A switching self-exciting jump diffusion process for stock prices

Abstract

This study proposes a new Markov switching process with clustering effects. In this approach, a hidden Markov chain with a finite number of states modulates the parameters of a self-excited jump process combined to a geometric Brownian motion. Each regime corresponds to a particular economic cycle determining the expected return, the diffusion coefficient and the long-run frequency of clustered jumps. We study first the theoretical properties of this process and we propose a sequential Monte-Carlo method to filter the hidden state variables. We next develop a Markov Chain Monte-Carlo procedure to fit the model to the S&P 500. We find that self-exciting jumps occur mainly during economic recession and nearly disappear in periods of economic growth. Finally, we analyse the impact of such a jump clustering on implied volatilities of European options.

Appendices

Appendix A: Proofs

Proof of proposition 2.1

As $\mathcal {F}_{0}\subset \mathcal {F}_{0}\vee \mathcal {G}_{t}$, the expectation of $\lambda _{t}$ is nested as follows:

$$\begin{aligned} \mathbb {E}(\lambda _{t}|\mathcal {F}_{0})= & {} \mathbb {E}\left( \mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0} \vee \mathcal {G}_{t}\right) |\mathcal {F}_{0}\right) \end{aligned}$$

and if we remember the expression (9) for the intensity, we infer that

$$\begin{aligned} \mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)= & {} \lambda _{0}-\alpha \int _{0}^{t}e^{\alpha (s-t)}\left( \lambda _{0} -\theta _{s}\right) ds+\mathbb {E}\left( \int _{0}^{t}\eta e^{\alpha (s-t)}dL_{s}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) . \end{aligned}$$

Let us denote by $\chi (t,l)$, the random measure of $L_{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}$. $\chi (t,l)$ is such that $L_{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}=\int _{0}^{t}\int _{0}^{\infty }\chi (ds,dl)$ and $d\left( L_{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) =\int _{0}^{\infty }\chi (ds,dl)$. Given that the integrand $\eta e^{\alpha (s-t)}$ is time integrable on $\mathbb {R}^{+}$, independent from the random measure $\chi$, we infer that

$$\begin{aligned} \mathbb {E}\left( \int _{0}^{t}\eta e^{\alpha (s-t)}dL_{s}|\mathcal {F}_{0} \vee \mathcal {G}_{t}\right)= & {} \int _{0}^{t}\int _{0}^{\infty }\eta e^{\alpha (s-t)} \chi (ds,dl)\\= & {} \int _{0}^{t}\eta e^{\alpha (s-t)}\int _{0}^{\infty }\chi (ds,dl)\\= & {} \int _{0}^{t}\eta e^{\alpha (s-t)}\mathbb {E}\left( dL_{s}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) . \end{aligned}$$

According to the Itô’s lemma for semi-martingale and by construction of $L_{s}$, $L_{s}$ is solution of the following SDE:

$$\begin{aligned} dL_{s}&=\int _{0}^{\infty }(L_{s}+|z|-L_{s})\chi (ds,dl)=|J|dN_{s}. \end{aligned}$$

Given that the jump J is independent from $N_{t}$, the expectation of $dL_{s}$ with respect to $\mathcal {F}_{0}\vee \mathcal {G}_{t}$ is then

$$\begin{aligned} \mathbb {E}\left( dL_{s}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)&=\mathbb {E}\left( \left| J\right| \right) \times \mathbb {E}\left( dN_{s}\,|\, \mathcal {F}_{0}\vee \mathcal {G}_{s}\right) \end{aligned}$$

(33)

for $s\le t$. Furthermore, conditionally to the filtration of $\lambda _{t}$ denoted $\left( \mathcal {L}_{t}\right) _{t\ge 0}$, $\left( N_{t}\right) _{t\ge 0}$ is a Poisson random variable of intensity and expectation equal to $\int _{0}^{t-}\lambda _{u}du$. Using nested expectations allows us to infer that

$$\begin{aligned} \mathbb {E}\left( dN_{s}\,|\,\mathcal {F}_{0}\vee \mathcal {G}_{s}\right) \,= & {} \mathbb {E}\left( \mathbb {E}\left( dN_{s}\,|\,\mathcal {F}_{0} \vee \mathcal {G}_{s}\vee \mathcal {L}_{s}\right) \,|\,\mathcal {F}_{0} \vee \mathcal {G}_{s}\right) \nonumber \\= & {} \mathbb {E}\left( \lambda _{s-}\,|\,\mathcal {F}_{0}\vee \mathcal {G}_{s}\right) . \end{aligned}$$

(34)

Combining (33) and (34) leads then to

$$\begin{aligned} \mathbb {E}\left( dL_{s}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)&=\mathbb {E}\left( \left| J\right| \right) \times \mathbb {E}\left( \lambda _{s-}\,|\,\mathcal {F}_{0}\vee \mathcal {G}_{s}\right) ds \end{aligned}$$

and

$$\begin{aligned} \mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)= & {} \lambda _{0}-\alpha \int _{0}^{t}e^{\alpha (s-t)}\\&\times \,\left( \lambda _{0}-\theta _{s}\right) ds+\eta \mathbb {E}\left( |J|\right) \int _{0}^{t}e^{\alpha (s-t)}\mathbb {E}\left( \lambda _{s-}\,|\,\mathcal {F}_{0}\vee \mathcal {G}_{s}\right) ds \end{aligned}$$

If we derive this last expression with respect to time, we find that $\mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)$ is solution of an ordinary differential equation (ODE):

$$\begin{aligned} \frac{\partial }{\partial t}\mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)= & {} -\,\alpha \left( \lambda _{0}-\theta _{t}\right) +\alpha ^{2}\int _{0}^{t}e^{\alpha (s-t)}\left( \lambda _{0}-\theta _{s}\right) ds+\eta \mathbb {E}\left( |J|\right) \mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \nonumber \\&-\,\alpha \eta \mathbb {E}\left( |J|\right) \int _{0}^{t}e^{-\alpha (t-s)}\mathbb {E}\left( \lambda _{s-}\,|\,\mathcal {F}_{0}\vee \mathcal {G}_{s}\right) ds\nonumber \\= & {} \left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) \mathbb {E}\left( \lambda _{t}| \mathcal {F}_{0}\vee \mathcal {G}_{t}\right) +\alpha \theta _{t}. \end{aligned}$$

(35)

The solution of this ODE is the following function:

$$\begin{aligned} \mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)= & {} \int _{0}^{t}\alpha \theta _{s}e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) (t-s)}ds+\lambda _{0}\,e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}, \end{aligned}$$

and the $\mathcal {F}_{0}$ conditional expectation is given by:

$$\begin{aligned} \mathbb {E}\left( \mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) |\mathcal {F}_{0}\right)= & {} \mathbb {E}\left( \int _{0}^{t}\alpha \theta _{s}e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) (t-s)}ds+\lambda _{0}\,e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}|\mathcal {F}_{0}\right) \nonumber \\= & {} \alpha \int _{0}^{t}e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) (t-s)}\mathbb {E}\left( \theta _{s}|\mathcal {F}_{0}\right) ds+\lambda _{0}\,e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}.\nonumber \\ \end{aligned}$$

(36)

The expected level of mean reversion at time s is the sum of $\bar{\theta }_{j}$ for $j=1$ to N weighted by the probabilities of transition:

$$\begin{aligned} \mathbb {E}\left( \theta _{s}|\mathcal {F}_{0}\right)= & {} \delta (0)'\,\exp \left( Q_{0}s\right) \,\bar{\theta } \end{aligned}$$

and if I is the $N\times N$ identity matrix, the integral in (36) may be rewritten as follows:

$$\begin{aligned}&\int _{0}^{t}e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) (t-s)}\mathbb {E}\left( \theta _{s}|\mathcal {F}_{0}\right) ds=e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}\int _{0}^{t}e^{-\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) s}\delta (0)'\,\exp \left( Q_{0}s\right) \,\bar{\theta }ds\\&\qquad =e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}\int _{0}^{t}\delta (0)'\,\exp \left( \left( Q_{0}-I\,(\eta \mathbb {E}\left( |J|\right) -\alpha )\right) s\right) \,\bar{\theta }ds\\&\qquad =e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}\left[ \delta (0)'\,\left( Q_{0}-I(\eta \mathbb {E}\left( |J|\right) -\alpha )\right) ^{-1}\exp \left( \left( Q_{0}-I(\eta \mathbb {E}\left( |J|\right) -\alpha )\right) s\right) \,\bar{\theta }\right] _{s=0}^{s=t}\\&\qquad =\delta (0)'\,\left( Q_{0}-I(\eta \mathbb {E}\left( |J|\right) -\alpha )\right) ^{-1}\left[ \exp \left( Q_{0}t\right) -\exp \left( I\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t\right) \right] \,\bar{\theta } \end{aligned}$$

$\square$

Proof of Corollary 2.2

Conditionally to $\mathcal {G}_{t}$, this expectation is equal to

$$\begin{aligned}&\mathbb {E}\left( \theta _{t}\lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) =\theta _{t}\mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\&\quad =\theta _{t}\int _{0}^{t}\alpha \theta _{s}e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) (t-s)}ds+\lambda _{0}\theta _{t}\,e^{\left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) t}. \end{aligned}$$

We infer from this relation that the derivative with respect to time is given by

$$\begin{aligned} \frac{\partial }{\partial t}\mathbb {E}\left( \theta _{t}\lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right)= & {} \theta _{t}\frac{\partial }{\partial t}\mathbb {E}\left( \lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) \\= & {} \left( \eta \mathbb {E}\left( |J|\right) -\alpha \right) \mathbb {E}\left( \theta _{t}\lambda _{t}|\mathcal {F}_{0}\vee \mathcal {G}_{t}\right) +\alpha \left( \theta _{t}\right) ^{2}. \end{aligned}$$

and we can conclude that $\mathbb {E}\left( \theta _{t}\lambda _{t}|\mathcal {F}_{0}\right)$ is well given by Eq. (12). $\square$

Proof of proposition 2.3

Let us introduce the following notations: $f(t,N_{t},\lambda _{t},\delta _{t})=\lambda _{t}^{2}$, according to the Itô’s lemma for semi-martingale, the infinitesimal generator of this function is given by:

$$\begin{aligned} \mathcal {A}f(t,N_{t},\lambda _{t},\delta _{t})= & {} \alpha (\theta _{t}-\lambda _{t})\,2\lambda _{t}+\lambda _{t} \int _{-\infty }^{+\infty }\left( \,\lambda _{t}+\eta \,|z|\right) ^{2} -\lambda _{t}^{2}\,d\nu (z)\\= & {} \alpha (\theta _{t}-\lambda _{t})\,2\lambda _{t}+\lambda _{t} \int _{-\infty }^{+\infty }\left( \,\lambda _{t}^{2}+2\eta \,|z| \lambda _{t}+\eta ^{2}|z|^{2}\right) -\lambda _{t}^{2}\,d\nu (z)\\= & {} 2\alpha \theta _{t}\lambda _{t}+2\left( \eta \,\mathbb {E}\left( |J|\right) -\alpha \right) \lambda _{t}^{2}+\eta ^{2}\lambda _{t}\mathbb {E}\left( |J|^{2}\right) \end{aligned}$$

On the other hand, we can prove that

$$\begin{aligned} \mathbb {E}\left( f(t,N_{t},\lambda _{t},\delta _{t})|\mathcal {F}_{0}\right)= & {} f(0,N_{0},\lambda _{0},\delta _{0})+\mathbb {E}\left( \int _{0}^{t} \mathcal {A}f(s,N_{s},\lambda _{s},\delta _{s})ds|\mathcal {F}_{0}\right) \nonumber \\= & {} f(0,N_{0},\lambda _{0},\delta _{0})+\int _{0}^{t}\mathbb {E} \left( \mathcal {A}f(s,N_{s},\lambda _{s},\delta _{s})|\mathcal {F}_{0}\right) ds. \end{aligned}$$

(37)

The derivative of this expectation with respect to time is equal to its expected infinitesimal generator:

$$\begin{aligned} \frac{\partial }{\partial t}\mathbb {E}\left( f(t,N_{t},\lambda _{t},\delta _{t})|\mathcal {F}_{0}\right)= & {} \mathbb {E}\left( \mathcal {A} f(t,N_{t},\lambda _{t},\theta _{t})|\mathcal {F}_{0}\right) , \end{aligned}$$

(38)

that allows us to infer Eq. (13). It is easy to check that the right term of Eq. (38) is linear and Lipschitz. Then, according to the theorem of Cauchy-Lipschitz the solution exists and is unique. $\square$

Proof of proposition 2.4

If we note $f(t,N_{t},\lambda _{t},\delta _{t})=\mathbb {E}\left( \omega ^{N_{s}}\,|\,\mathcal {F}_{t}\right)$, f is solution of an Itô’s equation for semi martingale. If $\delta _{t}=e_{i}$ then:

$$\begin{aligned} 0= & {} f_{t}+\alpha (\bar{\theta }_{i}-\lambda _{t})\,f_{\lambda }+\lambda _{t}\int _{-\infty }^{+\infty }f\left( t,N_{t}+1,\,\lambda _{t}+\eta \,|z|,\,e_{i}\right) -f(.)\,d\nu (z)\nonumber \\&+\sum _{j\ne i}^{N}q_{i,j}\left( f(t,N_{t},\lambda _{t},e_{j})-f(t,N_{t},\lambda _{t},e_{i})\right) \end{aligned}$$

(39)

If we assume that f is an exponential affine function of $\lambda _{t}$ and $N_{t}$:

$$\begin{aligned} f= & {} exp\left( A(t,s,e_{i})+B(t,s)\lambda _{t}+C(t,s)N_{t}\right) , \end{aligned}$$

where $A(t,s,e_{i})$ for $i=1$ to N, B(t, s), C(t, s) are time dependent functions, the partial derivatives of f are given by:

$$\begin{aligned}&f_{t} = \left( \frac{\partial }{\partial t}A(t,s,e_{j})+\frac{\partial }{\partial t}B(t,s)\lambda _{t}+\frac{\partial }{\partial t}C(t,s)N_{t}\right) f,\\&f_{\lambda }=B(t,s)f \end{aligned}$$

The integral in Eq. (39) is rewritten as follows:

$$\begin{aligned}&\int _{-\infty }^{+\infty }f(t,N_{t}+1,\lambda _{t}+\eta |z|,e_{j})-f(t,N_{t},\lambda _{t},e_{i})\,d\nu (z)\\&\quad =\int _{0}^{+\infty }exp\left( A(t,s,e_{j})+B(t,s)\lambda _{t}+C(t,s)N_{t}\right) \left( e^{B(t,s)\eta |z|+C(t,s)}-1\right) \,d\nu (z)\\&\quad =f\left[ e^{C}\psi \left( 0\,,\,B(t,s)\eta \right) -1\right] . \end{aligned}$$

As $q_{ii}=-\sum _{i\ne j}^{N}q_{i,j}$, the last term of Eq. (39) becomes

$$\begin{aligned} \sum _{j\ne i}^{N}q_{i,j}\left( f(t,N_{t},\lambda _{t},e_{j})-f(t,N_{t},\lambda _{t},e_{i})\right)= & {} \sum _{j=1}^{N}q_{i,j}f(t,N_{t},\lambda _{t},e_{j}). \end{aligned}$$

Then

$$\begin{aligned}&0=\left( \frac{\partial }{\partial t}A+\frac{\partial }{\partial t}B\,\lambda _{t}+\frac{\partial }{\partial t}C\,N_{t}\right) e^{A(t,s,e_{i})}+\alpha (\bar{\theta }_{i}-\lambda _{t})\,B\,e^{A(t,s,e_{i})}\\&\qquad +\lambda _{t}\left[ e^{C}\psi \left( 0\,,\,B(t,s)\eta \right) -1\right] e^{A(t,s,e_{i})}+\sum _{j=1}^{N}q_{i,j}e^{A(t,s,e_{j})}\,, \end{aligned}$$

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

$$\begin{aligned}&0=\left( \frac{\partial }{\partial t}A\right) e^{A(t,s,e_{i})}+\alpha \,\bar{\theta }_{i}\,B\,e^{A(t,s,e_{i})}+\sum _{j=1}^{N}q_{i,j}e^{A(t,s,e_{j})}\\&\qquad +\lambda _{t}\left( \frac{\partial }{\partial t}B-\alpha B+\left[ \omega \psi \left( 0\,,\,B(t,s)\eta \right) -1\right] \right) e^{A(t,s,e_{i})} \end{aligned}$$

or that

$$\begin{aligned} \frac{\partial }{\partial t}B= & {} \alpha B-\left[ \omega \psi \left( 0\,,\,B(t,s)\eta \right) -1\right] \\ \left( \frac{\partial }{\partial t}A\right) e^{A(t,s,e_{i})}= & {} -\alpha \,\bar{\theta }_{i}\,B\,e^{A(t,s,e_{i})}-\sum _{j=1}^{N}q_{i,j}e^{A(t,s,e_{j})} \end{aligned}$$

If we define $\tilde{A}(t,s,e_{i})=e^{A(t,s,e_{i})}$, the first equations can finally be put in matrix form as:

$$\begin{aligned} \frac{\partial \tilde{A}(t)}{\partial t}+\left( \text {diag}\left( \alpha \bar{\theta }B\right) +Q_{0}\right) \tilde{A}(t)=0, \end{aligned}$$

(40)

$\square$

Proof of proposition 2.5

If we note $f(t,X_{t},\lambda _{t},\delta _{t})=\mathbb {E}\left( e^{\omega _{1}X_{s}+\omega _{2}\lambda _{s}}\,|\,\mathcal {F}_{t}\right)$, f is solution of an Itô’s equation for semi-martingale. If $\delta _{t}=e_{i}$ then:

$$\begin{aligned}&0=f_{t}+f_{X}\left( \bar{\mu }_{i}-\frac{\bar{\sigma }_{i}^{2}}{2}-\lambda _{t}\,\mathbb {E}\left( e^{J}-1\right) \right) +f_{XX}\frac{\bar{\sigma }_{i}^{2}}{2}+\alpha (\bar{\theta }_{i}-\lambda _{t})\,f_{\lambda }\nonumber \\&\qquad + \lambda _{t}\int _{-\infty }^{+\infty }f\left( t,X_{t} +z,\,\lambda _{t}+\eta \,|z|,\,e_{i}\right) -f(.)\,\nu (dz)\nonumber \\&\qquad +\sum _{j\ne i}^{N}q_{i,j}\left( f(t,X_{t},\lambda _{t},e_{j})-f(t,X_{t},\lambda _{t},e_{i})\right) \end{aligned}$$

(41)

If we assume that f is an exponential affine function of $\lambda _{t}$ and $X_{t}$:

$$\begin{aligned} f= & {} exp\left( A(t,s,e_{i})+B(t,s)\lambda _{t}+C(t,s)X_{t}\right) , \end{aligned}$$

where $A(t,s,e_{i})$ for $i=1$ to N, B(t, s), C(t, s) are time dependent functions, the partial derivatives of f are given by:

$$\begin{aligned} f_{t}= & {} \left( \frac{\partial }{\partial t}A(t,s,e_{j})+\frac{\partial }{\partial t}B(t,s)\lambda _{t}+\frac{\partial }{\partial t}C(t,s)X_{t}\right) f,\\ f_{X}= & {} C(t,s)f\,f_{XX}=C(t,s)^{2}f \; f_{\lambda }=B(t,s)f \end{aligned}$$

and the integrand in Eq. (41) is rewritten as follows:

$$\begin{aligned} \int _{-\infty }^{+\infty }f\left( t,X_{t}+z,\,\lambda _{t}+\eta \,|z|,e_{i}\right) -f(.)\,\nu (dz)= & {} f\left[ \psi \left( C(t,s)\,,\,B(t,s)\eta \right) -1\right] . \end{aligned}$$

As $q_{ii}=-\sum _{i\ne j}^{N}q_{i,j}$, the last term of the Itô equation is also equal to

$$\begin{aligned} \sum _{j\ne i}^{N}q_{i,j}\left( f(t,X_{t},\lambda _{t},e_{j})-f(t,X_{t},\lambda _{t},e_{i})\right)= & {} \sum _{j=1}^{N}q_{i,j}f(t,X_{t},\lambda _{t},e_{j}) \end{aligned}$$

injecting these expressions into Eq. (41), leads to the following relation:

$$\begin{aligned} 0= & {} \left( \frac{\partial }{\partial t}A+\frac{\partial }{\partial t}B\,\lambda _{t}+\frac{\partial }{\partial t}C\,X_{t}\right) e^{A(t,s,e_{i})}+C\,\left( \bar{\mu }_{i}-\frac{\bar{\sigma }_{i}^{2}}{2}-\lambda _{t}\mathbb {E}\left( e^{J}-1\right) \right) e^{A(t,s,e_{i})}\\&+C^{2}\,\frac{\bar{\sigma }_{i}^{2}}{2}e^{A(t,s,e_{i})}+\alpha (\bar{\theta }_{i}-\lambda _{t})Be^{A(t,s,e_{i})}+\lambda _{t} \left[ \psi \left( C\,,\,B\eta \right) -1\right] e^{A(t,s,e_{i})}+\sum _{j=1}^{N}q_{i,j}e^{A(t,s,e_{j})}\,, \end{aligned}$$

from which we deduce that $C(t,s)=\omega _{1}$. Regrouping terms allows to infer that A and B are solutions of a system of ODE’s:

$$\begin{aligned} 0= & {} \left( \frac{\partial }{\partial t}A+\frac{\partial }{\partial t} B\,\lambda _{t}\right) e^{A(t,s,e_{i})}+\omega _{1}\,\left( \bar{\mu }_{i} -\frac{\bar{\sigma }_{i}^{2}}{2}-\lambda _{t}\mathbb {E}\left( e^{J}-1\right) \right) e^{A(t,s,e_{i})}\\&+\omega _{1}^{2}\,\frac{\bar{\sigma }_{i}^{2}}{2}e^{A(t,s,e_{i})} +\alpha (\bar{\theta }_{i}-\lambda _{t})Be^{A(t,s,e_{i})} +\lambda _{t}\left[ \psi \left( \omega _{1}\,,\,B\eta \right) -1\right] e^{A(t,s,e_{i})}\\&+\sum _{j=1}^{N}q_{i,j}e^{A(t,s,e_{j})}\,, \end{aligned}$$

From this last relation, we deduce that

$$\begin{aligned} {\left\{ \begin{array}{ll} 0= &{} \frac{\partial }{\partial t}A\,e^{A(t,s,e_{i})}+\omega _{1}\,\left( \bar{\mu }_{i}-\frac{\bar{\sigma }_{i}^{2}}{2}\right) e^{A(t,s,e_{i})}+\omega _{1}^{2}\,\frac{\bar{\sigma }_{i}^{2}}{2}e^{A(t,s,e_{i})}\\ &{} +\alpha \bar{\theta }_{i}B\,e^{A(t,s,e_{i})}+\sum _{j=1}^{N}q_{i,j}e^{A(t,s,e_{j})}\;for\,i=1\ldots N,\\ 0= &{} \frac{\partial }{\partial t}B-\alpha B-\omega _{1}\,\mathbb {E}\left( e^{J}-1\right) +\left[ \psi \left( \omega _{1}\,,\,B\eta \right) -1\right] \end{array}\right. } \end{aligned}$$

If we define $\tilde{A}(t,s)=(e^{A(t,s,e_{i})})_{i=1,\ldots ,N}$, the first equations can finally be put in matrix form as:

$$\begin{aligned} \frac{\partial \tilde{A}(t,s)}{\partial t}+\left( \text {diag}\left( \omega _{1}\,\left( \bar{\mu }-\frac{\bar{\sigma }^{2}}{2}\right) +\omega _{1}^{2}\,\frac{\bar{\sigma }^{2}}{2}+\alpha \bar{\theta }B\right) +Q_{0}\right) \tilde{A}(t,s)=0, \end{aligned}$$

(42)

$\square$

Proof of proposition

From the previous results, we know that B(t, s) is solution of the ODE:

$$\begin{aligned} \frac{\partial }{\partial t}B=\alpha B+\omega _{1}\left( \psi \left( 1,0\right) -1\right) -\left[ \psi \left( \omega _{1}\,,\,B\,\eta \right) -1\right] \end{aligned}$$

with the terminal condition: $B(s,s)=\omega _{2}$. If we set $B(t,s)=C(s-t)$ and $\tau =s-t$. Then $\frac{\partial }{\partial t}B=\frac{\partial }{\partial \tau }C\frac{\partial \tau }{\partial t}=-\frac{\partial }{\partial \tau }C$ and

$$\begin{aligned} \frac{\partial }{\partial \tau }C(\tau )= & {} -\alpha C(\tau )+\psi \left( \omega _{1}\,,\,C(\tau )\,\eta \right) -\underbrace{\left[ \omega _{1}\left( \psi \left( 1,0\right) -1\right) +1\right] }_{\beta (\omega _{1})}\nonumber \\= & {} -\alpha C(\tau )+\psi \left( \omega _{1}\,,\,C(\tau )\,\eta \right) -\beta (\omega _{1}) \end{aligned}$$

(43)

The left hand side is denoted h(C). Due to the convexity of $\psi \left( .\right)$ there is only one point $u^{*}$ such that $h(u)=0$. This equation is indeed equivalent to

$$\begin{aligned} \psi \left( \omega _{1}\,,\,u\,\eta \right)= & {} \beta \left( \omega _{1}\right) +\alpha \,u \end{aligned}$$

We rewrite the Eq. (43) as follows,

$$\begin{aligned} \frac{dC}{-\beta -\alpha C+\psi \left( \omega _{1}\,,\,C\,\eta \right) }= & {} d\tau . \end{aligned}$$

As $C(0)=\omega _{2}$, by direct integration, we have that

$$\begin{aligned} \int _{\omega _{2}}^{C}\frac{du}{-\beta -\alpha u+\psi \left( \omega _{1}\,,\,\eta \,u\right) }= & {} \tau \end{aligned}$$

with $C\in [\omega _{2},u^{*})$ or $C\in [u^{*},\omega _{2})$. Remark that if $C=u^{*}$ then $\tau =+\infty$ as the numerator converges to zero. If we define the function on the left hand side as

$$\begin{aligned} F_{\omega _{1}}(C):= & {} \int _{\omega _{2}}^{C}\frac{du}{-\beta -\alpha u+\psi \left( \omega _{1}\,,\,\eta \,u\right) } \end{aligned}$$

then $F_{\omega _{1}}(C)=\tau$ and $C=F_{\omega _{1}}^{-1}(\tau )$ or $B(t)=F_{\omega _{1}}^{-1}(s-t)$. $\square$

Appendix B: Particle filter

The structure of the particle filter algorithm is the following:

• 1. Initial step: draw M values of $v_{0}^{(i)}$ for $i=1,\ldots ,M$, from an initial distribution $p(v_{0})$

• 2. For $j=1\,:\,T$

  Prediction step: draw a sample of $\Delta L_{j}^{(i)}$ and $\delta _{j}^{(i)}$ and update $\lambda _{j}^{(i)}$, $\mu _{j}^{b\,(i)}$$\sigma _{j}^{(i)}$ , $\theta _{j}^{(i)}$ using the relations

  $$\begin{aligned}&\lambda _{j}^{(i)}=\lambda _{j-1}^{(i)}+\alpha (\theta _{j-1}^{(i)}-\lambda _{j-1}^{(i)})\Delta +\eta {\varvec{\Delta }}L_{j}^{(i)}\\&\mu _{j}^{(i)}=\delta _{j}^{(i)}\bar{\mu }\quad \sigma _{j}^{(i)}=\delta _{j}^{(i)}\bar{\sigma }\quad \theta _{j}^{(i)}=\delta _{j}^{(i)}\bar{\theta } \end{aligned}$$

  Correction step: the particle $v_{j}^{(i)}$ has a probability of occurrence equal to $w_{j}^{(i)}=\frac{p(x_{j}\,|\,v_{j})}{\sum _{i=1:M}p(x_{j}\,|\,v_{j})}$ where

  $$\begin{aligned} p(x_{j}\,|\,v_{j}^{(i)})&\sim \mathcal {N}\left( \left( \mu _{j}^{(i)}-\frac{\sigma _{j}^{(i)2}}{2}-\lambda _{j}^{(i)}\mathbb {E}\left( e^{J}-1\right) \right) \Delta -{\varvec{\Delta }}L_{j}^{'(i)}\,,\,\sigma _{j}^{(i)}\sqrt{\Delta }\right) \end{aligned}$$

  Resampling step: resample with replacement M particles according to the importance weights $w_{j}^{(i)}$. The new importance weights are set to $w_{j}^{(i)}=\frac{1}{M}$.