Consider the hybrid model given by $$dS=(r-q) S dt + \sqrt{v} S dZ_1$$ $$dv = \kappa_v (\theta_v - v) dt + \sigma_v \sqrt{v} dZ_2$$ $$dr = \kappa_r (\theta_r - r) dt + \sigma_r r^p dZ_3$$

with correlation values $\rho_{S,v}$, $\rho_{S,r}$ and $\rho_{v,r}$, where $p=0$ for the Heston-Hull-White (HHW) and $p=1/2$ for the Heston-CIR (HCIR) models.

I am trying to find the derivation of the pricing PDE given in the paper

$$0 = \frac{\partial f}{\partial t} + \frac{1}{2} v S^2 \frac{\partial^2 f}{\partial S^2} +\frac{1}{2} \sigma_v^2 v \frac{\partial^2 f}{\partial v^2} +\frac{1}{2} \sigma_r^2 r^{2p} \frac{\partial^2 f}{\partial r^2} + \rho_{S,v} \sigma_v \sqrt{v} r^p S \frac{\partial^2 f}{\partial S \partial v} + \rho_{S,r} \sigma_r r S \frac{\partial^2 f}{\partial S \partial r} + \rho_{v,r} \sigma_v \sigma_r \sqrt{v} r^p \frac{\partial^2 f}{\partial v \partial r} + (r - q) S \frac{\partial f}{\partial S} + \kappa_v (\theta_v - v) \frac{\partial f}{\partial v} + \kappa_r (\theta_r - r) \frac{\partial f}{\partial r} -rf$$

This is equation 2.16 without the log-transform of $S$.

If there is another derivation for the HHW or the HCIR pricing PDE's (preferably in terms of a delta heding portfolio), please could you point me to such a paper. I can get the derivations for the Heston model pricing PDE easily, but not for the hybrid models. Maybe I am looking in the wrong places?

  • $\begingroup$ As mentioned in this answer, you can get every pricing PDE from the no-arbitrage principle $\mathbb{E}^\mathbb{Q}[\text{d}V]=rV\text{d}t$. All you need is Itô's Lemma in the three dimensions, $f=f(t,S,v,r)$. $\endgroup$
    – Kevin
    Jul 11, 2022 at 21:15

1 Answer 1


Let's consider the following dynamics under a risk-neutral measure:\begin{align} \text{d}S&=(r-q) S \text{d}t + \sqrt{v} S \text{d}Z_S,\\ \text{d}v &= \kappa_v (\theta_v - v) \text{d}t + \sigma_v \sqrt{v} \text{d}Z_v,\\ \text{d}r &= \kappa_r (\theta_r - r) \text{d}t + \sigma_r r^p \text{d}Z_r. \end{align} with correlations $\text{d}Z_S\text{d}Z_v=\rho_{Sv}\text{d}t$, $\text{d}Z_S\text{d}Z_r=\rho_{Sr}\text{d}t$ and $\text{d}Z_v\text{d}Z_r=\rho_{vr}\text{d}t$.

Itô's Lemma suggests \begin{align} \text{d}f =& f_t\text{d}t + f_S\text{d}S+f_v\text{d}v+f_r\text{d}r+\frac{1}{2}f_{SS}(\text{d}S)^2+\frac{1}{2}f_{vv}(\text{d}v)^2+\frac{1}{2}f_{rr}(\text{d}r)^2 \\ &+ f_{Sv}\text{d}S\text{d}v+ f_{Sr}\text{d}S\text{d}r+ f_{vr}\text{d}v\text{d}r. \end{align}

Note that \begin{align} (\text{d}S)^2 &= vS^2\text{d}t, \\ (\text{d}v)^2 &= \sigma_v^2v\text{d}t,\\ (\text{d}r)^2 &= \sigma_r^2r^{2p}\text{d}t,\\ \text{d}S\text{d}v &= vS\sigma_v\rho_{Sv}\text{d}t,\\ \text{d}S\text{d}r &= \sqrt{v}S\sigma_rr^p\rho_{Sr}\text{d}t,\\ \text{d}v\text{d}r &= \sigma_v\sqrt{v}\sigma_rr^p\rho_{vr}\text{d}t. \end{align} Going back to Itô's Lemma yields \begin{align} \text{d}f =& f_t\text{d}t + (r-q) Sf_S \text{d}t +\kappa_v (\theta_v-v)f_v \text{d}t +\kappa_r (\theta_r - r)f_r \text{d}t \\ &+ \sqrt{v} Sf_S \text{d}Z_S+ \sigma_v \sqrt{v} f_v\text{d}Z_v + \sigma_r r^p f_r\text{d}Z_r\\ &+\frac{1}{2}f_{SS}vS^2\text{d}t+\frac{1}{2}f_{vv}\sigma_v^2v\text{d}t+\frac{1}{2}f_{rr}\sigma_r^2r^{2p}\text{d}t \\ &+ f_{Sv}vS\sigma_v\rho_{Sv}\text{d}t+ f_{Sr}\sqrt{v}S\sigma_rr^p\rho_{Sr}\text{d}t+ f_{vr}\sigma_v\sqrt{v}\sigma_rr^p\rho_{vr}\text{d}t. \end{align} Because of their martingale property, Itô integrals have zero expectation. Thus, \begin{align} \mathbb{E}^\mathbb{Q}[\text{d}f] =& f_t\text{d}t + (r-q) Sf_S \text{d}t +\kappa_v (\theta_v-v)f_v \text{d}t +\kappa_r (\theta_r - r)f_r \text{d}t \\ &+\frac{1}{2}f_{SS}vS^2\text{d}t+\frac{1}{2}f_{vv}\sigma_v^2v\text{d}t+\frac{1}{2}f_{rr}\sigma_r^2r^{2p}\text{d}t \\ &+ f_{Sv}vS\sigma_v\rho_{Sv}\text{d}t+ f_{Sr}\sqrt{v}S\sigma_rr^p\rho_{Sr}\text{d}t+ f_{vr}\sigma_v\sqrt{v}\sigma_rr^p\rho_{vr}\text{d}t. \end{align} Finally, due to the absence of arbitrage, we have $\mathbb{E}^\mathbb{Q}[\text{d}f]=rf\text{d}t$. Thus, the pricing PDE is \begin{align} f_t &+ (r-q) Sf_S+\kappa_v (\theta_v-v)f_v +\kappa_r (\theta_r - r)f_r \\ &+\frac{1}{2}f_{SS}vS^2+\frac{1}{2}f_{vv}\sigma_v^2v+\frac{1}{2}f_{rr}\sigma_r^2r^{2p} \\ &+ f_{Sv}vS\sigma_v\rho_{Sv}+ f_{Sr}\sqrt{v}S\sigma_rr^p\rho_{Sr}+ f_{vr}\sigma_v\sqrt{v}\sigma_rr^p\rho_{vr}-rf=0. \end{align}

Setting $x=\ln(S)$, we have $\text{d}S=S\text{d}x$ which gives $Sf_S=f_x$ and $S^2f_{SS}=f_{xx}-f_x$. Thus, the pricing PDE turns into \begin{align} f_t &+ \left(r-q-\frac{1}{2}v\right) f_x+\kappa_v (\theta_v-v)f_v +\kappa_r (\theta_r - r)f_r \\ &+\frac{1}{2}vf_{xx}+\frac{1}{2}\sigma_v^2vf_{vv}+\frac{1}{2}\sigma_r^2r^{2p}f_{rr} \\ &+ \rho_{Sv}\sigma_vvf_{xv}+ \rho_{Sr}\sigma_r\sqrt{v}r^pf_{xr}+ \rho_{vr}\sigma_v\sigma_r\sqrt{v}r^pf_{vr}-rf=0. \end{align}

Regarding a hedging argument: You can't use delta hedging. You have three sources of uncertainty and thus you need three tradable assets to hedge the risks. Only using the stock to hedge delta is not sufficient. If you have enough tradable assets, you can set up a portfolio $\Pi$ and eliminate all risks such that $\text{d}\Pi=...\text{d}t$ (without any $\text{d}Z$ terms). Then, you can equate $\text{d}\Pi=r\Pi\text{d}t$ and get a PDE. Alternatively, assuming that your SDEs are given under a risk-neutral measure, you can directly take an expectation of $\text{d}f$ and thereby removing the $\text{d}Z$ terms. Equating this conditional expectation with $rf\text{d}t$ then gives you the pricing PDE. Both approaches are equivalent in the end.

  • $\begingroup$ @JanStuller I use $\text{d}Z_S\text{d}Z_v$ as abusive notation for the quadratic covariation $\text{d}\langle Z_S,Z_v\rangle$ which is a simple Riemann/Lebesgue integral (no Ito integral). Hence, no expectation. Intuitively, $\text{d}Z_S\text{d}Z_v$ is deterministic and of order $\text{d}t$ and no stochastic term remains. Remember that writing $\text{d}Z$ is always just (meaningless) notation and a short hand expression for integrals. All I am doing is using the short-hand notation $(\text dW)^2=\text dt$. $\endgroup$
    – Kevin
    Jul 20, 2022 at 13:10
  • $\begingroup$ Ok. When I have simply two Wiener processes and I wanna say that their corr is $\rho$, should I write $\mathbb{E}[dW_1(t)dW_2(t)]=\rho dt$ or $dW_1(t)dW_2(t)=\rho dt$ or $\mathbb{E}[W_1(t)W_2(t)]=\rho t$? $\endgroup$ Jul 20, 2022 at 14:55
  • 1
    $\begingroup$ @JanStuller To be honest, I have seen (and actually written) both ways, with and without the expectation. Whenever I write $\text{d}W$, I am using a short-hand notation to make my life easier by avoid writing down integrals. Thus, in the sense of simplicity, I am more than happy to write $\text{d}W_1\text{d}W_2=\rho\text{d}t$. But as I said, you'll also find $\mathbb{E}[\text{d}W_1\text{d}W_2]=\rho\text{d}t$ in my writings. I also noticed that other authors switch between the two ways in different papers. $\endgroup$
    – Kevin
    Jul 20, 2022 at 15:03

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