In Paul Wilmotts quantitative finance books he says that the the value of an American option satisfies the following

$$ \frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2 \frac{\partial^2V}{\partial S^2}+rS\frac{\partial V}{\partial S}-rV \leq 0 \quad \quad (1) $$ and by the no arbitrage condition we have

$$V(S,t) \geq P(S,t) \quad \quad (2)$$ where P is the time dependent payoff. Now to quote the justification he gives for the price of American and European call options to be same is that

"If we substitute the Black–Scholes European call solution, in the absence of dividends, into the inequality (1) then it is clearly satisfied; it actually satisfies the equality. If we substitute the expression into the constraint (2) with P (S, t) = max(S − E, 0) then this too is satisfied. The conclusion is that the value of an American call option is the same as the value of a European call option when the underlying pays no dividends "

My question is why cant the same be said about American put options .As far as I understand the European put also satisfies (1) with equality .

  • 1
    $\begingroup$ The problem with the American put is the European solution can fail to satisfy (2), in which case the solution involves early exercise to restore (2). $\endgroup$
    – nbbo2
    Sep 10, 2021 at 17:12
  • $\begingroup$ @noob2 why can European option fail to satisfy 2?. Cant we take P(S,t)=max(E-S,0)? $\endgroup$
    – smbch
    Sep 10, 2021 at 17:17

1 Answer 1


For a very ITM put (i.e. $S \ll E$), when interest rates are positive, you can have $V \left(S, t\right) < \left(E - S\right)^+$ (see figure 2.8 p. 34 in the book). That would correspond roughly to the situation where, as time passes, the loss of time value is virtually null while the discounting effect (present value increases as time goes by) gives a positive theta to your option.

On an American option, this cannot be true as it would give rise to arbitrage opportunities (buy the put and exercise it immediately).


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