QuoteOriginally posted by: AlanOK, I'm am tired of this, so here is my last version of the argument.I am saying the IID Gaussian interpretation of the DeltaT->0 limit is a _consequence_ of things you agree with.Here is my argument for that.We start with(*) \delta x = delta Z'(t,x), where delta Z'(t,x) is some density.The draws from delta Z'(t,x) are NOT IID draws; they involve a distribution that depends on t and x.Then, to recover a diffusion, as DeltaT -> 0, you need (need means *must have*)E[delta Z'(t,x)] -> b(x,t) DeltaT and E[(delta Z'(t,x) - b(x,t) DeltaT)^2] -> a^2(x,t) DeltaT These last two eqns are short-hand for saying that there exist functions b(x,t) and a(x,t) > 0 such that(L1) lim(DeltaT->0) E[delta Z'(t,x)]/DeltaT = b(x,t) (L2) lim(DeltaT->0) E[(delta Z'(t,x) - b(x,t) DeltaT)^2]/DeltaT = a^2(x,t) You have agreed with this, so far, in previous posts.Now, we just introduce some rescaling:Define delta Z(t,x) = (delta Z'(t,x) - b(x,t) DeltaT)/a(x,t)This is just a notation change. In the new notation, (*) reads\delta x = b(x,t) DeltaT + a(x,t) delta Z(t,x).In the new notation, (L2) says(L2*) lim(DeltaT->0) E[(delta Z(t,x))^2]/DeltaT = 1So, (L2*) follows from things you have agreed with. In particular, the right-hand-side must be 1, not some function of t, not anything else: 1 It looks like we agree on everything up to here. In fact it is very straightforward and never been in doubt by either of us except I fail to see what you gain by introducing Z'. That the first two moments of delta Z divided by DeltaT become 0 and 1 as DeltaT->0 has always been a fundamental requirement of mine right from the start and had no need of derivation from your Z'. The question I have regards possible t-dependence in the higher moments n > 2. (Sorry I didn't make this clearer; I thought it was obvious that is what I intended.) Given that the first two moments are 0 and 1, any t-dependence in the distribution has to come from the higher moments does it not?I really do not understand why you introduced Z' at all, unless it stems from your moment conditions on delta x, whereas mine were on delta Z (equivalent to yours, assuming dZ is independent of x).QuoteThe aren't any other 'hidden assumptions'. There was never any assumption made that the draws fromdelta Z'(t,x) were IID.My question about identical distributions (no t-dependence in the infinitessimal limit) were about delta Z not your delta Z'.Quote I think it is fine to interpret the result (L2*) as saying the 'infinitesimal draws' fromthe limiting rescaled distribution delta Z are 'IID infinitesimal Gaussian draws'. So how does (L2*) get interpreted that way as Gaussian? I see nothing particularly Gaussian about (L2*) unless you impose the Levy characterization. In fact, if delta Z has any t-dependence in the infinitessimal limit, then I don't see how the higher moments could be independent of t (as they would be in a Gaussian).QuoteBut this result was never assumed. It is a _consequence_ of relations that you have already agreed to.Since I am done, feel free to have the last word. I'm sure we'll re-hash the argument in a couple years. I am not interested in having the last word, nor in arguing with you. My only desire here is to understand what the properties of a non-Gaussian delta Z in the limit deltaT->0 are that enable Fokker-Planck. I'd rather not wait another two years. You wrote earlier "you could have a sequence of discrete-time random walks, where each step is drawn from some weird density p(.) with continuous support. If these sequences tends to a limit, as DeltaT->0, where the first two moments match the moment relation I posted, then the result is the same Fokker-Planck equation". I took this as an acknowledgement that a Gaussian delta Z in the limit DeltaT->0 was not required for FPE. Was I wrong? If not and it is not the t-dependence of the distribution (which disappears in the steady-state limit where it does indeed become Gaussian) that enables FPE in the non-Gaussian case, then where is it?