covariance between 2 ts over time

I havea matrix(X) with 3 vectors with 380 elements each. I want to see how the covariance of 1st vector(A) and 2(B) with TS3 (C) evolves over time (25 periods used to compute the cov)

I supposed this should be done by means of a circular loop.

The problem is that the Cov(function) gives a matrix as a result .. and this fact translates into an error in the loop procedure. does someone know if there is a fromula to compute the COV without getting all the cov matrix?

the loop code i've written is :

for i = 26:380
  for u = 1:3
        covaariances(i,u)= ((cov(X(i,u),x(i,3))))
    end
end

this results in an error. could someone of you advise me a way to get what I am aiming for??? thank u for ur valuable time

[Newuser]

I made an error ... the code I have is:

for i = 26:380

for u = 1:3

covaariances(i,u)= ((cov(X(i:(i.25),u),x(i:(i-25),3))))

end

end

[Walter Roberson]

You must still have an error: the (i.25) would not be valid. Perhaps (i+25) ?

[Walter Roberson]

x(i:(i-25),3) would be the empty vector, as i is always going to be greater than i-25.

Answer 1

[Oleg Komarov]

EDIT

X = rand(1500,50);

% Window size
ws = 25;

% Preallocate
szX = size(X);
covC = cell(szX(1)-24,1);
covM = zeros(szX(1)-24,sx(2));

for n = 1:szX(1)-24
    covC{n} = cov(X(0+n:24+n,:));
    covM(n,:) = covC{n}(1,:);
end

You can also save on computations by implementing your own covariance calculation for the first series against the other 49 to get a column of values.

Or you can simply save on memory (even though covC is not more than 30 mb) by substituting the "covM line" inside the loop with:

covC{n} = covC{n}(1,:);

and calling outside of the loop:

covC = cat(1,covC{:});

[Newuser]

Thanks for your help.

The end result I am aiming at is a matrix containing the covariances (evolving over time) of all the 50 vectors with respect to the first vector.

This would mean that the first column would be of ones (1,1,1,1,1,1,)

[Newuser]

I mean ... of the fist matrix covC(1,1) ... i need the first line (this would be the first row of my new matric containig the covariances)

of the secon matrix covC(2,1) ... i need the first line (this would be the second row of my new matric containig the covariances)

and so on

Answer 2

[Teja Muppirala]

The COV function does return a matrix. But this is not a problem since you can just extract out the relevant pieces.

I think there are running covariance algorithms out there that do this calculation very efficiently, but even just using a plain old loop is very fast (this code is only slow because I'm plotting it).

t = 0.01*(0:379)';
X = [sin(20*t.^2) sin(5*t.^3) sin(2*t.^4)];
figure;
a1 = subplot(2,1,1);
plot(X);
legend({'X1' 'X2' 'X3'});
title('X');
blk = 25;
C = zeros(size(X,1)-blk+1,2);
a2 = subplot(2,1,2);
h = plot(C);
set(h(1),'color',[0 0.5 0]);
set(h(2),'color','r');

title('Running Covariance');

for n = 0:size(X,1)-blk
    c = cov(X(n + (1:blk),:));
    C(n+1,:) = c(2:3,1);
    set(h(1),'Ydata',C(:,1));
    set(h(2),'Ydata',C(:,2));
    drawnow;
end
legend({'cov(X1 , X2)' 'cov(X1 , X3)'});
linkaxes([a1 a2],'x');

If you're really against calculating the 3x3 covariance matrix, then you could do it using the formula for covariance which you can find on Wikipedia.

[Newuser]

Thank u for the help. Could u please suggest me some of the efficient running cov algorithms u were referring to? ...

I'm using the one u posted ... but with a X matrix of dimension (1581,500). It's taking hours for the operation

Answer 3

[Walter Roberson]

Covariance is inherently an operation that returns a matrix. It measures the correlation of every component of the first vector with every component of the second vector.

If you are looking for a single value that tells you how "similar" the two vectors are, then covariance is the wrong measure. Possibly you wish to use kstest2()

Answer 4

[Newuser]

Thank you all for the prompt and useful advices. Here is another questions for you.

If instead of 3 "assets" (vectors of the Main matrix" I were to have 350 "assets" and would like to compute the correlation/covariance between them? at each point in time, how could I do it? ... would it be possible using the algorithm postedy by Teja?

Thanks again

Answer 5

[Newuser]

Hallo!

I tried to use the Teja script ... with 50 colums vector containing each 1500 obs.

I've started the calculation 5 minutes ago ... and it's still "busy"

could someone suggest me a more efficient way ?

[Oleg Komarov]

It takes an instant to generate all the covariance matrices! Then you can plot all at once.

[Newuser]

cosa intendi precisamente?