On Sep 18, 3:12 pm, Herman Rubin <hru...@skew.stat.purdue.edu> wrote: > On 2012-09-17, 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 chi-squared > 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?