I'm reading Sinclair's Option Pricing and am confused by the proof for the maximum value of a call. It makes sense logically that a call can't be worth more than the underlying, and so:

c <= S

The proof the book uses however is as follows. Say there's a call trading for more than the underlying. Then, I will sell the call, and buy the underlying. At expiration (time T), our profit is:

c-(S_0 -S_T)

I don't understand why we subtract the S_T at the end? Doesn't this imply a huge profit? Say the call was worth 110, S=100, and S_T = 105. Then the profit would be 110-(100-105) = 115. That doesn't make sense.

ALSO, does this just assume that the option that we sold expires OTM, so it's not exercised??


1 Answer 1


I will try to clarify. Imagine that a European call option is written on $S$ and have maturity $T$. Moreover, imagine it is worth $c_t$ at time $t$. Can $c_t > S_t$? Under the no arbitrage assumption, it can't. To see why, let us build the following trading strategy:

  • We sell the call option and gain $c_t$
  • We buy the underlying $S_t$

because $c_t > S_t$, we have a sure profit in $t$ equal to $\pi_t = c_t - S_t > 0$.

What happens at expiration time $T$? It can be the case that the option is exercised by the buyer (if he is rational only if $S_T > K$):

  • Since we are the seller, we must give him the stock. But we already have the stock in our pocket (recall, we bought it in $t$). So, at time $T$ we will have another profit of $K = S_T + K - S_T$.

It can also happen that the option is not exercised. In this case nothing will happen, and we will hold the stock.

The crucial intuition is then the following: we have a SURE ($\pi_t$) profit and this profit is RISKLESS (we will not lose money in any state of the world). This is indeed arbitrage: since we want to price contingent claims under the assumption that no arbitrage exists, $c_t \leq S_t$.

  • 1
    $\begingroup$ @noob2: $K < S_T$. So, there's some positive amount that makes them equal ( if you add it to the LHS ) and he defined this positive amount as $S_t - k$. Atleast that's my understanding. A very nice proof indeed. $\endgroup$
    – mark leeds
    Commented May 27, 2021 at 14:53
  • $\begingroup$ @noob2: I'm not sure about your typo theory ( which could be correct ) but I also don't think my comment is correct because of the following: (c_t - S_t) is the initial profit of the seller of the call who then owns the stock. But, how can one be sure that $S_T$ is greater than $S_t$. Couldn't $S_T$ be less than $S_t$ in which case the seller of the call could lose money ? So, I'm still a little confused and not even clear on why part 2) needs to be considered anyway. The initial profit at time $t$ would seem to seal the deal ? Yes, you have call exposure but that call could expire OTM.. $\endgroup$
    – mark leeds
    Commented May 28, 2021 at 1:14
  • 1
    $\begingroup$ Sorry guys, the small $k$ was just a typo. I intended the strike $K$. Also, in the same inequality, I fixed $S_T$ in place of $S_t$. Even if we bought the stock in $t$, it will have a value of $S_T$ in $T$. $\endgroup$ Commented May 28, 2021 at 8:13
  • $\begingroup$ Very good. The point is clear now. You receive $c_t−S_t$ at time $t$ and then (instead of having to pay anything!) you have a chance to receive another amount $K$ at time $T$ and no chance of loss. Which is an arbitrage. $\endgroup$
    – nbbo2
    Commented May 28, 2021 at 10:47
  • $\begingroup$ @Yoda And Friends or noob2: It makes more sense now. So, if $S_T$ ends up less than $S_t$, then there's no exercise and owner still owns the stock. But couldn't the stock owner then have lost ( in stock price value ) more than he made initially ? Sorry to be such a pest but it's interesting so I just want to get it. Thanks. $\endgroup$
    – mark leeds
    Commented May 28, 2021 at 19:00

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