New answers tagged options
2
Here is a paper by the infamous Mark Rubinstein that should get you started.
http://www.haas.berkeley.edu/groups/finance/WP/rpf232.pdf
And here the trinomial tree version:
http://www.ederman.com/new/docs/gs-implied_trinomial_trees.pdf by no lesser than Derman and Kani.
This may also help with the actual computations:
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I tried to answer this in the comments but it got too long. simplest approach would be to guess a low and high volatility that is guaranteed to envelope the one to solve for. then compute the corresponding options prices at each of these guesses using your pricer. then while the difference between your guesses (the low/high volatility) is greater than some ...
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You don't need an algorithm to solve that - just program a simple BS option calculator using standard BS with dividend in Excel and fix all the inputs except the volatility. Then use goal seek/solver to change the volatility to get the given price and as a result you will have the implied volatility of the price.
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There are essentially two approaches you can take:
Approximate changes in IV by establishing a relationship between IV and option prices through a function of IV solely dependent on option price. While it is computationally very convenient it introduces huge estimation errors in certain cases. As pointed out in my comment above one such case is a slide in ...
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If you can get anywhere close to the same open-interest and volume using European options as the corresponding American ones, you'll have a much easier time just using them.
American options with high probability of early exercise don't contain information about that back end of the vol surface, and it's kind of hard to decide just what to do with their ...
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As already answered, your first question is call-put parity and this is an arbitrage relation independent of model assumption.
Your second question (under zero rates and dividends, in the Black-Scholes model) relates to call-put symmetry :
$$Call(spot=S_0,strike=K)=Put(spot=S_0,strike= \frac{S_0^2}{K})\times \frac{K}{S_0}$$
It can be easily derived from the ...
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