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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. \begin{equation} dS(t) = \alpha S(t)dt + \sigma_{2}S(t)dW(t). \end{equation}

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)?


Inspired by AFK, I try to answer this question by myself in mathematical way.

Firstly, the main idea of generating the portfolio with arbitrage opportunity is to buy a call option and at $\sigma_{1}$, and sell a call priced at $\sigma_{2}$. i.e. \begin{equation} X(t) = c(t,S(t)) - c^{\sigma_{2}}(t,S(t)) \end{equation} where $X(t)$ denote the value of portfolio, $c(t,S(t))$ is the value of the option at time $t$, and $c^{\sigma_{2}}$ is the value of the option priced in $\sigma_{2}$.

Actually, $c^{\sigma_{2}}$ was not exit in the market, but it doesn't matter since you can replicate it by hedging, which means, \begin{equation} X(t) = c(t,S(t)) - c_{x}(t,S(t))S(t) - \Gamma(t)M(t) \end{equation} Now, we want to show that X(t) has arbitrage opportunity.

It is trivial to see that X(0) = 0, then we want to show that dX(t) > 0 (Actually, we finally prove that de^{-rt}X(t) > 0). By Ito formula, we find that $$ dc(t,S(t)) = c_{t}dt + c_{x}dS(t) + 1/2c_{xx}d[S,S](t) $$ and $$ dX(t) = dc(t,S(t)) - c_{x}dS(t) - r(c - X(t) -c_{x}S(t))dt. $$ Then, substitute dc into this equation, we get $$ dX(t) = (c_{t}+1/2c_{xx}\sigma_{2}^{2}S(t)^{2}-rc+rc_{x}S(t))dt + rX(t) $$ Note that c(t,S(t)) follows the Black-scholes equation with $\sigma_{1}$, so we have $$ dX(t) - rX(t) = (c_{t}+1/2c_{xx}\sigma_{1}^{2}S(t)^{2}-rc+rc_{x}S(t))dt + 1/2c_{xx}(\sigma_{2}^{2} - \sigma_{1}^{2})S(t)^{2}dt $$ i.e. $$ de^{-rt}X(t) = 1/2c_{xx}(\sigma_{2}^{2} - \sigma_{1}^{2})S(t)^{2}dt $$ It is always positive ($\sigma_{2}>\sigma{1}$, and $c_{xx}>0$).

In summary, X(t) is a portfolio with X(0) = 0, and de^{-rt}X(t) is always positive, s.t. it has arbitrage opportunity.

If any problem in my idea and my proof, please let me know.

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  • $\begingroup$ essentially you are asking does the BS model enforce a price via no arbitrage. The answer is yes. See eg my book "concepts..." for the details. $\endgroup$ – Mark Joshi Apr 14 '15 at 9:06
  • $\begingroup$ @MarkJoshi Thank you for your answer. Do you mean the book "The Concepts and Practice of Mathematical Finance"? $\endgroup$ – Ben Dai Apr 14 '15 at 14:36
  • $\begingroup$ yes that is my book. $\endgroup$ – Mark Joshi Apr 14 '15 at 20:30
  • $\begingroup$ Price of the call increases with volatility. To arbitrage you want to buy the call priced at $\sigma_1$ and sell a call priced at $\sigma_2$. Obviously no one is going to buy you a call at a higher price but it doesn't matter since you can replicate it by a $\Delta$-hedge with the true volatility $\sigma_2$. $\endgroup$ – AFK Apr 14 '15 at 21:46
  • $\begingroup$ @AFK Your idea really inspire me. I want to do it in mathematical way tomorrow. Thank you very much. $\endgroup$ – Ben Dai Apr 15 '15 at 10:35
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I think the delta-replicating of $\sigma_2$ call is just a fancy way of saying "hedging the call option bought at $\sigma_1$ volatility, with deltas based on $\sigma_2$ volatility". This is full arbitrage in case the hedging/replicating is optimal, and just a statistical arbitrage in real life. You probably do not need such sophisticated proof of why this is an arbitrage, because in case of perfect replication, you replication portfolio will earn just the fair value of the option (based on $\sigma_2$ volatility). And whenever you buy the option at lower price, you have an arbitrage.

This question is probably a duplicate of: Volatility arbitrage - how is the profit extracted? (link is to my answer, see also valuable info in comments).

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  • $\begingroup$ Thank you very much, your answer is very useful to me especially the idea of statistical arbitrage. However, I do the mathematics not for finding out why is an arbitrage, but for understanding the model. $\endgroup$ – Ben Dai Apr 16 '15 at 12:09

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