[PairTheBoard]

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Just to be clear. In the real market where the Black-Scholes theoretical modeling after brownian motion applies, if the stock price is $60 would a Call with Strike 62 be in parity - ie. equally valued - with a Put with strike $58? Or not?

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Not. In the simplest setting, we'll take a unit bond (so interest is zero) and a unit volatility. Also assume our expiration time is 1. Then a call with strike price K is valued at

60*F(ln(60/K) + 1/2) - K*F(ln(60/K) - 1/2),

where F is the cdf of a normal. Take K=62 and get 22.37. Take K=58 and get 23.60.

So the call with strike price 62 has value 22.37. By put-call parity, the put with strike price 58 has value

23.60 + 58 - 60 = 21.60.

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Really? I'll take your word on it, but doesn't that seem a bit strange? With a stock price of $60 and strike prices of $60 for both the put and call they are valued equally by parity. But if you move the Strike price up $2 for the call and down $2 for the put, the Call becomes worth more than the Put. Do you have an intuitive explanation for this?


So in general Put-Call parity doesn't mean what I thought but actually means:

jason1990 --
"The put-call parity can be expressed as

put - call = strike - stock."

where they have the same Strike.


I'm now thinking that what Black-Scholes did was construct the theoretical Equivalent Trading Strategy for an option and calculate the theoretical cost of implemening the strategy - thus providing a rigorous pricing method for the option. Was Black-Scholes even needed then for the Put-Call parity you express above? Or does that follow automatically from the kind of arguments mosta has been making?

This is very cool stuff.

PairTheBoard