Reciprocal Class of Jump Processes

Abstract

Processes having the same bridges as a given reference Markov process constitute its reciprocal class. In this paper we study the reciprocal class of compound Poisson processes whose jumps belong to a finite set $\mathcal {A}\subset \mathbb {R}^{d}$. We propose a characterization of the reciprocal class as the unique set of probability measures on which a family of time and space transformations induces the same density, expressed in terms of the reciprocal invariants. The geometry of $\mathcal {A}$ plays a crucial role in the design of the transformations, and we use tools from discrete geometry to obtain an optimal characterization. We deduce explicit conditions for two Markov jump processes to belong to the same class. Finally, we provide a natural interpretation of the invariants as short-time asymptotics for the probability that the reference process makes a cycle around its current state.

Appendix

1.1 Proof of Step 1 in Theorem 4.2

We first observe that it is sufficient to prove that

$$\begin{aligned} \mathbb {Q}(.|\mathbf {N}_1=\mathbf {n}) << \mathbb {P}_{\nu }(.|\mathbf {N}_1=\mathbf {n}) \, \text { for all } \mathbf {n}\text { such that }\mathbb {Q}(\mathbf {N}_1=\mathbf {n})>0. \end{aligned}$$

To this aim, we use an approximation argument.

Let us fix $\mathbf {n}$ and construct a discrete (dyadic) approximation of the jump times. For $m \ge \max _{j=1,\ldots ,A} \log _{2} (n^j)+1:=\bar{m}$ , $\mathcal {D}^m$ is composed by A ordered sequences of dyadic numbers, the jth sequence having length $n^j$:

$$\begin{aligned} \mathcal {D}^{m}\!:= \!\left\{ \mathbf {k}= (k^j_i)_{j \le A, i \le n^j} : \, k^j_i \in 2^{-m} \mathbb {N}, 0 < k^{j} _{i-1} < k^{j}_{i} \le 1, \quad \forall j \le A, \quad \forall i \le n^j \right\} \end{aligned}$$

For $\mathbf {k}\in \mathcal {D}^m$, we define the subset of trajectories whose jump times are localized around $\mathbf {k}$:

$$\begin{aligned} O^{m}_{\mathbf {k}} = \left\{ \mathbf {N}_1= \mathbf {n}\right\} \cap \bigcap _{\mathop { i \le n^j}\limits ^{j \le A }} \left\{ 0 \le k^j_i - T^j_i < 2^{-m} \right\} \end{aligned}$$

(6.4)

Moreover, as a final preparatory step, we observe for every $m \ge \bar{m}$, $\mathbf {k},\mathbf {k}' \in \mathcal {D}^{m}$ one can easily construct $u \in \mathcal {U}$ such that:

$$\begin{aligned} u(j,t) = t+k'^j_i-k^j_i , \quad \forall j \le A, i \le n^j \ \mathrm{and} \ t \ \mathrm{s.t.} \ 0 \le k^j_i - t < 2^{-m} \end{aligned}$$

(6.5)

We can observe that (6.5) ensures $\dot{u}(j,T^j_i) =1$ over $O^{m}_{\mathbf {k}}$, and that $O_{\mathbf {k}'}^m = \pi _u^{-1}(O^{m}_{\mathbf {k}})$. We choose $F= \mathbbm {1}_{ O^{\mathbf {n}}_{\mathbf {k}'} } \mathbbm {1}_{\{\mathbf {N}_1=\mathbf {n}\}}/\mathbb {Q}(\mathbf {N}_1= \mathbf {n})$ and u as in (6.5) and apply (3.1) to obtain :

$$\begin{aligned} \begin{aligned} \mathbb {Q}\left( O^{m}_{\mathbf {k}'} \Big | \mathbf {N}_1= \mathbf {n}\right)&= \mathbb {Q}\left( \left\{ \omega : \pi _{u}(\omega ) \in O^{m}_{\mathbf {k}}\right\} | \mathbf {N}_1=\mathbf {n}\right) \\ \!&= \!\mathbb {Q}\left( \mathbbm {1}_{O^{m}_{\mathbf {k}}} \exp \left( \sum _{j=1}^{A}\int _{0}^{1} \log \varXi ^{\nu }(j,t,u(j,t)) \ \dot{u}(j,t) \mathrm{d}N_t^j \right) \Big | \mathbf {N}_1\!=\! \mathbf {n}\right) \\&\ge C \ \mathbb {Q}\left( O^{m}_{\mathbf {k}}\Big | \mathbf {N}_1= \mathbf {n}\right) , \end{aligned} \end{aligned}$$

where

$$\begin{aligned} C:= \Big ( \inf _{s,t \in [0,1], j \le A } \varXi ^{\nu } (j,s,t) \Big )^{\sum _{j} \mathbf {n}_j} >0 \end{aligned}$$

(6.6)

since $\nu \in \mathcal {J}$. With a simple covering argument, we obtain, for all $m \ge \bar{m}$ and $\mathbf {k}\in \mathcal {D}^m$,

$$\begin{aligned}&\sharp \mathcal {D}^m \min \left\{ 1,\frac{1}{C} \right\} \mathbb {Q}\left( O^{m}_{\mathbf {k}} | \mathbf {N}_1=\mathbf {n}\right) \\&\quad \le \mathbb {Q}\left( O^{m}_{\mathbf {k}} | \mathbf {N}_1=\mathbf {n}\right) + \sum _{\mathop {\mathbf {k}' \ne \mathbf {k}}\limits ^{\mathbf {k}' \in \mathcal {D}^m}} \mathbb {Q}\left( O^{m}_{\mathbf {k}'} |\mathbf {\mathbf {N}_1} = \mathbf {n}\right) \le 1 . \end{aligned}$$

It can be shown with a direct computation that $\frac{1}{|\mathcal {D}^m|} \le C' \mathbb {P}_{\nu }( O^m_{\mathbf {k}} | \mathbf {\mathbf {N}_1}=\mathbf {n})$ for some $C'>0$ uniformly in $m, \mathbf {k}\in \mathcal {D}^m$ (the proof is given in Lemma 6.6). Therefore there exists a constant $C^{''}>0$ such that:

$$\begin{aligned} \mathbb {Q}\left( O^{m}_{\mathbf {k}}|\mathbf {N}_1=\mathbf {n}\right) \le C^{''} \ \mathbb {P}_{\nu }\left( O^{m}_{\mathbf {k}}|\mathbf {N}_1=\mathbf {n}\right) ,\quad \forall m \ge \bar{m} , \mathbf {k}\in \mathcal {D}^{m} . \end{aligned}$$

With a standard approximation argument, using the fact that $\mathbb {Q}(\varOmega ) =1$, we can extend the last bound to any measurable set. This completes the proof of the absolute continuity.

We are left with the proof of the following Lemma.

Lemma 6.6

Let $\mathcal {D}^m$ and $\mathbb {P}_{\nu }$ as before. Then there exists a constant $C^{'}$ such that for m large enough,

$$\begin{aligned} C^{'} \ \mathbb {P}_{\nu }\left( O^{m}_{\mathbf {k}}|\mathbf {N}_1= \mathbf {n}\right) \ge \frac{1}{\sharp \mathcal {D}^{m}} \end{aligned}$$

Proof

We want to prove that for $\mathbf {n}\in \mathbb {N}^A$:

$$\begin{aligned} \frac{1}{\sharp \mathcal {D}^m} \le C' \mathbb {P}_{\nu }\left( O^{m}_{\mathbf {k}}| \mathbf {N}_1=\mathbf {n}\right) , \quad \forall \ m \ge \max _{j\le A} \log (n^j)+1 , \, \mathbf {k}\in \mathcal {D}^m \end{aligned}$$

(6.7)

We can first compute explicitly $\sharp \mathcal {D}^m$ with a simple combinatorial argument: Each $\mathbf {k}\in \mathcal {D}^m$ is constructed by choosing $n^j$ dyadic intervals, $j \le A$, and ordering them. Therefore

$$\begin{aligned} \sharp \mathcal {D}^m = \prod _{j=1}^{A} \left( {\begin{array}{c}2^{m}\\ n^j\end{array}}\right) . \end{aligned}$$

(6.8)

On the other hand, we observe that defining $\tilde{\nu }(\mathrm{d}x\mathrm{d}t) =\sum _{j=1}^{A}\delta _{a^j}(\mathrm{d}x) \otimes \mathrm{d}t$, $\mathbb {P}_{\tilde{\nu }}$ is equivalent to $\mathbb {P}_{\nu }$, and therefore, we can prove (6.7) replacing $\mathbb {P}_{\nu }$ with $\mathbb {P}_{\tilde{\nu }}$. To do this, for each $\mathbf {k}\in \mathcal {D}^m$, we define the function:

$$\begin{aligned} \delta : \{ 1,\ldots , 2^{m} \} \times \{1,\ldots ,A \} \longrightarrow \{ 0,1 \} \\ \delta ( i,j) := {\left\{ \begin{array}{ll} 1, &{} \quad \text{ if } i \in \left\{ 2^{m}\mathbf {k}^j_1,\ldots ,2^{m} \mathbf {k}^j_{n^j} \right\} \\ 0 , &{} \quad \text{ otherwise } \text{. }\end{array}\right. } \end{aligned}$$

Then, using the explicit distribution of $\mathbb {P}_{\tilde{\nu }}$,

$$\begin{aligned}&\mathbb {P}_{\tilde{\nu }}\left( O^m_{\mathbf {k}} | \mathbf {N}_1= \mathbf {n}\right) \\&\quad = \mathbb {P}_{\tilde{\nu }} \left( \bigcap _{(i,j) \in \{1,..,2^m \} \times \{1,..,A\}} \left\{ N^j_{\frac{i}{2^m}} - N^j_{\frac{i}{2^m}} = \delta (i,j) \right\} \Big | \mathbf {N}_1=\mathbf {n}\right) \\&\quad = \exp (A)\exp (-2^{-m})^{2^{m}A} (2^{-m})^{\left( \sum _j n^j \right) }\prod _{j=1}^A n^j! = \prod _{j=1}^A 2^{-mn^j} n^j! \end{aligned}$$

It is now easy to see that there exists a constant $C_0>0$ such that:

$$\begin{aligned} \left( {\begin{array}{c}2^m\\ n^j\end{array}}\right) \ge C_0 \frac{2^{m n^j}}{ n^j!} , \quad \forall \ j \le A , \ m \ge \max _{j=1,\ldots ,A} \log (n^j)+1, \ \mathbf {k}\in \mathcal {D}^m \end{aligned}$$

from which the conclusion follows. $\square$

1.2 Proof of Proposition 5.4

• (i) Let $\mathbf {n}\in \mathbb {N}^A, \mathbf {m} \in \mathfrak {F}_{\mathbf {A},\mathbf {n}}$. Since $\ker ^{*}_{\mathbb {Z}}(\mathbf {A})$ is a lattice basis, there exists $\mathbf {c}_1,\ldots ,\mathbf {c}_K \subseteq (\ker ^{*}_{\mathbb {Z}}(\mathbf {A})\cup -\ker ^{*}_{\mathbb {Z}}(\mathbf {A}))^K$ such that, if we define recursively

  $$\begin{aligned} w_0 = \mathbf {n}, \quad w_k = \theta _{\mathbf {c}_k} w_{k-1} \end{aligned}$$

  then we have that $w_K = \mathbf {m}$. Let us consider l large enough such that

  $$\begin{aligned} l \ \min _{j=1,\ldots ,A} \bar{c}^j \ge \left| \min _{\mathop {k=1,\ldots ,K}\limits ^{j=1\ldots ,A}}w^j_k\right| . \end{aligned}$$

  (6.9)

  We then consider the sequence $w'_k$, $k = 0,\ldots ,K+2l$ defined as follows:

  $$\begin{aligned} w'_k= {\left\{ \begin{array}{ll} \theta _{\bar{\mathbf {c}}}w'_{k-1}, \quad &{} \text{ if } \,\,1 \le k \le l\\ \theta _{\mathbf {c}_{k-l}} w'_{k-1}, \quad &{} \text{ if } \,\, l+1 \le k \le K+l \\ \theta _{-\bar{\mathbf {c}}} w'_{k-1} \quad &{} \text{ if } \,\, K+l+1 \le k \le K+2l .\end{array}\right. } \end{aligned}$$

  It is now easy to check, thanks to condition (6.9) that

  $$\begin{aligned} \ w'_k \in \mathfrak {F}_{\mathbf {A},\mathbf {n}} \quad \forall \ k \le K+2l . \end{aligned}$$

  Since all the shifts involved in the definition of $w'_k$ are associated with vectors in $\ker ^{*}_{\mathbb {Z}}(\mathbf {A})\cup - \ker ^{*}_{\mathbb {Z}}(\mathbf {A})$, we also have that $w'_k \in \mathfrak {F}_{\mathbf {A},\mathbf {n}}$ and $(w'_{k-1},w'_k)$ are an edge of $\mathcal {G}(\mathfrak {F}_{\mathbf {A},\mathbf {n}} ,\ker ^{*}_{\mathbb {Z}}(\mathbf {A})), k \le K+2l$.

  Moreover we can check that

  $$\begin{aligned} w'_{K+2l} = \mathbf {n}+ l \bar{c} + \sum _{k \le K} c_k - l \bar{c} = \mathbf {m} \end{aligned}$$

  Therefore $\mathbf {n}$ and $\mathbf {m}$ are connected in $\mathcal {G}(\mathfrak {F}_{\mathbf {A},\mathbf {n}} ,\ker ^{*}_{\mathbb {Z}}(\mathbf {A}))$, and the conclusion follows since the choice of $\mathbf {m}$ is arbitrarily in $\mathfrak {F}_{\mathbf {A},\mathbf {n}}$ and $\mathbf {n}$ any point in $\mathbb {N}^A$.

• (ii) Let $\mathbf {n}\in \mathbb {N}^A, \mathbf {m} \in \mathfrak {F}_{\mathbf {A},\mathbf {n}}$. Since $\ker ^{*}_{\mathbb {Z}}(\mathbf {A})$ is a lattice basis, there exists $K < \infty$ and $\mathbf {c}_1,\ldots ,\mathbf {c}_K \subseteq (\ker ^{*}_{\mathbb {Z}}(\mathbf {A})\cup - \ker ^{*}_{\mathbb {Z}}(\mathbf {A}))^K$ such that if we define recursively :

  $$\begin{aligned} w_0 = \mathbf {n}, \quad w_k = \theta _{\mathbf {c}_k} w_{k-1} \end{aligned}$$

  (6.10)

  then we have that $w_K = \mathbf {m}$

Observe that w.l.o.g there exists $K^+$ s.t. $\mathbf {c}_k \in \ker ^{*}_{\mathbb {Z}}(\mathbf {A})$ for all $k \le K^+ \$ and $\mathbf {c}_k \in \ -\ker ^{*}_{\mathbb {Z}}(\mathbf {A})\ ,k \in \{ K^+ +1,\ldots ,A\}$. Applying the hypothesis, one can check directly that $\{ w_{k} \}_{0 \le k \le K}$ is a path which connects $\mathbf {n}$ to $\mathbf {m}$ in $\mathcal {G}(\mathfrak {F}_{\mathbf {A},\mathbf {n}} ,\ker ^{*}_{\mathbb {Z}}(\mathbf {A}))$.