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Topic: Derivation of Ito Lemma
Replies: 12   Last Post: Feb 7, 2013 1:37 PM

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Paul

Posts: 263
Registered: 2/23/10
Re: Derivation of Ito Lemma
Posted: Sep 23, 2012 12:45 AM
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On Sep 21, 12:24 pm, Herman Rubin <hru...@skew.stat.purdue.edu> wrote:
> On 2012-09-20, Paul <paul.domas...@gmail.com> wrote:
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> > 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 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?

>
> >> 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 time-step 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^2-delT)^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.



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