Wednesday, July 18, 2018

option pricing - 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





  1. 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.





  2. 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'.




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