# 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.