# derivation of heston pde in gatheral

Following Gather (the volatility surface, chapter 2) we assume the following process:

$$dS_t = S_t(\mu_t dt+\sqrt{\nu_t}dZ^1_t)$$ $$d\nu_t= -\lambda(\nu_t-\bar{\nu})dt+\eta\sqrt{\nu_t}dZ^2_t$$

where $Z^1,Z^2$ are two brownian mortion such that $d\langle Z^1,Z^2\rangle_t= \rho dt$. Using the general valuation pde for a stochastic volatility model we get for this process the following pde:

$$\frac{\partial V}{\partial t} +\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\nu S^2+\rho\eta\nu S \frac{\partial^2 V}{\partial \nu \partial S} + \frac{1}{2}\eta^2\nu\frac{\partial^2 V}{\partial \nu^2} + rS \frac{\partial V}{\partial S}-rV=\lambda(\nu-\bar{\nu})\frac{V}{\partial \nu}$$

Now by introducing $F_{t,T}$ the time $T$ forward of the stock index, $x:=\log{(\frac{F_{t,T}}{K})}$, where $K$ denotes the strike space, $\tau:=T-t$ and $C$ the future value to expiration of the European option prices (rather than its value today, $V$) the above pde should transform to

$$-\frac{\partial C}{\partial \tau}+\frac{1}{2}\nu C_{11}-\frac{1}{2}\nu C_1+\frac{1}{2}\eta^2\nu C_{22}+\rho\eta\nu C_{12} - \lambda(\nu-\bar{\nu})=0$$

where the subscripts $1,2$ refer to differentiation w.r.t $x$ and $\nu$.

We have $V(S,\nu,t)=C(f(S),\nu,g(t))$, where $g(t):=\tau=T-t$. About the form of $f$ I'm unsure. Using this we get for the first term:

$$\frac{\partial V}{\partial t} = \frac{\partial C}{\partial \tau} \frac{\tau}{t}=-\frac{\partial C}{\partial \tau}$$

For $f$ we know $f(S)=\log{\frac{F_{t,T}(S)}{K}}$ ( I suppress the time subsctript on $t$). I've tried $F_{t,T}=S_t\exp{\int_t^T\mu_sds}$, with $\mu_s\equiv 0$. However I do not see how we can get this PDE in terms of $C$. It would be great if someone could explain the following two:

1. What is meant by future value to expiration?
2. how is $C$ related to $V$ in functional form?

Consider an asset $A$. You need to hold $A$ at time $T$ but since you don't need it now you don't want to buy it now. Instead you enter a forward contract with someone that says that at time $T$ you will pay the amount $K$ and get the asset in exchange. What should be the strike $K$ for the deal to be fair to both parties? By definition this is the $T$-forward price of $A$.
$$F^T_t = S_te^{r(T-t)}$$ Indeed the seller in the forward contract can borrow $S_t$ at the risk free rate, buy the asset and then at $T$, receive $K$, deliver the asset and payback what he borrowed + interest $S_te^{r(T-t)}$. So he started with 0 and ends up with $K - S_te^{r(T-t)}$. So by absence of arbitrage we must have $K = S_te^{r(T-t)}$. (In the case of dividend yield $q$, you just have to replace $r$ by $r-q$).
In general, for any $t\leq T$, $$\textrm{present value} = E^{\mathbb{Q}}_t[e^{-\int_t^Tr_s ds}g(S_T)] = P(t,T)E^{\mathbb{Q}_T}_t[g(S_T)] = P(t,T) \times \textrm{forward value}$$ where $P(t,T) = E^{\mathbb{Q}}_t[e^{-\int_t^Tr_s ds}]$ is the price of the ZCB with maturity $T$. With deterministic interest rate this is just $P(t,T) = e^{-r(T-t)}$.
2) Since $C$ corresponds to the undiscounted price of the call: $$V(t,S,\nu) = e^{-r(T-t)}C(T-t,\log(Se^{(r-q)(T-t)}/K),\nu)$$ Reasonning in forward terms allows to separate interest rate considerations from the rest and often simplify PDE's by making discounting terms disappear.
• many thanks for the clarification. Where did you find that $C$ is the undiscounted price? So $V$ is always discounted in the book? – math Dec 26 '14 at 19:18