# Gatheral's change of variables for stochastic volatility PDE

This is taken from Gatheral's book "The Volatility Surface", where he tries to go from equation 2.3 to equation 2.4.

We have the following PDE,

$$\frac{\partial V}{\partial t}+\frac{1}{2}vS^2\frac{\partial ^2 V}{\partial S^2} + \rho\eta vS\frac{\partial ^2 V}{\partial S\partial v}+\frac{1}{2}\eta^2v\frac{\partial ^2 V}{\partial v^2} + rS\frac{\partial V}{\partial S}-rV-\lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0$$

By using a change of variables,

$$x=\ln{\frac{Se^{r\tau}}{K}}, \tau = T-t$$

show that it reduces to

$$-\frac{\partial C}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 C}{\partial x^2} -\frac{1}{2}v\frac{\partial C}{\partial x} +\frac{1}{2}\eta^2v\frac{\partial ^2 C}{\partial v^2} + \rho\eta v\frac{\partial ^2 C}{\partial x\partial v} - \lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0$$

My working are as follows...

We have $C(x,v,\tau)=V(Ke^{x-r\tau}, v, T-\tau)$. The partial derivatives are,

\begin{aligned} \frac{\partial V}{\partial t} &= \frac{\partial V}{\partial S}\frac{\partial S}{\partial t} + \frac{\partial V}{\partial t}\\ &=\frac{\partial V}{\partial x}\frac{\partial x}{\partial S}\frac{\partial S}{\partial \tau}\frac{\partial \tau}{\partial t}+\frac{\partial V}{\partial \tau}\frac{\partial \tau}{\partial t}\\ &=\frac{\partial V}{\partial x}\frac{1}{S}(-rS)(-1)-\frac{\partial V}{\partial \tau}\\ &= r\frac{\partial V}{\partial x}-\frac{\partial V}{\partial \tau}\\ \frac{\partial V}{\partial S} &= \frac{\partial V}{\partial x}\frac{\partial x}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial x} \\ \frac{\partial^2 V}{\partial S^2}&= \frac{1}{S}\frac{\partial }{\partial x}\frac{\partial x}{\partial S}\frac{\partial V}{\partial x}-\frac{1}{S^2}\frac{\partial V}{\partial x} \\ &=\frac{1}{S^2}\left(\frac{\partial^2 V}{\partial x^2}-\frac{\partial V}{\partial x} \right)\\ \frac{\partial V}{\partial S\partial v} &= \frac{\partial }{\partial S}\frac{\partial V}{\partial v}=\frac{\partial }{\partial x}\frac{\partial x}{\partial S}\frac{\partial V}{\partial v} = \frac{1}{S}\frac{\partial V}{\partial x\partial v} \end{aligned}

Substituting into the original PDE, I unfortunately get,

$$r\frac{\partial C}{\partial x}-\frac{\partial C}{\partial \tau}+\frac{1}{2}v\frac{\partial ^2 C}{\partial x^2} -\frac{1}{2}v\frac{\partial C}{\partial x} +\frac{1}{2}\eta^2v\frac{\partial ^2 C}{\partial v^2} + \rho\eta v\frac{\partial ^2 C}{\partial x\partial v} +r\frac{\partial C}{\partial x}-rC- \lambda(v-\bar{v})\frac{\partial V}{\partial v} = 0$$

I'm not sure how to get rid of the $r\frac{\partial C}{\partial x}$ and $rC$ terms.

I think perhaps I left of crucial steps involving his remark that "Further, suppose that we consider only the future value to expiration C of the European option price rather than its value today and define $\tau = T − t$."

Can someone help? Thanks!

First note that you have a typo in the definition of the moneyness. It should be

$$x = \ln \left( F_{t, T} / K \right) = \ln \left( S e^{r \tau} / K \right).$$

Following the remark that you cited, we then define

$$e^{-r \tau} C(x, \nu, \tau) = V(S, \nu, t).$$

Note that $C$ is a function of $\tau$ and not $t$ - this seems strange in your notation. The corresponding partial derivatives are given by

\begin{eqnarray} \frac{\partial V}{\partial t} & = & e^{-r \tau} \left( r C - r \frac{\partial C}{\partial x} - \frac{\partial C}{\partial \tau} \right), \\ \frac{\partial V}{\partial S} & = & e^{-r \tau} \frac{1}{S} \frac{\partial C}{\partial x}, \\ \frac{\partial^2 V}{\partial S^2} & = & e^{-r \tau} \frac{1}{S^2} \left( \frac{\partial^2 C}{\partial x^2} - \frac{\partial C}{\partial x} \right). \end{eqnarray}

Substituting back yields Equation (2.4) in Gatheral's book. I would also recommend you clearly distinguish between $C$ and $V$ and don't have $V$ on both sides of your partial derivatives.

• thanks for pointing out my several typos, I have fixed them up. – Danny Jun 17 '17 at 14:00
• oh so what the paragraph is saying is that $C$ is defined to be the undiscounted price, so we have to add the discounting term back before we are able to equate that to $V$. thank you very much! – Danny Jun 17 '17 at 14:05