American Call Value, Intrinsic Value, and Early Exercise
Summary
The note explains the lower bound on an American call’s continuation value: it must be at least the payoff available from exercising immediately. The comparison is made using the stock price at the exercise time, not the stock’s eventual price. If immediate exercise yields less than retaining the option, a rational holder would keep it, which helps explain the bound.
For a non-dividend-paying stock with positive interest rates and volatility, the discussion gives a strict-value argument: an American call is worth at least as much as its European counterpart, whose value exceeds discounted intrinsic value, which in turn exceeds immediate intrinsic value. This supports the conclusion that early exercise is not optimal under those assumptions. The note contrasts the case with American puts, for which the same conclusion need not apply, but does not give a full put analysis.
Key ideas
- An American option’s value cannot be less than its immediate exercise payoff.
- The relevant stock price is the price when exercise is considered, not the terminal price.
- A holder should continue holding when the option is worth more than immediate exercise value.
- For a non-dividend-paying stock with positive rates and volatility, an American call is not optimally exercised early.
- The early-exercise conclusion differs for American puts.
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# Answer by Kurt G. (score 2)
# 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)?
"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 by Kurt G. (score 2)
https://quant.stackexchange.com/a/76487
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.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.