Deriving the Bates Model Jump Characteristic Function

# Characteristic function of the Bates model


# Characteristic function of the Bates model












I have a misunderstanding concerning the derivation of the SVJ model :

Firsty,I understand how to reach the final differential equation from :

\begin{gather} dS_t = (r - q - \lambda t (e^{m-\frac{\nu}{2}} - 1)) S_{t^-} dt + \sqrt{V_t} S_{t^-} dW^{Q,1}_t + S_{t^-} dJ_t \nonumber \\ dV_t = \kappa \left( \theta - V_t \right) dt + \xi \sqrt{V_t} dW^{Q,2}_t \nonumber\\ dJ_t = d\Big(\sum\limits_{i=1}^{N_t}(H_i-1)\Big)\,,\,dW^{Q,1}_t dW^{Q,2}_t = \rho dt \nonumber\\ \log(H_i - 1) \sim N(m,\nu)\,,\,N_t \sim Poisson(\lambda t) \nonumber \end{gather}

We first end up with :

\begin{gather} rC dt = \frac{\partial C}{\partial t}dt+\frac{1}{2}\frac{\partial C^2 }{\partial S_{-t}^2} d\langle S_{-t},S_{-t} \rangle + \frac{\partial ^2 C}{\partial S_{-t}\partial V_t} d\langle S_{-t},V_t \rangle +\frac{1}{2}\frac{\partial ^2 C}{\partial V_t^2}\langle V_t,V_t \rangle \label{eq:eq17}\\ + \frac{\partial C}{\partial S_{-t}}dS_{-t} + \frac{\partial C}{\partial V_t} dV_t + \lambda dt \int_{0}^{\infty}(C(t,S_{-t}(H_i - 1),V_t) - C(t,S_{-t},V_t)) p_{logN}(H_i - 1) d(H_i - 1) \nonumber \end{gather}

If we do a change of variable such $x_{-t} = \log{S_{-t}}$ we reach the step :

\begin{gather} 0 = -\frac{\partial D_j(x_{-t}, V_t, \tau)}{\partial \tau}+\frac{1}{2}V_t \frac{\partial^2 D_j(x_{-t}, V_t, \tau)}{\partial x_{-t}^2}-(\frac{1}{2}-j)V_t\frac{\partial D_j(x_{-t}, V_t, \tau)}{\partial x_{-t}} \\ +\frac{1}{2}V_t\xi^2\frac{\partial^2 D_j(x_{-t}, V_t, \tau)}{\partial V_t^2}+\rho\xi V_t\frac{\partial^2 D_j(x_{-t}, V_t, \tau)}{\partial V_t \partial x_{-t}}+(a-b_j V_t)\frac{\partial D_j(x_{-t}, V_t, \tau)}{\partial V_t}\\ +\lambda\int_{-\infty}^{\infty}(D_j(t,x_{-t} + \log{(H_i - 1)},V_t) - D_j(t,x_{-t},V_t)) p_{N}(\log{(H_i - 1)}) d(\log{(H_i - 1)}) \label{eq:eq1} \end{gather}

Where $r = 0$ and $ q = 0 $ and :

\begin{equation} a=\kappa \theta, b_{j}=\kappa-j \rho \xi \end{equation}

and :

\begin{gather*} P(x_t, V_t, \tau)= e^{r\tau}C(\tau,x_t,V_t) \\ \end{gather*} We can deduce : \begin{gather*} P(x_t, V_t, \tau)=K\{e^{x_t} D_1(x_{t}, V_t, \tau)-D_0(x_t, V_t, \tau)\} \end{gather*}

My question here is how to retrieve the characteristic function $\widehat{\Xi}$ form the PIDE such :

\begin{align} \widehat{\Xi}_{\lambda}(u,\tau) &= \mathbb{E^Q}( E(e^{ i u \log{J_t}}|N_t = k) )\nonumber \iff \sum_{k = 1}^{\infty} E(e^{ i u \log{(H_i-1)}}|N_t = k)^k P(N_t = k) \nonumber\\ &= \sum_{k = 1}^{\infty} \frac{e^{-\lambda \tau}}{k!}\Big(\lambda \tau \int_{-\infty}^{\infty}e^{i u \log{(H_i-1)}}p_N(\log{(H_i-1)}) d\log{(H_i-1)}\Big)^k \nonumber\\ \end{align} Set $j = \log{(H_i-1)}$ \begin{align} &= e^{-\lambda \tau}e^{\lambda \tau \int_{-\infty}^{\infty}e^{i u j}p_N(j) dj} \nonumber\\ &= e^{-\lambda \tau}e^{\lambda \tau \int_{-\infty}^{\infty} e^{i u j}\frac{e^{\frac{-(j - m)^2}{2\nu}}}{2\sqrt{\pi\nu}}dj} \nonumber \\ &= e^{\lambda \tau(e^{i u m - \frac{v u^2}{2}} - 1)} \nonumber \end{align}

In other words,I don't understand how we used the characteristic function of the last terms in order to find $\widehat{\Xi}$ : \begin{align} \lambda \int_{-\infty}^{\infty}(\widehat{D_j}(t,x_{-t} + j,V_t) - \widehat{D_j}(t,x_{-t},V_t))p_{N}(j) dj \end{align}

knowing that for a standard Heston Model: \begin{align} \widehat{D_j}(x_t, V_t, \tau,u) = \frac{e^{C(u, \tau) \theta+L(u, \tau) V_t}}{i u } \end{align}

And the final characteristic function should be if we add jumps : \begin{align} \widehat{D_j}(x_t, V_t, \tau,u) = \frac{e^{C_{SVJ}(u, \tau) \theta+L(u, \tau) V_t}\widehat{\Xi}_{\lambda}(u,\tau)}{i u } \end{align}

with :

\begin{align*} C_{SVJ}(u, \tau) = - i u \lambda \tau (e^{m + \frac{vu^2}{2}} - 1) + C \end{align*}

Thank you for your help

PS : Using intuition,I was wondering whether applying distributivity on $D_j(x_t + \log{(H_i - 1)} , V_t, \tau)$ could lead to an answer : \begin{equation} D_j(x_t + \log{(H_i - 1)} , V_t, \tau) = D_j(x_t , V_t, \tau)D_j(\log{(H_i - 1)} , V_t, \tau) \end{equation}