Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

the expected value of the option given the next period up, down values is:

$ Pexp = (p Price_{next, up} + (1 - p) Price_{next, down})/R$

where p is defined as $p = \frac{\exp(-r \times \Delta t) - d}{u - d}$ w/o a dividend yield and $p = \frac{\exp(-(r - q) \times \Delta t) - d}{u-d}$ with a dividend yield

Now I know from here that R is something like $\exp(-r \times \Delta t)$ however with a continuous dividend yield of q would it be $\exp(-(r-q) \times \Delta t)$ changing in a similar way that p changes?

Wikipedia says that it souldn't be, but trying both ways the second one gives the same result as the example here, slide 22, so I think that R should change in a similar way that p changes and effectively the new risk free rate is adjusted by q.


share|improve this question
shouldn't it be $exp(r \Delta t)$ and $exp((r-q)\Delta t)$ respectively ? (Confer the wikipedia-page for example) – Probilitator Feb 19 '14 at 8:58
up vote 1 down vote accepted

No the discounting factor that you use for backward induction won't change. (confer here Chapter IV)

This is only seems confusiong due to the mathematical formulation. Introducing continuous dividends basically adjusts your stock price (down) by discoutning the divididend (for it is paid out and thus dicreases the stock value). Your "risk-free" stock value at $t$ becomes $S_0 e^{r\Delta t}e^{-q \Delta t}$ instead of $S_0 e^{r\Delta t}$.

This leads to the following equation $$ S_0 e^{r\Delta t}e^{-q \Delta t}=pS_0u+(1-p)S_0d$$

Solving for $p$ gives your the desired result

Also note that there are several approaches to modelling dividends in a binomial model setting. Same document as above (Chapter IV)

share|improve this answer
2 comments, first I think you should have a plus sign at the addition of up/down? and secondly shouldn't it be $e^{q}$ instead of $e^{-q}$? – evan54 Feb 18 '14 at 23:33
thank you - I eddited the answer accordingly. The plus was a copy paste typo. The $-q$ is still correct but has to be incorporated differently. – Probilitator Feb 19 '14 at 7:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.