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frolloos
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Zassenhaus formula

July 11th, 2019, 3:20 pm

Ah, what the heck.

So I've been mulling this over for quite some time now: let [$]B[$] be the Black-Scholes generator, and [$]A[$] be SV generator terms, such as the vega, volga, vanna differential operators.

From semi-group theory it is justified to write the formal solution of an SV PDE as [$]C(\tau) = e^{\tau (A + B)} C(0)[$]. Kind of justified as [$] C(0) [$] is not smooth, but let's ignore that for the moment.

We know the Baker-Campbell-Hausdorff formula, but there is another (lesser known) formula called the Zassenhaus formula (see for instance https://arxiv.org/pdf/1702.04681.pdf) which expresses [$] e^{\tau(A+B)} [$] as a product of exponentiated differential operators, in particular one could write [$] e^{\tau(A+B)} = \cdots e^{\tau B} [$]. But [$] e^{\tau B} C(0) = C^{BS} (\tau)[$], which means that if the explicit form of " [$] \cdots [$] " is known then *maybe* a general perturbative solution could be found for any SV model.

Now I think the general form of "[$] \cdots [$]" can be found or is actually known (see link above to ArXiv paper), but I am not sure what the "best" way is to cut-off or Taylor expand the "[$] \cdots [$]" part, i.e. to what order in [$] \tau [$] or other parameters.

Thoughts? My thinking at the moment is that this approach could be too complex, not worth it, but maybe somebody else has some better insights.
 
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Cuchulainn
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Re: Zassenhaus formula

July 11th, 2019, 7:35 pm

There is a very close relationship between Yanenko/Marchuk/Strang FDM spltting and  BCH.

https://en.wikipedia.org/wiki/Strang_splitting
https://www.researchgate.net/profile/Ke ... ograms.pdf

Do you want an analytical solution using BCH? Difficult ...

https://www.cs.cornell.edu/cv/ResearchPDF/19ways+.pdf


http://wiredspace.wits.ac.za/bitstream/ ... sequence=1
 
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katastrofa
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Re: Zassenhaus formula

July 11th, 2019, 8:43 pm

I don't understand the math-fin applications you are writing about, but why not just expand the exp(tau*(A+B)) in a Taylor series and look at it. Looking at the form of the BS operator, I'd be worried if the functions on which it acts are differentiable many enough times to mathematically justify the higher expansion terms (or if the exp(tau (A+B)) is bounded and defined on a dense domain from which it can be extended onto those functions if they are not differentiable)...

I may be totally wrong, but I instinctively wouldn't look to BCH and Zassenhaus formulas here from physics, where operators, call it H, are self-adjoint and e^iH are bounded... while some terms of the BS operator, call it L, look suspicious enough to make me question if exp(tau*A) makes sense. I'm betting that it doesn't.

Just a brain dump of what's left of my physics mind.

PS, I'm scared to say it, but Zassenhaus formula is not an alternative but a dual to BCH (, dude :-P)
 
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Re: Zassenhaus formula

July 11th, 2019, 8:50 pm

expand the exp(tau*(A+B)) in a Taylor series 

Physicists and CS love TS, so they do. Pandora's Box.

Nope^2. See method THREE In the (above) Moler and Van Loan paper.
 
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katastrofa
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Re: Zassenhaus formula

July 11th, 2019, 10:25 pm

I don't know about CS guys, but self-respecting physicists use TS only when it converges. Otherwise, we use spectral theory. What's the spectrum of the BS oeprator? In the above case the Taloyr series expansion only led me to questioning the whole reasoning, but as I wrote - not my circus, not my acrobats.
 
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Alan
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Re: Zassenhaus formula

July 12th, 2019, 3:47 am

 then *maybe* a general perturbative solution could be found for any SV model.

Now I think the general form of "[$] \cdots [$]" can be found or is actually known (see link above to ArXiv paper), but I am not sure what the "best" way is to cut-off or Taylor expand the "[$] \cdots [$]" part, i.e. to what order in [$] \tau [$] or other parameters.

Thoughts? My thinking at the moment is that this approach could be too complex, not worth it, but maybe somebody else has some better insights.

There is already a well-established small-[$]\tau[$] expansion for SV models: the heat kernel expansion. It leads to an expansion for the implied volatility 
[$]\Sigma(\tau,k,V_0)[$],where [$]k = \log K/S_0[$] and [$]V_0[$] is the initial volatility

(*) [$]\Sigma(\tau,k,V_0) = a_0(k,V_0) + \tau \, a_1(k,V_0) + \tau^2 a_2(k,V_0) + O(\tau^3) [$]

Although (*) is almost certainly just an asymptotic expansion, asymptotic expansions (when they exist) are unique. So any putative small-[$]\tau[$] expansion derived from BCH or whatever is going to produce the same result.

As a practical matter, if one also expands in powers of k, the resulting double power series can be automated symbolically and taken to very high order. Again, this expansion is unique. 

See, for example, Paulot on (*) for SABR to  [$]O(\tau^2)[$]. Also, for the double expansion of the transition density, see Ait-Sahalia & KImmel (2007), "Maximum likelihood estimation of stochastic volatility models" JFE 83. Using Mathematica, for example, I have done the double expansion for the GARCH diffusion to something like [$]O(\tau^{11})[$] in unpublished work. Once you get it automated, it's easy to just insert arbitrary SV models, sit back, and let the computer grind it out.
Last edited by Alan on July 12th, 2019, 4:29 am, edited 2 times in total.
 
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frolloos
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Re: Zassenhaus formula

July 12th, 2019, 5:04 am

Thanks for all the inputs. First of al, I don't have a clear idea at all about this and what I am really looking for, so might be wasting your time.

It's not the BCH but its dual (Zassenhaus), and more specifically the left dual (the right dual is more commonly written down) that I've been looking at:

[$] e^{\tau(A + B)} = (\prod_2^\infty e^{\tau^n C_n} ) e^{\tau A} e^{\tau B} [$]

where the [$] C_n [$] involve commutators of [$] A [$] and [$] B [$].

So now suppose that [$] B [$] is the Black-Scholes differential operator, and [$] A [$] as in my first post the vega,vanna,volga operators. What I'd like to do is to work on

[$] (\prod_2^\infty e^{\tau^n C_n} ) e^{\tau A} C^{BS}(\tau) [$]

since [$] e^{\tau B} [$]  applied to the initial condition gives the Black-Scholes formula with the instantaneous volatility as implied volatility.

My main question for now is if I want the SV solution to order [$] \tau^2 [$] only, how should I expand the above expression? I wouldn't even know what the correct way is to do this. Do I Taylor expand all the exponentials up to order [$] \tau [$] and let the expansion work on the BS price formula successively thereby generating higher order Greeks? That seems to "easy" to be right.

I'm not looking for an analytical solution, that is probably very difficult or impossible, but as a first step just trying to understand how to work with the above formula. There is the heat kernel expansion and other established methods. But again, just curious about the left Zassenhaus formula. Is it linked to heat-kernel expansion?

Yes there are very technical issues/questions surrounding convergence. Is BCH / Zassenhaus well-defined for differential operators or only for finite dimensional lie-algebras? I don't know.

Alan, you did an expansion to O(\tau^11) for Garch. That's using heat kernel expansion? And heat kernel expansion works for most SV models, it's a very general method? I've always thought, maybe mistakenly, that the difficult part of heat kernel expansion is finding the geodesic distance.
 
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Re: Zassenhaus formula

July 12th, 2019, 1:28 pm

Are you looking for qualitative insights into the PDE and/or computational quantitative results? The latter follows from the former but using the former to avoid the latter does not work in general.

This thread is fuzzy. The main question is what can be done with these formula? i.e. what's the next step?
 
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frolloos
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Re: Zassenhaus formula

July 12th, 2019, 1:45 pm

I'm looking to compute the expression [$] (\prod_2^\infty e^{\tau^n C_n} ) e^{\tau A} C^{BS}(\tau) [$]. Not an exact computation to all orders, but let's say a first order perturbation.

I can't make it less fuzzy at the moment as am still groping my way around. But fair enough, let's close the topic for now until I can frame it better.
 
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Re: Zassenhaus formula

July 12th, 2019, 2:29 pm

This formula looks horrendous to compute. It;s a can of worms in its current form. BTW what does it represent?

Again, it does no harm to look into the bespoke Moler and van Loan. Numerical analysts have been doing this stuff for > 50 years. 

Idea: reduce the scope, see if you can it for the 2d heat equation to gain insights, This is a well-known approach in mathematics. At the moment we have few handles to get a grip on.

Not an exact computation to all orders, but let's say a first order perturbation.
How can we know if a given trick is 1st order or not? The maths is not there imo.
 
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Re: Zassenhaus formula

July 12th, 2019, 3:21 pm

 then *maybe* a general perturbative solution could be found for any SV model.

Now I think the general form of "[$] \cdots [$]" can be found or is actually known (see link above to ArXiv paper), but I am not sure what the "best" way is to cut-off or Taylor expand the "[$] \cdots [$]" part, i.e. to what order in [$] \tau [$] or other parameters.

Thoughts? My thinking at the moment is that this approach could be too complex, not worth it, but maybe somebody else has some better insights.

There is already a well-established small-[$]\tau[$] expansion for SV models: the heat kernel expansion. It leads to an expansion for the implied volatility 
[$]\Sigma(\tau,k,V_0)[$],where [$]k = \log K/S_0[$] and [$]V_0[$] is the initial volatility

(*) [$]\Sigma(\tau,k,V_0) = a_0(k,V_0) + \tau \, a_1(k,V_0) + \tau^2 a_2(k,V_0) + O(\tau^3) [$]

Although (*) is almost certainly just an asymptotic expansion, asymptotic expansions (when they exist) are unique. So any putative small-[$]\tau[$] expansion derived from BCH or whatever is going to produce the same result.

As a practical matter, if one also expands in powers of k, the resulting double power series can be automated symbolically and taken to very high order. Again, this expansion is unique. 

See, for example, Paulot on (*) for SABR to  [$]O(\tau^2)[$]. Also, for the double expansion of the transition density, see Ait-Sahalia & KImmel (2007), "Maximum likelihood estimation of stochastic volatility models" JFE 83. Using Mathematica, for example, I have done the double expansion for the GARCH diffusion to something like [$]O(\tau^{11})[$] in unpublished work. Once you get it automated, it's easy to just insert arbitrary SV models, sit back, and let the computer grind it out.
Sounds like a good process: set up the maths and then let MM take over. 
 
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ISayMoo
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Re: Zassenhaus formula

July 12th, 2019, 8:28 pm

 then *maybe* a general perturbative solution could be found for any SV model.

Now I think the general form of "[$] \cdots [$]" can be found or is actually known (see link above to ArXiv paper), but I am not sure what the "best" way is to cut-off or Taylor expand the "[$] \cdots [$]" part, i.e. to what order in [$] \tau [$] or other parameters.

Thoughts? My thinking at the moment is that this approach could be too complex, not worth it, but maybe somebody else has some better insights.

There is already a well-established small-[$]\tau[$] expansion for SV models: the heat kernel expansion. It leads to an expansion for the implied volatility 
[$]\Sigma(\tau,k,V_0)[$],where [$]k = \log K/S_0[$] and [$]V_0[$] is the initial volatility

(*) [$]\Sigma(\tau,k,V_0) = a_0(k,V_0) + \tau \, a_1(k,V_0) + \tau^2 a_2(k,V_0) + O(\tau^3) [$]

Although (*) is almost certainly just an asymptotic expansion, asymptotic expansions (when they exist) are unique. So any putative small-[$]\tau[$] expansion derived from BCH or whatever is going to produce the same result.

As a practical matter, if one also expands in powers of k, the resulting double power series can be automated symbolically and taken to very high order. Again, this expansion is unique. 

See, for example, Paulot on (*) for SABR to  [$]O(\tau^2)[$]. Also, for the double expansion of the transition density, see Ait-Sahalia & KImmel (2007), "Maximum likelihood estimation of stochastic volatility models" JFE 83. Using Mathematica, for example, I have done the double expansion for the GARCH diffusion to something like [$]O(\tau^{11})[$] in unpublished work. Once you get it automated, it's easy to just insert arbitrary SV models, sit back, and let the computer grind it out.
How do we know these expansions converge at all?
 
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Alan
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Re: Zassenhaus formula

July 12th, 2019, 10:56 pm

Alan, you did an expansion to O(\tau^11) for Garch. That's using heat kernel expansion? And heat kernel expansion works for most SV models, it's a very general method? I've always thought, maybe mistakenly, that the difficult part of heat kernel expansion is finding the geodesic distance.
@frolloos,
Yes, using the "double power series" I mentioned. Yes, it's very general: any SV model with time-independent generator coefficients (a property implied by your notations above for A and B). The geodesic distance is not too hard, and gets you the first term. The main difficulties are in the higher order corrections when each correction is done exactly. But, the automated double power series bypasses those difficulties. 

@ISayMoo,
The expansions are almost certainly asymptotic and likely not convergent. This does not stop them from being (sometimes) quite helpful. For example, suppose you want to estimate parameters of an SV model using daily data and maximum likelihood. Then an approximate (heat kernel) transition density is actually going to be more helpful than an exact expression, say from a spectral expansion. That's because spectral expansions typically have a very hard time being evaluated at small times. On the other hand, sometimes the lack of convergence does indeed cause problems. For example, there were huge threads here about  SABR and impossible negative implied volatilities, probably not accessible any more (but see here under 'Arbitrage problem'). 
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