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Alan
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Tricky integral question

April 15th, 2010, 6:42 pm

Consider the following (real-valued) integral, with (y,t) both positive reals: Here K is the modified Bessel function, sometimes called the Macdonald function.It is real-valued for the arguments shown.Ideally, I would like an analytic expression, but I doubt there is much hope for t> 0. My real question is below.Some background. If you picture a lognormal density, you'll have a rough picture of f(y,t).Like the lognormal, f(y,t) steepens to a DiracDelta(y - y0) as t -> 0. (up to a constant of proportionality)In this case, DiracDelta(y-1), which you can deduce from the relations here; [substitute for g(y) in the 2nd integral]For t > 0, f(y,t) is a smooth function of y. What I am really after, and should be provable, is the behavior as y -> 0.I have extremely strong numerical evidence thatf(y,t) ~ exp{- (log y)^2/(2t)} as y -> 0,where the precise meaning of the last eqn is that(*) I am looking for a proof of (*). In words, (*) says the lognormal notion becomes *exact* as y -> 0.My question: does anyone have any ideas for a proof of (*)? =================================================================Here is some of the numerical evidence. For example, with t = 9/4, and various y, I compute y ............. (-2 t)/(log y)^2 log f(y,t)10^(-05): 0.99127110^(-10): 0.99774910^(-19): 0.999369
Last edited by Alan on April 15th, 2010, 10:00 pm, edited 1 time in total.
 
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GoGoFa
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Tricky integral question

April 16th, 2010, 5:28 am

Have you tried the Kasahara's exponential Tauberian or a related theorem?
 
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mynetself
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Tricky integral question

April 16th, 2010, 9:20 am

Alan -> Just a quick idea. Since you need y->0, consider a Taylor expansion of the modified Bessel function about 0 and you only consider the linear approximation as the O(y^2) terms will be small. You should end up with a much simpler expression for your y(t,y) which should be easier to manipulate and make estimates about (your f(t,y) will look like f(t,y) = g(t) + y*h(t)). Not too sure it would help though as it doesn't look much like you can get an exponential... Some other expansion for the modified Bessel function perhapse?
 
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Alan
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Tricky integral question

April 16th, 2010, 12:35 pm

GoGoFa,Thanks for the suggestion; found some literature and will investigate.mynetself,Thank you also; considered this, but unfortunately BesselK[I nu, y] behaves kind of like sin(1/y) as y->0.BTW, 1. It looks like the Re[...] qualification in my definition is superfluous and can be droppped,as BesselK[I nu, y] is real for real (nu, y), assuming y>0. So, I have edited my original post to reflect this. 2. It looks like the Wikipedia page has a typo and the first integral there should read BesselK[I y,x] not the BesselK[I x,y] shown.With that correction, you can deduce the DiracDelta(y-1) result I mentioned.
Last edited by Alan on April 15th, 2010, 10:00 pm, edited 1 time in total.
 
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AVt
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Tricky integral question

April 17th, 2010, 5:04 pm

I 'played' with the task, however not with the transform suggested by GoGoFa(dito: I did not try asymptotically methods for integrals), but only with a firstorder approximation w.r.t. y of BesselK(I*x,y/2).Where I am aware, that this may not give the desired result 'analytical', butit may do numerical. Set C:=Pi*Re(1/GAMMA(x*I) * BesselK(x*I,1/2) * exp(Y*x*I)) * exp(-1/2*x^2*t)where Y := ln(y/4) and Re denotes the Real part.Now compare the integrands for y ~ 0+ small by plotting (ok,ok, ...) theirdifference. Since t is a kind of damping factor one can choose it to be 1 forthat hand waving 'proof'.What one should see is an oscillating error curve with error <= precision - 2,i.e. working with 18 Digits should give at most 1e-16 as error and that willvanish for x ---> infinity (because of the Gauss factor).So at least for Numerics it should be ok.Now the integral of C over the Positives can be seen as a kind of (half sided)Fourier transform, Y the parameter of the transform and instead of y ---> 0now the asymptotic Y ---> - infinity is what one wants.Not that I do know it (integrating Bessel w.r.t. index is a nightmare), butI hope it helps (and if not I just had some fun ...), as it looks more simple.
 
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Antonio
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Tricky integral question

April 19th, 2010, 8:00 am

Hello Alan,This might lead nowhere, but do you know this paper: http://www.springerlink.com/content/u54 ... ext.pdfThe integral they decompose is not the same as yours, but involves the product of the two Bessel functions J and Y.You can always express your function K in terms of these two.At first sight, your sinh(Pi*x) will cancel out with the one arising from expressing one of the two functions K in terms of the Bessel function I.If you transform the other K into I, then Y, you will not have any more sinh, and I guess you will end up with an integral like the one in the paper.This is a very rough guess obviously.Best,
 
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Alan
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Tricky integral question

April 19th, 2010, 1:04 pm

Thanks Axel & Antonio,Regarding the integral in the "On a Bessel function integral", I agree you can get the sinh term to cancel.But from the abstract, the authors are integrating over the Bessel arguments, not the index. I have made a little progress with GoGoFa's suggestion to try a Tauberian theorem type argument.Here is what I am trying. Let's fix t and write p(y) = f(y,t), which is a probability density on (0,infty)up to a normalization constant. The small y behavior of p(y) should be determined by the large s behavior of E[y^(i s)]. Now E[y^(i s)] only requires int y^(i s) BesselK(i x,y/2) = 4^(i s) Gamma(1/2(1 - i x + i s)) Gamma(1/2(1 + i x + i s)).So, that leaves an integral that perhaps I can analyze more easily as s >> 1. Need to show the latter behaveslike exp(-s^2 t/2), which would be true, for example if there was a saddle point x*(s) ~ s. Anyway, that's as faras I've gotten with the idea. Update: the moment idea E[y^(i s)] doesn't seem to work. The numerics suggest you can't interchange theintegration orders here.
Last edited by Alan on April 18th, 2010, 10:00 pm, edited 1 time in total.
 
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PaperCut
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Tricky integral question

April 19th, 2010, 1:45 pm

Is "y" a cash dividend?
 
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Alan
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Tricky integral question

April 19th, 2010, 2:57 pm

No, it's a volatility.
 
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Alan
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Tricky integral question

April 26th, 2010, 5:05 pm

Since I am still stuck on this problem, let me rephrase in a completely different way, whichmay prompt some more suggestions.Consider a diffusion on (0,infty) which can be described as "GBM + quadratic killing".Specifically, suppose we have a transition density p(t,x,y) which follows the Kolmogorov backwards eqn.p_t = 0.5 x^2 p_xx - c x^2 p,with p(0,x,y) = DiracDelta(x - y), and c is a non-negative (real) number.If c = 0, we know the answer is a lognormal density.If c > 0, I want to prove that the answer tends `essentially' to a lognormal as x -> 0.Analogous to before, `essentially lognormal as x -> 0' means precisely thatAlthough the exact solution will generally be norm-defective (i.e., the y-integral will be less than 1), the equation just typsetshould be compatible with that fact. (In fact, the exact solution will be given by a version of the tricky integralthat I started the thread with -- but let's ignore that here as it unhelpfully leads us back to the original query.) Any thoughts on how to attack the problem from this pde angle?
Last edited by Alan on April 25th, 2010, 10:00 pm, edited 1 time in total.
 
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mynetself
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Tricky integral question

April 26th, 2010, 6:57 pm

Alan -> What do you know about your "c" ? If it is small (<< 1), you could try a perturbative expansion of p in terms of c. What I mean is that you write p as a power series in c. (If c is small it will hopefully converge.) Then you plug the power series in your equation and obtain an infinite nr of equations, one for each power in c. The lowest order equation is the one for c=0, the next one is for O(c^1), then for O(c^2) and so on. At any given order you should find that you only need the solutions to the lowest order approximations, so you can evaluate the first order correction in terms of the c=0 one, which you know. And then the O(c^2) correction in terms of the O(c^1) solution and so on. You prob won't be able to solve these equations explicitely, but it might give you some insight on the x->0 behaviour.
Last edited by mynetself on April 25th, 2010, 10:00 pm, edited 1 time in total.
 
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Tricky integral question

April 26th, 2010, 7:20 pm

Here's some trickery with transformations, will have another check tomorrow when I am more awakeI. u_t = x^2 u_xx - cx^2 u (I leave out that 1/2)y = log(x)II. u_t = u_yy - b(y)u where b(y) = cexp(2y)III. u_t + b(y)u = u_yyUse integrating factor on LHSIV. (I(y,t)u)_t = I(y,t)u_yy where I(y) = exp(b(y)t)V. v_t = I(y,t)(1/I(y,t) v)_yy where 1/I(y,t) == exp(-b(y)t), and v(y,t) == I(y,t)u.now (not sure) integrate V between -INF and INF and integrate by parts, maybe something will come out of the woodwork at -INF otherwise the energy does not decrease??VI. Sanity check; if c = 0 then b(y) = 0 (and I(y,t) = 1) and PDE V is the 'Gaussian kernel PDE'. Maybe the bespoke asymptotic analysis from mynetself can be used here as c->0?Plan B: Numerically solve PDE and test the hypothesis on a fine mesh near x = 0 (y = -INF) (but i see you've done something similar)///Q. Your x in range (0, INF)?? edit: yes, OK (question answered)
Last edited by Cuchulainn on April 26th, 2010, 10:00 pm, edited 1 time in total.
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Alan
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Tricky integral question

April 27th, 2010, 2:56 pm

Thanks, guys, for the suggestions.The perturbation idea is a natural suggestion, but I couldn't see how to control it.The behavior I am looking to prove should hold for arbitrary large c.The integrating factor idea is novel and certainly something to mull over. As y -> -infty, b(y) -> 0, and l(y,t) -> 1, so you can see the pde turning into the pde for bm again.How to go from there, I don't see it. If you solved the pde numerically to very high precision, you should get something similar to my first post (bottom of post).
Last edited by Alan on April 26th, 2010, 10:00 pm, edited 1 time in total.
 
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Cuchulainn
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Tricky integral question

April 27th, 2010, 2:57 pm

QuoteOriginally posted by: AlanSince I am still stuck on this problem, let me rephrase in a completely different way, whichmay prompt some more suggestions.Consider a diffusion on (0,infty) which can be described as "GBM + quadratic killing".Specifically, suppose we have a transition density p(t,x,y) which follows the Kolmogorov backwards eqn.p_t = 0.5 x^2 p_xx - c x^2 p,with p(0,x,y) = DiracDelta(x - y), and c is a non-negative (real) number.If c = 0, we know the answer is a lognormal density.If c > 0, I want to prove that the answer tends `essentially' to a lognormal as x -> 0.Analogous to before, `essentially lognormal as x -> 0' means precisely thatAlthough the exact solution will generally be norm-defective (i.e., the y-integral will be less than 1), the equation just typsetshould be compatible with that fact. (In fact, the exact solution will be given by a version of the tricky integralthat I started the thread with -- but let's ignore that here as it unhelpfully leads us back to the original query.) Any thoughts on how to attack the problem from this pde angle?Is the PDE - strictly speaking - not a KPE but a reaction-diffusion KPP
Last edited by Cuchulainn on April 26th, 2010, 10:00 pm, edited 1 time in total.
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