We show the proof for the case
\(t_1,t_2>s\) only. For the other cases the proof is similar. We are going to use a fixed-point argument. For any choice of
\(T\in (s,\infty )\), the set
$$\begin{aligned} BV_2^{|\mathcal {Z}|}:=\Big \{ f:[s,T]\times [s,T]&\rightarrow \mathbb {R}^{|\mathcal {Z}|^2}\Big |\text { there exist} \text { finite signed measures }\mu _{ik}\\&\text { with }f_{ik}(x,y)=\mu _{ik}([s,x]\times [s,y]),x,y\in [s,T]\times [s,T]\Big \} \end{aligned}$$
is a linear space. The Hahn-Jordan decomposition offers for any finite signed measure a unique decomposition into a difference of two finite measures, and this decomposition can be furthermore used to decompose also any
\(f \in BV_2^{|\mathcal {Z}|}\) into a difference
\(f=f^+-f^-\) of nonnegative mappings
\(f^+,f^- \in BV_2^{|\mathcal {Z}|}\). Based on this unique construction of
\(f^+\) and
\(f^-\), we then define
\(|f|:= f^++f^-\). By equipping
\(BV_2^{|\mathcal {Z}|}\) with the norm
$$\begin{aligned} \Vert (f_{ik})_{i,k\in \mathcal {Z}}\Vert :=\sum _{i,k\in \mathcal {Z}}\;\int \displaylimits _{[s,T]\times [s,T]}|f_{ik}|(\textrm{d}t_1,\textrm{d}t_2)+\sum _{i,k\in \mathcal {Z}}| f_{ik}(s,s)| \end{aligned}$$
we obtain a metric space. On this metric space, we define an operator
\(O:BV_2^{|\mathcal {Z}|}\rightarrow BV_2 ^{|\mathcal {Z}|}\) as follows:
$$\begin{aligned} \left( O((f_{jl})_{j,l\in \mathcal {Z}})\right) _{ik}(t_1,t_2):&=\sum _{{j,l\in \mathcal {Z}}}\quad \;\int \displaylimits _{(s,t_1]\times (s,t_2]}f_{jl}(u_1^-,u_2^-) \Lambda _{jilk}(\textrm{d}u_1,\textrm{d}u_2) \end{aligned}$$
for
\(t_1,t_2\in [s,T]\). We want this operator to be a contraction, but unfortunately this is not true, so we need to replace our norm for
\(BV_2^{|\mathcal {Z}|}\) by another equivalent norm. Let
$$\begin{aligned} \nu (t_1,t_2):=4\sum _{\begin{array}{c} i,j,k,l\in \mathcal {Z}\\ i\ne j\\ k\ne l \end{array}}\Lambda _{ijkl}(t_1,t_2), \quad t_1,t_2 \in [s,T]. \end{aligned}$$
Because of assumption (
3.3), each
\(\Lambda _{ijkl}\),
\(i\ne j\),
\(k\ne l\), has an associated measure
\({\mu }_{\Lambda _{ijkl}}\) that is almost surely finite on
\([s,T]^2\). Hence, there exists a finite measure
\(\mu\) that satisfies
$$\begin{aligned} \nu (t_1,t_2)=\mu ([s,t_1]\times [s,t_2]),\quad t_1,t_2\in [s,T]. \end{aligned}$$
(4.1)
We moreover define
$$\begin{aligned} \nu _1(u):&=\mu ([s,u]\times [s,T]),\\ \nu _2(u):&=\mu ([s,T]\times [s,u]), \quad u \in [s,T], \end{aligned}$$
which are càdlàg by construction. For each
\(K \in (0,\infty )\) the mapping
\(\Vert \cdot \Vert _K\) defined by
$$\begin{aligned} \Vert (f_{ij})_{i,j\in \mathcal {Z}}\Vert _K:=\sum _{i,j\in \mathcal {Z}} \;\int \displaylimits _{[s,T]\times [s,T]}e^{-K(\nu _1(u_1)+\nu _2(u_2))}|f_{ij}|(\textrm{d}u_1,\textrm{d}u_2)+\sum _{i,k\in \mathcal {Z}}| f_{ik}(s,s)| \end{aligned}$$
is a norm on
\(BV_2^{|\mathcal {Z}|}\) that is equivalent to the norm
\(\Vert \cdot \Vert\). This construction of
\(\Vert \cdot \Vert _K\) is inspired by [
4] but is here extended to the two-dimensional case. For any
\(f \in BV_2^{|\mathcal {Z}|}\),
\((a,b]\times (c,d]\subset [s,T]\times [s,T]\) and
\(k,i\in \mathcal {Z}\), the definitions of operator
O and the two-dimensional transition rates
\((\Lambda _{ijkl})_{i,j,k,l\in \mathcal {Z}}\) in conjunction with the triangle inequality yield that
$$\begin{aligned} \int \displaylimits _{(a,b]\times (c,d]}|O(f)_{ik}|(\textrm{d}u_1,\textrm{d}u_2)&\le \sum _{\begin{array}{c} l:l\ne k\\ j:j\ne i \end{array}}\bigg (\;\int \displaylimits _{(a,b]\times (c,d]}|f_{jl}(u_1^-,u_2^-)|\Lambda _{jilk}(\textrm{d}u_1,\textrm{d}u_2)\\&\quad +\;\int \displaylimits _{(a,b]\times (c,d]}|f_{jk}(u_1^-,u_2^-)|\Lambda _{jikl}(\textrm{d}u_1,\textrm{d}u_2)\\&\quad +\;\int \displaylimits _{(a,b]\times (c,d]}|f_{il}(u_1^-,u_2^-)|\Lambda _{ijlk}(\textrm{d}u_1,\textrm{d}u_2)\\&\quad +\;\int \displaylimits _{(a,b]\times (c,d]}|f_{ik}(u_1^-,u_2^-)|\Lambda _{ijkl}(\textrm{d}u_1, \textrm{d}u_2)\bigg ). \end{aligned}$$
Summation over
i,
k and a reordering of some of the resulting sums lead to the inequality
$$\begin{aligned} \sum _{i,k} \int \displaylimits _{(a,b]\times (c,d]}|O(f)_{ik}|(\textrm{d}u_1,\textrm{d}u_2)&\le 4\sum _{\begin{array}{c} i,j,k,l\\ i \ne j, k \ne l \end{array} } \;\int \displaylimits _{(a,b]\times (c,d]}|f_{ik}(u_1^-,u_2^-)| \Lambda _{ijkl}(\textrm{d}u_1,\textrm{d}u_2)\\&\le 4 \sum _{i,k}\;\int \displaylimits _{(a,b]\times (c,d]}|f_{ik}(u_1^-,u_2^-)|\, \nu (\textrm{d}u_1,\textrm{d}u_2). \end{aligned}$$
Suppose now that
f(
s,
s) is zero. Then it holds that
$$\begin{aligned} \sum _{i,k} \int \displaylimits _{(a,b]\times (c,d]}|O(f)_{ik}|(\textrm{d}u_1,\textrm{d}u_2)&\le 4 \sum _{k,i}\;\int \displaylimits _{(a,b]\times (c,d]}\;\int \displaylimits _{[s,u_1)\times [s,u_2)}|f_{ik}|(\textrm{d}r_1,\textrm{d}r_2)\,\nu (\textrm{d}t_1,\textrm{d}t_2). \end{aligned}$$
As a consequence, the norm
\(\Vert \cdot \Vert _{K}\) of
O(
f) has an upper bound of
$$\begin{aligned} \begin{aligned} \Vert O(f)\Vert _{K}&=\sum _{i,k} \;\int \displaylimits _{[s,T]\times [s,T]}e^{-K(\nu _1(u_1)+\nu _2(u_2))}|O(f)_{ik}|(\textrm{d}u_1,\textrm{d}u_2)\\&\le \sum _{i,k} \;\int \displaylimits _{[s,T]\times [s,T]}e^{-K(\nu _1(u_1)+\nu _2(u_2))}\;\int \displaylimits _{[s,u_1)\times [s,u_2)}|f_{ik}|(\textrm{d}r_1,\textrm{d}r_2)\,\nu (\textrm{d}u_1,\textrm{d}u_2)\\&=\sum _{i,k} \;\int \displaylimits _{[s,T]\times [s,T]}\;\int \displaylimits _{(r_1,T]\times (r_2,T]}e^{-K(\nu _1(u_1)+\nu _2(u_2))}\nu (\textrm{d}u_1,\textrm{d}u_2))|f_{ik}|(\textrm{d}r_1,\textrm{d}r_2), \end{aligned}\end{aligned}$$
(4.2)
where the last equality uses Fubini’s theorem. Moreover, for arbitrary but fixed
\(u_1,u_2 \in [s,T]\) let
$$\begin{aligned} \tilde{\nu }(r_1,r_2)&:=\frac{\mu ((u_1,r_1]\times (u_2,r_2])}{\mu ((u_1,T]\times (u_2,T])}, \quad r_1 \in (u_1,T], \, r_2 \in (u_2,T], \end{aligned}$$
for the same
\(\mu\) is in equation (
4.1). Without loss of generality we assume that
\(\mu ((u_1,T]\times (u_2,T]) >0\). Otherwise the conclusion that we want to draw is trivial. Then
$$\begin{aligned} \tilde{\nu }_1(r_1)&:=\tilde{\nu }(r_1,T),\\ \tilde{\nu }_2(r_2)&:=\tilde{\nu }(T,r_2),\quad (r_1,r_2)\in (u_1,T]\times (u_2,T], \end{aligned}$$
correspond to cumulative distribution functions. Let
\(C:=\tilde{\nu }(u_1,u_2)>0\) and let (
A,
B) be a random vector that has
\(\tilde{\nu }\) as its two-dimensional cumulative distribution function. Then, by applying Sklar’s theorem, see e.g. Theorem 2.3.3 in [
13], we can show that
$$\begin{aligned} \begin{aligned}&\int \displaylimits _{(u_1,T]\times (u_2,T]}e^{-K(\nu _1(r_1)+\nu _2(r_2))}\nu (\textrm{d}r_1,\textrm{d}r_2)\\&\quad \le Ce^{-K(\nu _1(u_1)+\nu _2(u_2)}\;\int \displaylimits _{(u_1,T]\times (u_2,T]}e^{-CK(\tilde{\nu }_1(r_1)+\tilde{\nu }_2(r_2))}\tilde{\nu }(\textrm{d}r_1,\textrm{d}r_2)\\&\quad =Ce^{-K(\nu _1(u_1)+\nu _2(u_2)}\,\mathbb {E}\left[ e^{-CK(\tilde{\nu }_1(\tilde{\nu }_1^{-1}(U))+\tilde{\nu }_2(\tilde{\nu }_2^{-1}(V)))}\right] \end{aligned}\end{aligned}$$
(4.3)
for a suitable random vector (
U,
V) whose components are uniformly distributed on (0, 1) and such that (
A,
B) and
\((\tilde{\nu }_1^{-1}(U),\tilde{\nu }_2^{-1}(V))\) have the same distribution. Note here that the copula of (
U,
V) may be non-trivial. The inverse functions
\(\tilde{\nu }_1^{-1}\) and
\(\tilde{\nu }_2^{-1}\) are here defined as
\(\tilde{\nu }_n^{-1}(t):=\inf \{x:\tilde{\nu }_n(x)\ge t\}\),
\(n=1,2\) for
\(t\in (0,1)\). Since
\(\tilde{\nu }_1(\tilde{\nu }_1^{-1}(t)) \ge t\) for
\(t \in (0,1)\), see e.g. Theorem 3.1 in [
17], the inequality (
4.3) has an upper bound of
$$\begin{aligned}&\int \displaylimits _{(u_1,T]\times (u_2,T]}e^{-K(\nu _1(r_1)+\nu _2(r_2))}\nu (\textrm{d}r_1,\textrm{d}r_2) \le Ce^{-K(\nu _1(u_1)+\nu _2(u_2)}\,\mathbb {E}\left[ e^{-CK(U+V)}\right] . \end{aligned}$$
By applying Theorem 10.6.4 in [
10], we can show that the latter expectation has an upper bound of
$$\begin{aligned} \mathbb {E}\left[ e^{-CK(U+V)}\right] \le \mathbb {E}\left[ e^{-CK(2U)}\right] \end{aligned}$$
since
\(x \mapsto \exp \{ - C Kx\}\) is a convex function. As
U is uniformly distributed on (0, 1), we moreover have
$$\begin{aligned} \mathbb {E}\left[ e^{-CK(2U)}\right]&= \int \displaylimits _{(0,1)}e^{-CK(2u)}\textrm{d}u\\&= \frac{1}{2KC}\left( 1-e^{-2KC}\right) \\&\le \frac{1}{2KC}. \end{aligned}$$
We set
\(K=1\). Then, all in all we can conclude that the inequality (
4.2) has an upper bound of
$$\begin{aligned} \Vert O(f)\Vert _{K}&\le \frac{1}{2K} \sum _{i,k} \;\int \displaylimits _{[s,T]\times [s,T]}e^{-K(\nu _1(u_1)+\nu _2(u_2))}|f_{ik}|(\textrm{d}u_1,\textrm{d}u_2)\\&= \frac{1}{2} \Vert f \Vert _{K} \end{aligned}$$
whenever
f(
s,
s) equals zero. Suppose now that we have two solutions
\(R=(P_{ik})_{ik}\) and
\(R'=(P'_{ik})_{ik}\) of (
3.6) for given
\((P_{i})_{i}\). Then
\(R-R' \in BV_2^{|\mathcal {Z}|}\) is a fixed point of the operator
O and
\((R-R')(s,s)\) is zero. So we have
$$\begin{aligned} \Vert R-R' \Vert _{K} \le \frac{1}{2} \Vert R-R' \Vert _{K}, \end{aligned}$$
which necessarily implies that
\(R-R'\) is zero on
\([s,T]\times [s,T]\). In an analogous way it is possible to proof
\(R=R'\) also on the three rectangles
\([0,s]^2\),
\([0,s]\times (s,T]\) and
\((s,T]\times [0,s]\). Since
\(T \in (s,\infty )\) was arbitrary, we can expand the uniqueness property to infinity.