Why Stochastic Volatility PDE Terms Must Equal a Shared Function

# Derivation of the Stochastic Vol PDE


# Derivation of the Stochastic Vol PDE












I'm trying to follow the derivation of the stochastic vol pde for an option price - as given in Gatheral (The vol surface), Wilmott on Quant Finance and many other places. As usual one starts off with a portfolio $\Pi = C + \Delta S + \Delta_1 V$, from which we want to derive a PDE which the option price $C(t,S,\sigma)$ satisfies. Apply Ito's formula, and choosing the hedge ratios appropriately we can make this portfolio riskless. After some algebra we end up with something like

$$g(S, C,C_t, C_S, C_{SS}, C_\sigma, C_{\sigma,s})=g(S,V,V_t, V_S, V_{SS}, V_\sigma, V_{\sigma,s})$$

i.e. a relationship $g$ between the derivatives of $V$ and $C$ (in particular I'm referring to the first equation on page 6 here). Now this is the part I don't understand, right under that equation the author says something along the lines, "Since the left hand side is a function of $C$ only and the right hand side is a function of $V$ only, the only way the equality can be true is if both sides equal to some function $f$ of the independent variables $S$, $\sigma$ and $t$". Can someone explain why this is so exactly?

And also why without loss of generality can we assume this function takes the form $f(t,S,\sigma)=\alpha-\phi \beta \sqrt{\sigma}$?

The complete derivation can be found here.

## Answer by olaker (score 7, accepted)

https://quant.stackexchange.com/a/10186

- So we have the identity $$g(S,\sigma, t, C,C_t,C_S,...)=g(S, t,\sigma, V,V_t,V_S,...)$$ where $S$, $\sigma$, and $t$ are independent variables and $V=V(S,\sigma,t)$, $C=C(S,\sigma,t)$ are some unknown functions. But we can also treat the above identity formally and assume that the functions $C,C_t,C_S,...,V,V_t,V_S,... $ are themselves independent parameters or variables. Now, if we fix all variables except for $C$ and compute the partial derivative $\partial g/\partial C$ we will find that $$\frac{\partial g}{\partial C}=0$$ for all values of $S,\sigma, t,C_t,C_S,...,V,V_t,V_S,...$ since the right-hand side of the identity does not depend on $C$. This imples that $g$ is constant with respect to $C$. Similarly, one can get that $$\frac{\partial g}{\partial V}=0$$ since the left-hand side of the identity does not depend on $V$. Hence, $g$ is constant with respect to $V$ as well. We can repeat this argument for $C_t,C_S,...,V_t,V_S,... $, which would imply that $g$ does not depend on any of those formal variables. However, we cannot apply the same argument to any of the variables $S$, $\sigma$ or $t$, since both sides of the identity depend on each of the them. In other words, $$ g(S,\sigma, t, C,C_t,C_S,...)=g(S, t,\sigma, V,V_t,V_S,...)=f(S,\sigma, t)$$ for some function $f=f(S,\sigma, t)$. The above argument is not dissimilar to the standard trick of separation of variables which is used to solve simple partial differential equations.

- Now, we can simply express the unknown function $f$ in terms of another arbitrary function $\phi=\phi(S,\sigma, t)$: $$f= -(\alpha-\phi\beta).$$ We do not know the function $\phi$ explicitly, of course. However, it is just slightly more convenient to deal with $\phi$ rather than $f$, and, besides, the former can be interpreted as the 'market price of risk'.