I am having a look to Glasserman Book "Monte Carlo Methods in Financial Engineering". I think that it covers this topic.

Alan, I think now that is possible. I email you at your option city address.

- Cuchulainn
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QuoteOriginally posted by: AlanBetter grammar: I will be interested to hear if some map works for the full 2D Heston problem forputs or calls, including a cross term for correlation. Other cases (esp. 1D) seem much less interesting. The BC are whatever is appropriate for the (unknown at this point) numerical method.It's a lot of mathematical stuff (mostly differentiation and pages and pages!!). In order to reduce the scope I would like to take the following essential model problem (x == S, y == v) by neglecting drift and mixed for the moment.U_t + 1/2 y x^2 U_xx + 1/2 sig^2 y U_yy = 0It if works, I will continue. what do you think?

Last edited by Cuchulainn on November 4th, 2008, 11:00 pm, edited 1 time in total.

"Compatibility means deliberately repeating other people's mistakes."

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

This seems like an excellent special case to start with.The general case stand-alone volatility process pdeis dp/dt = (1/2) sig^2 v d^2 p/dv^2 + (a - b v) dp/dv,But, now you are taking a = b = 0. This special case model describes a volatility process trapped when it hits v=0,unlike the (usual full) Heston model. So the v=0 BC is different -- you canset option values to their (Black-Scholes) v=0 limit, which is generally wrong for Heston but right here.This limit, since r=0, is simply the intrinsic value.But, the large v behavior of p(t,v) is similar to the full model.Since the large v behavior is the problem at issue, I think your special case should indeed tell us if you can map v=infinity successfully to v=1 in a FDM scheme.

Last edited by Alan on November 4th, 2008, 11:00 pm, edited 1 time in total.

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: AlanBetter grammar: I will be interested to hear if some map works for the full 2D Heston problem forputs or calls, including a cross term for correlation. Other cases (esp. 1D) seem much less interesting. The BC are whatever is appropriate for the (unknown at this point) numerical method.It's a lot of mathematical stuff (mostly differentiation and pages and pages!!). In order to reduce the scope I would like to take the following essential model problem (x == S, y == v) by neglecting drift and mixed for the moment.U_t + 1/2 y x^2 U_xx + 1/2 sig^2 y U_yy = 0It if works, I will continue. what do you think?BTW, there is a good numerical example of this case in my book pg. 116, namelyT = sig= y =1, x = strike = 100 => Call Value(T,x,y) = 36.4791

- Cuchulainn
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QuoteOriginally posted by: AlanThis seems like an excellent special case to start with.The general case stand-alone volatility process pdeis dp/dt = (1/2) sig^2 v d^2 p/dv^2 + (a - b v) dp/dv,But, now you are taking a = b = 0. This special case model describes a volatility process trapped when it hits v=0,unlike the (usual full) Heston model. So the v=0 BC is different -- you canset option values to their (Black-Scholes) v=0 limit, which is generally wrong for Heston but right here.This limit, since r=0, is simply the intrinsic value.But, the large v behavior of p(t,v) is similar to the full model.Since the large v behavior is the problem at issue, I think your special case should indeed tell us if you can map v=infinity successfully to v=1 in a FDM scheme.I arrive at the same conclusions. For the original problem (no transformation, yet), I looked at it and 1. The boundaries B1 {x=0} and B2 {y= 0} are degenerate (quadratic form == 0) so here we must calculate the Fichera Function2. B1: FF = 0B2: FF = - 1/2 sig^2 (!! Heston would have kappa*theta and Feller ==> FF > 0 ==> no BC)Thus we must specify a BC on B2. How?3. On B1 the PDE becomes du/dt = 0 ==> u is constant right up to (0,0). So4. based on a 'analytic continuation'?? (heuristic?) from x = 0 outward we get the BC on x = 0 to be the intrinsic value, indeed.//////Q. It is WLOG (without loss of generality) to take a barrier option so that I don't have to transform x as well, for the moment? It would alow me to focus on v only, for the moment?It's trivial to put in -ru term, but no big deal? 5. The quadratic form is not degenerate on Xmax, Ymax (assuming they were part of the problem) then BC would need to be described, but that's another discussion.

Last edited by Cuchulainn on November 4th, 2008, 11:00 pm, edited 1 time in total.

"Compatibility means deliberately repeating other people's mistakes."

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

QuoteOriginally posted by: CuchulainnQ. It is WLOG (without loss of generality) to take a barrier option so that I don't have to transform x as well, for the moment? It would alow me to focus on v only, for the moment?A one-sided KO barrier on the 'small' side of the put or call payoff seems fine.A two sided KO barrier would make me worry that the method doesn't really workfor a regular put or call, as this grossly changes the payoff.

QuoteOriginally posted by: FaridMoussaouiHi,If you have a "deep" look to the paper on matheon.de, he is not proposing a standard TBC but a "non-standard" one.He first remarked that the discretised TBCs of Mayfied and Hu are somewhat "unstables".Instead, he managed to work directly on the numerical scheme (CN FDM), use a Z-transform. Knowing the behaviorof the solution at infinity, he derived the discrete TBC for his scheme (pages 9-10).F.I have looked at a few papers now on this TBC idea; for example "Discrete Transparent Boundary Conditions forParabolic Systems" Zisowsky & Ehrhardt. So far, I have to say the idea looks useless for practical PDE problems with non-constant coefs as the authors seem torequire constant coefs. What am I missing?

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I have now transformed the 1/4 plane: (x,y) -> (z,w) in [1,-1]^2The transformed PDE is degenerate on the whole boundary ==>For 3 of the 4 boundaries I get a PDE or the intrinsic value as before; there is complete compatibility at all vertices as well!!! (no blow up)//Now, when w = -1 (v = INFINITY) I also have a degenerate boundary but one of the coefficients (the 'y' in 1/2 y x^ U_xx) is unbounded only when w = -1. So, Fichera PDE there is not applicable as such (maybe). However, the end-point (-1,-1) and (1,-1) have the IV as Dirichlet BC (when z = -1 and +1).So, what's happening in -1 <= z <= 1?I don't yet want to hand wave but there should be a mathematical or financial value...If we can determine this BC I think the mathematical problem has been solved.

Last edited by Cuchulainn on November 5th, 2008, 11:00 pm, edited 1 time in total.

"Compatibility means deliberately repeating other people's mistakes."

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

- Cuchulainn
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QuoteOriginally posted by: AlanCall value = SWhat's the motivation? Why is the intrinsic value not good? Is C = S in the closed, open or other interval? I meanC = S, Smin <= S <= Smax etc.

Last edited by Cuchulainn on November 5th, 2008, 11:00 pm, edited 1 time in total.

David Wheeler

http://www.datasimfinancial.com

http://www.datasim.nl

The intuition comes from the Black-Scholes model, where v is simply a parameter.To actually prove it for specific stoch. vol. models (like Heston), see Ch 10 of my book. For other sv models (see Ch 10), there are counter-examples to C -> S, so you have to be careful.This is all for S in (0, infinity) p.s. There is very short 'proof' based on my Ch. 10 argument for your specific PDE.If you want me to post it, let me know.

Last edited by Alan on November 5th, 2008, 11:00 pm, edited 1 time in total.

Alan: I was reading the Quantlib code, and found a reference to you and to a Wilmot thread "QuantLib code is very high quality" concerning ExactVariance sampling for Heston.I can not find it into Wilmott site. This thread has been deleted or am I a bad searcher ?

I guess a bad searcher as my first try turned up this

I would like to precise : "a bad wilmott forum" searcher if you don't mind. Thank you very much for the link.