When does American and European call have the same price?

We will prove the following fact arising from finance.

  • American call and European call with the same strike and maturity has the same price given that their underlying assets are the same non-dividend stock.

Of course, this fact shall not be generalized to put or some other scenarios.

We denote

  • the stock price by random process {X},
  • and the strike by {K},
  • the maturity by {T},
  • interest rate by {r}.

The mathematical counterpart of the claim, which we want to prove below, is that

\displaystyle \mathbb E [ e^{-rT} (X_T - K)^+] = \sup_{\tau \le T} \mathbb E[ e^{-r\tau} (X_\tau - K)^+].

We set {\hat K = e^{-rT}K}. Because

  • {x\mapsto (x - \hat K)^+} is convex;
  • {t \mapsto e^{-rt} X_t} is a martingale;

we have

\displaystyle t\mapsto (e^{-rt} X_t - \hat K)^+

being a submartingale. This implies that

\displaystyle \mathbb E [ (e^{-rT} X_T - \hat K)^+] \ge \mathbb E[ (e^{-r\tau} X_\tau - \hat K)^+]

which in turn gives that

\displaystyle \mathbb E [ e^{-rT} (X_T - K)^+] \ge \mathbb E[ e^{-r\tau} (X_\tau - K)^+]

for any {\tau \le T}. The other direction of the inequality is obvious from {\sup} operator.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s