QuoteOriginally posted by: AlanBut, under the same circumstances, things are quite different for the put.An American-style put is worth strictly more thanits Euro-style counter-part. To prove it, assume otherwise. Then,since a perpetual Euro-style put is easily shown to be worth zero, a perpetual American-style put must be worth zero. But this is a nonsense conclusion since an American style put must not decrease in value as the time to expiration increases.Interesting proof, but it is wrong, or at least has so many holes...1) If p->0 for t->oo, it's also the same for c: c->0 for t->oo (why? because you can't exercise the euro-call option with a strike at t=oo) 2) much more interesting, the method to prove P>p uses an hypothesis that is unreal (t.exp=oo), while the P-C parity is an arbitrage relation, and useable only for "real" cases (eg: you can't arbitrage a theoric irrealistic case). It's equivalent to consider a system where t=vector t = [0,t] + [1,0] for any t3) The P-C parity is an arbitrage relation, that means you must impose it, when you think you're able to impose it for a profit, but you can't expect that others will impose it for you when you need it. And if P>p you can impose it always for less risk than imposing the one for p /early exercise on the short side is only a favor to you/4) The P-C parity represents the convenience to enter in a synthetic portfolio, for real cases, and for real times (arbitrages times are of the order of months). In these cases the P-C parity stands especially for American options, while it is also valid for European Options (that are less good for arbitrages, given their long times to close). Eg: it's convenient to enter in the synthetic portfolio with american options instead of european.So, what are the results from the above?1) Is that the P-C parity always stands for american and european REAL options.2) It's always convenient to use american options to impose the P-C parity, why because the european options will follow or will create other arbitrage opportunities.