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I'm trying to confirm my understanding of the 2 models. It is my understanding that the black-scholes is a special case of a binomial model with infinite steps.

Does this mean that if I were to start with a Binomial model with 1 step and increase steps towards infinity I would approach the same value concluded by the black-scholes?

If so does this mean I could use the implied volatility from Black-scholes formula derived from the market price of an option with the rest of the values (r, t, K, S, σ(IV) ) and approach the same market price from the black-scholes as # of steps approaches infinity? Would this only be the case for a European call with more disagreement on the value of American options with early exercise?

Thanks!

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After some more research... I found this article that shows the connections between the two models.. epublications.bond.edu.au/cgi/… that shows that prices do converge as N Periods increases. Also they provide all the Excel formulas to recreate their work if anyone is interested. –  Andromeda Apr 18 at 8:07
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Make it an answer and score some points ;) –  Bob Jansen Apr 18 at 9:56

2 Answers 2

As anticlimactic as this may be, I'm going to answer my own question here..

I found this article that shows the connections between the two models..

http://epublications.bond.edu.au/cgi/viewcontent.cgi?article=1126&context=ejsie that shows

that prices do converge as N Periods increases. Also they provide all the Excel formulas to recreate their work if anyones interested.

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FYI, the binomial distribution converges to normal as n goes to infinity, which is a nice way of thinking about the relationship between BS and the binomial tree models.

see here

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