Deriving the Heston-Hull-White PDE

Question

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?

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.