Then, we state that this sequence converges to
Xt in the following sense,
that the integral from a to b,
mathematical expectation of the difference Xtn minus Xt squared Tt, converges to zero.
And the theorem basically means that one can take exactly this sequence Xtn as
a sequence of step processes to construct sequence on the stage 2.
First of all, let me show that the process Xt is continuous in the mean squared sense.
Basically for take mathematical expectation of Xt minus Xs squared,
we can represent this difference as
mathematical expectation Xt squared minus 2 mathematical expectation XtXs,
plus mathematical expectation Xs squared.
And now, I would like to represent all objects of this sum by
the functions M and K. What we have here is actual is that
mathematical expectation of Xt squared is equal to the variance of
Xt plus mathematical expectation of Xt in squared.
And therefore, it is equal to Ktt plus m of t squared.
The second object is equal to 2 Kts plus mtms.
And the third object is equal to Kss,
plus m squared of s. Well,
and if you look attentively,
because of this is a presentation,
you will immediately conclude that since m and t are continuous,
then almost the whole sum converges to zero as s goes to t. And therefore,
the process Xt is continuos as a mean squared sense.
So in particular, we conclude that Xtn converges to Xt.
As a mean square senses as when n goes to infinity.
And therefore, if we have here the limit as n goes to infinity,
this limit is equal to zero.
But here, we shall proof,
not that the integrals of limits turns to zero.
But the limit of the integral goes to zero.
The question is whether we can change the places of the limit and the integral.
This is a very good question and it can be solved by the so-called Leibnitz theorem.
Is a basic mathematical concept which will short employ here and
now I would like to show you how you can apply this result here.
So Leibnitz theorem tells us there's a limit of a function fnt.
When n goes to infinity,
integral dt is equal to the limit,
if n goes to infinity,
integral f, n, t, dt.
If there is a function M,
which doesn't depend on n,
such that f n of t is bounded
by M of t and the integral M,
t, dt is finite.
Now, we should find this function M,
which maturates with medical expectations of the difference xt and minus xt squared.
This can be done by the following argument,
the mathematical expectation of
xtn minus xt squared is less or equal the mathematical expectation of
xtn squared multiplied by two plus two mathematical expectations of xt squared.
Well, this is just because of the inequality,
a minus b squared less or equal than two a squared plus two b squared.
Nothing more is applied here.
Well, and what we can say here that both of these mathematical expectations
are bounded by the maximum value of xt squared when t is from a to b.
Actually, it is less or equals than
four a maximal values or the mathematical expectation of xt squared,
when t is from a to b.
And this mathematical expectation can be represented as Ktt plus mt squared.
T is from a to b.
And since the function K is continuous and M is also continuous,
what we have here is a continuous function which is absorbed on the interval from a to b.
And we know that any continuous function takes his maximum on the closed interval.
And therefore, this maximum is actually a finite number.
So, we can use this constant as a function M of t. A constant is definitely integrable.
And therefore, we have found
the function M which is used inside the bound dominated theorem.
So finally, we conclude that the limit and the integral here can be changed.
And according to what we consider before,
we immediately realize that this limit is equal to zero.
This observation completes a proof.
And now, I would like to provide an example how can we use this theorem in
order to calculate the integrals from some given stochastic processes.
We can consider the following example.
I would like to take a Brownian motion at process xt and consider the
integral from zero to t Ws ds.
With no doubt, this Brownian motion satisfies the conditions of the theorem.
Because as mathematical expectations equal to zero.
Therefore, this continues.
As the coherence function is equal to the minimum between T and S. Therefore,
it is also continuous.
According to the theorem,
this integral can be defined as a limit as n
goes to infinity of the integrals of the following types.
So, as integrals of the step processes.
Right here we have Brownian motion since the point ti minus one.
We know that this integral are equal to the sum
Wti minus one multiplied by Wti minus Wti minus one.
So, the question is how to calculate this sum and the limit of this sum.
It's an easy exercise to show that this limit is equal to
the limit as n goes to infinity of minus one-half,
sum Wti minus Wti minus one,
squared plus one-half,
sum Wti squared minus Wti minus one squared.
Okay. The first summond is exactly as a quadratic equation of the process Wt.
One of our previous lectures we have shown that this quadratic variation is
equal to t and therefore this limit conversions to t.
As for the second summond,
here almost all summonds vanish.
And what we have here is just a Brownian motion t squared.
Finally, we conclude that this integral is equal to minus t divided by
two plus Brownian motion squared divided by two.
This result is as interesting because if here we had
a deterministic function f of s, the this integral,
Integral from zero to t F of s,
df of s should be equal to the F squared of two divided by two,
provided that F is zero at zero.
And here, you see we have one more summond.
This minus t divided by two.
Just to the fact that here we have
a stochastic process instead of using the derivaristic function.
The most crucial problem is the context of the stochastic integral of
this type is that your can apply this theorem only in few situations.
The point is that this sums can be constructed by the integrals related to this sums.
Sometimes, very difficult to analyze.
And in this situation,
we managed to do this work,
but in many other situations this work may be much more complicated.
But there is one method which is based on the Ito formula which
actually helps to calculate almost any integral of this type,
I mean, of this type.
And in what follows,
we will present this method.
But before we do this,
let us study this so-called Ito formula.
One of the most important formulas in this course.
