What is the effect of dividend yield being greater than the risk-free rate to American options pricing?

Even though dividends are discrete, literature often makes the assumption of continuous dividends (mostly in the case of indices but the individual stocks as well).

The dividend yield denoted by q is often considered as an adjustment to the risk free rate (i.e. r-q).

My question is, what happens to American Call options if r-q < 0? Is it now possible to exercise before maturity so it can no longer be calculated as a European option? Logic says you can early exercise but I am not sure.

Some footnote: In discrete dividend case we know that we should only exercise American Calls before maturity if the excess value of the option is less than the dividend. Otherwise value of the American Option will always be greater than the exercise price. This is mainly due to r > 0, and in the rare case of r < 0 American Puts become equivalent to European Puts.

I think that for any $q>0$ it becomes optimal to exercise an American call for a sufficiently high spot price $S$: if the spot increases enough, the dividend yield corresponds to sufficient cash dividend to render exercise optimal.
This would happen irrespective of the value of $r$ or the sign of $r-q$. What matters is that, for a given strike $K$, the price of a European call is of the order $$C \sim S\ e^{-qT}-K\ e^{-rT} \text{ for large }S$$ as both cumulative Normals go to one. This can become negative for large enough $q$, even though $S>>K$. The holder of an American option would not allow for this intrinsic value to become negative, and therefore would exercise early.
• Plus, now I see it is kind of a silly question since I made the mistake of confusing the effect of the sign of $r$ to the options with the effect of $q$ as a dark brother of $r$ missing the step they do not go together (hence $K$ is unaffected by $q$). – berkorbay Aug 4 '14 at 12:41