Why American Calls on Non-Dividend Stocks Are Not Exercised Early
Summary
The document explains why an American call on a stock that pays no dividends has the same value as an otherwise comparable European call. Early exercise gives the holder the stock price less the strike, while retaining the option preserves the right to buy later and delays payment of the strike. The response uses put-call parity to show that the European call's value exceeds the immediate exercise value when the interest rate is positive, so early exercise sacrifices value.
A second explanation cites a lower bound for the European call price based on the stock price and the discounted strike. This bound supports the comparison with immediate exercise and clarifies why the American option's early-exercise privilege adds no value in this setting. The reasoning depends on the non-dividend-paying assumption and the stated interest-rate condition; it does not address dividends or other situations where early exercise may be rational. The original question also distinguishes the option's exercise payoff from its market value before maturity.
Key ideas
- Exercising an American call early yields the stock price minus the strike when the option is in the money.
- Put-call parity implies a non-dividend European call is worth more than its immediate exercise value under the stated rate condition.
- The holder benefits from preserving optionality and delaying payment of the strike.
- Without dividends, the American call's early-exercise right does not increase its value.
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Full text
# Price American call equal to price European call (non-dividend-paying stock)
# Price American call equal to price European call (non-dividend-paying stock)
Let $\tilde{C}_K(t,T)$ be the value (price) of an American call option at strike $K$ and maturity $T$, and $C_K(t,T)$ the value (price) of a European call option at same parameters.
For a non-dividend-paying stock, $\tilde{C}_K(t,T) = C_K(t,T)$. Why?
My textbook says:
$\textbf{Proof}$: $\tilde{C}_K(t,T) \geq C_K(t,T)$ is obvious. (Why is this obvious?)
To prove $\tilde{C}_K(t,T) \leq C_K(t,T)$ consider
- The American is not exercised before $T$. Then $\tilde{C}_K(t,T) = C_K(t,T)$. (since then the American is the same as the European, this makes sense to me.)
- Suppose the American is exercised at $t<T$. Then
$$\tilde{C}_K(t,T) = S_t - K \leq C_K(t,T). \hspace{14cm}\blacksquare$$
I don't understand. I would think if the American $\tilde{C}_K$ is exercised at $t<T$, then $\tilde{C}_K(t,T)$ is the payoff of the call, hence $(S_t-K)^+$... And why would $S_t - K \leq C_K(t,T)$?
Reference: An introduction to quantitative finance by Stephen Blyth.
## Answer by user39119 (score 5, accepted)
https://quant.stackexchange.com/a/50231
From put-call parity we have $C_t =P_t +S_t - K e^{-r(T-t)},$ so $$C_t \geq S_t - K e^{-r(T-t)} > S_t - K.$$ This means that the price of the call $C_t$ at any time $0 < t<T$ is always greater than the value of exercising the call which is $S_t - K.$ Therefore, the optionality of exercising an American call option (with no dividends) before $T$ has no value.
## Answer by Idonknow (score 1)
https://quant.stackexchange.com/a/50235
From the same source (Introduction to Quantitative Finance by Stephen Blyth), before proving that both American and European call options on non-dividend paying stock have the same value, the author proves the following bound for non-dividend European call option value at page $57.$
> Result: The European call price on a non-dividend paying stock satisfies $$\max(0, S_t -K e^{-r(T-t)}) \leq C_K(t,T)\leq S_t$$ where $C_K(t,T)$ is the value of European call option at time $t$ with $K$ be the strike price and expire at time $T$ and $r$ is interest rate.
The left inequality can be proven using no-arbitrage argument whereas the right inequality can be proven using intrinsic value formula for European call option.
By equipping ourselves with result above, it is easy to prove the following statement:
> Suppose the American is exercised at $t<T$. Then $$\tilde{C}_K(t,T) = S_t - K \leq C_K(t,T).$$
Indeed, since $e^{-r(T-t)} \leq 1,$ so $$\tilde{C}_K(t,T) = S_t - K \leq S_t -K e^{-r(T-t)} \leq \max(0, S_t -K e^{-r(T-t)}) \leq C_K(t,T).$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.