A switching microstructure model for stock prices

Abstract

This article proposes a microstructure model for stock prices in which parameters are modulated by a Markov chain determining the market behaviour. In this approach, called the switching microstructure model (SMM), the stock price is the result of the balance between the supply and the demand for shares. The arrivals of bid and ask orders are represented by two mutually- and self-excited processes. The intensities of these processes converge to a mean reversion level that depends upon the regime of the Markov chain. The first part of this work studies the mathematical properties of the SMM. The second part focuses on the econometric estimation of parameters. For this purpose, we combine a particle filter with a Markov chain Monte Carlo algorithm. Finally, we calibrate the SMM with two and three regimes to daily returns of the S&P 500 and compare them with a non switching model.

Appendix

Proof of Lemma 2.1

To prove this relation, we differentiate the expression of $\lambda _{t}^{i}$ to retrieve its dynamic:

$$\begin{aligned} d\lambda _{t}^{i}= & {} \kappa _{i}c_{i,t}-\kappa _{i}\left( \lambda _{0}^{i}-\kappa _{i}\int _{0}^{t}e^{\kappa _{i}(s-t)}\left( \lambda _{0}^{i}-c_{i,s}\right) ds\right. \\&+\left. \int _{0}^{t}\delta _{i,1}e^{\kappa _{i}(s-t)}dL_{s}^{1}dt+\int _{0}^{t}\delta _{i,2}e^{\kappa _{i}(s-t)}dL_{s}^{2}\right) +\delta _{i,1}dL_{t}^{1}+\delta _{i,2}dL_{t}^{2}\\= & {} \kappa _{i}(c_{i,t}-\lambda _{t}^{i})dt+\delta _{i,1}dL_{t}^{1}+\delta _{i,2}dL_{t}^{2}\qquad i=1,2. \end{aligned}$$

To prove the positivity, we first remind that $\int _{0}^{t}\delta _{i,1}e^{\kappa _{i}(s-t)}dL_{s}^{1}$ and $\int _{0}^{t}\delta _{i,2}e^{\kappa _{i}(s-t)}dL_{s}^{2}$ are positive by construction. According to Eq. (12), the process $\lambda _{t}^{i}$ admit the following lower bound:

$$\begin{aligned} \lambda _{t}^{i}> & {} \lambda _{0}^{i}+\left( \min \left( c_{i}\right) -\lambda _{0}^{i}\right) \kappa _{i}\int _{0}^{t}e^{\kappa _{i}(s-t)}ds. \end{aligned}$$

(46)

Given that $\kappa _{i}\int _{0}^{t}e^{\kappa _{i}(s-t)}ds=\left( 1-e^{-\kappa _{i}t}\right) >0$, we conclude that

$$\begin{aligned} \lambda _{t}^{i}> & {} \lambda _{0}^{i}e^{-\kappa _{i}t}+\min \left( c_{i}\right) \left( 1-e^{-\kappa _{i}t}\right) >0. \end{aligned}$$

$\square$

Proof of Proposition 3.1

As $\mathcal {F}_{s}\subset \mathcal {F}_{s}\vee \mathcal {G}_{t}$, using nested expectations leads to the following expression for the expected intensity:

$$\begin{aligned} \mathbb {E}(\lambda _{t}^{i}|\mathcal {F}_{s})= & {} \mathbb {E}\left( \mathbb {E}\left( \lambda _{t}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) |\mathcal {F}_{s}\right) . \end{aligned}$$

If we remember the expression (13) of the intensity, using the Fubini’s theorem leads to the following expression for the expectation of $\lambda _{t}^{i}$, conditionally to the augmented filtration $\mathcal {F}_{s}\vee \mathcal {G}_{t}$ :

$$\begin{aligned} \mathbb {E}\left( \lambda _{t}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right)= & {} \lambda _{s}^{i}-\kappa _{i}\int _{s}^{t}e^{\kappa _{i}(u-t)}\left( \lambda _{s}^{i}-c_{i,u}\right) du\nonumber \\&+\int _{s}^{t}\delta _{i,1}e^{\kappa _{i}(u-t)}\mathbb {E}\left( dL_{u}^{1}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) +\int _{s}^{t}\delta _{i,2}e^{\kappa _{i}(u-t)}\mathbb {E}\left( dL_{u}^{2}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) .\nonumber \\ \end{aligned}$$

(47)

Using the same approach as in Errais et al. [17], $dL_{u}^{i}$ is rewritten as follows:

$$\begin{aligned} dL_{u}^{i}&={\left\{ \begin{array}{ll} O^{i} &{} \quad { if}\quad dN_{u}^{1}=1\\ 0 &{} \quad { otherwise} \end{array}\right. }. \end{aligned}$$

The order size $O_{i}$ being independent from all processes and then from $\mathcal {F}_{s}\vee \mathcal {G}_{t}$, we infer that

$$\begin{aligned} \mathbb {E}\left( dL_{u}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right)&=\mathbb {E}\left( O^{i}\right) \times \mathbb {E}\left( dN_{u}^{1}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) \\&=\mu _{i}\times \mathbb {E}\left( dN_{u}^{1}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{u}\right) . \end{aligned}$$

Using nested expectations and conditioning with respect to the sample path of $\lambda _{u}^{1}$ contained in the subfiltration $\mathcal {H}_{u}$ of $\mathcal {F}_{u}$ leads to equality for $u\le t$

$$\begin{aligned} \mathbb {E}\left( dL_{u}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right)&=\mu _{i}\times \mathbb {E}\left( \mathbb {E}\left( dN_{u}^{1}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{u}\vee \mathcal {H}_{u}\right) \,|\,\mathcal {F}_{s}\vee \mathcal {G}_{u}\right) . \end{aligned}$$

Conditionally to the sample path of $\lambda _{u}^{1}$, $N_{u}^{1}$ is a non-homogeneous Poisson process,

$$\begin{aligned} \mathbb {E}\left( dN_{u}^{1}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{u}\vee \mathcal {H}_{u}\right) =\lambda _{u-}^{i} \end{aligned}$$

therefore, we infer that

$$\begin{aligned} \mathbb {E}\left( dL_{u}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right)&=\mu _{i}\times \mathbb {E}\left( \lambda _{u-}^{i}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{u}\right) du\quad \forall \,u\le t, \end{aligned}$$

If we derive Eq. (47) with respect to time, we find that $\mathbb {E}\left( \lambda _{t}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right)$ is solution of an ordinary differential equation (ODE):

$$\begin{aligned} \frac{\partial }{\partial t}\mathbb {E}\left( \lambda _{t}^{i}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right)= & {} -\kappa _{i}\left( \lambda _{s}^{i}-c_{i,t}\right) +\kappa _{i}^{2}\int _{s}^{t}e^{\kappa _{i}(u-t)}\left( \lambda _{s}-c_{i,u}\right) du\\&+\,\delta _{i,1}\mu _{1}\mathbb {E}\left( \lambda _{t}^{1}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) -\kappa _{i}\delta _{i,1}\mu _{1}\int _{s}^{t}e^{-\kappa _{i}(t-u)}\mathbb {E}\left( \lambda _{u-}^{1}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) du\\&+\,\delta _{i,2}\mu _{2}\mathbb {E}\left( \lambda _{t}^{2}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) -\kappa _{i}\delta _{i,2}\mu _{2}\int _{s}^{t}e^{-\kappa _{i}(t-u)}\mathbb {E}\left( \lambda _{u-}^{2}\,|\,\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) du. \end{aligned}$$

Using Eq. ((47)), allows us to rewrite these ODE’s as follows:

$$\begin{aligned} \left( \begin{array}{c} \frac{\partial }{\partial t}\mathbb {E}\left( \lambda _{t}^{1}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) \\ \frac{\partial }{\partial t}\mathbb {E}\left( \lambda _{t}^{2}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) \end{array}\right)= & {} \left( \begin{array}{c} \kappa _{1}c_{1,t}\\ \kappa _{2}c_{2,t} \end{array}\right) +\left( \begin{array}{c@{\quad }c} \delta _{1,1}\mu _{1}-\kappa _{1} &{} \delta _{1,2}\mu _{2}\\ \delta _{2,1}\mu _{1} &{} \delta _{2,2}\mu _{2}-\kappa _{2} \end{array}\right) \left( \begin{array}{c} \mathbb {E}\left( \lambda _{t}^{1}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) \\ \mathbb {E}\left( \lambda _{t}^{2}|\mathcal {F}_{s}\vee \mathcal {G}_{t}\right) \end{array}\right) .\nonumber \\ \end{aligned}$$

(48)

Solving this system of equation requires to determine eigenvalues $\gamma$ and eigenvectors $(v_{1},v_{2})$ of the matrix present in the right term of this system:

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} (\delta _{1,1}\mu _{1}-\kappa _{1}) &{} \delta _{1,2}\mu _{2}\\ \delta _{2,1}\mu _{1} &{} (\delta _{2,2}\mu _{2}-\kappa _{2}) \end{array}\right) \left( \begin{array}{c} v_{1}\\ v_{2} \end{array}\right) =\gamma \left( \begin{array}{c} v_{1}\\ v_{2} \end{array}\right) . \end{aligned}$$

We know that eigenvalues cancel the determinant of the following matrix:

$$\begin{aligned} \det \left( \begin{array}{c@{\quad }c} (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma &{} \delta _{1,2}\mu _{2}\\ \delta _{2,1}\mu _{1} &{} (\delta _{2,2}\mu _{2}-\kappa _{2})-\gamma \end{array}\right) =0, \end{aligned}$$

and are solutions of the second order equation:

$$\begin{aligned} \gamma ^{2}-\gamma \left( (\delta _{1,1}\mu _{1}-\kappa _{1})+(\delta _{2,2}\mu _{2}-\kappa _{2})\right) +(\delta _{1,1}\mu _{1}-\kappa _{1})(\delta _{2,2}\mu _{2}-\kappa _{2})-\delta _{1,2}\delta _{2,1}\mu _{1}\mu _{2}=0. \end{aligned}$$

Roots of the last equation are $\gamma _{1}$ and $\gamma _{2}$, as defined by the Eq. (14). One way to find an eigenvector is to note that it must be orthogonal to each rows of the matrix:

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma &{} \delta _{1,2}\mu _{2}\\ \delta _{2,1}\mu _{1} &{} (\delta _{2,2}\mu _{2}-\kappa _{2})-\gamma \end{array}\right) \left( \begin{array}{c} v_{1}\\ v_{2} \end{array}\right) =0, \end{aligned}$$

then necessary,

$$\begin{aligned} \left( \begin{array}{c} v_{1}^{i}\\ v_{2}^{i} \end{array}\right) =\left( \begin{array}{c} -\delta _{1,2}\mu _{2}\\ (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{i} \end{array}\right) \quad for\,i=1,2. \end{aligned}$$

If we note $D:=diag(\gamma _{1},\gamma _{2})$, the matrix in the right term of Eq. (48) admits the decomposition:

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} \delta _{1,1}\mu _{1}-\kappa _{1} &{} \delta _{1,2}\mu _{2}\\ \delta _{2,1}\mu _{1} &{} \delta _{2,2}\mu _{2}-\kappa _{2} \end{array}\right)= & {} VDV^{-1}, \end{aligned}$$

where V is the matrix of eigenvectors, as defined in Eq. (16). Its determinant, $\Upsilon$, and its inverse are respectively provided by Eqs. (18) and (17). If two new variables are defined as follows:

$$\begin{aligned} \left( \begin{array}{c} u_{1}\\ u_{2} \end{array}\right)= & {} V^{-1}\left( \begin{array}{c} m_{1}\\ m_{2} \end{array}\right) . \end{aligned}$$

The system (48) is decoupled into two independent ODEs:

$$\begin{aligned} \frac{\partial }{\partial t}\left( \begin{array}{c} u_{1}\\ u_{2} \end{array}\right)= & {} V^{-1}\left( \begin{array}{c} \kappa _{1}c_{1,t}\\ \kappa _{2}c_{2,t} \end{array}\right) +\left( \begin{array}{c@{\quad }c} \gamma _{1} &{} 0\\ 0 &{} \gamma _{2} \end{array}\right) \left( \begin{array}{c} u_{1}\\ u_{2} \end{array}\right) . \end{aligned}$$

(49)

And introducing the following notations

$$\begin{aligned} V^{-1}\left( \begin{array}{c} \kappa _{1}c_{1,t}\\ \kappa _{2}c_{2,t} \end{array}\right)= & {} \left( \begin{array}{c} \epsilon _{1}(t)\\ \epsilon _{2}(t) \end{array}\right) , \end{aligned}$$

leads to the solutions for the system (49):

$$\begin{aligned} \left( \begin{array}{c} u_{1}(t)\\ u_{2}(t) \end{array}\right)= & {} \left( \begin{array}{c} \int _{s}^{t}\epsilon _{1}(u)e^{\gamma _{1}\left( t-u\right) }du\\ \int _{s}^{t}\epsilon _{2}(u)e^{\gamma _{2}\left( t-u\right) }du \end{array}\right) +\left( \begin{array}{l@{\quad }l} e^{\gamma _{1}(t-s)} &{} 0\\ 0 &{} e^{\gamma _{2}(t-s)} \end{array}\right) V^{-1}\left( \begin{array}{c} \lambda _{s}^{1}\\ \lambda _{s}^{2} \end{array}\right) \end{aligned}$$

that allows us to infer the expression (15) for moments of $\lambda _{t}^{i}$. Notice that the determinant $\Upsilon$ is always real and if parameters of mutual excitations $\delta _{1,2},\,\delta _{2,1}$, are positive. As $\mu _{1},\mu _{2}>0$, the determinant is also strictly positive and the matric V is invertible. Finally, Eq. (15) states that conditionally to the sample path of the Markov chain $\theta _{t}$, processes $\lambda _{t}^{1}$ and $\lambda _{t}^{2}$ are Markov given that their $\mathcal {F}_{s}\vee \mathcal {G}_{t}$-expectations only depend on the pair $\left( \lambda _{t}^{1},\lambda _{t}^{2}\right)$. $\square$

Proof of Proposition 3.2

From the previous proposition, we infer that the unconditional expectations of OAI are the solutions of the following system

$$\begin{aligned} \left( \begin{array}{c} \mathbb {E}\left( \lambda _{t}^{1}\,|\,\mathcal {F}_{s}\right) \\ \mathbb {E}\left( \lambda _{t}^{2}\,|\,\mathcal {F}_{s}\right) \end{array}\right)= & {} V\int _{s}^{t}\left( \begin{array}{l@{\quad }l} e^{\gamma _{1}(t-u)} &{} 0\\ 0 &{} e^{\gamma _{2}(t-u)} \end{array}\right) V^{-1}\left( \begin{array}{c} \kappa _{1}\mathbb {E}\left( c_{1,u}\,|\,\mathcal {F}_{s}\right) \\ \kappa _{2}\mathbb {E}\left( c_{2,u}\,|\,\mathcal {F}_{s}\right) \end{array}\right) du\nonumber \\&+V\left( \begin{array}{l@{\quad }l} e^{\gamma _{1}(t-s)} &{} 0\\ 0 &{} e^{\gamma _{2}(t-s)} \end{array}\right) V^{-1}\left( \begin{array}{c} \lambda _{s}^{1}\\ \lambda _{s}^{2} \end{array}\right) . \end{aligned}$$

(50)

Given that $\theta _{t}$ is a finite state Markov chain of generator $Q_{0}$ and if we remember that $c_{i}=\left( \begin{array}{c} c_{i,1}\\ \vdots \\ c_{i,l} \end{array}\right)$ for $i=1,2$ are l- vectors, the expected level of mean reversion at time u is equal to:

$$\begin{aligned} \mathbb {E}\left( c_{i,u}|\mathcal {F}_{s}\right)= & {} \theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{i} \end{aligned}$$

then expectation of intensities, conditionally to $\mathcal {F}_{s}$:

$$\begin{aligned} \left( \begin{array}{c} \mathbb {E}\left( \lambda _{t}^{1}\,|\,\mathcal {F}_{s}\right) \\ \mathbb {E}\left( \lambda _{t}^{2}\,|\,\mathcal {F}_{s}\right) \end{array}\right)= & {} V\int _{s}^{t}\left( \begin{array}{c@{\quad }c} e^{\gamma _{1}(t-u)} &{} 0\\ 0 &{} e^{\gamma _{2}(t-u)} \end{array}\right) V^{-1}\left( \begin{array}{c} \kappa _{1}\theta _{s}^{\top }\,\exp \left( Q_{0}(u-s)\right) \,c_{1}\\ \kappa _{2}\theta _{s}^{\top }\,\exp \left( Q_{0}(u-s)\right) \,c_{2} \end{array}\right) du\nonumber \\&+V\left( \begin{array}{c@{\quad }c} e^{\gamma _{1}(t-s)} &{} 0\\ 0 &{} e^{\gamma _{2}(t-s)} \end{array}\right) V^{-1}\left( \begin{array}{c} \lambda _{s}^{1}\\ \lambda _{s}^{2} \end{array}\right) . \end{aligned}$$

(51)

If we replace $V^{-1}$ by its definition (17), we obtain that

$$\begin{aligned}&V^{-1}\left( \begin{array}{c} \kappa _{1}\theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{1}\\ \kappa _{2}\theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{2} \end{array}\right) \\&\qquad =\frac{1}{\Upsilon }\left( \begin{array}{c} \left( \begin{array}{c} \kappa _{1}\left( (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{2}\right) \theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{1}\\ +\,\kappa _{2}\delta _{1,2}\mu _{2}\theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{2} \end{array}\right) \\ \left( \begin{array}{c} \kappa _{1}\left( \gamma _{1}-(\delta _{1,1}\mu _{1}-\kappa _{1})\right) \theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{1}\\ -\,\kappa _{2}\delta _{1,2}\mu _{2}\theta _{s}^{\top }\,\exp \left( Q_{0}\left( u-s\right) \right) \,c_{2} \end{array}\right) \end{array}\right) . \end{aligned}$$

The integrand in Eq. (51) becomes then:

$$\begin{aligned}&\left( \begin{array}{c@{\quad }c} e^{\gamma _{1}(t-u)} &{} 0\\ 0 &{} e^{\gamma _{2}(t-u)} \end{array}\right) V^{-1}\left( \begin{array}{c} \kappa _{1}\theta _{s}^{\top }\,\exp \left( Q_{0}(u-s)\right) \,c_{1}\\ \kappa _{2}\theta _{s}^{\top }\,\exp \left( Q_{0}(u-s)\right) \,c_{2} \end{array}\right) \\&\quad =\frac{1}{\Upsilon }\left( \begin{array}{c} \left( \begin{array}{c} e^{\gamma _{1}t}\kappa _{1}\left( (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{2}\right) \theta _{s}^{\top }\,\exp \left( \left( Q_{0}-\gamma _{1}I\right) (u-s)\right) \,c_{1}\\ +\,e^{\gamma _{1}t}\kappa _{2}\delta _{1,2}\mu _{2}\theta _{s}^{\top }\,\exp \left( \left( Q_{0}-\gamma _{1}I\right) (u-s)\right) \,c_{2} \end{array}\right) \\ \left( \begin{array}{c} e^{\gamma _{2}t}\kappa _{1}\left( \gamma _{1}-(\delta _{1,1}\mu _{1}-\kappa _{1})\right) \theta _{s}^{\top }\,\exp \left( \left( Q_{0}-\gamma _{2}I\right) (u-s)\right) \,c_{1}\\ -\,e^{\gamma _{2}t}\kappa _{2}\delta _{1,2}\mu _{2}\theta _{s}^{\top }\,\exp \left( \left( Q_{0}-\gamma _{2}I\right) (u-s)\right) \,c_{2} \end{array}\right) \end{array}\right) . \end{aligned}$$

and we can conclude by direct integration that expected value of $\lambda _{t}^{i}$ are given by Eq. (19). This result also states processes $\lambda _{t}^{1}$ and $\lambda _{t}^{2}$ are Markov given that their $\mathcal {F}_{s}$expectations only depend on the information available at time s: $\left( \lambda _{s}^{1},\lambda _{s}^{2},\theta _{s}^{1},\theta _{s}^{2}\right)$. $\square$

Proof of Corollary 3.3

To prove this statement, it is sufficient to show that the conditional expectation of these processes with respect to $\mathcal {F}_{s}$ depends exclusively upon the information available at time s. Using the Tower property of conditional expectation, the expected number of supply order conditionally to $\mathcal {F}_{s}$ is then equal to the following product:

$$\begin{aligned} \mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\right)= & {} \mathbb {E}\left( \mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\vee \mathcal {H}_{t}\right) |\mathcal {F}_{s}\right) . \end{aligned}$$

By construction, the compensator of process $N_{t}^{1}$ is an $\mathcal {H}_{t}$-adapted process $\int _{0}^{t}\lambda _{u}^{1}du$ such that the compensated process $M_{t}^{1}=N_{t}^{1}-\int _{0}^{t}\lambda _{u}^{1}du$ is a martingale. Given that $\mathbb {E}\left( M_{t}^{1}|\mathcal {F}_{s}\vee \mathcal {H}_{t}\right) =M_{s}^{1}$, we deduce that $\mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\vee \mathcal {H}_{t}\right) =N_{s}^{1}+\int _{s}^{t}\lambda _{u}^{1}du$. Using the Fubini’s theorem, we infer that

$$\begin{aligned} \mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\right)= & {} \left( N_{s}^{1}+\mathbb {E}\left( \int _{s}^{t}\lambda _{u}^{1}du|\mathcal {F}_{s}\right) \right) \nonumber \\= & {} N_{s}^{1}+\int _{s}^{t}\mathbb {E}\left( \lambda _{u}^{1}|\mathcal {F}_{s}\right) du. \end{aligned}$$

(52)

According to Proposition 3.2, $\mathbb {E}\left( \lambda _{u}^{1}|\mathcal {F}_{s}\right)$ depends only upon $\lambda _{s}^{1}$, $\lambda _{s}^{2}$ and $\theta _{s}$. From Eq. (52), we immediately deduce that $\mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\right)$ is exclusively a function of ($\lambda _{s}^{1}$, $\lambda _{s}^{2}$ ,$\theta _{s}$, $N_{s}^{1}$). The same holds for $N_{t}^{2}$. By definition, $L_{t}^{1}$ is a sum of independent random variables:

$$\begin{aligned} \mathbb {E}\left( L_{t}^{1}|\mathcal {F}_{s}\right)= & {} \mathbb {E}\left( \sum _{n=1}^{N_{t}^{1}}O_{n}^{1}|\mathcal {F}_{s}\right) \\= & {} \mu _{1}\mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\right) . \end{aligned}$$

As $\mathbb {E}\left( N_{t}^{1}|\mathcal {F}_{s}\right)$ is a function of ($\lambda _{s}^{1}$, $\lambda _{s}^{2}$, $\theta _{s}$, $N_{s}^{1}$), the same conclusion holds for $\mathbb {E}\left( L_{t}^{1}|\mathcal {F}_{s}\right)$. A similar reasoning for $L_{t}^{2}$ and Proposition 3.2 allows to end the proof. $\square$

Proof of Proposition 3.4

If we remember the Eq. (48), we infer that the expectations of $c_{j,t}\lambda _{t}^{i}$ for $i,j=1,2$ are solution of ordinary differential equations (ODE):

$$\begin{aligned}&\underbrace{\left( \begin{array}{c} \frac{\partial }{\partial t}\mathbb {E}\left( c_{1,t}\lambda _{t}^{1}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\ \frac{\partial }{\partial t}\mathbb {E}\left( c_{2,t}\lambda _{t}^{1}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\ \frac{\partial }{\partial t}\mathbb {E}\left( c_{1,t}\lambda _{t}^{2}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\ \frac{\partial }{\partial t}\mathbb {E}\left( c_{2,t}\lambda _{t}^{2}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \end{array}\right) }_{:=dE(t)}=\underbrace{\left( \begin{array}{c@{\quad }c@{\quad }c} \kappa _{1} &{} 0 &{} 0\\ 0 &{} 0 &{} \kappa _{1}\\ 0 &{} 0 &{} \kappa _{2}\\ 0 &{} \kappa _{2} &{} 0 \end{array}\right) }_{:=K}\underbrace{\left( \begin{array}{c} c_{1,t}^{2}\\ c_{2,t}^{2}\\ c_{1,t}c_{2,t} \end{array}\right) }_{:=C_{t}^{2}}\\&\quad +\underbrace{\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \delta _{1,1}\mu _{1}-\kappa _{1} &{} 0 &{} \delta _{1,2}\mu _{2} &{} 0\\ 0 &{} \delta _{1,1}\mu _{1}-\kappa _{1} &{} 0 &{} \delta _{1,2}\mu _{2}\\ \delta _{2,1}\mu _{1} &{} 0 &{} \delta _{2,2}\mu _{2}-\kappa _{2} &{} 0\\ 0 &{} \delta _{2,1}\mu _{1} &{} 0 &{} \delta _{2,2}\mu _{2}-\kappa _{2} \end{array}\right) }_{WFW^{-1}}\underbrace{\left( \begin{array}{c} \mathbb {E}\left( c_{1,t}\lambda _{t}^{1}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\ \mathbb {E}\left( c_{2,t}\lambda _{t}^{1}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\ \mathbb {E}\left( c_{1,t}\lambda _{t}^{2}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\ \mathbb {E}\left( c_{2,t}\lambda _{t}^{2}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \end{array}\right) }_{E(t)}. \end{aligned}$$

We summarize this system of ODE as follows

$$\begin{aligned} dE(t)= & {} K\,C_{t}^{2}+W\,F\,W^{-1}E(t). \end{aligned}$$

If we note $U(t)=W^{-1}E(t)$, we rewrite this last system:

$$\begin{aligned} dU(t)= & {} W^{-1}K\,C_{t}^{2}+F\,U(t), \end{aligned}$$

that admits the following solution:

$$\begin{aligned} U(t)= & {} \int _{0}^{t}\exp \left( F\,s\right) \,W^{-1}K\,C_{s}^{2}\,ds+\exp \left( F\,t\right) U(0), \end{aligned}$$

and we can conclude. $\square$

Proof of Proposition 3.5

If we remember Eq. (27), we can develop it as follows

$$\begin{aligned}&\exp \left( F\,s\right) W^{-1}\left( \begin{array}{c} \kappa _{1}c_{1,s}^{2}\\ \kappa _{1}c_{1,s}c_{2,s}\\ \kappa _{2}c_{1,s}c_{2,s}\\ \kappa _{2}c_{2,s}^{2} \end{array}\right) \\&\quad =\frac{1}{\Upsilon }\left( \begin{array}{c} \kappa _{1}\left( (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{2}\right) \left( e^{\gamma _{1}s}c_{1,s}^{2}\right) +\delta _{1,2}\mu _{2}\kappa _{2}\left( e^{\gamma _{1}s}c_{1,s}c_{2,s}\right) \\ \kappa _{1}\left( \gamma _{1}-(\delta _{1,1}\mu _{1}-\kappa _{1})\right) \left( e^{\gamma _{2}s}c_{1,s}^{2}\right) -\delta _{1,2}\mu _{2}\kappa _{2}\left( e^{\gamma _{2}s}c_{1,s}c_{2,s}\right) \\ \kappa _{1}\left( (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{2}\right) \left( e^{\gamma _{1}s}c_{1,s}c_{2,s}\right) +\delta _{1,2}\mu _{2}\kappa _{2}\left( e^{\gamma _{1}s}c_{2,s}^{2}\right) \\ \kappa _{1}\left( \gamma _{1}-(\delta _{1,1}\mu _{1}-\kappa _{1})\right) \left( e^{\gamma _{2}s}c_{1,s}c_{2,s}\right) -\delta _{1,2}\mu _{2}\kappa _{2}\left( e^{\gamma _{2}s}c_{2,s}^{2}\right) \end{array}\right) , \end{aligned}$$

and its expectation is given by

$$\begin{aligned}&\mathbb {E}\left( \exp \left( F\,s\right) W^{-1}\left( \begin{array}{c} \kappa _{1}c_{1,s}^{2}\\ \kappa _{1}c_{1,s}c_{2,s}\\ \kappa _{2}c_{1,s}c_{2,s}\\ \kappa _{2}c_{2,s}^{2} \end{array}\right) \right) \\&\quad =\frac{1}{\Upsilon }\left( \begin{array}{c} \kappa _{1}\left( (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{2}\right) \left( \theta _{0}e^{\left( Q_{0}+\gamma _{1}I\right) s}\bar{c}_{1}^{2}\right) +\delta _{1,2}\mu _{2}\kappa _{2}\left( \theta _{0}e^{\left( Q_{0}+\gamma _{1}I\right) s}\bar{c}_{1,2}\right) \\ \kappa _{1}\left( \gamma _{1}-(\delta _{1,1}\mu _{1}-\kappa _{1})\right) \left( \theta _{0}e^{\left( Q_{0}+\gamma _{2}I\right) s}\bar{c}_{1}^{2}\right) -\delta _{1,2}\mu _{2}\kappa _{2}\left( \theta _{0}e^{\left( Q_{0}+\gamma _{2}I\right) s}\bar{c}_{1,2}\right) \\ \kappa _{1}\left( (\delta _{1,1}\mu _{1}-\kappa _{1})-\gamma _{2}\right) \left( \theta _{0}e^{\left( Q_{0}+\gamma _{1}I\right) s}\bar{c}_{1,2}\right) +\delta _{1,2}\mu _{2}\kappa _{2}\left( \theta _{0}e^{\left( Q_{0}+\gamma _{1}I\right) s}\bar{c}_{2}^{2}\right) \\ \kappa _{1}\left( \gamma _{1}-(\delta _{1,1}\mu _{1}-\kappa _{1})\right) \left( \theta _{0}e^{\left( Q_{0}+\gamma _{2}I\right) s}\bar{c}_{1,2}\right) -\delta _{1,2}\mu _{2}\kappa _{2}\left( \theta _{0}e^{\left( Q_{0}+\gamma _{2}I\right) s}\bar{c}_{2}^{2}\right) \end{array}\right) . \end{aligned}$$

Integrating this last equation allows us to conclude. $\square$

Proof of Proposition 3.6

If we remember the expression (24) of the infinitesimal generator, we have

$$\begin{aligned} \mathcal {A}\left( \left( \lambda _{t}^{1}\right) ^{2}\right)= & {} 2\kappa _{1}\left( c_{1,t}-\lambda _{t}^{1}\right) \lambda _{t}^{1}+\lambda _{t}^{1}\int _{-\infty }^{+\infty }\left( \lambda _{t}^{1}+\delta _{1,1}z\right) ^{2}-\left( \lambda _{t}^{1}\right) ^{2}\nu _{1}(dz)\\&+\,\lambda _{t}^{2}\int _{-\infty }^{+\infty }\left( \lambda _{t}^{1}+\delta _{1,2}z\right) ^{2}-\left( \lambda _{t}^{1}\right) ^{2}\nu _{2}(dz),\\ \\ \mathcal {A}\left( \left( \lambda _{t}^{2}\right) ^{2}\right)= & {} 2\kappa _{2}\left( c_{2,t}-\lambda _{t}^{2}\right) \lambda _{t}^{2}+\lambda _{t}^{1}\int _{-\infty }^{+\infty }\left( \lambda _{t}^{2}+\delta _{2,1}z\right) ^{2}-\left( \lambda _{t}^{2}\right) ^{2}\nu _{1}(dz)\\&+\,\lambda _{t}^{2}\int _{-\infty }^{+\infty }\left( \lambda _{t}^{2}+\delta _{2,2}z\right) ^{2}-\left( \lambda _{t}^{2}\right) ^{2}\nu _{2}(dz),\\ \\ \mathcal {A}\left( \lambda _{t}^{1}\lambda _{t}^{2}\right)= & {} \kappa _{1}\left( c_{1,t}-\lambda _{t}^{1}\right) \lambda _{t}^{2}+\kappa _{2}\left( c_{2,t}-\lambda _{t}^{2}\right) \lambda _{t}^{1}\\&+\,\lambda _{t}^{1}\int _{-\infty }^{+\infty }\left( \lambda _{t}^{1}+\delta _{1,1}z\right) \left( \lambda _{t}^{2}+\delta _{2,1}z\right) -\lambda _{t}^{1}\lambda _{t}^{2}\nu _{1}(dz)\\&+\,\lambda _{t}^{2}\int _{-\infty }^{+\infty }\left( \lambda _{t}^{1}+\delta _{1,2}z\right) \left( \lambda _{t}^{2}+\delta _{2,2}z\right) -\lambda _{t}^{1}\lambda _{t}^{2}\nu _{2}(dz). \end{aligned}$$

And given that $\frac{\partial }{\partial t}g=\mathbb {E}\left( \mathcal {A}g\,|\,\mathcal {F}_{0}\right)$, we can conclude. $\square$

Proof of Proposition 3.9

Let us assume that $\theta _{t}=e_{i}$. If we denote by $g(\lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},\theta _{t})=\mathbb {E}\left( \omega ^{N_{T}^{k}}\,|\,\mathcal {F}_{t}\right)$, g is solution of the following Itô’s equation for semi martingale :

$$\begin{aligned} 0= & {} g_{t}+\kappa _{1}\left( c_{1,t}-\lambda _{t}^{1}\right) g_{\lambda ^{1}}+\kappa _{2} \left( c_{2,t}-\lambda _{t}^{2}\right) g_{\lambda ^{2}}\nonumber \\&+\,\lambda _{t}^{1}\int _{-\infty }^{+\infty }g \left( \lambda _{t}^{1}+\delta _{1,1}z,J_{t}^{1}+(z,1)^{\top }, \lambda _{t}^{2}+\delta _{2,1}z,J_{t}^{2},e_{i}\right) \nonumber \\&-\quad g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{i} \right) d\nu _{1}(z)\nonumber \\&+\,\lambda _{t}^{2}\int _{-\infty }^{+\infty }g \left( \lambda _{t}^{1}+\delta _{1,2}z,J_{t}^{1},\lambda _{t}^{2}+\delta _{2,2}z, J_{t}^{2}+(z,1)^{\top },e_{i}\right) \nonumber \\&-\,g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{i}\right) d \nu _{2}(z)\nonumber \\&+\,\sum _{j\ne i}^{N}q_{i,j}\left( g(\lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{j})-g(\lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{i})\right) . \end{aligned}$$

(53)

Next, we assume that g is an exponential affine function of $\lambda _{t}^{1}$, $\lambda _{t}^{2}$ and $N_{t}^{i}$ :

$$\begin{aligned} g= & {} \exp \left( A(t,T,e_{j})+B_{1}(t,T)\lambda _{t}^{1}+B_{2}(t,T)\lambda _{t}^{2}+C(t,T)N_{t}^{k}\right) , \end{aligned}$$

where $A(t,T,e_{i})$ for $i=1$ to l, $B_{1}(t,T)$, $B_{2}(t,T)$ and C(t, T) are time dependent functions. The partial derivatives of g are then given by:

$$\begin{aligned} g_{t}= & {} \left( \frac{\partial }{\partial t}A(t,T,e_{j})+\frac{\partial }{\partial t}B_{1}(t,T)\lambda _{t}^{1}+\frac{\partial }{\partial t}B_{2}(t,T)\lambda _{t}^{2}+\frac{\partial }{\partial t}C(t,T)N_{t}^{k}\right) g, \\ g_{\lambda ^{1}}= & {} B_{1}(t,T)g \quad and \quad g_{\lambda ^{2}}=B_{2}(t,T)g. \end{aligned}$$

And the integrands in Eq. (53) are rewritten with the notations $A:=A(t,T,e_{i}),\,B_{1}:=B_{1}(t,T),$$B_{2}:=B_{2}(t,T)$ and $C:=C(t,T)$ as follows:

$$\begin{aligned}&\int _{-\infty }^{+\infty }g\left( \lambda _{t}^{1}+\delta _{1,1}z,J_{t}^{1}+\left( z,1\right) ^{\top },\lambda _{t}^{2}+\delta _{2,1}z,J_{t}^{2},e_{i}\right) -g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{i}\right) \,d\nu _{1}\left( z\right) \\&\qquad =g\left[ e^{1_{k=1}C}\psi _{1}\left( B_{1}\delta _{1,1}+B_{2}\delta _{2,1}\right) -1\right] , \\&\int _{-\infty }^{+\infty }g\left( \lambda _{t}^{1}+\delta _{1,2}z,J_{t}^{1},\lambda _{t}^{2}+\delta _{2,2}z,J_{t}^{2}+\left( z,1\right) ^{\top },e_{j}\right) -g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{i}\right) \,d\nu _{2}\left( z\right) \\&\qquad =g\left[ e^{1_{k=2}C}\psi _{2}\left( B_{1}\delta _{1,2}+B_{2}\delta _{2,2}\right) -1\right] . \end{aligned}$$

As the sum of instantaneous probabilities is null, $q_{ii}=-\sum _{i\ne j}^{l}q_{i,j}$, we have that

$$\begin{aligned} \sum _{j\ne i}^{l}q_{i,j}\left( g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{j} \right) -g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{i}\right) \right)= & {} \sum _{j=1}^{l}q_{i,j}g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{j}\right) . \end{aligned}$$

Then the Eq. (53) becomes:

$$\begin{aligned} 0= & {} \left( \frac{\partial }{\partial t}A+\frac{\partial }{\partial t}B_{1}\,\lambda _{t}^{1}+\frac{\partial }{\partial t}B_{2}\,\lambda _{t}^{2}+\frac{\partial }{\partial t}C\,N_{t}^{i}\right) e^{A\left( t,T,e_{i}\right) }\nonumber \\&+\,\kappa _{1}\left( c_{1,t}-\lambda _{t}^{1}\right) B_{1}\,e^{A\left( t,T,e_{i}\right) }+\kappa _{2}\left( c_{2,t}-\lambda _{t}^{2}\right) B_{2}\,e^{A\left( t,T,e_{i}\right) }\nonumber \\&+\,\lambda _{t}^{1}\left( e^{1_{k=1}C}\psi _{1}\left( B_{1}\delta _{1,1}+B_{2}\delta _{2,1}\right) -1\right) e^{A\left( t,T,e_{i}\right) }\nonumber \\&+\,\lambda _{t}^{2}\left( e^{1_{k=2}C}\psi _{2}\left( B_{1}\delta _{1,2}+B_{2}\delta _{2,2}\right) -1\right) e^{A\left( t,T,e_{i}\right) }\nonumber \\&+\sum _{j=1}^{l}q_{i,j}g\left( \lambda _{t}^{1},J_{t}^{1},\lambda _{t}^{2},J_{t}^{2},e_{j}\right) , \end{aligned}$$

(54)

from which we guess that $C(t,s)=\ln \omega$. Regrouping terms allows to infer that

$$\begin{aligned} 0= & {} \frac{\partial }{\partial t}Ae^{A(t,T,e_{i})}+\kappa _{1}c_{1,t}\,B_{1}\,e^{A(t,T,e_{i})}+\kappa _{2}c_{2,t}\,B_{2}\,e^{A(t,T,e_{i})}+\sum _{j=1}^{l}q_{i,j}e^{A(t,T,e_{j})}\\&+\,\lambda _{t}^{1}\left( \frac{\partial }{\partial t}B_{1}-\kappa _{1}B_{1}+\left[ 1_{k=1}\omega \psi _{1}\left( B_{1}\delta _{1,1}+B_{2}\delta _{2,1}\right) -1\right] \right) e^{A(t,T,e_{i})}\\&+\,\lambda _{t}^{2}\left( \frac{\partial }{\partial t}B_{2}-\kappa _{2}B_{2}+\left[ 1_{k=2}\omega \psi _{2}\left( B_{1}\delta _{1,2}+B_{2}\delta _{2,2}\right) -1\right] \right) e^{A(t,T,e_{i})}. \end{aligned}$$

Given that $\lambda _{t}^{1}$ and $\lambda _{t}^{2}$ are random quantities, this equation is satisfied if and only if

$$\begin{aligned} \frac{\partial }{\partial t}B_{1}= & {} \kappa _{1}B_{1}-\left[ 1_{k=1}\omega \psi _{1}\left( B_{1}\delta _{1,1}+B_{2}\delta _{2,1}\right) -1\right] \\ \frac{\partial }{\partial t}B_{2}= & {} \kappa _{2}B_{2}-\left[ 1_{k=2}\omega \psi _{2}\left( B_{1}\delta _{1,2}+B_{2}\delta _{2,2}\right) -1\right] \\ \left( \frac{\partial }{\partial t}A\right) e^{A(t,T,e_{i})}= & {} -\kappa _{1}c_{1,t}\,B_{1}\,e^{A(t,T,e_{i})}-\kappa _{2}c_{2,t}\,B_{2}\,e^{A(t,T,e_{i})}-\sum _{j=1}^{l}q_{i,j}e^{A(t,T,e_{j})}. \end{aligned}$$

If we define $\tilde{A}(t,T)=\left( e^{A(t,T,e_{1})},\ldots ,e^{A(t,T,e_{l})}\right)$ , the last equations can finally be put in matrix form as:

$$\begin{aligned} \frac{\partial \tilde{A}}{\partial t}+\left( \text {diag}\left( \kappa _{1}c_{1,t}\,B_{1}+\kappa _{2}c_{2,t}\,B_{2}\right) +Q_{0}\right) \tilde{A}=0. \end{aligned}$$

$\square$

Proof of Proposition 3.11

From previous results, we know that $B_{k}(t,T)$ is solution of the following ODE

$$\begin{aligned} \frac{\partial }{\partial t}B_{k}=\kappa _{k}B_{k}+\left( -1\right) ^{k}\omega _{1}\alpha _{k}\mu _{k}-\left[ \psi _{k}\left( B_{1}\delta _{1,k}+B_{2}\delta _{2,k}+C_{k}\right) -1\right] ,\quad k=1,2 \end{aligned}$$

with terminal condition $B_{k}(T,T)=\omega _{k+1}$. If we set $B_{k}(t,T)=D_{k}(T-t)$ and $\tau =T-t$. Then

$$\begin{aligned} \frac{\partial B_{k}}{\partial t}=\frac{\partial D_{k}}{\partial \tau }\frac{\partial \tau }{\partial t}=-\frac{\partial D_{k}}{\partial \tau }. \end{aligned}$$

Thus we obtain

$$\begin{aligned} \frac{\partial D_{k}}{\partial \tau }= & {} -\kappa _{k}B_{k}(\tau )-\left( -1\right) ^{k}\omega _{1}\alpha _{k}\mu _{k}+\left[ \psi _{k}\left( D_{1}(\tau )\delta _{1,k}+D_{2}(\tau )\delta _{2,k}+C_{k}\right) -1\right] \nonumber \\= & {} -\kappa _{k}D_{k}(\tau )+\psi _{k}\left( D_{1}(\tau )\delta _{1,k}+D_{2}(\tau )\delta _{2,k}+C_{k}\right) -\left[ \left( -1\right) ^{k}\omega _{1}\alpha _{k}\mu _{k}+1\right] \nonumber \\= & {} -\kappa _{k}D_{k}(\tau )+\psi _{k}\left( D_{1}(\tau )\delta _{1,k}+D_{2}(\tau )\delta _{2,k}+C_{k}\right) -\beta _{k}(\omega _{1}). \end{aligned}$$

(55)

The left hand side is then denoted $h_{k}(D_{1},D_{2})$. Due to the convexity of $\psi _{k}$ there is only one point $\left( u_{1}^{*},u_{2}^{*}\right)$ such that $h_{k}(u)=0$ for $k=1,2$. These equations are indeed equivalent to

$$\begin{aligned} \psi _{k}\left( u_{1}\delta _{1,k}+u_{2}\delta _{2,k}+C_{k}\right) =\beta _{k}(\omega _{1})+\kappa _{k}u_{k}. \end{aligned}$$

We rewrite the Eq. (55) as follows,

$$\begin{aligned} \frac{dD_{k}}{-\kappa _{k}D_{k}+\psi _{k}\left( D_{1}\delta _{1,k}+D_{2}\delta _{2,k}+C_{k}\right) -\beta _{k}(\omega _{1})}=d\tau . \end{aligned}$$

As $D_{k}(0)=\omega _{k+1}$ for $k\in \{1,2\}$ by direct integration, we have that

$$\begin{aligned} \int _{\omega _{2}}^{D_{1}}\frac{du_{1}}{-\kappa _{1}u_{1}+\psi _{1}\left( u_{1}\delta _{1,1}+D_{2}\delta _{2,1}+C_{1}\right) -\beta _{1}(\omega _{1})}=\tau , \\ \int _{\omega _{3}}^{D_{2}}\frac{du_{2}}{-\kappa _{2}u_{2}+\psi _{2}\left( D_{1}\delta _{1,2}+u_{2}\delta _{2,2}+C_{2}\right) -\beta _{2}(\omega _{1})}=\tau . \end{aligned}$$

with $D_{k}\in [\omega {k+1},u_{k}^{*})$ or $D_{k}\in [u_{k}^{*},\omega {k+1})$.

We can remark that if $\left( D_{1},D_{2}\right) =\left( u_{1}^{*},u_{2}^{*}\right)$ then $\tau =+\infty$ as the numerator converges to zero. If we define the functions $F_{\omega _{1}}^{1}(x,y)$ and $F_{\omega _{1}}^{2}(x,y)$ from $\mathbb {R}^{2}$ to $\mathbb {R}^{+}$ by Eq. (36), $D_{1}$ and $D_{2}$ are such that $F_{\omega _{1}}^{k}(D_{1},D_{2})=\tau$. If $\left( F_{\omega _{1}}^{1}\right) ^{-1}(\tau \,|\,y)$ and $\left( F_{\omega _{1}}^{2}\right) ^{-1}(\tau \,|\,x)$ are respectively the inverse functions of $F_{\omega _{1}}^{1}(.,y)$ and $\,F_{\omega _{1}}^{2}(x,.)$, then $D_{1}$ and $D_{2}$ satisfy the following system

$$\begin{aligned} D_{1}&=\left( F_{\omega _{1}}^{1}\right) ^{-1}(\tau \,|\,D_{2}),\\ D_{2}&=\left( F_{\omega _{1}}^{2}\right) ^{-1}(\tau \,|\,D_{1}), \end{aligned}$$

or $B_{1}(t,T)=\left( F_{\omega _{1}}^{1}\right) ^{-1}(T-t\,|\,B_{2}(t,T))$ and $B_{2}(t,T)=\left( F_{\omega _{1}}^{2}\right) ^{-1}(T-t\,|\,B_{1}(t,T))$ . $\square$