Why American Calls on Non-Dividend-Paying Stocks Need Not Be Exercised Early
Summary
The document discusses the result that an American call on a non-dividend-paying stock has the same value as a European call under the model considered. The quoted explanation relates this to the discounted immediate-exercise payoff behaving as a submartingale under risk-neutral probabilities, so exercising early does not improve expected value. It contrasts calls with puts, where receiving the strike price earlier can make early exercise valuable.
The question challenges whether the put payoff, (K − s)+, is convex. The accepted response correctly says it is convex: it is the maximum of two affine functions, K − s and zero. Thus Jensen’s inequality can apply to the payoff, although that fact alone does not prove the discounted put intrinsic value is a submartingale; discounting the strike payment can counter the convexity effect. The passage is a conceptual discussion rather than a full proof, and the call result depends on assumptions such as no dividends and the specified pricing model.
Key ideas
- The discussed result equates American and European call values under the stated model setting.
- The call argument uses the submartingale property of discounted immediate-exercise value under risk-neutral probabilities.
- The put payoff (K − s)+ is convex because it is the maximum of two affine functions.
- Put early exercise may be optimal when receiving the strike earlier outweighs the effects described by convexity.
- Convexity alone does not establish a submartingale property after discounting.
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Full text
# Value of European Call equals Value of American Call, Question on Explanation/Proof
# Value of European Call equals Value of American Call, Question on Explanation/Proof
I am reading S. Shreve, Stochastic Calculus for Finance, Vol. I. There he proves that American Call Options have the same value as European Call Options. In the proof he uses that for a Call option the payoff function $g(s) = (s - K)^+$ is convex, and then shows that $$ \frac{1}{(1+r)^n} g(S_n) $$ is a submartingale (he calles this process the discounted intrinsic value process, where the intrinsic value process is the process which gives the immediate payout at each time and for each event). Because it is a submartingale under the risk-neutral measure by the optional sampling theorem for each stopping time $\tau : \Omega \to \mathbb N \cup \{\infty\}$ the stopped process considered at the $N$-th time step $$ \frac{1}{(1+r)^{N \wedge \tau}} g(S_{N\wedge \tau}) $$ has a lower expectation, i.e. we can not get better by stopping some time before $N$, which shows the claim.
After the proof, there are some explanations I do not understand:
> [The Theorem] shows that the early exercise feature of the American call contributes nothing to its value. An examination of the proof of the theorem indicates that this is because the discounted intrinsic value of the call is a submartingale (i.e. has a tendency to rise) under the risk-neutral probabilities. The discounted intrinsic value of an American put is not a submartingale. If $g(s) = (K - s)^+$, then the Jensen inequality still holds but [here he refers to an equation in the proof] does not. Jensen's inequality says that the convex payoff of the put imparts to the discounted intrinsic value of the put a tendency to rise over time, but this may be overcome by a second effect. Because the owner of the put receives $K$ upon exercise, she may exercise early in order to prevent the value of this payment from being discounted away. For low stock prices, this second effect becomes more important than the convexity, and early exercise becomes optimal.
In his explanations he seems to say that $(K - s)^+$ is convex too ("[...] Jensen's inequality says that the convex payoff of the put [...]"), but it is not (indeed it is concave...) and so his explanations and Jensen's inequality does not apply, so I can not follows his explanations, maybe I have understood something wrong? Could someone please explain?
EDIT: The relevant part could be accessed with google books.
## Answer by emcor (score 1, accepted)
https://quant.stackexchange.com/a/18309
A convex function is when the line between two points on the graph always lies above the graph. And this does hold for the put, its also sometimes called a sublinear function.
Also see http://en.wikipedia.org/wiki/Convex_function
So the author is correct in saying that $(K-s)^+$ is convex.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.