Paul
Posts:
493
Registered:
2/23/10


Re: Derivation of Ito Lemma
Posted:
Sep 20, 2012 6:44 PM


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.

