Option Hedging with a Stochastic Portfolio Diffusion

# Black and Scholes equation for portfolio **with** arbitrage


# Black and Scholes equation for portfolio **with** arbitrage












I am well aware of how the ordinary Black and Scholes equation is derived, under the assumption of an arbitrage free portfolio, $V=G-hS$. Here $S$ is the price of the underlying and $G$ is the option price. Suppose now that the portfolio is stochastic, and follows the process $$(1): dV=rVdt+f(V,t)VdW,$$ where $r$ is a constant parameter. The change of the portfolio also satisfies $dV=dG-h \cdot dS$. In the standard Black and Scholes, we have $h = \partial G/\partial S$. Assuming that the underlying follows a GBM, we have $$dS = \mu S dt + \sigma SdW,$$ where $\mu$ and $\sigma$ are constant parameters. Itô's lemma then provides $$dG = \left(\frac{\partial G}{\partial t}+\mu S\frac{\partial G}{\partial S}+\frac{1}{2}\sigma^2S^2\frac{\partial G^2}{\partial S^2} \right)dt + \sigma S\frac{\partial G}{\partial S}dW,$$ so that $$(2): dV=dG-hdS = \left(\frac{\partial G}{\partial t}+\mu S\left(\frac{\partial G}{\partial S}-h\right)+\frac{1}{2}\sigma^2S^2\frac{\partial G^2}{\partial S^2} \right)dt + \sigma S\left(\frac{\partial G}{\partial S}-h\right)dW $$

Matching 1) and 2), we see that we must have the following implicit relation for $h$ $$(3): f(G-hS,t)(G-hS)=\sigma S \left(\frac{\partial G}{\partial S}-h\right). $$ In the standard Black and Scholes we would have $f=0$, thus recovering $h = \frac{\partial G}{\partial S}$. If $f$ only depends on its second argument ($t$), then 3) can also easily be solved for $h$, yielding $$(5):h =\frac{Gf(t)-\sigma S \frac{\partial G}{\partial S}}{S(f(t)-\sigma)}, \ f \ \mathrm{independent \ of} \ V$$

Combining 1) and 2) would then give the Black and Scholes like equation $$(6):\frac{\partial G}{\partial t}+\mu S\frac{\partial G}{\partial S}+\frac{1}{2}\sigma^2S^2\frac{\partial G^2}{\partial S^2}+(r-\mu)\frac{Gf(t)-\sigma S \frac{\partial G}{\partial S}}{f(t)-\sigma}-rG=0, \ f \ \mathrm{independent \ of} \ V $$

EDIT

The above model is considered in the following paper by Contreras et al.. They write the stochastic portfolio 1) as $$(1'): dV=rVdt+f(t,S(t))VdW. $$ Then $h$ can again be determine as in equation 5), yielding the Black and Scholes like equation 6). See section 2 of the cited paper.

However, I would like to know what could we do if $f$ is not independent of $V$, i.e. if we retain the formulation in 1).

EDIT

I edit the question to clarify what I mean. Suppose we retain the dependence of $V$ in $f$. With $V=G-hS$, we can write equation 3) as $$(7):f(V,t)-\sigma V = \sigma S \frac{\partial G}{\partial S} - \sigma G.$$ Suppose the function $V \mapsto f(V,t)V-\sigma V$ is invertible for each $t$ and $\sigma$, and call the inverse $\phi$, so that $\phi(x,t) = (V \mapsto f(V,t)V-\sigma V)^{-1}(x)$. We can then solve for $h$ to get $$(8):h=\frac{G-\phi(\sigma S \frac{\partial G}{\partial S} - \sigma G,t)}{S},$$ from which we get the modified Black and Scholes equation $$(9): \frac{\partial G}{\partial t}+\mu S\frac{\partial G}{\partial S}+\frac{1}{2}\sigma^2S^2\frac{\partial G^2}{\partial S^2}+(r-\mu)\left[G-\phi(\sigma S \frac{\partial G}{\partial S} - \sigma G,t) \right]-rG=0.$$ If $f$ is independent of $V$ then $\phi(x,t) = x/(f(t)-\sigma)$, so we recover the previous equation 6).

However, for a general function $f$ the map $V \mapsto f(V,t)V-\sigma V$ is not invertible. Consider, for example, $f(V,t)=V$, for which $V \mapsto f(V,t)V-\sigma V = V^2 - \sigma V$ is not invertible. Then we cannot uniquely determine $h$. Can we say anything about the equation for $G$ in this case?