On Sep 19, 3:39 pm, Dave <divergent.tser...@gmail.com> wrote: > I am not meaning to be nosy, but why do you need to know? If it is for finance, I recommend you readhttp://papers.ssrn.com/sol3/papers.cfm?abstract_id=1678726 > > If it is for physics then ignore the above. > > As to dB^2 and dt, the above is correct, but it is important to remember that dt is deterministic and dB is not. The expectation is the limiting form of the problem. Gauss provides a great warning that you can only discuss ill posed problems in their limiting form. Ito's lemma is a method of discussing this ill posed problem by solving the limiting form. I do not know of a proof that shows this limiting form is the unique method to solve this problem. It is possible a different limiting form exists, although I very seriously doubt it unless it arrives at this same place.
It is for physics. A colleague went through it with me. dB^2 needs to be written as delB^2 = (SNRV^2)delt, where SNRV is a value from a standard normal random variable, delt is a small increment of time, of which there are T of in the interval t=0 to t=T. Hence, N.dt=T. The SNRV and delB values really need to be enumerated as SNRVi and delBi, i=1 to N. The sum of delBi^2 over T is then T(1/N)(sum of SNRVi^2, i=1 to N). As N increases toward infinity, (1/N)(sum of SNRVi^2) becomes the expectation of SNRVi^2, which is simply the variance of SNRVi, which by definition is unity. Hence T(1/N)(sum of SNRV^2) integrates to T, which means that delB^2 integrates over T to T. Since dt also integrates over T to T, the differential delB^2 is treated as dt, even though the equivalence only holds in integral form. I suspect that this shorthand equivalence can only be taken for granted in differential form if the delB^2 term is not multiplied by anything that changes with the variable of integration (I think) and that it occurs as a linear additive term in whatever expression is being evaluated.