I'm trying to derive the Heston-Hull-White PDE. The correct backwards PDE is equation (1.3) of this paper on page (2). I will begin deriving the forward PDE, but switching between the two is trivial.

The model I am working with is the Heston-Hull-White model, given below:

$$\mathrm{d}S = rS\mathrm{d}t + \sqrt{v}S\mathrm{d}W_1$$ $$\mathrm{d}v = \kappa (\bar{v}-v)\mathrm{d}t+\omega\sqrt{v}\mathrm{d}W_2$$ $$\mathrm{d}r = \lambda(\theta(t)-r)\mathrm{d}t + \eta\mathrm{d}W_3\text{.}$$

I assume that $S$ and $v$ have correlation $\rho_{S,v}$, $S$ and $r$ have correlation $\rho_{S,r}$, and $v$ and $r$ have correlation $\rho_{v,r}$.

My approach is to apply the Feynman-Kac theorem. This is standard, so I will skip most steps.

  1. Let $h(S(T))$ be the payoff function of the option. For a vanilla call, $$h(S(T))=S(T)-K\text{.}\tag{1}$$

  2. Let $$g(t, S(t), v(t), r(t)) =\tilde{\mathbb{E}}\Big(e^{-\int_{u=t}^{u=T}r(u)\mathrm{d}u}h(X(T))\Big)\tag{2}$$ be the price of the option. My goal is to find the PDE (implied by Feynman-Kac) for $g$.

  3. $g$ is not a martingale, so we want to first make a transformation to get a martingale. If we follow a Black-Scholes example, we may try to do something like this: $$f(t, S(t), v(t), r(t)) = e^{-\int_{u=0}^{u=t}r(u)\mathrm{d}u}g(t, S(t), v(t), r(t))\text{.}\tag{3}$$ so that $f$ becomes a martingale. The problem here is that $r$ is a random variable, and we can't pull it outside of the expected value. I continue now as if the definition of $f$ makes sense.

  4. I now apply Itô's lemma and set the coefficient of $\mathrm{d}t$ equal to $0$. That gives the following PDE for $f$: $$f_t + rSf_S +\kappa(\bar{v}-v)f_v+\lambda(\theta(t)-r)f_r + \rho_{S,v}Sv\omega f_{s,v} + \rho_{S, r}\eta S\sqrt{v}f_{S, r} + \rho{v, r}\omega \sqrt{v}\eta f_{v, r}+\frac{1}{2}vS^2f_{S, S} + \frac{1}{2}v\omega^2f_{v, v} + \frac{1}{2}f_{r, r}\eta^{2}=0\text{.}\tag{4}$$

The PDE in (4) is the PDE for $f$, but I need the PDE for $g$. Following a Black-Scholes example, I get the PDE of $g$ by computing the partial derivatives of $f$ in terms of $g$ using (3).

For example, I compute

$$f_t = e^{-\int_{u=0}^{u=t}r(u)\mathrm{d}u}(-r(t) g + g_t)\text{.}\tag{5}$$

I then substitute this back into equation (4).

Now I need to compute $f_r$, but again, $r$ is a random variable. Step (3) was indeed a mistake. How do I continue?


1 Answer 1


(Just what I think is the right start)

The pricing PDE comes out of the dynamics of a self-financing portfolio, $\Pi$, hedged against the movements of stock, $S$, its volatility, $v$, and interest rate $r$.

With $V$ the target path-independent derivative, $U$ vanilla European option (different from $V$), and $P$ zero-coupon bond, the portfolio would be:

$$ \Pi = V+\alpha S + \beta U + \gamma P,$$ $$ d\Pi = dV+\alpha dS + \beta dU + \gamma dP, $$

and we would exploit the fact that $\Pi$ is riskless (set the integrands of $dS$, $dv$ and $dr$ to $0$) and also the fact that it returns $r$ :

$$ d\Pi = r \Pi dt. $$

We note that, under HW dynamics, $P(t,T)$ is just a deterministic function of $r_t$ only (so we don't need to worry about the integrated short rate) and it has its own PDE (in $r$ variable).

This would be an extension of the detailed derivation of Heston PDE by Rouah here.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.