## how to compute running correlation coefficient

### Kathleen (view profile)

on 3 Apr 2015
Latest activity Answered by David J. Mack

### David J. Mack (view profile)

on 21 Dec 2017
Hi, I'd like to calculate a 'running' correlation coefficient of two variables. t seems that my problem is that the corrcoef(x,y) function writes a (2,2)-Matrix but I only wanna save the actual coefficient, i.e. (1,2) or (2,1), into my resuls. How can I do this? Any help is very much appreciated! > > > I've tried the following:
Here some fake data:
x = [1 3 4 8 9 12 13 14 40 30];
y = [2 3 4 7 8 9 13 14 41 32];
window=2; % window size for the correlation coefficient

### the cyclist (view profile)

on 3 Apr 2015

If you have the Statistics Toolbox, you can use the function corr, which will only return the single value of the correlation.

Kathleen

### Kathleen (view profile)

on 3 Apr 2015
Thanks for your reply and help. But the syntax corr cannot be used to facilitate this.

### Roger Stafford (view profile)

on 3 Apr 2015
Edited by Roger Stafford

### Roger Stafford (view profile)

on 3 Apr 2015

By "running correlation coefficient" I assume you mean you want the correlation computed for each possible length of x and y from 1:1, 1:2, ..., 1:N for some large maximum number N. If N is very large it becomes inefficient to repeatedly compute using 'corrcoef'. Fortunately it is possible to express the correlation in terms of five running sums which can greatly reduce the total computation required. You can see this form in the Wikipedia site
http://en.wikipedia.org/wiki/Correlation_and_dependence
near the end of the first section.
The following code carries out that computation in terms of the Wikipedia formula.
sx = x(1); sy = y(1);
sx2 = x(1)^2; sxy = x(1)*y(1); sy2 = y(1)^2;
c(1) = 1; % Avoid a NaN on the first step
for n = 2:N
sx = sx + x(n);
sy = sy + y(n);
sx2 = sx2 + x(n)^2;
sxy = sxy + x(n)*y(n);
sy2 = sy2 + y(n)^2;
c(n) = (n*sxy-sx*sy)/sqrt((n*sx2-sx^2)*(n*sy2-sy^2));
end
Note that for each step in the loop there are only about twenty operations performed for updating the correlation value c(n), whereas with a large value for N, each step would involve some multiple of n of such operations.
You should not necessarily take the above code literally, but its method may serve to give you a procedure for performing the computation more efficiently for your "running" type situation.

Kathleen

### Kathleen (view profile)

on 3 Apr 2015
Dear Roger,
Thanks for your reply and help! I want to compute the sliding or running window correlation coefficient. I have read related papers, the formula is as following:
t=n,n+1,n+2,n+3，......。 n means the length of silding or running window.
Could you translate this formula into Matlad codes? Any help is very much appreciated! Many many thanks!
Roger Stafford

### Roger Stafford (view profile)

on 3 Apr 2015
Just change the way the five "running" sums are generated.
t = 10; % <-- You choose the width of the window
sx = cumsum(x); sx = sx(t+1;end)-sx(1:end-t);
sy = cumsum(y); sy = sy(t+1;end)-sy(1:end-t);
sx2 = cumsum(x.^2); sx2 = sx2(t+1;end)-sx2(1:end-t);
sxy = cumsum(x.*y); sxy = sxy(t+1;end)-sxy(1:end-t);
sy2 = cumsum(y.^2); sy2 = sy2(t+1;end)-sy2(1:end-t);
c = (t*sxy-sx.*sy)./sqrt((t*sx2-sx.^2).*(t*sy2-sy.^2));