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.