Can someone provide an example of how arbitrage would be used when an american call option can be bought for less than max(final stock - strike,0)? [closed]

Question

"Final stock" means the stock price at expiration, and "strike" means strike price. If a call option had to be purchased for more than the max(final stock - strike,0)then you would never make money off buying the call option right? I found the no-arbitrage assumption in Merton's Theory of Rational Option Pricing.

Answer 1

Merton writes

Further it follows from conditions of arbitrage that $$\tag{3}F(S,\tau;E)\ge{\rm Max}(0,S-E)\,.$$ In general , a relation like (3) need not hold for a European warrant.

This is the well-known relationship between continuation value $F(S,\tau;E)$ of the American option and the payoff you get from it when you exercise at time $t=T-\tau\,.$ In (3), $S$ is not the final stock but the price of the stock at $t\,.$

Now we have a decent QSE question we can ask: Why does (3) hold?

Hint: Will you as the holder of the American option be rational when you exercise at $t$ in the case that the payoff you get is less than the value of the option you could keep by not exercising?

As a further exercise highly recommend to try to find the simple proof that when the stock does not pay dividends and interest rates and volatility are strictly greater than zero then (3) always holds strictly. In other words: We never exercise that American call before maturity.

Proof:

First, the American call is worth more than the European call $G(S,\tau;E)$ and that is strictly worth more than its intrinsic value ${\rm Max}(0,S-\color{red}{e^{-r\tau}}E)\,.$ That intrinsic value is strictly worth more than ${\rm Max}(0,S-E)\,.$

Finally, it is instructive to think about the question why this does not hold for the American put.