When American Calls Match European Calls and When Early Exercise Matters

# American Call: when it's European?


# American Call: when it's European?












It is a rather well-spread fact that in Black-Scholes (BS) model for a stock with no dividends that follows Geometric Brownian Motion (GBM), the price of American call coincides with that of its European counterpart. Regarding that I have some questions:

- When I wanted to check the proof of that, even though I have quite a library of FinMath books, I have only found the proof in Shreve's "Stochastic Calculus for Finance Vol II". The proof there is probabilistic, and just based on the fact that the price of the call is submartingale, hence by optional sampling theorem it's not optimal to early exercise it. However, I was not able to find a proof which is more in the spirit of the PDE approach to pricing. For example, I did not see it in any books of Wilmott, which surprised me a lot. So the first question: what are other sources with the proof, and are there proofs different from that in Shreve?

- Based on the proof by Shreve, it does not seem that the underlying must follow GBM, the only important part seems to be that we shall be able to take discounting outside of expectation, so in particular for deterministic interest rates. I guess that implies that many other models: local volatility, stochastic volatility etc. also imply the price of American call being equal to the European call price. Next thought was about the models with jumps - but wait... stock paying dividends is a particular case of the jump model, and we know for sure that in such case American call worth strictly more than European one. Is it due to the fact that there does not exist probability measure under which the (discounted) price of such stock is a martingale?

In general: what are sufficient (and perhaps necessary) conditions for American call price being equal to European call price. This question is purely theoretical: namely, for which models does that hold. Does that also hold for futures on stocks, Fixed Income and FX products being the underlying? From the first glance, at least in case of futures it seems that as long as its price can be modeled by GBM, same theory shall apply (that's nothing more than yet another underliyng, we don't care whether it's a stock or futures on stock), however I've heard somewhere that it may be strictly suboptimal to hold American calls on futures till expiry.

## Answer by Mark Joshi (score 10, accepted)

https://quant.stackexchange.com/a/17678

it's a model-free result. The conditions are $d\leq 0, r\geq 0.$

The proof is that for a european $$ C_t > S_t - Ke^{-r(T-t)} \geq S_T - K $$ and the American is worth at least as much so you never early exercise. So it's worth the same as European.

To prove the inequality, observe if $B_T =1,$ take $K$ units of $B_t$ and one of $C_t$ to get something worth $$ \max(S_T,K) \geq S_T $$ at time $T$. So you must have $$ C_t + K B_t > S_t $$ before.

(you need to get my book "Concepts" in your library!)

## Answer by user9403 (score 7)

https://quant.stackexchange.com/a/17679

It can also be proved by Jenson's inequality. It can only be optimal to exercise the American option if the option is below its intrinsic value; but since the "max" function is convex, the European price satisfies the following inequality: $$c(S_t, t)=e^{-rT}\mathbb{E}[(S_T-K)^+]>=e^{-rT}\left(\mathbb{E}[S_T]-\mathbb{E}[K]\right)^+=S_t-Ke^{-rT} $$ The expectations are taken with respect to the filtration generated by $S_t$ and are in the risk-neutral measure. The rest of the proof follows Joshi's answer.

## Answer by dm63 (score 3)

https://quant.stackexchange.com/a/26067

For options on futures, the American call is worth more than the European when rates >0. Imagine an option that is deep in the money. When you exercise, you get delivered a futures position, which is subject to immediate variation margin (ie you collect the intrinsic immediately). The comparable deep in the money European option is worth the PV of the intrisic.