Why a Call Option Cannot Be Worth More Than Its Underlying
Summary
The note explains the upper bound on a European call’s value: under the stated no-arbitrage assumption, its price cannot exceed the price of the underlying asset. The proposed argument is to sell an overpriced call and buy the underlying, creating a positive cash balance at the outset. At expiration, if the call is exercised, the seller can deliver the stock already held; if it is not exercised, the seller retains the stock. The key point is that the terminal stock position covers the delivery obligation, so the strategy does not depend on the call expiring out of the money.
The explanation is qualitative and focuses on the payoff logic rather than a full, time-value-adjusted arbitrage calculation. In particular, it describes the initial surplus as riskless but does not detail financing, dividends, or other contract and market assumptions. Its central lesson is the coverage argument behind the call-price bound, not a numerical valuation method.
Key ideas
- Selling a call while buying its underlying creates an initial surplus if the call costs more than the asset.
- The stock held in the strategy can be delivered if the call is exercised.
- If the call is not exercised, the seller keeps the underlying asset.
- The argument for the upper bound does not require the call to expire out of the money.
- The proof relies on a no-arbitrage setting and gives a simplified payoff explanation.
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Full text
# Maximum value of a call option proof # Maximum value of a call option proof 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?? ## Answer by Yoda And Friends (score 1) https://quant.stackexchange.com/a/64250 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, $c_t \leq S_t$.
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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.