Machine learning in trading: theory, models, practice and algo-trading - page 2522

[Aleksey Nikolayev]

secret #:
The multiplication table in the market is non-stationary)

Like a sine with a wartime value of four)

[Valeriy Yastremskiy]

Aleksey Nikolayev #:

Yes.

No. Let j<k, then COV(Yj,Yk)= COV(Yj,Yj+X(j+1)+...+Xk)= COV(Yj,Yj)+ COV(Yj, X(j+1)+...+Xk)=.

Well then you just have to substitute.

ACF = COV(Yj,Yk)/sqrt(COV(Yj,Yj)*COV(Yk,Yk))= (COV(Yj,Yj)+COV(Yj,X(j+1)+...+Xk)) / sqrt(COV(Yj,Yj)*COV( Yj,Yj+X(j+1)+..+Xk, Yj,Yj+X(j+1)+..+Xk)) = (COV(Yj,Yj)+COV(Yj,X(j+1)+...+Xk)) / sqrt(COV(Yj,Yj)*COV(X(j+1)+...+Xk,X(j+1)+...+Xk)) = ...

I'm getting confused with the rules of the COV function

[Aleksey Nikolayev]

Valeriy Yastremskiy #:

Well, then you just have to substitute.

ACF = COV(Yj,Yk)/sqrt(COV(Yj,Yj)*COV(Yk,Yk))= (COV(Yj,Yj)+COV(Yj,X(j+1)+...+Xk)) / sqrt(COV(Yj,Yj)*COV( Yj,Yj+X(j+1)+..+Xk, Yj,Yj+X(j+1)+..+Xk)) = (COV(Yj,Yj)+COV(Yj,X(j+1)+...+Xk)) / sqrt(COV(Yj,Yj)*COV(X(j+1)+...+Xk,X(j+1)+...+Xk)) = ...

I get further confused with the rules of the COV function

No, substitute early) There one of two terms is zero (which one?) COV(Yj,Yj)==0 orCOV(Yj,X(j+1)+...+Xk)==0? We need to take advantage of the linearity of covariance for each argument and remember the definition of SB)

[Dmytryi Nazarchuk]

Valeriy Yastremskiy #:

Why are you feeding him?

It is clear that he is a pure theorist, it is clear that the ACF SB has no practical value - no one just needs to dig through textbooks and recall long forgotten lectures....

[Aleksey Nikolayev]

Dmytryi Nazarchuk #:

Why are you feeding him?

It is clear that he is a pure theorist, it is clear that the ACF SB has no practical meaning - no one just needs to dig through textbooks and recall long forgotten lectures....

green grapes)

[Evgeny Dyuka]

Ivan Butko #:
Gentlemen, regulars of this thread, tell me if there are any successes in machine learning, ready-made products that work?

The thread has grown fiercely, it is probably the most active.

https://www.mql5.com/ru/blogs/post/746398#comment_26491969

[Valeriy Yastremskiy]

Aleksey Nikolayev #:

No, substitute early) There one of the two summands is zero (which one?) COV(Yj,Yj)==0 orCOV(Yj,X(j+1)+...+Xk)==0? We need to take advantage of the linearity of covariance on each argument and remember the definition of SB)

The second equals zero if Yj is not equal to the remaining sum of x's. But it can also be equal. In the singular case.

And the first equals one.

[Aleksey Nikolayev]

Valeriy Yastremskiy #:
The second equals zero if Yj is not equal to the remaining sum of x's. But it can also be equal. In the singular case.

Yes, from the definition of SB that all increments in the future are independent with values in the present and past and therefore the covariances are all zero.

Valeriy Yastremskiy #:

And the first is equal to one.

Not one, it is the variance increasing with time j. If we denote by d the variance of white noise Xi, thenCOV(Yj,Yj)=j*d^2. To do this, we have to represent Yj as the sum of X1+...+Xj and calculate, taking into account the properties of white noise.

As a result, after substitution, ACF=sqrt(min(j,k)/max(j,k)). If I didn't mess up something, of course).

I propose to close here the subject of ACF SB, so as not to make especially impressionable practical practitioners nervous)

[Valeriy Yastremskiy]

Aleksey Nikolayev #:

Yes, from the definition of SB, that all increments in the future are independent with values in the present and past and hence the covariances are all zero.

Not unity, it is the variance increasing with time j. If we denote by d the variance of white noise Xi, thenCOV(Yj,Yj)=j*d^2. To do this, we have to represent Yj as the sum of X1+...+Xj and calculate, taking into account the properties of white noise.

As a result, after substitution, ACF=sqrt(min(j,k)/max(j,k)). If I didn't mess up something, of course).

I propose to close here the subject of ACF SB, so as not to make particularly impressionable practical practitioners nervous)

(Yes, I mixed up) was not my favorite thing to do in the terver formulas). On Schrödinger the complexity of understanding is exponential) although it is not even a theorem.)

[mytarmailS]

Evgeny Dyuka # :
https://www.mql5.com/ru/blogs/post/746398#comment_26491969

how do you do your broadcast is a third party service or what? how does it work? how can I do this?