Let's say we have a system we know all about, such as a geometric Brownian motion. We know its density is log-normal.But let's pretend naively that we don't know this, and we construct an eigenfunction solution to the relevant Kolmogorov backward equation. So we have this series solution, let's say.Then my (invisible) friend walks in and says, "no it's just log-normal." And I say "but my solution should be equivalent."How do we establish this equivalence?

Let's use log coordinates and suppose there is no drift and unit volatility.Then, given some arbitrary complicated looking p(t,x), there are various ways to establish it is a std BM density:1. Show [$]\int e^{i z x} p(t,x) \, dx = e^{-z^2 t/2}[$]2. Show [$]p_t = \frac{1}{2} p_{xx}[$] with [$]p(0,x) = \delta(x)[$]3. Show all the moments are correct.4. Analytically evaluate your sum or integral representation. This may involve an appeal to the Poisson summation formulaPerhaps there are more. But, since you apparently started from (2), really you're done and don't have to show anything, since I believe the soln is unique. Nevertheless, I suspect 1. and/or 4. might not be too hard to carry out for your eigenfunction solution, at least in this standardized case.

You can write the transition density function as an infinite sum over some family of orthogonal polynomials. These polynomials are the eigenfunctions of the infinitesimal generator of your particular diffusion. So say you have a Vasicek model. Then, the associated orthogonal polynomials are (maybe not surprisingly) the Hermite polynomials. For a CIR process, the polynomials would be the Laguarre polynomials... Then, there are some "cool" summation formulae in Erdelyi, A. (1953). Higher Transcendental Functions, Vol. 2. McGraw-Hill, New York.) that show that these infinite sums equal to the well known closed-form expressions. E.g. for Vasicek, you use Mehler's formula to map sum over Hermite polynomials to the known Gaussian density. For CIR's Laguerre polynomials, you use Hille-Hardy formula to obtain the modified Bessel function. These formulae exist not for any processes obviously. You can find more details about the usage of spectral methods in Linetsky, V. (2007). Spectral methods in derivatives pricing. Hand-book of Financial Engineering, Handbooks in Operations Research and Management Sciences Vol. 15. Elsevier, Amsterdam.

QuoteOriginally posted by: AlanLet's use log coordinates and suppose there is no drift and unit volatility.Then, given some arbitrary complicated looking p(t,x), there are various ways to establish it is a std BM density:1. Show [$]\int e^{i z x} p(t,x) \, dx = e^{-z^2 t/2}[$]2. Show [$]p_t = \frac{1}{2} p_{xx}[$] with [$]p(0,x) = \delta(x)[$]3. Show all the moments are correct.4. Analytically evaluate your sum or integral representation. This may involve an appeal to the Poisson summation formulaPerhaps there are more. But, since you apparently started from (2), really you're done and don't have to show anything, since I believe the soln is unique. Nevertheless, I suspect 1. and/or 4. might not be too hard to carry out for your eigenfunction solution, at least in this standardized case.Alan, thank you so much. Yes as you point out, I think my (hypothetical) problem would come from (2). The Fourier inversion, (1) is a great idea. Not sure about the moments because I'm not sure some distributions are really defined by their moments, i.e. hitting time problems?The Poisson summation formula looks wild, this is new to me. Thanks for that.

QuoteOriginally posted by: overkill112358You can write the transition density function as an infinite sum over some family of orthogonal polynomials. These polynomials are the eigenfunctions of the infinitesimal generator of your particular diffusion. So say you have a Vasicek model. Then, the associated orthogonal polynomials are (maybe not surprisingly) the Hermite polynomials. For a CIR process, the polynomials would be the Laguarre polynomials... Then, there are some "cool" summation formulae in Erdelyi, A. (1953). Higher Transcendental Functions, Vol. 2. McGraw-Hill, New York.) that show that these infinite sums equal to the well known closed-form expressions. E.g. for Vasicek, you use Mehler's formula to map sum over Hermite polynomials to the known Gaussian density. For CIR's Laguerre polynomials, you use Hille-Hardy formula to obtain the modified Bessel function. These formulae exist not for any processes obviously. You can find more details about the usage of spectral methods in Linetsky, V. (2007). Spectral methods in derivatives pricing. Hand-book of Financial Engineering, Handbooks in Operations Research and Management Sciences Vol. 15. Elsevier, Amsterdam.Thank you overkill. I am familiar with Linetsky's stuff, although he can be a bit challenging to read. I will also check out Erdelyi too. Thank you.

QuoteOriginally posted by: OrbitQuoteOriginally posted by: AlanLet's use log coordinates and suppose there is no drift and unit volatility.Then, given some arbitrary complicated looking p(t,x), there are various ways to establish it is a std BM density:1. Show [$]\int e^{i z x} p(t,x) \, dx = e^{-z^2 t/2}[$]2. Show [$]p_t = \frac{1}{2} p_{xx}[$] with [$]p(0,x) = \delta(x)[$]3. Show all the moments are correct.4. Analytically evaluate your sum or integral representation. This may involve an appeal to the Poisson summation formulaPerhaps there are more. But, since you apparently started from (2), really you're done and don't have to show anything, since I believe the soln is unique. Nevertheless, I suspect 1. and/or 4. might not be too hard to carry out for your eigenfunction solution, at least in this standardized case.Alan, thank you so much. Yes as you point out, I think my (hypothetical) problem would come from (2). The Fourier inversion, (1) is a great idea. Not sure about the moments because I'm not sure some distributions are really defined by their moments, i.e. hitting time problems?The Poisson summation formula looks wild, this is new to me. Thanks for that.You're welcome. Re 3, good point -- I wasn't thinking about the moment problem. Looking it up, say here, it seems for a Gaussian, the moments *do* determine the distribution because the characteristic function (1) is analytic about z=0 -- in fact, it is entire.However, for a lognormal, this is not the case. So, in general, you are right that 3. is an unreliable guide. BTW, I second overkill's good comments. As long as you are doing eigenfunction expansions, it is a wonderfulexercise to create the series expansion for one of the fundamental solutions to the KBE,(*) [$]p_t = x \, p_{xx} + (a - b \, x) \, p_x[$] with [$](a,b) > 0[$] and state space [$]x \in (0,\infty)[$],and then to see that it can be exactly summed (Hille-Hardy) to a nice well-known closed-form.The fundamental soln I refer to is the one with [$]p(t=0,x) = \delta(x-y)[$], with [$]y > 0[$] and [$]\int p(t,x,y) \, dy = 1[$].This is the unique soln to (*) for [$]a > 1[$] with Dirac IC, but the soln extends to any [$]a > 0[$].

Another simple way is to expand your solution (log-normal or whatever) into the basis of eigenfunctions. If all such obtained expansions coefficients are identical to the ones you have in your series, there you go.This is so to say the other direction of Alan's alternative number 4 (--or, in other words, the inverse unitary transformation).

QuoteOriginally posted by: davidhighAnother simple way is to expand your solution (log-normal or whatever) into the basis of eigenfunctions. If all such obtained expansions coefficients are identical to the ones you have in your series, there you go.This is so to say the other direction of Alan's alternative number 4 (--or, in other words, the inverse unitary transformation).davidhighThanks for your response. Actually my original thought was that my solution would already be an eigenfunction expansion.Anyways it's all hypothetical. I want to try some of the suggestions on a hitting time density and see how it shakes out.

QuoteOriginally posted by: OrbitActually my original thought was that my solution would already be an eigenfunction expansion.Mine as well. You have two ingredients here: the closed function [$]p(x)[$] and some solution [$]g(x)[$] in terms of eigenfunctions (--or any other complete orthonormal basis),$$g(x) = \sum_k \lambda_k \,\phi_k(x)$$However, in the last equation you only know the [$]\lambda_k,\phi_k(x)[$], but not the closed form [$]g(x)[$].The question is now, whether [$]g(x) \equiv p(x)[$]?As far as I understand, Alan (in his alternative 4) and similarly overkill suggest to analytically sum up the series to obtain the closed form [$]g(x)[$], after which comparison with [$]p(x)[$] is simple (--at least hopefully, as there might be bad hypergeometric function identities or similar stuff involved). Intuitively, I would consider the calculations to be quite hard sometimes, unless you find some clear analytic relations in the literature (I guess you wouldn't want to derive them by yourself ... I wouldn't) -- EDIT: in fact, by using the following approach one effectively is doing such a derivation.On the other hand, to do it the other way round seems to me like a more straightforward approach. Just project your given function [$]p(x)[$] onto the eigenbasis, which leads to the expansion$$p(x) = \sum_k \underbrace{\Bigg(\int \phi_k^\ast(z) \, p(z)\, dz \Bigg)}_{=:c_k} \phi_k(x)$$With this your functions are identical iff [$]c_k=\lambda_k[$] for all [$]k[$]. Note that with this approach you "only" need to calculate integrals here, alas an infinite number of them ... thus some analytic tractability is required ... still, a Gaussian plus Hermite-eigenfunctions should be easy to handle (--even without special references).EDIT: to mention the series summation again: note that once you showed the equivalence, you also established another summation formula of the kind suggested earlier.

@davidhighThanks!