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mghiggins
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Exact calibration in implied trees ?

February 24th, 2002, 12:21 am

but we can also argue that a local vol model is only as good as the stochastic vol model of which it is a conditional expectation

Except that the Dupire vol is a function of current traded option prices only, not of any particular choice of stochastic vol model. If there are no gaps, it is the risk-neutral expectation of the instantaneous vol^2 no matter *what* model you choose to fit to the current option prices.
 
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reza
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Exact calibration in implied trees ?

February 24th, 2002, 12:32 am

you're right, thanks for clarifying this for me
 
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Pat
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Exact calibration in implied trees ?

February 25th, 2002, 3:48 pm

Pat, I'm not sure I understood your argument. I was referring to "knowing" the risk-neutral transition density P(T,S | t0,S0) >>

If you know P(Tf,Sf | t0,S0) for t0 = today and T = exercise date, this may not be enough information to use in hedging. To hedge, you need to know P(Tf,Sf | t,S) for all dates t (and possible states S(t)) from today to T. In a 1-D model, the transisiton density P(Tf,Sf | t,S) for all dates t can be predicted soley from the value S(t)=S. In a 2-D model, the transition densities depend on two variables: P(Tf,Sf,Xf | t,S,X). You can always write today's 1-D transition density as
P(Tf,Sf | t0, S0) = int{ove Xf}P(Tf,Sf,Xf | t0,S0,X0)
since you know today's value of X0, but this isn't enough to get the intermediate transition densities
 
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Pat
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Exact calibration in implied trees ?

February 25th, 2002, 3:51 pm

but we can also argue that a local vol model is only as good as the stochastic vol model of which it is a conditional expectation >>

But the local vol model gives the wrong hedges, which can make it unusable.
 
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scholar
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Joined: October 17th, 2001, 8:03 pm

Exact calibration in implied trees ?

February 25th, 2002, 5:29 pm


If you know P(Tf,Sf | t0,S0) for t0 = today and T = exercise date, this may not be enough information to use in hedging. To hedge, you need to know P(Tf,Sf | t,S) for all dates t (and possible states S(t)) from today to T. >>
This is certainly true. I was not precise in my notation, sorry. I had in mind "knowing" the transition probabilities P(T,Sf | t,S) for ALL times 0 < t,T < T_0 (T_0 is a time horizon). Do you agree that this will be sufficient to price an instrument whose payoff depends on S only ?
 
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Pat
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Exact calibration in implied trees ?

February 25th, 2002, 5:31 pm

Oh yes.
 
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pafein
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Exact calibration in implied trees ?

February 27th, 2002, 2:22 am

the Alan Lewis book is very good >>

The three books that seem to cover stoch vol models are Lewis [Option Valuation under Stoch. Vol.], PWQF, & Foque [Derivatives in Financial Markets with Stoch. Vol.]. I've been unable to find any of these in a store (it's a shame the Barnes & Noble near the CBOT closed a while back). Any one of these preferred, particularly for someone who, though mathematically inclined & not completely ignorant, has a somewhat limited background (relative to you all )? I.e., if the text takes the "and then one applies the 3rd Laplace-Fou*^$%%& transform" approach, I'll use it to heat my apartment. TIA

--Pete
 
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reza
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Joined: August 30th, 2001, 3:40 pm

Exact calibration in implied trees ?

February 27th, 2002, 2:26 am

the Alan Lewis book is very good*

--------------------
* I've heard it gets pretty cold in Chicago, you do need something to heat your place.
 
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pafein
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Exact calibration in implied trees ?

February 27th, 2002, 2:45 am

the Alan Lewis book is very good* >>

Yeah, I caught that part. Though I think Hull is good, my mom wouldn't get it at all. We're still working on the "No, not the currency" Eurodollar issue after 3+ years. Nevermind.

* I've heard it gets pretty cold in Chicago, you do need something to heat your place. >>

True. Perhaps I'll start with Sklansky & Malmuth.
 
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reza
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Joined: August 30th, 2001, 3:40 pm

Exact calibration in implied trees ?

February 27th, 2002, 11:32 am

but seriously I recommend the Lewis book over the Fouque book, it is more pedagogical and explains things step-by-step
as for PWQF, it is indispensable, just like the Hull book
... there are a lot of good threads on the Book Forum, take a look
Omar is the specialist on books
 
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scholar
Topic Author
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Joined: October 17th, 2001, 8:03 pm

Exact calibration in implied trees ?

February 27th, 2002, 12:43 pm

Oh yes. >>
which leads me back to my question. Given that we had a good model for transition probabilities P(T,Sf | t,S) for all times 0 < t,T < T_0 (T_0 is a time horizon), would we really need any stochastic vol model to price a derivative whose payoff depends only on Sf ?
 
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Pat
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Joined: September 30th, 2001, 2:08 am

Exact calibration in implied trees ?

February 27th, 2002, 11:30 pm

No. You wouldn't need it, but you have described an impossible situation (unless it is a linear payoff like a fwd contract).

If the vol is stochastic, the transition densities P(Tf,Sf | t,S) for intermediate times t depend on what the then-current vol is.
 
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scholar
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Joined: October 17th, 2001, 8:03 pm

Exact calibration in implied trees ?

February 28th, 2002, 7:42 am

No. You wouldn't need it, but you have described an impossible situation (unless it is a linear payoff like a fwd contract). >>
Why ? I thought the knowledge of all transition probabilities would let me price e.g. a Bermudan option with payoff max(K - S_T,0) and a discrete series of exercise dates, or American option. Is it wrong ?
 
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Pat
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Joined: September 30th, 2001, 2:08 am

Exact calibration in implied trees ?

February 28th, 2002, 2:29 pm

In general, in a two state variable system, you cannot know the transition denisty
P(Tf,Sf|t,S) for intermediate times t

Suppose we are at state X0,Y0, and consider the transitions
X0,Y0 -> X1,Y1 -> Xf,Yf

Suppose we integrate the transition density of X0,Y0 -> X1,Y1 over all Y1; we then get the probability of going from X0,Y0 to X1; since we know the starting value of Y, namely Y0, this is also the probability of ending up at X1.

Now suppose we integrate the transitions density of X1,Y1 -> Xf,Yf over all Yf; this gives the probability of ending up in state Xf given that we started at X1,Y1. To reduce this to an "X only" model, we would have to assume that this "reduced" probability ddensity did not depend on the value Y1; ie we would have to assume that X and Y are independent random variables.

In a stochastic vol model, if Y is anything like a vol for the other variable X, it will affect the transition densities of the variable X. Ie, no independence
 
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scholar
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Exact calibration in implied trees ?

February 28th, 2002, 7:32 pm

Pat, what you say is formally true, but I think if this were the whole story we would be in a deep s**t. It seems to me that the way out is simply not to condition the transition densities on values of Y at intermediate times! Let us assume, as in your example, that we have two stochastic factors X and Y, while the payoff function depends on X only, and there is a series {T_i} i= 1,..,N of exercise dates. Then all we need are transitional densities p(X_{i+1},T_{i+1}| X_{i},T_{i} Y_0) where Y_0 = Y(t=0). There is no need at all to try to calculate
densities of the form p(X_{i+1}T_{i+1}| X_{i}T_{i}Y_{t}), just to run into troubles you described.
My feeling is that a model should be formulated in terms of observables, which are X's, not Y's. If we have a reliable method that allows one to calculate quantities p(X_{i+1},T_{i+1}| X_{i},T_{i} Y_0), we don't have to resort to any modeling of Y's.