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Week 5.7: Moving-average filters-2

Upon completing this week, the learner will be able to determine whether a given stochastic process is stationary and ergodic; determine whether a given stochastic process has a continuous modification; calculate the spectral density of a given wide-sense stationary process and apply spectral functions to the analysis of linear filters.

The purpose of this course is to equip students with theoretical knowledge and practical skills, which are necessary for the analysis of stochastic dynamical systems in economics, engineering and other fields. More precisely, the objectives are 1. study of the basic concepts of the theory of stochastic processes; 2. introduction of the most important types of stochastic processes; 3. study of various properties and characteristics of processes; 4. study of the methods for describing and analyzing complex stochastic models. Practical skills, acquired during the study process: 1. understanding the most important types of stochastic processes (Poisson, Markov, Gaussian, Wiener processes and others) and ability of finding the most appropriate process for modelling in particular situations arising in economics, engineering and other fields; 2. understanding the notions of ergodicity, stationarity, stochastic integration; application of these terms in context of financial mathematics; It is assumed that the students are familiar with the basics of probability theory. Knowledge of the basics of mathematical statistics is not required, but it simplifies the understanding of this course. The course provides a necessary theoretical basis for studying other courses in stochastics, such as financial mathematics, quantitative finance, stochastic modeling and the theory of jump - type processes.

Visualizar o programa do curso

Ministrado por

Vladimir Panov
Assistant Professor

Transcrição

The following the theorem holds.

Let xt be a weakly stationary,

Process with 0 mean,

And some spectral density, which I will denote by gx.

And assume that Yt is a linear filter, so

it is equal to integral over R, rho(s) xt- s ds.

Then, First of all, the process Y,

so an image of the process X with respect to this filter,

is also a weakly stationary process.

And secondly, it turns out that the spectral

density of the process Y at the point x is equal

to the spectral density of the process x at the point

x multiplied by the modulus of the Fourier transform

of the function rho at the point x squared.

Here, a Fourier transform of the function

rho is equal to the integral exponent iux,

rho (u)du, Integral is taken over R.

So the most important part of the theorem is,

of course, this statement that Y is also stationary.

Because this means that filters of this type

transfer stationary processes to stationary processes.

And the second part is also very important,

because it tells us what is the spectral density of the image of

a process x with respect to this transformation.

Then later on I will show that this result is very

helpful in practice in many situations.

Let me now give you intuition how one can prove the theorem.

Let me start with the first part, so we shall prove that Yt is stationary.

I just would like to recall that many properties of such integrals

were considered previously in this lecture.

This is exactly a stochastic integral of the type when we have some

stochastic process, and to integrate according to.

And as you know from that part of our lecture,

mathematical expectation of the process

Y is equal to the integral over R rho(s)

mathematical expectation xt-s ds.

And you know that since this is equal to 0,

we conclude that mathematical expectation of Yt is also equal to 0.

Here we essentially use the fact that mathematical expectation and

integral can be changed, and I have discussed this sometime before.

As for the covariance function of the process Y,

let me take two time moments, t1 and t2.

Of course, as covariance function is

equal to the mathematical expectation

of the product of one integral rho (s1)

x(t1- s1) ds multiplied by another integral over R,

rho(s2) x(t2- s2) ds2.

So you know that you can combine these two integrals and

you will get a double integral ds1, ds2.

And moreover, according to the properties of stochastic integration,

it turns out that mathematical expectation of these integrals can be changed.

And therefore we will get,

this is equal to the doubled integral, RR, rho s1 rho s2.

And here, we will put mathematical expectation

of the difference x(t1- s1) x(t2- s2),

and here it will be ds1 ds2.

Now, we will use the properties of the process x.

So we know that it's a stationary process.

Therefore, there exists an outer covariance function.

And actually here, we have covariance of the process xt and

the point t1- s1 and t2- s2.

Therefore it is equal to gamma as a time

point (t2- t1)- (s2- s1).

And finally, we conclude that the covariance function

of the process Y can be represented as some function

which depends on the difference between t2 and t1.

This function can be written in the closed form, and

this is actually a covariance function of the process Y.

So this function is equal to the doubled

integral, rho s1 rho s2 gamma

(x- (s2- s1)) ds1 ds2.

So we have proven the first part of the theorem.

Because we have shown the mathematical expectation is a constant,

then the covariance function depends on the difference between t2 and t1.

Now, I'm going to prove the second part of this theorem.

So as we have already seen, the outer covariance function

of the process Y can be represented as some doubled integral.

Let me now rewrite this integral in the following form.

It is actually an integral rho s1

multiplied by the integral, and

here I will take outer covariance

function of the process x at the point x

+ s1- s2 multiplied by rho of s2.

Here I should write ds2 and afterwards ds1.

This formula was recently shown, but

now I represent it as a doubled integral in the following form.

And this, you see, what we have in the first integral

is exactly the convolution between the functions gamma x and the function rho.

And here's a convolution, understand, in the sense of densities.

And this convolution is actually taken in the time moment at point x + s1.

So continuing to the line of reasoning,

we get that outer covariance function

of the process y is equal to the integral

over R convolution between gamma x and

rho at the point x + s1,

multiplied by rho at the point s1 ds1.

Now, let me introduce the following notation.

I will write a rho with circle of x for

the function which is equal to rho as a point minus x.

And if you look at density of this integral,

you can just change this sign here.

So you can change s1 to minus s1, and therefore,

we will get here minus and here instead of plus s1, we get minus s1.

And this function rho of minus s1, according to our new notation,

is equal to rho with a circle with a point s1.

And what we have here is also a convolution between

the function gamma x convoluted with rho and rho circle.

So finally, we conclude that this outer covariance function

of Y is equal to the convolution of three functions.

These are gamma x,

rho, and rho circle.

Now we shall take the Fourier transform from both parts of this equality,

so from this part and this part.

Before we'll do this, let me shortly mention

the spectral density ofthe process Y and

the point x is actually equal to 1 divided by 2

pi Fourier transform of the function gamma y at the point minus x.

This directly follows from the definition of spectral density.

So let me Fourier transforms of both parts, and

I will also divide both parts by 2 pi.

Moderation of this division is exactly this formula.

So what I will finally get is 1 divided

by 2 pi Fourier transform of the function gamma Y at the point x.

Is equal to 1 divided by 2 pi Fourier transform

of the function gamma X at the point x,

multiplied by Fourier transform of rho,

and Fourier transform of rho circle.

Now let me replace x to minus x in this equality here,

should be here minus x, minus x, minus x, minus x.

And let me finally conclude that what happens to the left hand

part is exactly the outer covariance functions of process Y.

What we have here is exactly outer covariance function of the process X.

And these two objects are complex conjugates to each other.

This is a simple outcome from the definition of the function rho circle.

Therefore, we conclude that this equality is fulfilled.

And in the next sub section,

I will show you an example how one can use it in practical tasks.

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