# Method for finding a arbitrage opportunity when market price of call is incorrect

The solution of the Black-scholes equation is the price of a European call. And the option price assumes the underlying stock is a geometric Brownian motion with volatility $\sigma_{1}>0$.

Suppose, however, the underlying asset is really a geometric Brownian motion with volatility $\sigma_{2} > \sigma_{1}$, i.e. $$dS(t) = \alpha S(t)dt + \sigma_{2}S(t)dW(t).$$

Consequently, the market price of the call is incorrect.

Can we set up a portfolio which has an arbitrage opportunity in the market? Furthermore, if there any methods to generate a portfolio arbitrage opportunity (how to consider this problem)?

• Could you clarify what you mean by the second question? Commented Apr 13, 2015 at 15:53
• @user223097 That equation means that the stock price is a Geometric Brownian Motion. W(t) is a Brownian motion. Commented Apr 13, 2015 at 16:05

This is an excellent question! You can! This is basically the result called "The fundamental theorem of derivatives trading" see link. We assume $$dS(t) =S(t) (\alpha dt + \sigma_2 dW(t) )$$ but some derivative is priced at implied volatility $\sigma_1<\sigma_2$, assume it has payoff function $h(S_T)$ and has value $C_h(S_t,r,\sigma_2,t)=C(t,S_t)$ consider a last $\hat{\sigma}^2$, this is the volatility we will use for hedging and can be either or something else. Assume we have constant interest rate $r$. The derivative price function satisfies the Black-Scholes PDE $$\frac{\partial C}{\partial t} + rs \frac{\partial C}{\partial s}+ 1/2 \sigma_2^2 s^2 \frac{\partial^2 C}{\partial s^2}- rC = 0$$ and C(T,s) = h(s). To make the argument similar to usual hedging strategies we will assume we sell the underlying and hedge it - in fact to enjoy the arbitrage we need to exactly reverse the strategy described. Say we sell the derivative and delta hedge it using $\hat{\sigma}$ we will continuously hold $\phi_t = \frac{\partial C}{\partial s}$ (determined by the BS equation with $\hat{\sigma}^2$) of the risky asset and make the hedging portfolio self-financing in the risk free we get $$dX_t = \frac{\partial C}{\partial s}(S_t) dS_t + (X_t -\frac{\partial C}{\partial s}(S_t) \cdot S_t ) r dt$$ while the actual price process $Y$ (Ito on C from BS satisfies) $$dY_t = \frac{\partial C}{\partial s} dS_t + (\frac{\partial C}{\partial t}(S_t) +1/2 \sigma_2^2 S_t^2 \frac{\partial^2 C}{\partial s^2}(S_t) )dt$$ so the hedge error $Z = X - Y$ is (use BS equation) $$dZ_t = r X_t + 1/2 S_t^2 \frac{\partial^2 C}{\partial s^2}(S_t) (\hat{\sigma}^2-\sigma^2_2 ) dt$$ this yields that (solve it as a differential equation) $$Z_T = X_T-h(S_T) = \int_0^T e^{r(T-s)} 1/2 S_t^2 \Gamma_t ^2 (\hat{\sigma}^2 -\sigma_2 ^2) dt$$ note all terms in the integral are positive so if we hedge using $\hat{\sigma}=\sigma_1 <\sigma_2$ the (combined) position is free but will surely loose money, while in the case $\hat{\sigma}=\sigma_1$ the integral cancels, but the position will initially cost us money (we initiate a replication of a more expensive derivative). Ergo reversing the strategy yields an arbitrage :)
• Thank for your answer, it is really useful for me. I guess your main idea for this solution is to check the "hedge error" i.e. Z in your post. Do you think the portfolio $X_{t}$ could hedge the option priced by $\hat{\sigma}$? Commented May 7, 2015 at 13:57
• @BenDAI I'm glad - it is a very good question. The X does exactly hedge the option with $\hat{\sigma}^2$, Z is the result of the hedge. Do I understand your question correctly?