I'm trying to derive an approximation for the zero-rebate barrier option under the Heston model: $$dS_t=\mu S_tdt+\sqrt{v_t}S_tdW^S_t$$ $$dv_t=\kappa(\bar{v}-v_t)dt+\eta\sqrt{v_t}dW^v_t,\quad d\langle W^S,W^v\rangle_t=\rho dt$$ The payoff of a down-and-out option is: $$\mathcal{C}_T=(S_T-k)\mathbb{I}_{\{S_T\geq K\}}\mathbb{I}_{\{(\max_{0\leq t\leq T}S_t=:)m_T\geq H\}}$$

Under the Black-Scholes dynamics we have the closed-form solution: $$\mathcal{C}_t=S_te^{r\tau}\left(\Phi\left(\frac{\ln(S_t/K)+\nu\tau}{\sigma\sqrt{\tau}}\right)-\left(\frac{H}{S_t}\right)^{1+2r/\sigma^2}\Phi\left(\frac{\ln(H^2/(S_tK))+\nu\tau}{\sigma\sqrt{\tau}}\right)\right) - K\left(\Phi\left(\frac{\ln(S_t/K)+(\nu-\sigma^2)\tau}{\sigma\sqrt{\tau}}\right)-\left(\frac{H}{S_t}\right)^{-1+2r/\sigma^2}\Phi\left(\frac{\ln(H^2/(S_tK))+(\nu-\sigma^2)\tau}{\sigma\sqrt{\tau}}\right)\right)$$ where $\nu=r+\frac{\sigma^2}{2}$ Obviously, under the Heston dynamics we don't have a closed-form solution. However, I would like to know whether there exists approximations allowing to price the down-and-out call option in a similar fashion.

As a starting point, I can use the result of the hitting probability of Heston for $\rho$ and $\mu$ equal to 0, i.e. the probability that during the time interval $[0,t]$ the process $S_t$ was positive, with $x$ and $y$ are initial values, solved via the Fourier transform in $x$: $$\frac{2}{\pi}\int^\infty_0 \frac{\sin (\omega x)}{\omega}\bigg(\frac{\Delta(\omega)e^{-m_-(\omega)\kappa t}}{m_+(\omega)+m_-(\omega)e^{-\Delta(\omega)\kappa t}}\bigg)^{\frac{2\theta\kappa}{\xi^2}}e^{-\frac{2y\kappa}{\xi^2}B(\omega,\kappa t)}d\omega$$ where $B(\omega,\tau)$ is a solution of the Riccati equation $$\frac{\partial B}{\partial \tau}(\omega,\tau)=-B(\omega,\tau)-B(\omega,\tau)^2+\frac{\xi^2 \omega^2}{4\kappa^2}, B(\omega,0)=0;$$ $$\Delta(\omega)=\sqrt{1+(\xi\omega/\kappa)^2}, m_{\pm}(\omega)=\frac{\Delta(\omega)\pm 1}{2}$$

Any hints in this direction as much as completely different solutions are much appreciated.

  • $\begingroup$ If I can ask, is this just a mathematical challenge, or is there a real world motivation behind it? Pricing this with finite differences is straightforward and trivial to implement if you already have a vanilla Heston FD engine. $\endgroup$
    – Yian Pap
    Feb 5, 2019 at 23:13
  • $\begingroup$ @YianPap good point! It's mainly a math challenge actually, although I'm seeking an alternative to Monte Carlo pricing with importance sampling which works well for discretely monitored barriers but cannot reproduce the price for continuously monitored barriers. I don't have a vanilla Heston FD engine, but I guess it is also a valid approach worth being considered. Although it only works in low dimensions right? $\endgroup$
    – FunnyBuzer
    Feb 6, 2019 at 0:16
  • $\begingroup$ Right, if you want high dimensions then you only have MC. You can make your MC price continuously monitored barriers pretty accurately though. In the link below I talk about this a little and show some results for Heston: acenumerics.com/miscellaneous/… $\endgroup$
    – Yian Pap
    Feb 6, 2019 at 19:50
  • $\begingroup$ @YianPap so basically you mean to use the Brownian Bridge result for a GBM like the one I wrote above? Also, for comparison, would you please write an answer about the exact PDE-based solution with full-truncation Euler discretisation? $\endgroup$
    – FunnyBuzer
    Feb 6, 2019 at 21:23
  • 1
    $\begingroup$ This paper explains it in section 1: "Advanced Monte Carlo methods for barrier and related exotic options", by Emmanuel Gobet (hal.archives-ouvertes.fr/hal-00319947/document) I suggest you read that first and then my comments in the blog about applying this technique to Heston. $\endgroup$
    – Yian Pap
    Feb 10, 2019 at 14:59


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