I was wondering if the representation by Duffie, Pan, and Singleton (2000) is already accounting for the little Heston trap. DPS represent their 'general' discounted characteristic function as: $$ \begin{align} \psi(u,(y,v),t,T) = exp(\alpha(u,T-t) + yu + \beta(u,T-t)v), \end{align} $$ where

$$ \begin{align} \beta(\tau,u) &= -a\frac{1-\exp{(-\gamma\tau)}}{2\gamma-(b+\gamma)(1-\exp{(-\gamma\tau)})}\\ \alpha_{0}(\tau,u) &= -r\tau +(r-\xi)\tau u - \kappa\sigma\left(\frac{b+\gamma}{\sigma^{2}}\tau + \frac{2}{\sigma^{2}}\log\left(1 - \frac{b+\gamma}{2\gamma}\left(1-\exp{(-\gamma\tau)}\right)\right)\right)\\ \alpha(\tau,u) &= \alpha_{0}(\tau,u) - \bar{\lambda}\tau(1+\bar{\mu}u) + \bar{\lambda}\int^{\tau}_{0} \theta(u,\beta(s,u))ds,\\ a &= u(1-u),\\ b &= \sigma\rho u - \kappa,\\ \gamma &= \sqrt{b^{2} + a\sigma^{2}},\\ \bar{\lambda} &= \lambda_{y} + \lambda_{v} + \lambda_{c}. \end{align} $$ The transform $\theta(c_{1},c_{2})$ can be found in their paper. When comparing this discounted characteristic function to other variations of the original Heston characteristic function, they look quite different from each other. It is giving me a hard time to figure out if this representation already takes care of 'the little Heston trap'. In the case what would I need to change in here to handle the little Heston trap. Do I also need to change something in the integral?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.