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fashionmarina
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mean reversion characteristic function

December 22nd, 2007, 4:53 pm

does anyone know not too complex references about mean-reversion (OU) models with a martingale component driven by a lèvy process whose it is known characteristic function in closed forme?thanks
 
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fashionmarina
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mean reversion characteristic function

January 1st, 2008, 6:07 pm

SDE OUdS(t) = (a - kS(t))dt + dL(t) where L(t) is a generic Lévy processProcessS(t) = exp(-kt) S(0) + Int_0^t [ exp(-k(t -s)) (ads - dL(s)) ]Characteristic function of Sphi_{S(t)}(u) = exp[ iu exp(-kt) S(0) + iu (1- exp(-kt) a/k - Int_0^t [ psi_{L(t)}(u exp(-k(t-s)))ds ] ] where psi_{L(t)}(u) is the characteristic exponent of LQuestions1) How can I change this char. fun. of S to the one of logreturns x=lnS(0)/S(0) ? so I can use it to price usign classical fft algorithm, otherwise, how can I change fft algorithm to use the char fun of S?2) I am not sure that i shoulde use psi_{L(t)}(.) or psi_{L(1)}(.).3) How should I change the char function to use RN measure with drift (r-q)t ? (L is not in an exponential so i can't use convexity adjustment)thanks very much
 
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Alan
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mean reversion characteristic function

January 1st, 2008, 7:35 pm

Is S(t) supposed to be a non-negative underlying price? If so, then your firstequation doesn't make sense if L(t) is truly a "generic Levy process", since they can be negative.It only makes sense if L(t) is a subordinator (an increasing Levy process).Assuming that L(t) -is- a subordinator, then the way to evaluate the expectation of option-type payoffs is to -not- try to convert to log S(t).Instead, here's the put expectation: P = E[( K - S(T))^+], ignoring discounting, right? Next, convert this to a transform valuation using the c.f. that you already have: phi_{S(T)}(z) = E[exp(i z S(T))]. This is a reasonably straighforward exercise and you will get this formula for the put:(*) valid for every complex z-plane contour C running parallel to the real axis, with Im z >= 0 (important!)Note that you can take Im z = 0, which means integrating along the real z-axis if you want to (I wouldn't). I can write (*) immediately because I am using this same formula right now for something else.Personally, I would just take a contour, say Im z = 1, and run (*) through Mathematica's NIntegrate[] to get a number.It's a one-liner, there. But, I suppose you can use fft if you really want to.regards, -----------------------------------------------------------------------------------------p.s. (Q1) My (*) gives you a way to use the c.f. you have. I don't know about adapting the integral to fft. (Q2): I don't know(Q3):Clearly S(t) is not a martingale for -any- Levy process L(t).Given that, the put formula I wrote must be interpretted as just an expected value under the implied measure,which is the measure under which the process has the stated form.Now, what about converting to a RN measure? I'm not sure that's possible at all here ifL(t) is indeed a subordinator. This makes me think that this model may not make sense at all for what youwant to do with it, which you should probably explain in detail.
Last edited by Alan on December 31st, 2007, 11:00 pm, edited 1 time in total.
 
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fashionmarina
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mean reversion characteristic function

January 2nd, 2008, 7:51 am

Thanks for your help,What I try to do is only include in a generic way mean reversion in basic Lèvy models (VG, NIG, CGMY...) and using the char. fun. (as a generic function with argument the char fun (or exponent) of a specific Lèvy process) to calibrate price options by fft based pricing algorithms (as Carr Madan '99 and other).I don't find any (simple) reference about it. if I havedS(t) = (a - lnS(t))S(t)dt + S(t)dL(t)and x(t)=lnS(t)dx(t) = (a-x(t))dt + dL(t)what is the char. function?what is the right RN adjustment? It is sufficient to choose L(t)= (r-q+w)t + B(t) where B is a generic lévy process and w is its convexity correction?thanks again for any suggestions.Regards.
 
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Alan
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mean reversion characteristic function

January 2nd, 2008, 2:56 pm

Well, now you've changed the model, but it seems you already have the answers.You wrote the c.f of X(t) (before risk-adjustment) in your previous post.As you seem to expect, the mean-reversion disappears after risk-adjustment, and you are just left with the "standard" modelsyou already know.