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:
- What is meant by future value to expiration?
- how is $C$ related to $V$ in functional form?