Put-Call Symmetry by Peter Carr

quotes

Member
Hello everyone,

I heard a formula like C(S,K)=P(K,S) proved by Peter Carr.

Its simple proof is on foreign exchange market
where One Call on Euro denominated by Dollar is somehow equivalent to K Puts on Dollar denominated by Euro, considering the exchange effect,

Co(S0,K)=KSo Po(1/So,1/K)

where 1EUR=SoUSD now

Is there an elaboration on this issue? thanks.
 

quotes

Member
The argument is really intuitive.

If you get the right to buy 1 Euro by only K dollars, it is equivalent to K times the right to sell 1 dollar for 1/K euro.

Thus their current value shall be the same
Call(So, K) USD = K Put (1/So, 1/K) EUR = KSo Put(1/So, 1/K) USD
Call(So, K)= KSo Put(1/So, 1/K)

But note 1/S follows a different Brownian motion than S,
1/St=1/So exp{-(r-ss/2)t - sWt}
Use Bt=-Wt and it is the same as that of St except for the starting value 1/So and drift -r

A simpler way is to add special property of r=0
aPut(S,K)=Put(aS,aK) and a=SK

Call(S,K)=Put(K,S) (r=0)

This formula is easy to use since normally the derivative of S is also the derivative of discounted S, who has no drift.
And the symmetry works for American Options as well!
I wonder if they use this formula to price American Puts!
 

physEcon

Member
This is quite useful for pricing barrier options as it allows you to replicate the payoff with a static portfolio.
 

quotes

Member
This is quite useful for pricing barrier options as it allows you to replicate the payoff with a static portfolio.
could you be more specific?

P.S. It seems flawed with the quadratic variation drift
1/St=1/So exp{-(r-ss/2)t - sWt}
=1/So exp{-r+sst/2 +sBt}
Note +sst/2 here who contaminates a Geometric Brownian Motion formula
 
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