})=\mathbb{P}\left(\mathsf{Y}(1)=\mathbf{y}^{\prime}|\mathsf{X}(1)=\mathbf{x},\mathsf{X}(0)=\mathbf{x},\mathsf{Y}(0)=\mathbf{y}\right).$

Our starting point is the following explicit expression for
$\mathcal{R}^{N,N+1}$.

######  Proposition 5.13.

We have, with
$(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{W}_{N,N+1}$,

$\displaystyle\mathcal{R}^{N,N++1}(\mathbf{y}^{\prime};\mathbf{x}^{\prime},\mathbf{x},\mathbf{y})=\mathsf{Blk}\left(y_{1}^{\prime},y_{1},x_{1}\right)\prod_{n=2}^{N}\mathsf{b}\left(x_{n},y_{n}^{\prime}\right)\mathsf{Psh}\left(y_{n}^{\prime},y_{n},x_{n-1}^{\prime}\right)\mathsf{Psh}\left(y_{N+1}^{\prime},y_{N+1},x_{N}^{\prime}\right),$

where the factors are defined by

$\displaystyle\mathsf{Blk}\left(y^{\prime},y,x\right)$
$\displaystyle=\mathsf{T}(y,y^{\prime})\mathsf{b}\left(x,y^{\prime}\right),$
$\displaystyle\mathsf{Psh}\left(y^{\prime},y,x^{\prime}\right)$
$\displaystyle=\mathsf{T}\left(y,y^{\prime}\right)\mathbf{1}_{x^{\prime}<y}+\mathsf{T}\left(x^{\prime}+1,y^{\prime}\right)\mathbf{1}_{x^{\prime}\geq
y},$ $\displaystyle\mathsf{b}\left(x,y^{\prime}\right)$
$\displaystyle=\mathbf{1}_{y^{\prime}<x}+(1+\beta
a_{y^{\prime}})\mathbf{1}_{y^{\prime}=x}.$

###### Proof.

The $\mathsf{Psh}$ factor is a direct translation in symbols of the pushing
mechanism. The fact that the $\mathsf{Blk}$ factor corresponds to blocking can
be seen by multiplying out $\mathsf{T}(y,y^{\prime})$ with
$\mathsf{b}(x,y^{\prime})$ and using the identity, with $y\geq x$,

$\sum_{m=x}^{\infty}\mathsf{T}(y,m)=\mathsf{T}(y,x)(1+\beta a_{x}).$

This follows from the trivial to check identity, for $y\leq x\leq m$,

$\frac{\mathsf{T}(y,m)}{\mathsf{T}(y,x)}=(1+\beta a_{x})\mathsf{T}(x,m).$

Similarly, the fact that the factor with subscript $n$ in the product
corresponds to the interaction of $\mathsf{Y}_{n}$ with $\mathsf{X}_{n-1}(1)$
by pushing and $\mathsf{X}_{n}(0)$ by blocking can be seen by multiplying out
$\mathsf{Psh}\left(y^{\prime},y,x^{\prime}\right)$ with
$\mathsf{b}\left(x,y^{\prime}\right)$ and the relation

$\displaystyle\sum_{m=x}^{\infty}\mathsf{Psh}\left(m,y,x^{\prime}\right)=\mathsf{Psh}\left(x,y,x^{\prime}\right)(1+\beta
a_{x}),$

which is again a consequence of the previous identity. ∎

We now go on to prove an analogue of Proposition 5.9 with
$\mathsf{Q}_{0,1}^{N,N+1}$ replaced by $\mathcal{G}^{N,N+1}$ (clearly, if we
knew equality (74) were true there would be nothing to show). The following
identity is the key ingredient in the computation.

######  Proposition 5.14.

We have

$\displaystyle\sum_{x_{i}=y_{i}^{\prime}}^{x_{i}^{\prime}\wedge(y_{i+1}-1)}\frac{1}{a_{x_{i}}}\mathsf{T}(x_{i},x_{i}^{\prime})\mathsf{Blk}(y_{i}^{\prime},y_{i},x_{i})$
$\displaystyle\mathsf{Psh}(y_{i+1}^{\prime},y_{i+1},x_{i}^{\prime})$
$\displaystyle=\frac{1}{a_{x_{i}}^{\prime}}\mathsf{T}(y_{i},y_{i}^{\prime})\mathsf{T}(y_{i+1},y_{i+1}^{\prime})\mathbf{1}_{y_{i}\leq
y_{i}^{\prime}<y_{i+1}\leq y_{i+1}^{\prime}}\mathbf{1}_{y_{i}^{\prime}\leq
x_{i}^{\prime}<y_{i+1}^{\prime}}.$

###### Proof.

Observe that, the left hand side is zero if we do not have
$y_{i}^{\prime}<y_{i+1}$ or if we do not have $y_{i}^{\prime}\leq
x_{i}^{\prime}<y_{i+1}^{\prime}$. Thus, we can restrict to this scenario in
which case the indicator functions on the right are both $1$. By using
$\mathsf{Blk}(y_{i}^{\prime},y_{i},x_{i})=\mathsf{T}(y_{i},y_{i}^{\prime})\mathsf{b}(x_{i},y_{i}^{\prime})$
and cancelling out $\mathsf{T}(y_{i},y_{i}^{\prime})$ it will suffice to
establish the following claim

$\displaystyle\sum_{x_{i}=y_{i}^{\prime}}^{x_{i}^{\prime}\wedge(y_{i+1}-1)}\frac{a_{x_{i}^{\prime}}}{a_{x_{i}}}\mathsf{T}(x_{i},x_{i}^{\prime})\left(\mathbf{1}_{y_{i}^{\prime}<x_{i}}+(1+\beta
a_{y_{i}^{\prime}})\mathbf{1}_{x_{i}=y_{i}^{\prime}}\right)\left(\mathsf{T}(y_{i+1},y_{i+1}^{\prime})\mathbf{1}_{x_{i}^{\prime}<y_{i+1}}+\mathsf{T}(x_{i}^{\prime}+1,y_{i+1}^{\prime})\mathbf{1}_{x_{i}^{\prime}\geq
y_{i+1}}\right)$ $\displaystyle=\mathsf{T}(y_{i+1},y_{i+1}^{\prime}).$

Let us denote by $\mathcal{E}(y_{i}^{\prime})$ the left hand side above as a
function of $y_{i}^{\prime}$. We will now show that
$\mathcal{E}(x_{i}^{\prime}\wedge(y_{i+1}-1))=\mathsf{T}(y_{i+1},y_{i+1}^{\prime})$
and that $\mathcal{E}(m)=\mathcal{E}(m-1)$ from which the claim follows.

Suppose $x_{i}^{\prime}<y_{i+1}$ then $x_{i}^{\prime}\wedge(y_{i+1}-1)$. It is
immediate then that
$\mathcal{E}(x_{i}^{\prime}\wedge(y_{i+1}-1))=\mathsf{T}(y_{i+1},y_{i+1}^{\prime})$.
If instead we suppose $x_{i}^{\prime}\geq y_{i+1}$ then
$x_{i}=y_{i}^{\prime}=y_{i+1}-1$. After some elementary manipulations using
the explicit formulae we again obtain
$\mathcal{E}(x_{i}^{\prime}\wedge(y_{i+1}-1))=\mathsf{T}(y_{i+1},y_{i+1}^{\prime})$.
We now show $\mathcal{E}(m)=\mathcal{E}(m-1)$. We observe that the sums
$\sum_{x_{i}=m+1}^{x_{i}\wedge(y_{i+1}-1)}$ on both sides of the desired
identity $\mathcal{E}(m)=\mathcal{E}(m-1)$ are equal (the summands are exactly
equal). We thus need to check that the sum $\sum_{x_{i}=m}^{m}$ on the left
hand side and the sum $\sum_{x_{i}=m-1}^{m}$ are equal. After some
manipultions this boils down to proving the identity

$\frac{1}{a_{m}}\mathsf{T}(m,x_{i}^{\prime})(1+\beta
a_{m})=\frac{1}{a_{m}}\mathsf{T}(m,x_{i}^{\prime})+\frac{1}{a_{m-1}}\mathsf{T}(m-1,x_{i}^{\prime})(1+\beta
a_{m-1}).$

Elementary manipulations readily establish this. ∎

We finally prove the required intertwinings.

######  Proposition 5.15.

We have the intertwinings

$\displaystyle\left[\mathcal{G}^{N,N+1}\Pi_{N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right)$
$\displaystyle=\left[\Pi_{N}\mathfrak{P}^{(N)}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right),\
$
$\displaystyle(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1},\mathbf{x}^{\prime}\in\mathbb{W}_{N},$
(75)
$\displaystyle\left[\mathfrak{P}^{(N+1)}\mathfrak{L}_{N+1,N}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right)$
$\displaystyle=\left[\mathfrak{L}_{N+1,N}\mathcal{G}^{N,N+1}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),\
$
$\displaystyle\mathbf{y}\in\mathbb{W}_{N+1},\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\in\mathbb{W}_{N,N+1}.$
(76)

###### Proof.

The proof of (75) is immediate from the definition of $\mathcal{G}^{N,N+1}$.
For identity (76) we make repeated use of Proposition 5.14. The right hand
side of (76) is equal to

$\frac{\mathfrak{h}_{N}(\mathbf{x}^{\prime})}{\mathfrak{h}_{N+1}(\mathbf{y})}\sum_{\mathbf{x}\prec\mathbf{y}}\prod_{i=1}^{N}\frac{1}{a_{x_{i}}}\prod_{i=1}^{N}\mathsf{T}(x_{i},x_{i}^{\prime})\mathbf{1}_{\mathbf{x}\prec\mathbf{x}^{\prime}}\mathcal{R}^{N,N+1}(\mathbf{y}^{\prime};\mathbf{x}^{\prime},\mathbf{x},\mathbf{y}).$

Now, recall that

$\mathsf{P}^{(N)}(\mathbf{x},\mathbf{x}^{\prime})=\det\left(\mathsf{T}(x_{i},x_{j}^{\prime})\right)_{i=1}^{N}=\prod_{i=1}^{N}\mathsf{T}(x_{i},x_{i}^{\prime})\mathbf{1}_{\mathbf{x}\prec\mathbf{x}^{\prime}}.$

Moreover, we can restrict the sum above to a sum over $x_{i}\in\llbracket
y_{i}^{\prime},x_{i}^{\prime}\wedge(y_{i+1}-1)\rrbracket$ for otherwise the
summand is zero and use the form of $\mathcal{R}^{N,N+1}$ from Proposition
5.13 to obtain that the right hand side of (76) is equal to

$\displaystyle\frac{\mathfrak{h}_{N}(\mathbf{x}^{\prime})}{\mathfrak{h}_{N+1}(\mathbf{y})}\sum_{x_{i}\in\llbracket
y_{i}^{\prime},x_{i}^{\prime}\wedge(y_{i+1}-1)\rrbracket}\prod_{i=1}^{N}\frac{1}{a_{x_{i}}}\prod_{i=1}^{N}\mathsf{T}(x_{i},x_{i}^{\prime})\mathsf{Blk}\left(y_{1}^{\prime},y_{1},x_{1}\right)$
$\displaystyle\times\prod_{n=2}^{N}\mathsf{b}\left(x_{n},y_{n}^{\prime}\right)\mathsf{Psh}\left(y_{n}^{\prime},y_{n},x_{n-1}^{\prime}\right)\mathsf{Psh}\left(y_{N+1}^{\prime},y_{N+1},x_{N}^{\prime}\right).$

Consider the factor

$\displaystyle\frac{1}{a_{x_{1}}}\mathsf{T}(x_{1},x_{1}^{\prime})\mathsf{Blk}(y_{1}^{\prime},y_{1},x_{1})\mathsf{Psh}(y_{2}^{\prime},y_{2},x_{1}^{\prime}).$

Summing over $x_{1}\in\llbracket
y_{1}^{\prime},x_{1}^{\prime}\wedge(y_{2}-1)\rrbracket$ using Proposition 5.14
we get that this is equal to

$\frac{1}{a_{x_{1}}^{\prime}}\mathsf{T}(y_{1},y_{1}^{\prime})\mathsf{T}(y_{2},y_{2}^{\prime})\mathbf{1}_{y_{1}\leq
y_{1}^{\prime}<y_{2}\leq y_{2}^{\prime}}\mathbf{1}_{y_{1}^{\prime}\leq
x_{1}^{\prime}<y_{2}^{\prime}}.$

We can then combine the factor $\mathsf{T}(y_{2},y_{2}^{\prime})$ with
$\mathsf{b}(x_{2},y_{2}^{\prime})$ to get
$\mathsf{Blk}(y_{2}^{\prime},y_{2},x_{2})$ and perform the sum over
$x_{2}\in\llbracket y_{2}^{\prime},x_{2}^{\prime}\wedge(y_{3}-1)\rrbracket$
using Proposition 5.14. If we keep iterating this procedure of using
Proposition 5.14 we obtain that the right hand side of (76) is equal to

$\displaystyle\frac{\mathfrak{h}_{N}(\mathbf{x}^{\prime})}{\mathfrak{h}_{N+1}(\mathbf{y})}\prod_{i=1}^{N}\frac{1}{a_{x_{i}^{\prime}}}\prod_{i=1}^{N+1}\mathsf{T}\left(y_{i},y_{i}^{\prime}\right)\mathbf{1}_{\mathbf{y}\prec\mathbf{y}^{\prime}}\mathbf{1}_{\mathbf{x}^{\prime}\prec\mathbf{y}^{\prime}}=\left[\mathfrak{P}^{(N+1)}\mathfrak{L}_{N+1,N}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),$

which concludes the proof. ∎

###  5.4 Continuous-time pure-birth chain coupling

We have the following continuous-time analogue for pure-birth chains of our
previous discrete-time results. This result was proven in [8].

######  Proposition 5.16.

Consider the continuous-time stochastic process
$\left(\mathsf{X}(t),\mathsf{Y}(t);t\geq 0\right)$ in $\mathbb{W}_{N,N+1}$.
The $\mathsf{X}$-component evolves autonomously according to the semigroup
$\left(\mathfrak{P}^{(N)}_{\exp(-tz)}\right)_{t\geq 0}$. The
$\mathsf{Y}$-component evolves as $N+1$ independent pure-birth chain with jump
rate $a_{x}$ at $x\in\mathbb{Z}_{+}$ with the following interactions with the
$\mathsf{X}$-component:

  * •

If the clock of $\mathsf{Y}_{i}$ rings for it to jump and
$\mathsf{Y}_{i}=\mathsf{X}_{i}$ then nothing happens (block).

  * •

If $\mathsf{X}_{i-1}=x$ and $\mathsf{Y}_{i}=x+1$ and $\mathsf{X}_{i-1}$ jumps
to $x+1$ then $\mathsf{Y}_{i}$ instantaneously moves to $x+2$ (push).

Then, the time-homogeneous transition probabilities of
$\left(\mathsf{X}(t),\mathsf{Y}(t);t\geq 0\right)$ from
$(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1}$ to
$(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\in\mathbb{W}_{N,N+1}$ in time $t$
are given by:

$\mathsf{Q}_{\exp(-tz)}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\frac{\mathfrak{h}_{N}(\mathbf{x}^{\prime})}{\mathfrak{h}_{N}(\mathbf{x})}\mathcal{Q}_{\exp(-tz)}^{N,N+1}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right],$
(77)

where $\mathcal{Q}_{f}^{N,N+1}$ is defined as in Definition 5.3 and Remark
5.4. Moreover, we have the intertwinings:

$\displaystyle\left[\mathsf{Q}_{\exp(-tz)}^{N,N+1}\Pi_{N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right)$
$\displaystyle=\left[\Pi_{N}\mathfrak{P}_{\exp(-tz)}^{(N)}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right),\
$
$\displaystyle(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1},\mathbf{x}^{\prime}\in\mathbb{W}_{N},$
(78)
$\displaystyle\left[\mathfrak{P}_{\exp(-tz)}^{(N+1)}\mathfrak{L}_{N+1,N}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right)$
$\displaystyle=\left[\mathfrak{L}_{N+1,N}\mathsf{Q}_{\exp(-tz)}^{N,N+1}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),\
$
$\displaystyle\mathbf{y}\in\mathbb{W}_{N+1},\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\in\mathbb{W}_{N,N+1}.$
(79)

######  Remark 5.17.

Observe that, in the definition of the stochastic process
$\left(\mathsf{X}(t),\mathsf{Y}(t);t\geq 0\right)$ above the interaction of
the $\mathsf{Y}$-components with the $\mathsf{X}$-components is exactly given
by the interaction between consecutive levels in the pure-birth push-block
dynamics of Definition 2.4.

######  Remark 5.18.

An analogous result holds for $\mathcal{Q}_{\exp(-tz)}^{N,N+1}$ without the
Doob-transform by $\mathfrak{h}_{N}$, except that now the autonomous
$\mathsf{X}$-component evolves as $N$ independent pure-birth chains killed
when they collide (the whole process is killed as well).

###  5.5 Couplings of Borodin-Ferrari-Olshanski

Given two Markov semigroups which are intertwined such as
$\mathfrak{P}_{f}^{(N)}$ and $\mathfrak{P}_{f}^{(N+1)}$ (with an appropriate
choice of $f$) there are different interesting ways to couple the
corresponding stochastic processes. We now briefly survey, and compare to our
setting, couplings developed181818We note that in these constructions the
intertwining from Proposition 4.11 between $\mathfrak{P}_{f}^{(N)}$ and
$\mathfrak{P}_{f}^{(N+1)}$ is taken as input. In the coalescing random walk
framework the intertwining can be arrived at without apriori knowledge, see
for example the discussion before Proposition 5.8. by Borodin, Ferrari and
Olshanski which have been made use of in many works [24, 20, 32, 21], see also
[35, 44, 43] for further developments and generalisations. Although the
origin191919These couplings have their origin in an idea of Diaconis and Fill
[59] from their study of strong stationary times [59, 67, 107, 108] for
convergence to equilibrium for Markov chains. of these constructions is very
different from the coalescing random walk framework we have developed, it
turns out that these couplings in the case of pure-birth and Bernoulli
dynamics coincide with ours. On the other hand, in the case of geometric jumps
their coupling is different from the Warren-Windridge geometric dynamics,
which are in some sense preferable since they have more Markovian projections.

####  5.5.1 Sequential-update

Consider the Markov kernel, see [24],

$\mathfrak{P}_{f}^{(N,N+1),\textnormal{Seq}}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\begin{cases}\frac{\mathfrak{P}_{f}^{(N)}(\mathbf{x},\mathbf{x}^{\prime})\mathfrak{P}_{f}^{(N+1)}(\mathbf{y},\mathbf{y}^{\prime})\mathfrak{L}_{N+1,N}(\mathbf{y}^{\prime},\mathbf{x}^{\prime})}{\left[\mathfrak{L}_{N+1,N}\mathfrak{P}_{f}^{(N)}\right](\mathbf{y},\mathbf{x}^{\prime})},&\left[\mathfrak{L}_{N+1,N}\mathfrak{P}_{f}^{(N)}\right](\mathbf{y},\mathbf{x}^{\prime})>0,\\\
0,&\textnormal{otherwise}.\end{cases}$ (80)

on the state space

$\mathbb{S}^{N,N+1}_{\textnormal{Seq}}=\left\\{(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N}\times\mathbb{W}_{N+1}\big{|}\mathfrak{L}_{N+1,N}(\mathbf{y},\mathbf{x})>0\right\\}=\mathbb{W}_{N,N+1}.$

By virtue of the intertwining from Theorem 4.11 it is immediate that this is
correctly normalised. Moreover, the conditional distribution of the projection
on $\mathbb{W}_{N}$ given the projection on $\mathbb{W}_{N+1}$ is given by
$\mathfrak{L}_{N+1,N}$, see [24]. It can be checked that in the case of
Bernoulli jumps, namely with $f(z)=1-\alpha z$, the dynamics arising from (80)
are exactly the sequential-update push-block Bernoulli dynamics we have been
considering. In particular, (80) with $f(z)=1-\alpha z$ is equal to (67) with
$s=0,t=1$ and $f_{0,1}(z)=f(z)$. In the case of geometric jumps however, the
coupling obtained from (80) is different from the Warren-Windridge type
coupling we have been considering in this work. It can be checked that its
projection on the right edge is still the geometric push-TASEP but the
projection on the left edge is not even Markovian (and the evolution of the
rest of the array is indeed different).

####  5.5.2 Parallel-update

Consider the Markov kernel, see [24],

$\mathfrak{P}_{f}^{(N,N+1),\textnormal{Par}}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]=\frac{\mathfrak{P}_{f}^{(N)}(\mathbf{x},\mathbf{x}^{\prime})\mathfrak{P}_{f}^{(N+1)}(\mathbf{y},\mathbf{y}^{\prime})\mathfrak{L}_{N+1,N}(\mathbf{y}^{\prime},\mathbf{x})}{\left[\mathfrak{L}_{N+1,N}\mathfrak{P}_{f}^{(N)}\right](\mathbf{y},\mathbf{x})}$
(81)

on the more complicated state space:

$\mathbb{S}^{N,N+1}_{\textnormal{Par}}=\left\\{(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N}\times\mathbb{W}_{N+1}\big{|}\left[\mathfrak{L}_{N+1,N}\mathfrak{P}_{f}^{(N)}\right](\mathbf{y},\mathbf{x})>0\right\\}.$

Again, by virtue of the intertwining from Theorem 4.11 it is immediate that
this correctly normalised. Also, the conditional distribution of the
projection on $\mathbb{W}_{N}$ given the projection on $\mathbb{W}_{N+1}$ is
now given by $\mathfrak{L}_{N+1,N}\mathfrak{P}_{f}^{(N)}$, see [24]. In the
case of Bernoulli jumps this gives rise, as the projection on the left edge,
to parallel-update Bernoulli TASEP. It would be possible to obtain analogous
results to the ones in this paper (show the existence of determinantal
correlations, compute the correlation kernel and so on) for the parallel-
update model but the situation is more involved and we will not pursue the
details here.

####  5.5.3 Continuous-time

For continuous time and countable state space an analogous framework for
coupling intertwined semigroups was developed202020In the earlier paper of
Borodin-Ferrari [24] the continuous-time dynamics were obtained as a scaling
limit from discrete-time. by Borodin and Olshanski in [32]. The two-level
dynamics are now defined through their explicit infinitesimal jump rates, see
[32], instead of their transition kernels which in continuous time are almost
never explicit (unlike the explicit kernels (80), (81) for discrete-time). It
can be checked that the dynamics coming from the coupling of [32], associated
to the intertwining from Theorem 4.11, actually match the pure-birth push-
block dynamics we have been considering in this paper. So in particular,
$\mathsf{Q}^{N,N+1}_{\textnormal{exp}(-tw)}$ is a rare example of an explicit
transition kernel for the coupling framework developed in [32].

##  6 Dynamics on arrays

###  6.1 The space-time inhomogeneous case

Combining our previous results we prove the following.

######  Proposition 6.1.

Let $N\geq 1$ be arbitrary. Consider an initial configuration in
$\mathbb{IA}_{N}$ distributed according to

$\mu_{N}\left(\mathbf{x}^{(N)}\right)\mathfrak{L}_{N,N-1}\left(\mathbf{x}^{(N)},\mathbf{x}^{(N-1)}\right)\cdots\mathfrak{L}_{2,1}\left(\mathbf{x}^{(2)},\mathbf{x}^{(1)}\right),$
(82)

for some probability measure $\mu_{N}$ on $\mathbb{W}_{N}$. Then, perform
$M_{1}$ steps of sequential-update Bernoulli dynamics with parameters
$\alpha_{1},\dots,\alpha_{M_{1}}$, $M_{2}$ steps of Warren-Windridge geometric
dynamics with parameters $\beta_{1},\dots,\beta_{M_{2}}$ and finally
continuous-time pure-birth dynamics for time $t$. The parameters
$\alpha_{i},\beta_{i}$ are assumed to satisfy
$0\leq\alpha_{i}\leq(\sup_{x\in\mathbb{Z}_{+}}a_{x})^{-1}$ and
$0\leq\beta_{i}<(\sup_{x\in\mathbb{Z}_{+}}a_{x}-\inf_{x\in\mathbb{Z}_{+}}a_{x})^{-1}$.
Then, the distribution of the resulting configuration in $\mathbb{IA}_{N}$ is
given by

$\left[\mu_{N}\mathfrak{P}^{(N)}_{\prod_{i=1}^{M_{1}}(1-\alpha_{i}w)\prod_{i=1}^{M_{2}}(1+\beta_{i}w)^{-1}\exp(-tw)}\right]\left(\mathbf{x}^{(N)}\right)\mathfrak{L}_{N,N-1}\left(\mathbf{x}^{(N)},\mathbf{x}^{(N-1)}\right)\cdots\mathfrak{L}_{2,1}\left(\mathbf{x}^{(2)},\mathbf{x}^{(1)}\right).$
(83)

Finally, consider the process $\left(\mathsf{X}_{i}^{(n)}(t);t\geq
0\right)_{1\leq i\leq n;1\leq n\leq N}$ in $\mathbb{IA}_{N}$ defined in
Theorem 2.14. Assume it is initialised according to (82). Then, for any $1\leq
n\leq N$, the projection on the $n$-th row evolves as a Markov process with
transition probabilities
$\mathfrak{P}_{s,t}^{(n)}=\mathfrak{P}_{f_{s,t}}^{(n)}$ from (12).

###### Proof.

The proof is by induction, making use of Propositions 5.9, 5.15 and 5.16, by
virtue of the Markov functions theory of Rogers-Pitman [126]. This type of
argument, and variations, have been documented in many places [134, 24, 114,
12, 6, 121, 5] and we do not repeat all details. The main point is that under
such dynamics, the Gibbs-type property of the initial distribution of the
array, namely the fact that the law of the first $N-1$ rows of the array given
the $N$-th row being equal to $\mathbf{y}\in\mathbb{W}_{N}$ is given by:

$\mathfrak{L}_{N,N-1}\left(\mathbf{y},\mathbf{x}^{(N-1)}\right)\mathfrak{L}_{N-1,N-2}\left(\mathbf{x}^{(N-1)},\mathbf{x}^{(N-2)}\right)\cdots\mathfrak{L}_{2,1}\left(\mathbf{x}^{(2)},\mathbf{x}^{(1)}\right),$

is preserved for all times $t>0$, and the projections on single levels are
Markovian with the desired transition probabilities
$\mathfrak{P}_{s,t}^{(n)}=\mathfrak{P}_{f_{s,t}}^{(n)}$. Then, the measure
$\mu_{N}$ on level $N$ is evolved accordingly,

$\displaystyle\left[\mu_{N}\mathfrak{P}^{(N)}_{(1-\alpha_{1}w)}\cdots\mathfrak{P}^{(N)}_{(1-\alpha_{M_{1}}w)}\mathfrak{P}^{(N)}_{(1+\beta_{1}w)^{-1}}\cdots\mathfrak{P}^{(N)}_{(1+\beta_{M_{2}}w)^{-1}}\mathfrak{P}^{(N)}_{\exp(-tw)}\right](\mathbf{x})=$
$\displaystyle\left[\mu_{N}\mathfrak{P}^{(N)}_{\prod_{i=1}^{M_{1}}(1-\alpha_{i}w)\prod_{i=1}^{M_{2}}(1+\beta_{i}w)^{-1}\exp(-tw)}\right](\mathbf{x}),$

by virtue of Proposition 4.4, which is where we need the (technical)
assumption on the $\beta_{i}$ (recall the assumption on the $\alpha_{i}$
parameters is there for the transition probabilities to be well-defined).
Observe that, the order in which we take the jumps is not important. ∎

We note that, the deterministic law of the fully-packed configuration is of
the form (82) by taking
$\mu_{N}(\mathbf{x})=\mathbf{1}_{(0,1,\dots,N-1)}(\mathbf{x})$.

###  6.2 The space-level inhomogeneous case

We now consider the space and level inhomogeneous setting of Theorem 2.24. Our
process evolves with either only continuous-time pure-birth dynamics, or only
discrete time Bernoulli or only discrete time geometric dynamics which are all
homogeneous in time. The extra new ingredient for the argument are the
functions $\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}$, which are
defined recursively in (87) and then shown to have a particular determinant
expression in Proposition 6.9 which reveals the desired parameter symmetry.
Other than that, most of the work has already been done.

Recall that, we write $\mathsf{T}_{t}^{\bullet}$ with
$\bullet\in\\{\textnormal{pb, B, g}\\}$, where the abbreviations stand for
pure-birth, Bernoulli and geometric, to denote $\mathsf{T}_{f}$ with the
following choices of function $f(w)$: if $\bullet=\textnormal{pb}$, then
$f(w)=\exp(-tw)$, if $\bullet=\textnormal{B}$, then $f(w)=(1-w)^{t}$ and if
$\bullet=\textnormal{g}$, then $f(w)=(1+w)^{-t}$.

######  Lemma 6.2.

We have the following eigenfunction relations

$\mathsf{T}_{t}^{\bullet}h_{\gamma}^{\bullet}(x)=c_{t,\gamma}^{\bullet}h_{\gamma}^{\bullet}(x),$

where the functions $h_{\gamma}^{\bullet}$ are given by

$h_{\gamma}^{\textnormal{pb}}(x)=p_{x}(-\gamma),\ \
h_{\gamma}^{\textnormal{B}}(x)=p_{x}\left(1-\gamma^{-1}\right),\ \
h_{\gamma}^{\textnormal{g}}(x)=p_{x}(\gamma^{-1}-1),$

and the constants $c_{t,\gamma}^{\bullet}$ by,

$c_{t,\gamma}^{\textnormal{pb}}=e^{t\gamma},\ \
c_{t,\gamma}^{\textnormal{B}}=\gamma^{-t},\ \
c_{t,\gamma}^{\textnormal{g}}=\gamma^{t}.$

For $\bullet=\textnormal{g}$ we require the technical condition
$\sup_{k\in\mathbb{Z}_{+}}a_{k}-\inf_{k\in\mathbb{Z}_{+}}a_{k}<1$ on
$\mathbf{a}$. Moreover, if the parameter $\gamma$ satisfies $\gamma\geq 0$ for
$\bullet=\textnormal{pb}$, $0<\gamma\leq 1$ for $\bullet=\textnormal{B}$,
$\gamma\geq 1$ for $\bullet=\textnormal{g}$, then $h_{\gamma}^{\bullet}$ is
strictly positive.

###### Proof.

The eigenfunction relation in the pure-birth and Bernoulli cases follows
directly from Lemma 3.7. For the geometric case, subject to the condition on
$\mathbf{a}$, it also follows from Lemma 3.7 but only for $\gamma$ close to
$1$. Instead we argue as follows. A direct computation using the explicit
formula for $\mathsf{T}_{1}^{\textnormal{g}}$ from Lemma 3.4 gives for any
$\gamma\in\mathbb{C}$,

$\mathsf{T}_{1}^{\textnormal{g}}h_{\gamma}^{\textnormal{g}}(x)=c_{1,\gamma}^{\textnormal{g}}h_{\gamma}^{\textnormal{g}}(x).$

Then, by virtue of Lemma 3.4, subject to the condition on $\mathbf{a}$, we
obtain the result for any $t\in\mathbb{Z}_{+}$. The final statement regarding
positivity of $h^{\bullet}_{\gamma}(x)$ is immediate from the fact that the
polynomial $p_{x}$ is strictly positive on $(-\infty,0]$. This gives a range
of $\gamma$ which is not optimal, but we restrict to it for simplicity. ∎

Recall Definition 2.27.

######  Proposition 6.3.

In the setting of Definition 2.27 we have,

$\mathsf{T}_{t,\gamma}^{\bullet}(x,y)=\frac{1}{c_{t,\gamma}^{\bullet}}\frac{h_{\gamma}^{\bullet}(y)}{h_{\gamma}^{\bullet}(x)}\mathsf{T}_{t}^{\bullet}(x,y).$

###### Proof.

A direct computation reveals this Doob transform relation between
$\mathsf{T}_{t,\gamma}^{\bullet}$ and $\mathsf{T}_{t}^{\bullet}$. ∎

Clearly, we have the relation

$\mathsf{T}_{t,{\gamma_{1}}}^{\bullet}(x,y)=\frac{c_{t,{\gamma_{2}}}^{\bullet}}{c_{t,{\gamma_{1}}}^{\bullet}}\frac{h_{\gamma_{1}}^{\bullet}(y)h_{\gamma_{2}}^{\bullet}(x)}{h_{\gamma_{2}}^{\bullet}(y)h_{\gamma_{1}}^{\bullet}(x)}\mathsf{T}_{t,\gamma_{2}}^{\bullet}(x,y).$
(84)

We then define the following kernels from $\mathbb{W}_{N+1}$ to
$\mathbb{W}_{N}$ by,

$\displaystyle\Lambda_{N+1,N}^{\gamma_{N+1},\bullet}\left(\mathbf{y},\mathbf{x}\right)$
$\displaystyle=\prod_{j=1}^{N}v_{\gamma_{N+1}}^{\bullet}(x_{j})\mathbf{1}_{\mathbf{x}\prec\mathbf{y}},$
(85)
$\displaystyle\Lambda_{N+1,N}^{\gamma_{N+1},\gamma_{N},\bullet}\left(\mathbf{y},\mathbf{x}\right)$
$\displaystyle=\prod_{j=1}^{N}v_{\gamma_{N+1}}^{\bullet}(x_{j})\frac{h_{\gamma_{N}}^{\bullet}(x_{j})}{h_{\gamma_{N+1}}^{\bullet}(x_{j})}\mathbf{1}_{\mathbf{x}\prec\mathbf{y}},$
(86)

where $v_{\gamma}^{\bullet}(x)$ is given by:

$v_{\gamma}^{\textnormal{pb}}(x)=(a_{x}+\gamma)^{-1},\
v_{\gamma}^{\textnormal{B}}(x)=(\gamma a_{x}+1-\gamma)^{-1},\
v_{\gamma}^{\textnormal{g}}(x)=(\gamma a_{x}-1+\gamma)^{-1}.$

######  Definition 6.4.

Define the strictly positive function
$\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}$ on $\mathbb{W}_{N}$ as
follows:

$\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}\left(\mathbf{x}\right)=\left[\Lambda_{N,N-1}^{\gamma_{N},\gamma_{N-1},\bullet}\Lambda_{N-1,N-2}^{\gamma_{N-1},\gamma_{N-2},\bullet}\cdots\Lambda_{2,1}^{\gamma_{2},\gamma_{1},\bullet}\mathbf{1}\right](\mathbf{x}).$
(87)

For the rest of this section it will be convenient to use the following
notation (note that $\gamma$ here is a scalar)

$\mathcal{P}_{t}^{\gamma,\bullet,N}(\mathbf{x},\mathbf{y})=\det\left(\mathsf{T}^{\bullet}_{t,\gamma}(x_{i},y_{j})\right)_{i,j=1}^{N}.$

Observe that $\mathcal{P}^{\gamma,\bullet,N}_{t}(\mathbf{x},\mathbf{y})$
coincides with $\mathsf{P}_{f}^{(N)}(\mathbf{x},\mathbf{y})$ from Definition
4.1 with the following choices of function $f$ and inhomogeneity $\mathbf{a}$:
if $\bullet=\textnormal{pb}$ then $f(w)=\exp(-tw)$ and the inhomogeneity is
$\mathbf{a}+\gamma$, for $\bullet=\textnormal{B}$, then $f(w)=(1-w)^{t}$ and
the inhomogeneity is $\gamma\mathbf{a}+1-\gamma$ and for
$\bullet=\textnormal{g}$ then $f(w)=(1-w)^{-t}$ and the inhomogeneity is
$\gamma\mathbf{a}+\gamma-1$. Here, for a scalar $c$, $\mathbf{a}+c$ is simply
the sequence $(a_{x}+c)_{x\in\mathbb{Z}_{+}}$. Thus, by virtue of Theorem 4.6
we have the intertwining, with $\mathbf{x}\in\mathbb{W}_{N}$,
$\mathbf{y}\in\mathbb{W}_{N+1}$,

$\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N+1}\Lambda_{N+1,N}^{\gamma_{N+1},\bullet}\left(\mathbf{y},\mathbf{x}\right)=\Lambda_{N+1,N}^{\gamma_{N+1},\bullet}\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N}\left(\mathbf{y},\mathbf{x}\right).$

From relation (84) we then obtain yet another intertwining:

$\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N+1}\Lambda_{N+1,N}^{\gamma_{N+1},\gamma_{N},\bullet}\left(\mathbf{y},\mathbf{x}\right)=\left(\frac{c^{\bullet}_{t,\gamma_{N}}}{c^{\bullet}_{t,\gamma_{N+1}}}\right)^{N}\Lambda_{N+1,N}^{\gamma_{N+1},\gamma_{N},\bullet}\mathcal{P}_{t}^{\gamma_{N},\bullet,N}\left(\mathbf{y},\mathbf{x}\right).$
(88)

Hence, by induction we obtain that
$\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}$ is a strictly positive
eigenfunction of $\mathcal{P}_{t}^{\gamma_{N},\bullet,N}$.

######  Lemma 6.5.

We have:

$\mathcal{P}_{t}^{\gamma_{N},\bullet,N}\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}=\left(c^{\bullet}_{t,\gamma_{N}}\right)^{-(N-1)}\prod_{j=1}^{N-1}c^{\bullet}_{t,\gamma_{j}}\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}.$

With all these preliminaries in place we arrive to the following definition.

######  Definition 6.6.

For all $N\geq 1$, define the Markov kernel
$\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}$ from $\mathbb{W}_{N+1}$ to
$\mathbb{W}_{N}$ by

$\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}\left(\mathbf{y},\mathbf{x}\right)=\prod_{j=1}^{N}v_{\gamma_{N+1}}^{\bullet}(x_{j})\frac{h_{\gamma_{N}}^{\bullet}(x_{j})}{h_{\gamma_{N+1}}^{\bullet}(x_{j})}\frac{\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}(\mathbf{x})}{\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N+1})}(\mathbf{y})}\mathbf{1}_{\mathbf{x}\prec\mathbf{y}},$
(89)

and define abusing notation the Markov kernel
$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}$ from $\mathbb{W}_{N}$ to
itself by

$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}\left(\mathbf{x},\mathbf{y}\right)=\left(c_{t,\gamma_{N}}^{\bullet}\right)^{N-1}\prod_{j=1}^{N-1}\frac{1}{c_{t,\gamma_{j}}^{\bullet}}\frac{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{y})}{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{x})}\det\left(\mathsf{T}_{t,\gamma_{N}}^{\bullet}(x_{i},y_{j})\right)_{i,j=1}^{N}.$
(90)

Here, we have abused notation since in principle we have already defined
$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}$ in (15). From expression
(90) the symmetry in the parameters $\gamma_{1},\gamma_{2},\dots,\gamma_{N}$
is not obvious at all due to the recursive definition of
$\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}$ from (87). Of course,
we will prove shortly in Proposition 6.10 below that (15) and (90) are one and
the same. The following intertwining is then immediate from (88), with
$\mathbf{y}\in\mathbb{W}_{N+1}$, $\mathbf{x}\in\mathbb{W}_{N}$,

$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N+1}\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}\left(\mathbf{y},\mathbf{x}\right)=\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}\left(\mathbf{y},\mathbf{x}\right),$
(91)

and we also note the following explicit formulae.

######  Lemma 6.7.

We have the expressions

$\displaystyle\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}\left(\mathbf{x},\mathbf{y}\right)$
$\displaystyle=\prod_{j=1}^{N}\frac{1}{c_{t,\gamma_{j}}^{\bullet}}\frac{h_{\gamma_{N}}^{\bullet}(y_{j})}{h_{\gamma_{N}}^{\bullet}(x_{j})}\frac{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{y})}{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{x})}\det\left(\mathsf{T}_{t}^{\bullet}(x_{i},y_{j})\right)_{i,j=1}^{N}$
$\displaystyle=\left(c^{\bullet}_{t,\gamma_{N+1}}\right)^{N}\prod_{j=1}^{N}\frac{1}{c_{t,\gamma_{j}}^{\bullet}}\frac{h_{\gamma_{N}}^{\bullet}(y_{j})h_{\gamma_{N+1}}^{\bullet}(x_{j})}{h_{\gamma_{N}}^{\bullet}(x_{j})h_{\gamma_{N+1}}^{\bullet}(y_{j})}\frac{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{y})}{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{x})}\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N}(\mathbf{x},\mathbf{y}).$

We now obtain an alternative expression for
$\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}$ which will give us a
final expression for $\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}$ from
which parameter symmetry will be obvious. The following little observation is
the key ingredient.

######  Lemma 6.8.

We have the identity, $x,y\in\mathbb{Z}_{+}$, $y>x$,

$\sum_{m=x}^{y-1}\frac{v_{\gamma_{2}}^{\bullet}(m)h_{\gamma_{1}}^{\bullet}(m)}{h_{\gamma_{2}}^{\bullet}(m)}=c^{\bullet}_{\gamma_{2};\gamma_{1}}\left[\frac{h_{\gamma_{1}}^{\bullet}(x)}{h_{\gamma_{2}}^{\bullet}(x)}-\frac{h_{\gamma_{1}}^{\bullet}(y)}{h_{\gamma_{2}}^{\bullet}(y)}\right],$
(92)

where the constant $c^{\bullet}_{\gamma_{2};\gamma_{1}}$ is given by

$c^{\textnormal{pb}}_{\gamma_{2};\gamma_{1}}=\frac{1}{\gamma_{2}-\gamma_{1}},\
c^{\textnormal{B}}_{\gamma_{2};\gamma_{1}}=\frac{\gamma_{1}}{\gamma_{1}-\gamma_{2}},\
c^{\textnormal{g}}_{\gamma_{2};\gamma_{1}}=\frac{\gamma_{1}}{\gamma_{2}-\gamma_{1}}.$

###### Proof.

This is proven by induction on $y$. For $y=x+1$ an elementary computation
gives

$\frac{h_{\gamma_{1}}^{\bullet}(x)}{h_{\gamma_{2}}^{\bullet}(x)}-\frac{h_{\gamma_{1}}^{\bullet}(x+1)}{h_{\gamma_{2}}^{\bullet}(x+1)}=\frac{1}{c^{\bullet}_{\gamma_{2};\gamma_{1}}}\frac{v_{\gamma_{2}}^{\bullet}(x)h_{\gamma_{1}}^{\bullet}(x)}{h_{\gamma_{2}}^{\bullet}(x)}.$

Then, we have a telescoping sum and the conclusion follows. ∎

######  Proposition 6.9.

We have

$\displaystyle\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}\left(x_{1},\dots,x_{N}\right)=\prod_{j=2}^{N}\prod_{i=1}^{j-1}c_{\gamma_{j};\gamma_{i}}^{\bullet}\prod_{j=1}^{N}\frac{1}{h_{\gamma_{N}}^{\bullet}(x_{j})}\det\left(h_{\gamma_{i}}^{\bullet}(x_{j})\right)_{i,j=1}^{N}.$

###### Proof.

This is obtained by induction. For the inductive step we can bring the sums
inside the determinant by multinearity and then use the identity (92) from
Lemma 6.8. ∎

By combining the above we obtain the final expression for
$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}$ matching (15).

######  Proposition 6.10.

We have

$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}(\mathbf{x},\mathbf{y})=\prod_{j=1}^{N}\frac{1}{c_{t,\gamma_{j}}^{\bullet}}\times\frac{\det\left(h_{\gamma_{i}}^{\bullet}(y_{j})\right)_{i,j=1}^{N}}{\det\left(h_{\gamma_{i}}^{\bullet}(x_{j})\right)_{i,j=1}^{N}}\det\left(\mathsf{T}_{t}^{\bullet}(x_{i},y_{j})\right)_{i,j=1}^{N}.$
(93)

In particular, $\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}$ is symmetric
in the parameters $\gamma_{1},\gamma_{2},\dots,\gamma_{N}$.

###### Proof.

Combine Lemma 6.7 and Proposition 6.9. ∎

We need one more definition.

######  Definition 6.11.

Define the sub-Markov kernel $\mathfrak{W}_{t}^{N,N+1,\bullet,\gamma_{N+1}}$
from $\mathbb{W}_{N,N+1}$ to itself as follows:

  1. 1.

If $\bullet=\textnormal{pb}$, then with $t\in\mathbb{R}_{+}$,
$\mathfrak{W}_{t}^{N,N+1,\textnormal{pb},\gamma_{N+1}}=\mathcal{Q}_{\exp(-tw)}^{N,N+1}$,
with the underlying inhomogeneity sequence required to define
$\mathcal{Q}_{\exp(-tw)}^{N,N+1}$ being $\mathbf{a}+\gamma_{N+1}$.

  2. 2.

If $\bullet=\textnormal{B}$, then with $t\in\mathbb{Z}_{+}$,
$\mathfrak{W}_{t}^{N,N+1,\textnormal{B},\gamma_{N+1}}=\mathcal{Q}_{(1-w)^{t}}^{N,N+1}$,
with the underlying inhomogeneity sequence required to define
$\mathcal{Q}_{(1-w)^{t}}^{N,N+1}$ being
$\gamma_{N+1}\mathbf{a}-\gamma_{N+1}+1$.

  3. 3.

If $\bullet=\textnormal{g}$, we first define the sub-Markov kernel
$\tilde{\mathcal{G}}^{N,N+1}$ by

$\tilde{\mathcal{G}}^{N,N+1}\left[\left(\mathbf{x},\mathbf{y}\right),\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\right]=\frac{\mathfrak{h}_{N}(\mathbf{x})}{\mathfrak{h}_{N}(\mathbf{x}^{\prime})}\mathcal{G}^{N,N+1}\left[\left(\mathbf{x},\mathbf{y}\right),\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\right],$

with the underlying inhomogeneity sequence defining $\mathcal{G}^{N,N+1}$
being $\gamma_{N+1}\mathbf{a}+\gamma_{N+1}-1$ and $\beta=1$. Then, with
$t\in\mathbb{Z}_{+}$,
$\mathfrak{W}_{t}^{N,N+1,\textnormal{g},\gamma_{N+1}}=\tilde{\mathcal{G}}^{N,N+1}\cdots\tilde{\mathcal{G}}^{N,N+1}$
convolved with itself $t$ times. .

######  Remark 6.12.

The slightly more involved definition for the geometric case is because
$\mathcal{G}^{N,N+1}$ is already Markov as it is the analogue of
$\mathsf{Q}_{f}^{N,N+1}$; in fact, as mentioned earlier, they should be equal
for $f(z)=(1+\beta z)^{-1}$. So we need to unnormalise it first to obtain
$\tilde{\mathcal{G}}^{N,N+1}$, the analogue of $\mathcal{Q}_{f}^{N,N+1}$, in
order to treat all three cases uniformly.

The following is simply a re-writing of Propositions 5.9, 5.15 and 5.16 in the
uniform notation of $\mathfrak{W}_{t}^{N,N+1,\bullet,\gamma_{N+1}}$ (but we
note again that the way the parameter $\gamma_{N+1}$ appears in the
inhomogeneity sequence is different in each case).

######  Proposition 6.13.

Let $N\geq 1$ and $t\geq 0$. Then, we have

$\displaystyle\left[\mathfrak{W}_{t}^{N,N+1,\bullet,\gamma_{N+1}}\Pi_{N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right)$
$\displaystyle=\left[\Pi_{N}\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right),$
$\displaystyle\left[\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N+1}\Lambda_{N+1,N}^{\gamma_{N+1},\bullet}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right)$
$\displaystyle=\left[\Lambda_{N+1,N}^{\gamma_{N+1},\bullet}\mathfrak{W}_{t}^{N,N+1,\bullet,\gamma_{N+1}}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),$

where we view $\Lambda_{N+1,N}^{\gamma_{N+1},\bullet}$ as a Markov kernel from
$\mathbb{W}_{N+1}$ to $\mathbb{W}_{N,N+1}$ as before.

Observe that, by virtue of Lemma 6.7 we can Doob transform the semigroup
$\mathcal{P}_{t}^{\gamma_{N+1},\bullet,N}$ to
$\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}$ using the eigenfunction

$\prod_{j=1}^{N}\frac{h_{\gamma_{n}}^{\bullet}(x_{j})}{h_{\gamma_{N+1}}^{\bullet}(x_{j})}\mathsf{H}^{\bullet}_{(\gamma_{1},\dots,\gamma_{N})}(x_{1},\dots,x_{N})$

having eigenvalue
$\left(c^{\bullet}_{t,\gamma_{N+1}}\right)^{-N}\prod_{j=1}^{N}c_{t,\gamma_{j}}^{\bullet}$.
Hence, we can correctly define the Markov kernel
$\mathsf{D}_{t}^{N,N+1,\bullet,\gamma_{N+1}}$ on $\mathbb{W}_{N,N+1}$ by,

$\displaystyle\mathsf{D}_{t}^{N,N+1,\bullet,\gamma_{N+1}}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right]$
$\displaystyle=\left(c^{\bullet}_{t,\gamma_{N+1}}\right)^{N}\prod_{j=1}^{n}\frac{1}{c_{t,\gamma_{j}}^{\bullet}}\frac{h_{\gamma_{N}}^{\bullet}(y_{j})h_{\gamma_{N+1}}^{\bullet}(x_{j})}{h_{\gamma_{N}}^{\bullet}(x_{j})h_{\gamma_{N+1}}^{\bullet}(y_{j})}\times$
$\displaystyle\times\frac{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{y})}{\mathsf{H}_{(\gamma_{1},\dots,\gamma_{N})}^{\bullet}(\mathbf{x})}\mathfrak{W}_{t}^{N,N+1,\bullet,\gamma_{N+1}}\left[(\mathbf{x},\mathbf{y}),(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right].$

Putting everything together we obtain, by virtue of Proposition 6.13:

######  Proposition 6.14.

Let $N\geq 1$ and $t\geq 0$. Then, we have

$\displaystyle\left[\mathsf{D}_{t}^{N,N+1,\bullet,\gamma_{N+1}}\Pi_{N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right)$
$\displaystyle=\left[\Pi_{N}\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N}\right]\left((\mathbf{x},\mathbf{y}),\mathbf{x}^{\prime}\right),\
(\mathbf{x},\mathbf{y})\in\mathbb{W}_{N,N+1},\mathbf{x}^{\prime}\in\mathbb{W}_{N},$
(94)
$\displaystyle\left[\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,N+1}\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right)$
$\displaystyle=\left[\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}\mathsf{D}_{t}^{N,N+1,\bullet,\gamma_{N+1}}\right]\left(\mathbf{y},(\mathbf{x}^{\prime},\mathbf{y}^{\prime})\right),\
\mathbf{y}\in\mathbb{W}_{N+1},\left(\mathbf{x}^{\prime},\mathbf{y}^{\prime}\right)\in\mathbb{W}_{N,N+1},$
(95)

where we view $\Lambda_{N+1,N}^{\boldsymbol{\gamma},\bullet}$ as a Markov
kernel from $\mathbb{W}_{N+1}$ to $\mathbb{W}_{N,N+1}$ as before.

Theorem 2.24 is then a direct consequence of the following proposition with
the choice $\mu_{N}(\mathbf{x})=\mathbf{1}_{(0,1,\dots,N-1)}(\mathbf{x})$.

######  Proposition 6.15.

Let $N\geq 1$ be arbitrary. Consider the process
$\left(\mathsf{X}_{i}^{(n),\bullet}(t);t\geq 0\right)_{1\leq i\leq n;1\leq
n\leq N}$ from Definition 2.23 in $\mathbb{IA}_{N}$. We assume that the
sequence $\boldsymbol{\gamma}$ satisfies (14) and if $\bullet=\textnormal{B}$
then $\sup_{x\in\mathbb{Z}_{+}}a_{x}\leq 1$ or if $\bullet=\textnormal{g}$
then $\sup_{x\in\mathbb{Z}_{+}}a_{x}-\inf_{x\in\mathbb{Z}_{+}}a_{x}<1$.
Suppose the initial condition in $\mathbb{IA}_{N}$ is of the form,

$\mu_{N}\left(\mathbf{x}^{(N)}\right)\Lambda_{N,N-1}^{\boldsymbol{\gamma},\bullet}\left(\mathbf{x}^{(N)},\mathbf{x}^{(N-1)}\right)\cdots\Lambda_{2,1}^{\boldsymbol{\gamma},\bullet}\left(\mathbf{x}^{(2)},\mathbf{x}^{(1)}\right),$
(96)

for some probability measure $\mu_{N}$ on $\mathbb{W}_{N}$. Then, for any
$1\leq n\leq N$, the projection on the $n$-th row
$\left(\mathsf{X}^{(n),\bullet}(t);t\geq 0\right)$ evolves as a Markov process
with transition probabilities
$\left(\mathcal{P}_{t}^{\boldsymbol{\gamma},\bullet,n}\right)_{t\geq 0}$ given
in (15) and the distribution of
$\left(\mathsf{X}^{(1),\bullet}(\mathfrak{t}),\dots,\mathsf{X}^{(N),\bullet}(\mathfrak{t})\right)$
in $\mathbb{IA}_{N}$ for fixed time $\mathfrak{t}$ is given by

$\left[\mu_{N}\mathcal{P}_{\mathfrak{t}}^{\boldsymbol{\gamma},\bullet,N}\right]\left(\mathbf{x}^{(N)}\right)\Lambda_{N,N-1}^{\boldsymbol{\gamma},\bullet}\left(\mathbf{x}^{(N)},\mathbf{x}^{(N-1)}\right)\cdots\Lambda_{2,1}^{\boldsymbol{\gamma},\bullet}\left(\mathbf{x}^{(2)},\mathbf{x}^{(1)}\right).$
(97)

###### Proof.

Similarly to the proof of Proposition 6.1 this is obtained by induction using
Proposition 6.14 now instead, see [134, 24] for completely analogous
arguments. As in Proposition 6.1, the extra condition for
$\bullet=\textnormal{B}$ is so that the transition probabilities are positive
and for $\bullet=\textnormal{g}$ so that we can use our convolution of kernels
result. ∎

##  7 Computation of the correlation kernels

By virtue of display (83) in Proposition 6.1, with
$\mu_{N}(\mathbf{x})=\mathbf{1}_{\mathbf{x}=(0,1,\dots,N-1)}$, a simple
computation gives that, after running the dynamics described in Theorem 2.9 we
can write out the resulting distribution of the array
$\left(\mathsf{X}_{i}^{(n)}\right)_{1\leq i\leq n;1\leq n\leq N}$, for all
$N\geq 1$, explicitly as a product of determinants. We then apply the Eynard-
Mehta theorem [66, 37, 82] in the form that can be found in Lemma 3.4 of [26]
(we do not recall this theorem explicitly as it has been documented many
times). This immediately gives the existence of a determinantal point process
structure. Finding an explicit expression for the correlation kernel
$\mathfrak{K}_{f}$ is our next task. We need to introduce some notation.

######  Definition 7.1.

Define the functions $\Psi_{N-i}^{(N)}$, for $i=1,\dots,N$, by

$\Psi_{N-i}^{(N)}(x)=-\frac{1}{a_{x}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{w^{N-i}f(w)}{p_{x+1}(w)}dw,\
x\in\mathbb{Z}_{+},$ (98)

with
$f(w)=\prod_{i=1}^{M_{1}}(1-\alpha_{i}w)\prod_{i=1}^{M_{2}}(1+\beta_{i}w)^{-1}\exp(-tw)$
as first defined in (11).

Up to a multiplicative constant,
$\mathfrak{P}_{f}^{(N)}((0,\dots,N-1),\mathbf{x})$ is equal to
$\det(\Psi_{N-j}^{(N)}(x_{i}))_{i,j=1}^{N}$. Note that, the choice of the
$\mathsf{C}_{\mathbf{a}}$ contour in the integral also gives the same
functions $\Psi$ (however the choice of $\mathsf{C}_{\mathbf{a},0}$ will be
important below). Define the kernel $\phi^{(n)}(x_{1},x_{2})$ as the
convolution of $\phi$ from Lemma 4.3 with itself $n$ times.

######  Definition 7.2.

For $1\leq n\leq N-1$ and $1\leq j\leq N$ define the functions
$\Psi_{n-j}^{(n)}$ by convolution with $\phi^{(N-n)}$:

$\Psi_{n-j}^{(n)}(y)=\left[\phi^{(N-n)}\Psi_{N-j}^{(N)}\right](y),\
y\in\mathbb{Z}_{+}.$

We will make use of the following intermediate lemmas.

######  Lemma 7.3.

Let $1\leq n\leq N$ and $1\leq j\leq N$. Then, we have, with $f$ as in (11),

$\Psi_{n-j}^{(n)}(x)=-\frac{1}{a_{x}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{w^{n-j}f(w)}{p_{x+1}(w)}dw.$
(99)

###### Proof.

We prove this by induction. We need to show

$\left[\phi\Psi_{n-j}^{(n)}\right](x)=\Psi_{n-1-j}^{(n-1)}(x).$

The left hand side is equal to

$-\frac{1}{a_{x}}\frac{1}{2\pi\textnormal{i}}\sum_{y>x}-\frac{1}{a_{y}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{w^{n-1-j}f(w)}{p_{y+1}(w)}dw.$

We deform the contour $\mathsf{C}_{\mathbf{a},0}$ to the contour
$\tilde{\mathsf{C}}_{\mathbf{a}}$ from the proof of Lemma 3.5. Then, we can
bring the sum inside the integral, use identity (40) and deform the contour
back to $\mathsf{C}_{\mathbf{a},0}$, without crossing any poles, which gives
$\Psi_{n-1-j}^{(n-1)}(x)$ and concludes the proof. ∎

######  Lemma 7.4.

Let $1\leq k\leq N$. Then, we have

$\phi^{(k)}(y,x)=-\frac{1}{a_{y}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{p_{x}(w)}{p_{y+1}(w)w^{k}}dw.$

###### Proof.

This is Lemma 3.5 in [8]. ∎

######  Lemma 7.5.

Let $1\leq k\leq N$. Then, we have

$\phi^{(k)}(\mathsf{virt},x)=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{0}}\frac{p_{x}(w)}{w^{k}}dw.$

###### Proof.

This is Lemma 3.6 in [8]. ∎

######  Definition 7.6.

Let $1\leq n\leq N$ and $1\leq j\leq n$. Define the following function, with
$f$ as in (11),

$\Phi_{n-j}^{(n)}(x)=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{0}}\frac{p_{x}(w)}{f(w)w^{n-j+1}}dw,\
x\in\mathbb{Z}_{+}.$ (100)

######  Proposition 7.7.

Let $1\leq n\leq N$. Then, we have

$\displaystyle\sum_{x=0}^{\infty}\Psi_{i}^{(n)}(x)\Phi_{j}^{(n)}(x)=\mathbf{1}_{i=j},$

for $0\leq i,j\leq n-1$.

###### Proof.

Observe that since $i$ is non-negative we can use the
$\mathsf{C}_{\mathbf{a}}$ contour in the definition of $\Psi_{i}^{(n)}$. Then,
by deforming, if required, the contour $\mathsf{C}_{0}$ to a contour inside
the region $\mathcal{U}$ defined in (35), we can make use of Lemma 3.3, to
compute with $g(w)=f(w)w^{i}\in\mathsf{Hol}\left(\mathbb{H}_{-R}\right)$ where
$R>R(\mathbf{a})$,

$\displaystyle\sum_{x=0}^{\infty}\Psi_{i}^{(n)}(x)\Phi_{j}^{(n)}(x)$
$\displaystyle=\sum_{x=0}^{\infty}-\frac{1}{a_{x}}\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{w^{i}f(w)}{p_{x+1}(w)}dw\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{0}}\frac{p_{x}(u)}{f(u)u^{j+1}}du$
$\displaystyle=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{0}}\frac{1}{f(u)u^{j+1}}\sum_{x=0}^{\infty}p_{x}(u)\mathsf{T}_{g}(x)du$
$\displaystyle=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C_{0}}}\frac{1}{f(u)u^{j+1}}f(u)u^{i}du=\mathbf{1}_{i=j},$

as desired. ∎

######  Lemma 7.8.

The functions $\left\\{\Phi_{j}^{(n)}(\cdot);0\leq j\leq n-1\right\\}$ span
the space

$\textnormal{span}\left\\{\phi^{(1)}(\mathsf{virt},\cdot),\phi^{(2)}(\mathsf{virt},\cdot),\dots,\phi^{(n)}(\mathsf{virt},\cdot)\right\\}.$

###### Proof.

Using the Cauchy integral formula we see that

$\textnormal{span}\left\\{\phi^{(1)}(\mathsf{virt},\cdot),\phi^{(2)}(\mathsf{virt},\cdot),\dots,\phi^{(n)}(\mathsf{virt},\cdot)\right\\}=\textnormal{span}\left\\{x\mapsto\frac{d^{k-1}}{dw^{k-1}}p_{x}(w)\bigg{|}_{w=0};1\leq
k\leq n\right\\}.$

Again, using the Cauchy integral formula, since $f$ has no zeroes in
$\mathsf{C}_{0}$, we have

$\Phi_{j}^{(n)}(x)=\frac{1}{j!}\frac{d^{j}}{dw^{j}}\left(\frac{p_{x}(w)}{f(w)}\right)\bigg{|}_{w=0}=\frac{1}{j!}\frac{d^{j}}{dw^{j}}p_{x}(w)\bigg{|}_{w=0}+\sum_{k=0}^{j-1}c_{k,j}\frac{d^{k}}{dw^{k}}p_{x}(w)\bigg{|}_{w=0}.$

Thus, we get

$\textnormal{span}\left\\{\Phi_{j}^{(n)}(\cdot);0\leq j\leq
n-1\right\\}=\textnormal{span}\left\\{x\mapsto\frac{d^{k-1}}{dw^{k-1}}p_{x}(w)\bigg{|}_{w=0};1\leq
k\leq n\right\\}$

as required. ∎

######  Lemma 7.9.

For $1\leq n\leq N$, we have

$\phi(\mathsf{virt},\cdot)=\Phi_{0}^{(n)}(\cdot)\equiv 1.$

###### Proof.

This is due to the fact that $f(0)=1$ and the fact that $f$ has no zeroes in
$\mathsf{C}_{0}$. ∎

###### Proof of Theorem 2.9.

We apply a variant of the Eynard-Mehta theorem, in exactly the form found in
Lemma 3.4 of [26], by virtue of all the preceding results in this section, to
obtain the explicit form of the correlation kernel $\mathfrak{K}_{f}$ as
follows:

$\mathfrak{K}_{f}\left[(n_{1},x_{1});(n_{2},x_{2})\right]=-\phi^{(n_{2}-n_{1})}(x_{1},x_{2})\mathbf{1}_{n_{2}>n_{1}}+\sum_{k=1}^{n_{2}}\Psi_{n_{1}-k}^{(n_{1})}(x_{1})\Phi_{n_{2}-k}^{(n_{2})}(x_{2}).$

We need to simplify the sum

$\displaystyle\sum_{k=1}^{n_{2}}\Psi_{n_{1}-k}^{(n_{1})}(x_{1})\Phi_{n_{2}-k}^{(n_{2})}(x_{2})$
$\displaystyle=-\frac{1}{a_{x_{1}}}\frac{1}{(2\pi\textnormal{i})^{2}}\sum_{k=1}^{n_{2}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{f(w)w^{n_{1}-k}}{p_{x_{1}+1}(w)}dw\oint_{\mathsf{C}_{0}}\frac{p_{x_{2}}(u)}{f(u)u^{n_{2}-k+1}}du$
$\displaystyle=-\frac{1}{a_{x_{1}}}\frac{1}{(2\pi\textnormal{i})^{2}}\oint_{\mathsf{C}_{\mathbf{a},0}}dw\oint_{\mathsf{C}_{0}}du\frac{p_{x_{2}}(u)f(w)}{p_{x_{1}+1}(w)f(u)}\sum_{k=1}^{\infty}\frac{w^{n_{1}-k}}{u^{n_{2}-k+1}}$
$\displaystyle=-\frac{1}{a_{x_{1}}}\frac{1}{(2\pi\textnormal{i})^{2}}\oint_{\mathsf{C}_{\mathbf{a},0}}dw\oint_{\mathsf{C}_{0}}du\frac{p_{x_{2}}(u)f(w)}{p_{x_{1}+1}(w)f(u)}\frac{w^{n_{1}}}{u^{n_{2}}}\frac{1}{w-u},$

where we have extended the sum over $k$ from $n_{2}$ to infinity. This is
allowed because there are no additional contributions for $k>n_{2}$ since
there are no poles in $u$ in $\mathsf{C}_{0}$ and thus the $u$-contour
integral vanishes. Moreover, note that since $|u|<|w|$ the geometric series
converges. This concludes the proof. ∎

######  Lemma 7.10.

The functions $\mathfrak{h}_{N}$ can be written as

$\mathfrak{h}_{N}\left(\mathbf{x}\right)=\det\left(\phi^{(N+1-i)}\left(\mathsf{virt},x_{j}\right)\right)_{i,j=1}^{N}=\det\left(\frac{1}{(i-1)!}\left(-\frac{d}{dw}\right)^{i-1}p_{x_{j}}(w)\big{|}_{w=0}\right)_{i,j=1}^{N}.$
(101)

###### Proof.

This easily follows by the formula for $\mathfrak{h}_{N}(\mathbf{x})$ from
Definition 4.7. ∎

######  Proposition 7.11.

Consider the process $\left(\mathsf{X}_{k}^{(n)}(t);t\geq 0\right)_{1\leq
k\leq n;n\geq 1}$ with discrete-time Bernoulli or geometric dynamics
determined by the functions $f_{i,i+1}(z)$ as in Theorem 2.14. Let $M\geq 1$
and $t_{1}<t_{2}<\cdots<t_{M}$ be arbitrary times. The joint distribution of
$\left(\mathsf{X}^{(N)}(t_{1}),\dots,\mathsf{X}^{(N)}(t_{M})\right)$ is then
given by

$\displaystyle\frac{1}{Z}\det\left(\tilde{\Phi}^{(t_{1})}_{j}\left(x_{j}^{(1)}\right)\right)_{i,j=1}^{N}\prod_{r=1}^{M-1}\det\left(\mathsf{T}_{f_{t_{r},t_{r+1}}}\left(x^{(r)}_{i},x^{(r+1)}_{j}\right)\right)_{i,j=1}^{N}\det\left(\tilde{\Psi}^{(t_{M})}_{j}\left(x_{j}^{(M)}\right)\right)_{i,j=1}^{N},$

where the functions $\tilde{\Psi}_{j}$ and $\tilde{\Phi}_{j}$ are given by:

$\displaystyle\tilde{\Phi}^{(t_{1})}_{j}(x)$
$\displaystyle=-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{x}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{f_{0,t_{1}}(w)w^{j-1}}{p_{x+1}(w)}dw,$
$\displaystyle\tilde{\Psi}^{(t_{M})}_{j}(x)$
$\displaystyle=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{0}}\frac{p_{x}(w)}{f_{0,t_{M}}(w)w^{j}}dw,$

and $Z$ is some normalisation constant. In the case of
$\left(\mathsf{X}_{k}^{(n)}(t);t\geq 0\right)_{1\leq k\leq n;n\geq 1}$
following the continuous-time pure-birth dynamics the conclusion is exactly
the same but with $t_{i}\in\mathbb{R}_{+}$ and
$f_{t_{1},t_{2}}(z)=e^{-(t_{2}-t_{1})z}$ instead.

###### Proof.

By virtue of Proposition 6.1, we also note Lemma 7.10,
$\left(\mathsf{X}^{(N)}(t);t\geq 0\right)$ evolves as a Markov process with
transition probabilities given by (12). Then, by using Markov property we can
write down the following formula for the distribution of
$\left(\mathsf{X}^{(N)}(t_{1}),\dots,\mathsf{X}^{(N)}(t_{M})\right)$,

$\displaystyle\frac{1}{\tilde{Z}}\det\left(\mathsf{T}_{f_{0,t_{1}}}\left(i-1,x^{(1)}_{j}\right)\right)_{i,j=1}^{N}\det\left(\mathsf{T}_{f_{t_{1},t_{2}}}\left(x^{(1)}_{i},x^{(2)}_{j}\right)\right)_{i,j=1}^{N}\cdots\det\left(\mathsf{T}_{f_{t_{M-1},t_{M}}}\left(x^{(1)}_{i},x^{(2)}_{j}\right)\right)_{i,j=1}^{N}\mathfrak{h}_{N}\left(x^{(N)}\right),$

for some normalisation constant $\tilde{Z}$. Then, by virtue of Lemma 7.8 (in
particular, since the $\tilde{\Psi}_{i}^{(t_{M})}(z)$ functions, which are
essentially the $\Phi$ functions from Definition 7.6, span the space spanned
by the $\phi^{(i)}(\mathsf{virt},z)$ functions) and simple row operations on
the determinants defining the first and last factors we obtain the desired
result. Finally, the exact same argument gives the conclusion in the
continuous-time case. ∎

###### Proof of Theorem 2.14.

It only remains to compute the correlation functions, as the probabilistic
statement that projections on single levels are Markovian follows from
Proposition 6.1. Define, for $1\leq k\leq M$, the functions
$\tilde{\Phi}_{i}^{(t_{k})}$ and $\tilde{\Psi}_{j}^{(t_{k})}$ by the
convolutions

$\displaystyle\tilde{\Phi}_{j}^{(t_{k})}\left(x\right)$
$\displaystyle=\left[\tilde{\Phi}_{j}^{(t_{1})}\mathsf{T}_{f_{t_{1},t_{2}}}\cdots\mathsf{T}_{f_{t_{k-1},t_{k}}}\right]\left(x\right),$
$\displaystyle\tilde{\Psi}_{j}^{(t_{k})}\left(x\right)$
$\displaystyle=\left[\mathsf{T}_{f_{t_{k},t_{k+1}}}\cdots\mathsf{T}_{f_{t_{M-1},t_{M}}}\tilde{\Psi}_{j}^{(t_{M})}\right]\left(x\right).$

Then, arguing as in the proof of Lemma 3.4 we obtain the following explicit
description for them, for any $1\leq k\leq M$:

$\displaystyle\tilde{\Phi}^{(t_{k})}_{j}(x)$
$\displaystyle=-\frac{1}{2\pi\textnormal{i}}\frac{1}{a_{x}}\oint_{\mathsf{C}_{\mathbf{a},0}}\frac{f_{0,t_{k}}(w)w^{j-1}}{p_{x+1}(w)}dw,$
$\displaystyle\tilde{\Psi}^{(t_{k})}_{j}(x)$
$\displaystyle=\frac{1}{2\pi\textnormal{i}}\oint_{\mathsf{C}_{0}}\frac{p_{x}(w)}{f_{0,t_{k}}(w)w^{j}}dw.$

Finally, for any $1\leq k\leq M$, we have by an obvious adaptation of
Proposition 7.7,

$\sum_{x=0}^{\infty}\tilde{\Phi}_{i}^{(t_{k})}(x)\tilde{\Psi}_{j}^{(t_{k})}(x)=\mathbf{1}_{i=j},$

for $1\leq i,j\leq N$. Thus, by virtue of Proposition 7.11 and the Eynard-