Alan, since the similarity reduction for Asians isn't necessarily that helpful I suspect that here it will be worse.Could you work your tranform magic on it? (Change to variables S, R=S/I, t first.)P

Played around a little. Here is a solvable case.Take mu = constant and sigma(x) = x^(1/2), where x = S(t)/I(t)(1) dS = mu S(t) dt + V(t)^(1/2) S(t) dW, where V(t) = [sigma(S(t)/I(t))]^2 = S(t)/I(t)If I haven't made any mistakes, Ito =>(2) dV(t) = [(mu+lambda) V(t) - V(t)^2] dt + V(t)^(3/2) dW,so (1) and (2) combined are the 3/2 sv-model with rho=1, solved in my book, Ch11 for any |rho| <= 1 [Warning, this is again very likely an example of a stricly local martingale when mu=0, so put-call parity fails, etc.Maybe it is possible to flip the sign of rho, which would be better; see my p.s.] p.s.Just tried sigma(x) = x^(-1/2) , which seems to lead to dV(t) = [1 - (mu + lambda) V(t) + V(t)^2] dt - V(t)^(3/2) dW. Now, pretty sure you have a perfectly good martingalefor S(t) when mu=0; also pretty sure S(t) hits 0 as V(t) explodes, which is fine and expected in this case as itis analogous to cev with beta=1/2. Is this one exactly solvable? I don't know -- but suspect the slight change in the drift is a spoiler. The natural next one to try is sigma(x) = x^(-1), but I will stop here.

Last edited by Alan on March 14th, 2012, 11:00 pm, edited 1 time in total.

QuoteWhy it's interesting ? Because you can mimick joint distribution.Any reference on THIS (and not other mimicking ala Gyongy or Krylov/Atlan)?QuoteI think it will become more and more popular.Why now, when this has been what people call an SDE since the 1940's :s

Dave, you right about return on 2 or 1 stocks. Indeed for given S ( t ) return [ S ( t ) - S ( 0 ) ] / S ( 0 ) = [NS ( t ) - NS ( 0 ) ] / NS ( 0 ) regardless of the model for S. What I said is related to dynamics. 2 stocks does not satisfy the same DE as one in contrast to linear case in which initial data 2S ( 0 ) will bring you to 2S ( t ) by solving the same equation for each t. I thought that DE that govern 1 or 2 stocks should be the same as it is in the linear case. If we do not expect that 1 or 2 stocks follow the same DE that is unusual then a nonlinear eq can be used for a stock price model. I do not see at least at this moment an error. I have thought that 2 stocks have to be govern by the same equation as one.

Last edited by list on March 14th, 2012, 11:00 pm, edited 1 time in total.

Alan,I haven't checked your maths, but all those square roots make it look plausible!In the case where everything is nice and therefore hedgeable, won't you be able to replace mu by r for valuation?Any scope for lambda being related to r? (And have you got all signs right?)Wouldn't you expect V to be a decreasing function of x anyway? Historically low stock leading to higher vol.P

Yes, mu=r for pricing.The general case of what I am trying is let V(t) = sigma(t)^2 = [I(t)/S(t)]^p.You would like p > 0 so that the vol increases as S decreases.Ito =>dV = p [-(mu + lambda) V + 0.5 (p+1) V^2 + V^{(p-1)/p}] dt - p V^(3/2) dWThe solvable 3/2 model hasdV = [a V + b V^2] dt + c V^(3/2) dWSo, the last term of the drift is the "spoiler". The only case where it 'works' is the unphysical p = -1, which I reported before.It's your lambda, so I suppose you can relate it to r if you have some rationale.

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QuoteOriginally posted by: AlanPlus, I dsicuss dS = S^p dW (p > 1) in Ch 9 of my stoch. vol. book. It is a classic example of a strictly local martingale.For p >= 1/2 pde theory concludes that C == 0.0 and P == K for all time.Is this consistent with martingale theory?

Last edited by Cuchulainn on March 14th, 2012, 11:00 pm, edited 1 time in total.

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All sounds plausible.Questions/comments:1. It's not (as) obvious from the valuation PDE in variables S, I and t that there is a problem. Which is a shame because S, I and t are more natural, since there's only one-dim diffusion and two time-like variables2. What constraints are there on the behaviour of V(S/I) that make the model acceptable? 3. I guess that we have left the world of closed-form solutions(?) and now have to do numerics or asymptoticsP

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: AlanPlus, I dsicuss dS = S^p dW (p > 1) in Ch 9 of my stoch. vol. book. It is a classic example of a strictly local martingale.For p >= 1/2 pde theory concludes that C == 0.0 and P == K for all time.Is this consistent with martingale theory?No, that can't be right.

QuoteOriginally posted by: PaulAll sounds plausible.Questions/comments:1. It's not (as) obvious from the valuation PDE in variables S, I and t that there is a problem. Which is a shame because S, I and t are more natural, since there's only one-dim diffusion and two time-like variables2. What constraints are there on the behaviour of V(S/I) that make the model acceptable? 3. I guess that we have left the world of closed-form solutions(?) and now have to do numerics or asymptoticsP2. Well, it's apparently very much like the cev model, so you have to decide what's acceptable to you there.If you don't want put/call parity ambiguities (I wouldn't), then you need V(x) bounded as x grows large. [x=S/I, your convention, the reverse of my earlier post]. For example, if V(x) ~ x^p for large x, then any -infty < p < = 0 is fine.In the other direction, is the behavior as x -> 0. You have to decide if you want to allow S to hit 0 or not.If you do, then you must make that boundary a trap to avoid an arbitrage opp for equities.Of course, if you are modelling something else then these criteria need not apply.3. yes

Last edited by Alan on March 14th, 2012, 11:00 pm, edited 1 time in total.

Or just have V being bounded both above and below, so V-> constant as S/I-> infty and V-> a larger constant as S/I ->0. If we are going numerical then might as well. But then what function makes modelling sense?P

Cuch,Let's have V being a sigmoidal function with a few parameters. We also have a lambda. Let's calibrate! Skews will be taken care of by V function and time behaviour (perhaps) by the lambda. (Obviously won't be a perfect calibration, but as a model it sure as hell beats most others for common sense!)P

Yes, you can do that. You can have a function that is "cev-like" for an intermediate range and then ultimatelythe vol. becomes "BS-like". There may be problems with this, just like there are problems with cev.Over time, given equity growth, the model evolution might be to the upper asymptote.So, even if you match a market skew today, the model may suggest that the smile willultimately evolve into a flat smile -- a basic problem with 'local vol'-type models. Maybe you need some Monte Carlo's to explore this.

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As you probably know, I don't like calibration anyway! I'm much more comfortable with a model that has some economic/behavioural basis and if the values are different from the market, so they should be...that's what markets are for...to bring together people with different values for things! And prices are governed by human behaviour (well, maybe not any longer!) and humans like anchoring!P

Sure, and one thing I like about the model is that people cause volatility and they will sell sometimes just becauseothers are selling and vice-versa. But I also know that you are more interested in "time-series" behavior than"point-in-time" calibration. And my point is that the time series behavior may not be what you are hoping for, atleast over very large times. With V sigmoidal, I wonder if ultimately your model will simply evolve into the BS model -- with thevolatility given by one of the two asymptotes.

Last edited by Alan on March 14th, 2012, 11:00 pm, edited 1 time in total.

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