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In Euan Sinclair's book, Option Trading, he writes that $c <= S$, the price of a European call must be lower than the price of the underlying stock. To prove it, he applies the principle of no arbitrage:

Imagine a European call is trading for more than the underlying. We then choose to sell the call and buy the underlying at a price of $S_0$. At expiration (time $T$) our profit will be $c - (S_0 - S_T)$. But the second term has to be less than $S_0$, and our assumption was that $c > S_0$, so this profit must be greater than zero. We have to make money. So for there to be no arbitrage we need to have $c <= S$.

When I put myself in the arbitrageur's shoes, I imagine myself selling the call, buying the underlying, invested the proceeds at $r$ until expiry, then selling the underlying, all of which gets me the following profit:

$$ PL = (c-S_0)\,e^{\,r\,T}+\text{min}\,(K,S_T) $$ I have looked in every book I can find in order to understand why Sinclair's $$ PL = c-(S_0-S_T) $$ is a better equation for arbitrage profits on an overpriced ($c>S_0$) European call than $$ PL = (c-S_0)\,e^{\,r\,T}+\min\,(K,S_T) $$ but most books do not even bother starting with the profit equation when deriving an upper bound for an overpriced European call, except Sinclair. I like his approach of systematically thinking in terms of arbitrage, and I can derive the other bounds, but this one has me flummoxed.

Why is my, more complicated profit equation wrong or inaccurate?

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I have looked in every book I can find in order to understand why $$ PL = c-S_0+S_T $$ is a better equation for arbitrage profits on an overpriced ($c>S_0$) European call than $$ PL = (c-S_0)\,e^{\,r\,T}+\min\,(K,S_T) $$ but most books do not even bother starting with the profit equation when deriving an upper bound for an overpriced European call, except Sinclair. I like his approach of systematically thinking in terms of arbitrage, and I can derive the other bounds, but this one has me flummoxed.

Why is the second equation wrong or inaccurate?

Is there perhaps a better forum for this question? I notice that most of the topics herein are vastly more complicated.

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  • $\begingroup$ Bah, I just realized I probably should not have repeated my question as an "Answer." Apologies. $\endgroup$ – Dürenand Mar 9 at 14:41

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