# Derivation for call option upper bound

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?

## 2 Answers

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.

• Bah, I just realized I probably should not have repeated my question as an "Answer." Apologies. – Dürenand Mar 9 at 14:41
• you can edit your question and put this part there as well – Cettt Jun 11 at 6:45

I literally was going through this exact same problem going over it for too long. I came up with what you had independently so I think we are right.

Just found this: 