Zimin,

 to generalize your initial comment, for any process dS = Mu(S,t)*S*dt + 
Sigma(S,t)*S*dz,
the delta-hedging argument leads to the Black-Scholes PDE.  
This is true for any arbitrary functions Mu and Sigma, and so includes GBM, 
Mean Reversion, and others.
There is no problem with this, because in the risk-neutral world, which is 
what you enter if you can hedge,
the drift of the "actual" process is irrelevant.

I believe your concern is that you would like to see a different option price 
for Mean Reversion process.  This can only happen if the asset is not 
hedgeable, and so the actual dynamics then need to be factored into the 
option pricing.   If you assume that the underlying is a non-traded factor, 
then the PDE will have to reflect the market price of risk, and the drift of 
the actual process is then reflected in the PDE.

Vasant





Zimin Lu
10/17/2000 05:20 PM
To: Vince J Kaminski/HOU/ECT@ECT, Stinson Gibner/HOU/ECT@ECT, Vasant 
Shanbhogue/HOU/ECT@ECT, Pinnamaneni Krishnarao/HOU/ECT@ECT, Alex 
Huang/Corp/Enron@ENRON, Kevin Kindall/Corp/Enron@ENRON, Tanya 
Tamarchenko/HOU/ECT@ECT
cc:  
Subject: Option Pricing Challenge


Dear All,

I have a fundamental question back in my mind since 95.    Hope you can give 
me a convincing answer .

Zimin

---------------------------------

In deriving BS differential equation, we assume the underlying follows GBM

ds= mu*s*dt + sigma*s*dz

where mu is the drift, sigma is the volatility, both can be a function of s.

Then we use delta hedging argument, we obtain the BS differential equation 
for the option price, regardless
of mu.  

With the BS PDE and boundary condition, we can derive BS formula.  Fine. No 
problem.

Question comes here.   Suppose the underlying is traded security and follows, 
say, mean-reverting process

ds=beta(alpha-s)dt + sigma*s*dz

Apparantly, this SDE leads to a different probability distribution. However, 
using the delta hedging argument,
we still get the same BS differential equation, with the same boumdary 
condition, we get the same BS formula.  
Not fair !

From another angle, I can derive the distribution from the BS PDE for the 
underlying, which is the lognormal distribution.
My thinking is: can I drive the distribution for any SDE from the option PDE 
?  The answer should be yes, but got to be
from a different PDE rather than BS PDE.  Then what we do about the 
delta-hedging argument ?

Thanks.