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Paul
Posts:
493
Registered:
2/23/10


Re: Derivation of Ito Lemma
Posted:
Sep 23, 2012 12:45 AM


On Sep 21, 12:24 pm, Herman Rubin <hru...@skew.stat.purdue.edu> wrote: > On 20120920, Paul <paul.domas...@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.
I see that your statement for the sum is true, and that's what the textbooks show. But as I tried to say (probably not all that clearly), it isn't obvious to me that this means the differential deltaB^2 approaches deltaT as the latter shrinks, i.e., the unintegrated differences. Apparently, it does because the variance on deltaB^2 shrinks faster than deltaT. In my last message, I posted the integral that is needed to solve th variance of deltaB^2...haven't done real math in many years, but it seems rather challenging.



