# 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

1. 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.)

2. Suppose the American is exercised at $$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 $$t, then $$\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.

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 $$0 < t is always greater than the value of exercising the call which is $$S_t - K.$$ Therefore, the optionality of exercising an American call option (with no dividends) before $$T$$ has no value.

• Is it conventional to assume $r>0$, always? Dec 11, 2019 at 13:25
• In B-S model yes.
– user39119
Dec 11, 2019 at 13:28
• @UBM is the second plus sign you wrote supposed to be a minus? Dec 11, 2019 at 13:35
• @Slade: Thanks, I edited it.
– user39119
Dec 11, 2019 at 13:39

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 $$t. Then $$\tilde{C}_K(t,T) = S_t - K \leq C_K(t,T).$$

Indeed, since $$e^{-r(T-t)} \leq 1,$$ so $$\tilde{C}_K(t,T) = S_t - K \leq S_t -K e^{-r(T-t)} \leq \max(0, S_t -K e^{-r(T-t)}) \leq C_K(t,T).$$