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The put-call symmetry states that C(S,t;X,r,q) = P(X,t;S,q,r), and that this works for American options. According to my notes, this is 'model dependent' because it depends on the assumption that the underlying price follows geometric brownian motion.

However, I don't understand why this would matter: Considering that the payoffs are the same, shouldn't the two options have the same value even if the stock price doesn't follow geometric brownian motion?

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Very simple, think in terms of the terminal payoff functions between call and put and setup two portfolios, one holding a call and K bonds, and the other a put and one unit of underlying. Verify that both portfolios are of equal value at expiration and you should easily be able derive PC-Parity. The reason why distributional assumptions matter is because the call and put prices underly the same assumptions. Make different distributional assumptions and you may not even be able to derive a closed form solution. –  Matt Wolf Sep 23 '13 at 13:20
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up vote 5 down vote accepted

American put-call symmetry relies on the observation that trading $S$ for $K$ is optimal when $\frac1K$ is optimally traded for $\frac1S$. So long as the dynamics of the inverse process $\frac1S$ are sufficiently tractable, you can derive the symmetry formula.

You don't have to have a "pure" GBM for this to work. For example, non-constant (and even price-dependent) volatility and rates are OK. But, given you need a good inverse process you probably do need a log-process of some kind.

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"You don't have to have a "pure" GBM for this to work". True, but the Put-Call relationship may then be different. –  Matt Wolf Sep 24 '13 at 2:01
    
Thanks Brian. If anyone would like more elaboration, it can be found here: quantnet.com/threads/put-call-symmetry-by-peter-carr.9540 –  Twilight Sparkle Sep 24 '13 at 3:40
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