The following proposition is very usefull for changing measure:
If $S_{t}=S_{0} \exp \left(r t+X_{t}\right)$ where $\left(X_{t}\right)_{t \in[0, T]}$ is a Lévy process with Lévy triplet $\left(\sigma^{2}, \nu, b\right) $ and if there exists a solution $\beta \in \mathbb{R}$ to the equation: $$ \begin{array}{l} b+\left(\beta+\frac{1}{2}\right) \sigma^{2}+\int_{-1}^{+1} \nu(d x)\left[\left(e^{x}-1\right) e^{\beta\left(e^{x}-1\right)}-x\right] \\ \left.+\int_{|x|>1} e^{x}-1\right) e^{\beta\left(e^{x}-1\right)} \nu(d x)=0 \end{array} $$ then the minimal entropy martingale $S_{t}^{*}$ is also an exponential-Lévy process $S_{t}^{*}=S_{0} \exp \left(r t+X_{t}^{*}\right)$ where $\left(Z_{t}\right)_{t \in[0, T]}$ is a Lévy process with Lévy triplet $\left(\sigma^{2}, \nu^{*}, b^{*}\right)$ given by: $$ \begin{aligned} b^{*} &=b+\beta \sigma^{2}+\int_{-1}^{+1} \nu(d x)\left[e^{\beta\left(e^{x}-1\right)} x\right] \\ \nu^{*}(d x) &=\exp \left[\beta\left(e^{x}-1\right)\right] \nu(d x) \end{aligned} $$
And now lets consider Affine Markov Processes: \begin{aligned} d S_{t} &=\mu S_{t} d t+\sqrt{V}_{t} S_{t} d Z_{t}^{(1)}+S_{t} d\left(Q_{t}^{(s)}-\lambda \mu_{J} t\right) \\ d V_{t} &=\kappa\left(\theta-V_{t}\right) d t+\sigma_{v} \sqrt{V_{t}}\left[\rho d Z_{t}^{(1)}+\sqrt{1-\rho^{2}} d Z_{t}^{(2)}\right]+d Q_{t}^{(v)} \end{aligned}
where $Z$ is Brownian Motions, $Q$ is jump-processes and $\rho$ is a correlation factor.
My questions is: How to derive the characteristic triplet of this model?