Deriving the Forward-Moneyness Transformation for a Stochastic Volatility PDE
Summary
The document explains a change of variables that converts a stochastic volatility pricing PDE from spot price and calendar time to forward log-moneyness and time to expiry. Its central correction is to define the option’s future-value function with a discount factor: the original price equals the transformed function multiplied by the discount factor. This distinction produces time and log-moneyness derivatives that cancel the interest-rate terms in the transformed equation.
The response lists the relevant derivative relationships and says that substituting them into the original PDE yields the desired form. It also highlights a notation issue: the transformed function depends on time to expiry, not calendar time, and should be kept distinct from the original price. The material gives an algebraic derivation rather than numerical validation, and the displayed target equation retains a variance-drift derivative whose notation should be handled consistently with the transformed function.
Key ideas
- Define the transformed option value as a discounted future value to eliminate the risk-free rate terms.
- Use forward log-moneyness and time to expiry as the new independent variables.
- Apply the chain rule carefully to time, spot, and mixed derivatives.
- Keep the original price and transformed value distinct in notation.
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# Gatheral's change of variables for stochastic volatility PDE
# 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!
## Answer by LocalVolatility (score 4, accepted)
https://quant.stackexchange.com/a/34744
First note that you have a typo in the definition of the moneyness. It should be
\begin{equation} x = \ln \left( F_{t, T} / K \right) = \ln \left( S e^{r \tau} / K \right). \end{equation}
Following the remark that you cited, we then define
\begin{equation} e^{-r \tau} C(x, \nu, \tau) = V(S, \nu, t). \end{equation}
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.