My understanding was that as you increase the time to expiry of an option, the value of the option increases. However, I have run a bunch of scenarios and have realized that if you assume a dividend yield > 0%, the value of the option starts decreasing after x number of years. In other words, in the first z years, the value of the option increases, and after x years, it starts to decrease. This is even the case if the risk free rate > dividend yield. Can someone please explain to me intuitively why the dividend yield causes the value of the option to decrease after a certain number of years and why the time value of the option doesn't outweigh the effects of the dividend?

Thanks very much.

  • $\begingroup$ Which option valuation model did you use? $\endgroup$ – emcor Jun 6 '15 at 21:16
  • $\begingroup$ He was obviously talking about the standard BS model with a non-zero dividend. $\endgroup$ – SmallChess Jun 7 '15 at 15:29


let $q t$ be big (t goes to infinity where q is the yield) and you will see why . The first part of the BS formula becomes zero.

Also in accordance to put call parity, the call must be worth zero if the entire stock price has been paid out in dividends:


Dividends cause the stock price to call so if you pay out enough of them the price will be zero, hence C=0. If r>d then you must discount your dividends. (A 1% dividend isn't so great when the risk-free interest is 10%) If d>r then the dividends erode the call price more then interest boosts stock prices. Eventually you have a situation where r=d(discounted dividend). If d>r you can intuitively see how dividends would make calls worthless. When d=r it's harder to visualize, but consider 'interest' causes the price to be boosted by p(1+r) and then dividends cause it to fall p(1-d) so you have p=p(1+r)(1-d) since d=r and if you repeat this 100 or so times the (1-d^2)^100 part goes to zero and the expected value of the stock as time goes to infinity is zero. If r>d the discounting process on the dividend makes the above relation hold.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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