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Re: Derivation of Ito Lemma
Posted:
Sep 21, 2012 12:24 PM


On 20120920, Paul <paul.domaskis@gmail.com> wrote: > On Sep 20, 6:44 pm, Paul <paul.domas...@gmail.com> wrote: >> On Sep 18, 11:19 pm, Paul <paul.domas...@gmail.com> wrote: >> > On Sep 18, 3:12 pm, Herman Rubin <hru...@skew.stat.purdue.edu> wrote: >> > > On 20120917, Paul <paul.domas...@gmail.com> wrote: >> > > > I'm looking athttp://en.wikipedia.org/wiki/It%C5%8D%27s_formula#Informal_derivation >> > > > where it says dB^2 tends to E(dB^2). I followed the link to the basic >> > > > properties for Wiener processes, but I can't find why dB^2 tends to >> > > > E(dB^2). I am guessing that it has to do with the limit as dt >> > > > approaches zero. The closest thing seems to be that the variance of a >> > > > Wiener process is t, but that's not quite the same thing. dB is a >> > > > sampling of a normal random variable, it is not a summary statistic. >> >> > > > For context, I am looking at the Ito Lemma for Geometric Brownian >> > > > motion (immediately above the Ito derivation link above). In the >> > > > second line, there is a (1/2)(sigma^2)dt. This is a direct result of >> > > > the fact that the random variable dB^2 gets replaced by dt. It seems >> > > > to be a pivotal change, so I'd like to understand it. >> >> > > > Thanks. >> >> > > One way to see this is to compute the mean and variance for >> > > a computation of the sum of deltaB^2 for a fine partition >> > > of an interval of length T. If the partition is of equal >> > > intervals on the t axis, the sum will be a chisquared >> > > distribution with the number N of intervals divided by N, >> > > which has variance 2T/N. >> >> > > Another way to look at this is to note that when X and Y >> > > are independent random variables with mean 0, >> >> > > E(X^2 + Y^2) = E((X+Y)^2). >> >> > > and as powers of X and Y are also independent, >> >> > > V(X^2 + Y^2) = V((X+Y)^2)  4 E(X^2) E(Y^2) >> >> > > Apply this to the normal case, and this shows that >> > > the variance decreases. Again, calculation will >> > > give the result that the variance converges to 0 >> > > in probability, and if one uses refinements, the >> > > convergence is even with probability one. >> >> > So the expression db^2 doesn't actually tend to dt as dt approaches >> > zero, right? It only has the effect of tending to dt if you consider >> > variances integrated over the interval T? >> >> I've looked at a number of textbooks, and they all prove that >> integrating dB^2 from t=0 to t=T yields T. Since T is also the result >> of integrating dt over the same interval, they conclude from this that >> dB^2 *is* dt. To me, going from integrals that evaluate to the same >> result is not sufficient basis to conclude that the integrands are >> equal at all points in the integration. I *do* understand the Central >> Limit Thereom, which in this context is a statement of the equivalence >> of the integral. But this does not extend down to integrand evaluated >> at points along the integration interval. Thanks if anyone can shed >> some light on the seemingly mysterious equivalence of dB^2 and dt > > By the way, one of the books made it seem easy to show that the > *variance* in dB^2 approached zero in the limit as dt approached > zero. In that way, dB^2 approached its mean of dt. The implication > was that var(dB^2)>0 was easily shown by using the PDF of dB^2. I > did that using two different ways (corroborated), but the expression > to integrate in order to get the variance of dB^2 seems like a tough > nut to crack. One basically has to integrate the following from > d_delW=infinity to d_delW=+infinity: > > [ exp( delW^2 / 2 delT ) ] ( delW^2  delT )^2 d_delW > > For the time being, this assumes that the timestep delT is small, but > has not been taken to the limit of 0. delW is the dB above (the book > I followed used different notation). The exponential is simply due to > the gaussian nature of delW. The (delW^2delT)^2 comes from the > definition of variance (of delW^2, not delW), which is just the square > of the residual, with the mean of delW^2 being delT (as per Weiner > process). Note that assuming very small delT and applying the > binomial approximation does seem to simplify the integrand for the > purpose of determining the symbolic integral, so I suspect that if > this was ever solved, it might have been capitalizing on the fact that > it is a definite integral from infinity to +infinity.
The proof shows that the Riemann sums for (deltaB)^2 converge in probability, and even almost surely if nested partitions are used. This is for the sums from 0 to s for ANY s. By monotonicity, it is enough to use a countable number of points, so the events of probability 0 end up with a total probability of 0. So this means that with probability 1 the limit of the sums of deltaB^2 from 0 to s is s for all s. Going to infinity still only brings in another countable process. Usually Brownian motion is not defined on the negative numbers, but if it is the increments which are of importance, this holds from infinity to infinity.
 This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558



