what I am puzzled about is, why dont we instead of having
\begin{equation} dX_t = \sqrt{V_t} dB_t - (\frac{1}{2} V_t^2-r-\lambda\Phi(\rho)) dt - \rho dZ_{\lambda t}\nonumber \end{equation}
we just have
\begin{equation} dX_t = V_t dB_t - (\frac{1}{2} V_t-r-\lambda\Phi(\rho)) dt - \rho dZ_{\lambda t}\nonumber \end{equation}
where \begin{equation} dV_t = -\lambda V_t dt + dZ_{\lambda t}\nonumber \end{equation}
I have been working on American put problem for this. Without the squre root, I think some things can be simplified in a much nicer manner. Though I have not done any computation, without the square root, $V$ is 'in the same dimension' as the log price. The equation has a nice interpretation that a jump in 'volatility' correspond to a jump in price rather than a jump in 'volatility squared' correspond to a jump in price?