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From a mathematical point of view,

Brownian motion is exactly the same as the Wiener process.

And according to this terminology,

this process is now treated as either Bt,

B from Brown or Wt, W from Wiener.

Let me shortly mention that it is called Brownian motion in honor of Robert Brown,

who considered movement of

small particles suspended in a fluid, around 1827.

The next scientist who worked with

a very similar scientific phenomena was actually Louis Bachelier,

the father of modern financial mathematics in 1900.

He considered serious speculations in finance,

and he wrote a PhD thesis about this topic.

And actually, his ideas was very close to the ideas of Albert Einstein,

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He introduced his theory in 1927.

And so, this process has two names;

Brownian motion, Wiener process.

And let me now explain what does this theorems mean.

Actually Brownian motion can be defined in two ways.

First of all, it can be defined as a Gaussian process with

zero mathematical expectations and covariance function

equal to the minimum between t and s. As explained to you previously,

this function is symmetric and positive.

So, I'm definite and therefore,

there exist a Gaussian process with this covariance function.

And exactly this process is called the Brownian motion.

Second definition, we can say that

the Brownian motion is a process such that property number zero,

it starts from zero almost surely,

then it has stationary and independent increments or non-independent increments.

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And stationarity follows from the second item.

Actually Bt minus Bs has a normal distribution

with zero mean and variance t minus s. In particular,

from here it follows that the Bt has

a normal distribution with means zero and variance equal to

t. Let me now show that these definitions are equivalent.

Let me first show that definition two follows for definition one.

That is, if Bt is a Gaussian process with

zero mathematical expectation and

covariance function equal to the minimum between arguments.

So, all of these three properties are fulfilled.

Let me start with property number zero.

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Well, you know that the mathematical expectation of

B0 is equal to zero because this is m at a point

zero and variance of B0 is equal to the covariance function at a point 00,

therefore, it is also equal to zero.

So, what we have is that the random variable B0

has zero mathematical expectation and zero variance.

From here, we immediately conclude that B to zero is equal to zero almost surely.

And therefore, the property number zero holds.

Next, let me show that process Bt has independent increments.

I will show that two increments are independent and this proof can be extended to

resolve any problems to arbitrary number of intervals.

So, let me consider two non-interceptable intervals from A to

B and from C to D. And let me

show that Bb minus

Ba and Bd minus Bc are independent.

Let me first show that these two increments are uncorrelated.

I will consider the covariance between Bb minus

Ba and Bd minus Bc.

Since covariance is a linear function,

it is equal to covariance between Bb and

Bd minus covariance between Ba and Bd,

minus covariance between Bb and Bc,

and plus covariance between Ba and Bc.

Now you know that covariance function is equal to minimum between the arguments.

Therefore, the first covariance function is equal to

the minimum between B and D. That this is equal to B.

The second covariance function is equal to A.

This covariance is equal to B,

and this covariance is equal to A.

So, what we have here is a B minus A minus B plus A,

that this is equal to zero.

So, we have that these increments are uncorrelated.

But as you know, it isn't sufficient to say that they are uncorrelated.

We shall prove that they are independent.

And according to the theorem which I showed you a little bit earlier,

we have to show that Bb minus Ba,

Bd minus Bc is a Gaussian vector.

This can be shown just by definition because if you

take any linear combination of the components.

So, lambda one Bb minus Ba plus lambda two Bd minus Bc,

and this linear combination can be considered as a linear combination of Bb,

Ba, Bd, and Bc.

Since the process Bt is a Gaussian process,

any linear combination of the final dimensional distribution Ba,

Bb, Bc, Bd has normal distribution.

And therefore, we get that this expression has

a normal distribution since lambda one and lambda two or any two numbers,

we get that vector Bb minus Ba and Bd minus Bc is a Gaussian vector.

And since we know that the components of these vectors are uncorrelated,

we immediately conclude that they're independent.

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So, the first property is also proven.

And now, we proceed with the second one.

Actually, there are three statements which are hidden here.

So first of all,

we should prove that this difference has a normal distribution.

Secondly, we should prove that the mean is equal to zero.

And thirdly, we shall prove that the variance is equal to the difference

between t and s. Let me start with the first part,

so that this difference has a normal distribution.

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Therefore, any linear combination of these components has normal distribution,

and in particular if I take combination with coefficients one and minus one.

We also get a random variable was normal distribution.

So, we concludes that Bt minus Bs are normally distributed.

Secondly, if we consider a mathematical expectation of the difference between Bt and Bs,

we can simply use that mathematical expectation in a linear function and get this

is equal to mathematical expectation of Bt minus mathematical expectation of Bs.

So, this is equal it is equal to mt minus ms and it's equal to zero.

And finally, let me consider as variance of Bt minus Bs.

We know that variance is exactly the same as

the covariance between Bt minus Bs and Bt minus Bs.

And now, we can use this as the covariance function as actually a linear function

and therefore it is equal to covariance between Bt and

Bt minus two covariances between Bt and

Bs plus covariance between Bs and Bs.

So, if you're looking at the first amount.

It is equal to a minimum between T and T,

so it is equal to T. The second amount is equal to two s,

here of course we assume that T is larger than S and is larger or equal than zero.

Otherwise, of course the statement doesn't have any sense,

and as for the last part of this expression,

it's equal to S. So we have T minus two s plus s. Finally we conclude that

this is equal to t minus s, so theorem one.

So, if a process is a Gaussian process with zero mean and this covariance function,

then all of these properties are satisfied and we

conclude that definition two follows from definition one.

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Let me now show that and draw a statement,

that definition one follows from definition two.

So, I'm going to prove that definition one follows from definition two.

Let me first show that if beta t is a process which satisfies these three properties,

then this is a Gaussian process.

In fact, if we can see that any linear combination of beta tk by t1,

t2 and tl are sometime points,

then I claim that this sum can be actually

represented as a linear combination of the increments process beta.

Let me assume that t1 is less than t2 and so on is less than

tn and let me take the largest tn from this sum.

So, I will just put this sum

outside the sum with this term outside of the sum and therefore,

I will get lambda n beta tn.

And now I would like to subtract from this from beta tn minus one.

Since I have subtracted lambda and beta tn minus one,

I should add this two into the sum and get lambda

n plus lambda n minus one beta tn minus one.

And two other terms remain,

so it's sum k runs from one to n minus two lambda k beta tk.

Then I can do exactly the same with beta tn minus one and so on.

And finally I will get that this sum was equal to the sum of k from one to

n. Sum numbers dk multiplied by

the difference between beta tn and beta tn minus one.

And here t0 is equal to zero.

So, we have represented the sum lambda k beta tk as

a linear combination of the increments of the process beta with some coefficients.

Now, you know that all of these increments have normal distribution,

and this means that here we have a sum of Gaussian random variables.

Moreover, the central variables are independent because of the first property.

So, as we know from the probability theory,

the sum also has a normal distribution.

And therefore, we have proven that

any linear combination of beta tks have normal distribution and we

conclude from here that beta t1 and so on beta tn is a Gaussian vector.

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And therefore, our process Bt is Gaussian.

So, we have proven that Bt is a Gaussian process and let me now proceed to raise

the proof of these two statements

that arithmetical expectation is equal to

zero and covariance function is equal to minimum.

Let me first note that the process Bt at point t has

a normal distribution with zero mean and variance equal to t. In fact,

this follows from the second property,

if you put s equal to zero here and here's is at B0 is equal to zero.

So, you immediately get this statement.

Therefore, function m(t) which is

equal to the mathematical expectation of Bt is equal to zero.

And as for the covariance function.

We have here covariance between Bt and Bs.

We have a very typical problem which very often arises as a context of

stochastic processes that here is these two random variables are

dependent but the increments are independent.

And therefore we should somehow consider

increments instead of the values of the process beta.

In this case, we can use the following.

We can subtract Bs and at Bs

here on the first place inside the covariance and if we use that sub-process,

the covariance function is linear regardless.

This is equal to covariance between Bt minus Bs and

Bs plus covariance between Bs and Bs.

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Now, you can just mention that here you have no Bs

but Bs minus B0 because B0 is equal to zero.

Therefore, you have increments of the process beta with respect to sum

non-acceptable intervals Sd and

zero S. Since you know that the increments are independent,

covariance is equal to zero.

And as for the second term,

you have here covariance between two random variables which are the same.

That is therefore it's equal to the variance of Bs and the variance of Bs equal to s

because distribution of Bt at any time point t

is normal with zero mean and variance equal to t. So,

all of these things were given under assumption

that t is larger than s. If you now consider t smaller than s,

so you will get that the covariance function is equal to t. And finally,

you conclude that the covariance function is equal to minimum between t and s. So,

all of the items which I included in definition one are proven,

so we have shown that the Bt is a Gaussian process,

that the mathematical expectation is equal to zero and as

the covariance function is equal to the minimum between t and s. Therefore,

look that I have shown that definitions one and two are equivalent.