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Covered Calls and the Stock Price Upper Bound for Call Options

Article Quant Q&A · Author: kirafa

Summary

The document examines why a European or American call option cannot be worth more than its underlying stock. The questioner attempts an arbitrage argument by buying the stock and writing a call, but incorrectly treats the stock’s terminal price gain as realized cash while continuing to hold the shares. The accepted answer clarifies that a covered call retains the stock so it can be delivered if the option is exercised; the proposed profit expression therefore does not describe the final cash position.

The discussion states that if the call premium exceeds the stock price, buying the stock and selling the call produces an initial excess of premium over stock cost, while the stock covers the exercise obligation. This supports the upper-bound intuition for both call styles. The explanation is brief and does not develop the full payoff accounting or address complications such as dividends, financing, or transaction costs. A second answer is incoherent and contributes no usable analysis.

Key ideas

  • A call option’s value is bounded above by the current price of its underlying stock.
  • A covered-call position combines long stock with a short call.
  • The stock’s terminal price change is not realized cash if the shares are retained to cover exercise.
  • The proposed arbitrage reasoning depends on accounting for the stock delivery obligation correctly.

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Full text
# Upper Bound on European/American Call Option (Hull)


# Upper Bound on European/American Call Option (Hull)












I recently began reading Hull's derivatives textbook, and found a line that he didn't expand on much. Let $c$ be the price of a European call, $C$ be the price of an American call, and $S_0$ be the current price of the stock underlying the call. Hull then says:

> "No matter what happens, the option can never be worth more than the stock. Hence, the stock price is an upper bound to the option price: $c \leq S_0$ and $C \leq S_0$. If these relationships were not true, an arbitrageur could easily make a riskless profit by buying the stock and selling the call option."

I had no clue how this would end up in an arbitrage, so I tried prove it myself for the American call case. But when I did, I got something different from what Hull claimed, so I must have made a mistake somewhere:

Suppose the call has strike $K$. Initially, we buy the stock for $S_0$ and write the call for $c$, so that we currently have $C-S_0$. At maturity, the stock's price has changed to $S_T$, so that so far, we have $C + (S_T - S_0)$.

Now, we either have $K < S_T$ or $K \geq S_T$. Consider the case where $K \geq S_T$. In this case, the owner of the call will not exercise, so our final profit is $C + (S_T - S_0)$.

Isn't the final results conditional on $S_T$? How does this make the profit riskless?

## Answer by KaiSqDist (score 0, accepted)

https://quant.stackexchange.com/a/79927

I don't agree that the final results are conditional on $S_T$. The whole point of a covered call (short call long underlying), is to hold the asset in the event the option counterparty exercises the call i.e. $S_T>K$ as you have mentioned. You are not expected to sell the underlying asset when it is worth $S_T$. Therefore, the expression $C+(S_T-S_0)$ is erroneous.

The riskless profit will therefore always be given by:

$$C-S_0$$

## Answer by Herberto Silva (score 0)

https://quant.stackexchange.com/a/81716

is simply in-free and oldGRADE for what spoil can afford of a poor one called down here this time Joes theory instead of twins one and say <the smarter eat up the fool ones» but defensive system abbreviates and enlongates both edges criteria for the suspention of a third menber entrance with Am profits and this words like ´soglasno´or other are meaningless,economics tailburning with the speed of light relatively1!

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.