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Now, I'd like to speak about the integrals of the following type.

We take the ministic function f of t,

and integrate it with the respect to d,

wt, where wt is a Brownian motion.

And we consider this is integral or the interval from A to B.

These kind of integrals is known as, Wiener Integrals.

And, we knows that Brownian motion and

Wiener process are two notions for exactly the same object.

Well, we will define these integrals for any function F,

forms a space l2.

Let me very shortly recall what does it mean.

So basically, on the set of functions,

you can define a so-called inner product.

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Inner product of two functions f and G,

is the integral f of X, g of X,

d x and is integral over the integral from A to B. I would note it as f,

g. And, basically this is an analog of this color product in the space of vectors.

This object has the following properties.

First of all, it is symmetric.

So, f, g is equal to g f. Secondly,

it is linear in the sense that if you take any two constant,

this object is linear in the sense that if you take two functions f1 and f2,

and two real numbers a1 and a2.

There is a product between a1,

f1 plus a2 f2,

and g is equal to a1 f1,

g, plus a2 f2,

g. And finally, if we consider a in a product,

between f and f, this inner product,

is a negative, and it is equal to zero if and only if the function f is equal to zero.

Since, all of these three properties hold,

then the space of functions with these inner product,

is called a Hilbert space.

This is a complete analog of the [inaudible] space of vectors.

So, we'll try to define this integral for any function from the space l2.

We says that this function is from this space,

if and only if the integral of f of x squared from A to B,

the X is infinite.

And we can also say that a sequence of functions f n

converges to f in the sense of space as a space l2.

If and only if their inner product

of f n minus f

with f n minus f converges to zero as n is turned into infinity.

In other words, this means that the integral f n of x minus f of x squared,

the x is turned into zero.

So, integral is considered also from A to B.

Let me now try to define this type of integrals,

and I will do this in two steps.

On the first step on the first stage,

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Function f is called a step function,

if and only if it can be represented in the following way.

So, f of x is equal to the sum of Lambda I,

I from one to N,

multiply it with the indicator that ti minus one is less or equals and X,

and is less than ti.

Here, t1. Excuse me,

t0, t1 and so on tn,

is a subdivision of the interval between N and B.

And for these functions,

we define the integral,

the linear integral f of ft d wt,

as the sum from I from 1 to n alpha I,

Brownian motions the moment ti minus Brownian motions and moment ti minus one.

I forgot to say, that alpha one,

alpha two, and so on,

alpha N are just a real numbers.

Let me now provide an example of a step function,

such as this integral can be computed directly.

Let the function f t be the following,

f of t is equal to

one if t is between zero and one.

It equal to 10 if t is from one to two,

is equals to 0 if t is larger or equals than two.

Of course, this function can be illustrated by a picture.

So, we have a function from zero to one,

that is equal to one.

Then from one to two,

it is equal to 10.

And afterwards, it is equal to zero,

as the plot is quite simple.

And the question is,

what is the value of the integral of f of t d wt,

if we can see this is integral from zero to

capital T. Of course the answer depends on the value of capital T. And basically,

if we just apply this scheme,

this definition, exactly for this example,

then we get the following answer.

This integral is equal to.

So, first of all, let me consider the case when capital T is less than one.

In this case, we have here the integral of

all the one constant one from zero to capital T. And therefore,

this integral is equal to capital W of T. Otherwise,

if T is between one and two,

then we have a function outside this form.

It's one multiplied with the indicator that x is between zero,

and one plus 10 multiplied by the indicator that d is

between one and capital T. And therefore,

the answer here is the following.

So, we should take one multiplied by capital W1 plus

10 multiplied by Wt minus W1.

And the third situation which is also possible,

when capital T is large or equal than two,

then we have the sum of three elements and the sum gives us W1 plus 10,

multiplied by W2, minus W1.

So, this is the answer of what is the value of this

integral for this function f. As you see from this example,

in any case for any capital T,

the value of this integral is a normally distributed random variable.

And this of course is done not by chance.

This is a kind of theorem,

which I would like to show right now.

The theorem is the full length.

Let me denote the integral from A to B,

f of T by I

of f. And the theorem tells us that if f is a step function,

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then, this integral I have f as a normal distribution,

which mean there is zero and variant equal to L to normal,

the function f is a power of two.

It is equal to the integral from A to B,

f squared of x, d x.

Let me proved the theorem,

actually this is a very a simple fact.

And, the proof is the following one.

So at first, why it is normal,

this is just because of our definition.

Here you sees that we have a set of independent random variables,

because you know that increments of a Brownian motion are independent.

And therefore, here we have a sum

of normally distributed random variables, which I independent.

And therefore, the sum has also normal distribution.

As for the parameters of this normal distribution,

let me check that the mathematical expectation is equal to zero.

Let's work simple exercise.

So, I take mathematical expectation of I of f. And,

due to the linearity of the mathematical expectation,

I can conclude that this expectation is equal to the sum of alpha I,

I from one to N,

mathematical expectation of W,

ti minus W, ti minus one.

And, you know that the Brownian Motion has a zero mathematical expectation.

And therefore, all mathematical expectations here are equal to zero.

And finally, we conclude that the sum is also equal to zero.

Now, what's about variants of this integral?

So, variants I ff.

Well, what happen exactly the same situation.

Here is the sum of independent random variables,

we knows that the variance of a sum of a dependent random variables,

is equal to the sum of variances.

And therefore, it is equal to the sum,

I from one to N, alpha I squared,

multiply it by the variance of the difference W ti minus W, ti minus one.

You know that one of the properties of the Brownian emotion is that this increment is

distributed according to a normal distribution with zero mean and variance

equal to ti minus ti minus one.

And therefore, here you have a sum,

I from one to N. Alpha I squared,

multiplied by ti, minus ti minus one.

And, this is exactly equal to the L to norm was the function f,

in the power of two.

So, all points of all items of this theorem are proven.

We have showed that this integral has a normal distribution,

with zero and mean at exactly this variance.

And therefore, the theorem is proven.