Assuming constant volatility across all strikes, how to use known premium of options to determine premium of options with another strike? e.g. suppose we know premium of \$40 call and put, \$50 call and put, how to determine premium of \$30 and \$60 options?


Under the Black-Scholes framework, you can calculate the implied volatility, given the option's price, underlying's price, time to maturity and the risk free rate. To calculate the implied volatility you have to use a root finding method, since there is not a closed form of the inverse of the B-S option pricing equation for volatility.

In the real world implied volatility varies across maturities (volatility smile). However, if you want to assume constant volatility as stated in the question you can use it to calculate option prices for other strikes.

| improve this answer | |
  • $\begingroup$ The volatility smile, in my experience, is nearly linear near the underlying's current value (but with different slopes on the two sides). With a sufficient number of data points, you could build a reasonably accurate model. $\endgroup$ – user59 Oct 20 '14 at 16:05
  • $\begingroup$ @barrycarter, could you please elaborate? It is either linear or exhibits different slopes on either side (in which case it is hardly linear). I would argue that the function of the smile highly depends on the exact type of underlying asset but rarely is it linear (except in special cases). $\endgroup$ – Matthias Wolf Oct 21 '14 at 4:48
  • $\begingroup$ @MattWolf I should have said "bilinear" or something. In other words, it looks like a line to the left of the strike price and a different line to the right of the strike price (forming sort of a V shape, but the two sides of the V have different slopes). If you still disagree, let me know, and I'll explain my position further. $\endgroup$ – user59 Oct 21 '14 at 13:28
  • $\begingroup$ Thanks for the clarification. I would still disagree with the described "shape". Especially currency option volatility smiles can hardly be described as linear on either side of the atm strike. $\endgroup$ – Matthias Wolf Oct 21 '14 at 16:16

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.