Assume I use some finite difference solver to solve for American type of exercise in BS framework where stock pays dividend discretely. Then at every time iteration, for call option, I firstly adjust for dividends and then account for early exercise by taking a max between intrinsic value and "value to hold".

Should I swap the order in a case of a put? Should I first adjust for early exercise and after for a dividend?

The reason I am in doubt is because with a call I better exercise right before the dividend and with a put right after. Is that correct?


1 Answer 1


I would recommend to do both.

Consider the situation where a single discrete dividend is paid at $t$. You use a Finite Difference (FD) scheme to price a European option. Starting from the terminal condition at $T$, by backward induction you manage to obtain the solution $$V(t^+, \mathcal{S})$$ for a discrete grid of spot levels $\mathcal{S}$ at time $t^+$. For the moment, it is as if you did not consider dividends.

Now, because the stock goes ex at $t$, the no-jump condition writes: $$V(t^-,\mathcal{S}) = V(t^+,\mathcal{S}-D) \tag{1}$$

When you talk about "accounting for the dividend payment", I assume you talk about the transformation $V(t^+,\mathcal{S}) \to V(t^-,\mathcal{S})$ you need to take care of before being able to resume your FD backward stepping up to time $t=0$. From $(1)$ you know that $V(t^+,\mathcal{S}) \to V(t^-,\mathcal{S})=V(t^+,\mathcal{S}-D)$. This means you can (for instance) perform an interpolation to find $V(t^+,\mathcal{S}-D)$ from $V(t^+,\mathcal{S})$ and set $V(t^-,\mathcal{S})$ equal to the result.

For American options I would advise to do, when you reach time step $t$

  • Once you get $V(t^+,\mathcal{S})$, do $V(t^+,\mathcal{S}) = \max( V(t^+,\mathcal{S}), (\phi(\mathcal{S}-K))^+)$ = check for optimal exercise opportunity after a dividend payment
  • Account for the dividend payment $V(t^+,\mathcal{S}) \to V(t^-,\mathcal{S})=V(t^+,\mathcal{S}-D)$ = account for dividend payment
  • Once you get $V(t^-,\mathcal{S})$, do $V(t^-,\mathcal{S}) = \max( V(t^-,\mathcal{S}), (\phi(\mathcal{S}-K))^+)$ = check for optimal exercise opportunity before a dividend payment
  • Move backward to previous time step $t-\Delta t$... and repeat.

with $\phi=\pm1$ for call/put respectively.

  • $\begingroup$ It certainly will find that max over all stopping times I am looking for but it is extra work that could be avoided. I just don't get why in all the books/papers only calls are considered! Or perhaps one can just make a comparison between the two orders(dividends, then American and vice versa) to see which one provides a greater number... $\endgroup$
    – Medan
    Aug 10, 2016 at 17:22
  • $\begingroup$ @Medan It cannot be done otherwise IMHO, except if you are willing to add an if clause to distinguish put vs call. But again what would you do with a more general payoff which is callable at a particular date which coincides with a dividend? If you want to have only 1 pricer this is the way to go $\endgroup$
    – Quantuple
    Aug 10, 2016 at 17:33
  • $\begingroup$ And this extra work is just adding a line after having dealt with dividends (simply taking the max), I don't think this constitutes some kind of bottleneck $\endgroup$
    – Quantuple
    Aug 10, 2016 at 17:33
  • $\begingroup$ I agree that this would be a more general pricer. Just though to find the most optimal solution here for the particular case of a put. $\endgroup$
    – Medan
    Aug 10, 2016 at 18:04
  • $\begingroup$ For a put it is indeed enough to just check exercise after the dividend payment. It should be easy to check also. $\endgroup$
    – Quantuple
    Aug 10, 2016 at 18:06

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