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

 Messages: [ Previous | Next ]
 Paul Posts: 517 Registered: 2/23/10
Re: Derivation of Ito Lemma
Posted: Sep 18, 2012 11:19 PM

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?

Date Subject Author
9/17/12 Paul
9/18/12 Herman Rubin
9/18/12 Paul
9/20/12 Paul
9/20/12 Paul
9/21/12 Herman Rubin
9/23/12 Paul
9/19/12 divergent.tseries@gmail.com
9/28/12 Paul
9/28/12 Paul
9/29/12 divergent.tseries@gmail.com
9/30/12 Paul
2/7/13 voyteg