Alan, this function indeed looks like the solution. It can be checked directly by taking d X ( t ) where X is represented by the formula you wrote above.

Yes indeed it is. The issue is not that solution, but the following generalization for d >= 2: Here the W_i are independent BM's.From Ito's lemma:(*) Since E[d beta]=0 and then beta(t) is a 1D BM and (*) is equivalent to the orig. Sqrt process sde I posted with omega = (\sigma^2)/4 d The Y_i's each have a closed-form OU soln in terms of their respective W_i's.Then X is given as the sum of the squares of those.The question is:Should this solution also be considered "closed-form" for the original sqrt process sde with d=2,3,4,... ??(I say 'no' for the reasons given in the last paragraph of my previous post)

Last edited by Alan on December 25th, 2011, 11:00 pm, edited 1 time in total.

Do we consider beta as a Wiener process?

Last edited by list on December 25th, 2011, 11:00 pm, edited 1 time in total.

QuoteOriginally posted by: AlanYes indeed it is. The issue is not that solution, but the following generalization for d >= 2: Here the W_i are independent BM's.From Ito's lemma:(*) Since E[d beta]=0 and then beta(t) is a 1D BM and (*) is equivalent to the orig. Sqrt process sde I posted with omega = (\sigma^2)/4 d The Y_i's each have a closed-form OU soln in terms of their respective W_i's.Then X is given as the sum of the squares of those.The question is:Should this solution also be considered "closed-form" for the original sqrt process sde with d=2,3,4,... ??(I say 'no' for the reasons given in the last paragraph of my previous post)Eq (*) has unique solution which can be represented in explicit form X = ...Whether it can be representation unique? It might be yes or not, it does not matter. In this contraction beta does not arbitrary Wiener process as it should be in benchmark theory. If we are interested in a problem related to calculation of probabilities some events the construction is sufficient for that. For applications we need first be given components X_i. They should be meaningfully defined.

Last edited by list on December 25th, 2011, 11:00 pm, edited 1 time in total.

- Cuchulainn
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QuoteOriginally posted by: EBalSubstitution makes itthen the solution is trivial.This looks like 1 = 4 * omega / sig^2 which corresponds to 1 dof in the nonCentral chi^2 distribution. Is that why it is a special case?

Last edited by Cuchulainn on December 25th, 2011, 11:00 pm, edited 1 time in total.

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In our problem the case the process Y takes two signs values. Therefore X = Y^2 does not equal to Y = sqrt(X) as far as sqrt is assumed to be positive.

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: EBalSubstitution makes itthen the solution is trivial.This looks like 1 = 4 * omega / sig^2 which corresponds to 1 dof in the nonCentral chi^2 distribution. Is that why it is a special case?Well, it's all related. The chi^2 dof are the dimensions d in my previous posts which are the integral # of independent BM'sof the underlying OU processes.

Quoteit explicitly contructs "a solution" X(t) as a functional of time and B(t). With that definition of closed-formI had a crack at it, but I'm not too good with ODEs.First, I think it would be more informative to rewrite the OU like this:Y_{t}=e^{-\lambda t} Y_{0} -\int_{0}^{t}c(s,t) W_{s} ds +W_{t},(sorry equation editor is not working for me. Is it firefox? You know the deterministic c(s,t) I just threw away my notes and don't want to introduce a mistake.)that shows a nice linear mapping f defined on C^{0} ( whereas the dW integral is unambiguously defined only up to a null set.) such that Y=f(W) a.s. . (So likewise in the d dim case the CIR is a nice "quadratic form" of the d underlying BM's. Smooth, explicit )Sticking to 1 BM, the classic pathwise solution method brings one to solve\dot{H}_{t}(H_{t}+k_{t})=(\omega - \frac{\sigma^{2}}{4})e^{2 \lambda t}H_{0}=0for any generic forcing term k (maybe such that k_{0}=0!). This is easy in the case of Ebal's solution, and seems harder in the other cases.Now to sort of dispute the purpose...If you are interested in the process X itself, knowing the transition seems like the best information you could have use for! (But maybe sticking a BM in there and using known facts on BM can help.)If you are interested in the smoothness of the solution map then "usually" some bounds can be established without explicit knowledge of the solution H. (But maybe you want to get H explicitly, aiming at a numerically useful description of the solution map.)

Last edited by croot on December 28th, 2011, 11:00 pm, edited 1 time in total.

QuoteOriginally posted by: crootQuoteNow to sort of dispute the purpose...If you are interested in the process X itself, knowing the transition seems like the best information you could have use for! (But maybe sticking a BM in there and using known facts on BM can help.)Yes, I agree completely. In my experience it is much easier to find transition densities. Of course, for thesqrt model, all the transition densities are known, for just about any boundary behavior you might specify at X=0.In fact, here is a good challenge. Can anybody show a single example of an SDE where we (i) have an explicit solution, and (ii) need this solution to contruct a transition density because otherwise we don't know how??(This is not a brainteaser -- I will guess nobody will have such an example.)So, what good are exact SDE solutions? Well, since this is the Brainteaser forum, maybe they are just good for amusement.

Last edited by Alan on December 28th, 2011, 11:00 pm, edited 1 time in total.

Here one for you maybe guys dX=sigma(Xm)*dW where Xm is the running minimum.One can show that the weak solution is simply the integral X=X0+int(sigma(Xm)*dW,[0,t])This is the exact solution. Now try to find the transition density explicitely (actually it's possible but not so simple). Usually SFDE (stochastic functionnal differential equations) of the form dX=sigma(functionnal of X)*dW with functionnal being : running min, running max, running sum, running quadradic variation or what you want; have weak solution as a simple stochastic integral but good luck to find explicitely the transition probability. Nevertheless I agree that 99.99% of the time exact solution is worthless, it's better to known the solution to the associated Kolmogorov equation.

Last edited by frenchX on December 28th, 2011, 11:00 pm, edited 1 time in total.

Well, since the running maximum _is_ a functional of {X}, this doesn't seem like a solution tome. It seems the same as saying the solution to dX = f(X_t) dW_t is X(t) = X(0) + int(0,t) f(X_s) dW_s,which is no solution at all, since a functional of {X} appears on the rhs. Ditto for having the running max on the rhs. Anyway, the challenge is not to show an sde solution. The challenge is to show an sde solution thatis used to get a transition density and where that explicit soln is the easiest (or better, the only) way to get that density.

Last edited by Alan on December 28th, 2011, 11:00 pm, edited 1 time in total.

Agree with you that Fokker Planck is easier if you want to obtain a transition probability. The difference is that the running maximum is a stochastic functionnal and not a deterministic one. It's a bit like a 2D process.Exact solution of SDE are good for mathematicians who only believe in EXACT solution

QuoteOriginally posted by: frenchXAgree with you that Fokker Planck is easier if you want to obtain a transition probability. The difference is that the running maximum is a stochastic functionnal and not a deterministic one. It's a bit like a 2D process.Exact solution of SDE are good for mathematicians who only believe in EXACT solution If the solution of 'running maximum' say exists. Then whether it is a Markov process?

The solution exists and it's called an Azema Yor martingale.Azema Yor process But well that's another story

Last edited by frenchX on December 28th, 2011, 11:00 pm, edited 1 time in total.

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