I have two vectors:
A_1 =
10
200
7
150
A_2 =
0.001
0.450
0.0007
0.200
I would like to know
Given:
A_1 = [10 200 7 150]';
A_2 = [0.001 0.450 0.007 0.200]';
(As others have already pointed out) There are tools to simply compute correlation, most obviously corr
:
corr(A_1, A_2); %Returns 0.956766573975184 (Requires stats toolbox)
You can also use base Matlab's corrcoef
function, like this:
M = corrcoef([A_1 A_2]): %Returns [1 0.956766573975185; 0.956766573975185 1];
M(2,1); %Returns 0.956766573975184
Which is closely related to the cov
function:
cov([condition(A_1) condition(A_2)]);
As you almost get to in your original question, you can scale and adjust the vectors yourself if you want, which gives a slightly better understanding of what is going on. First create a condition function which subtracts the mean, and divides by the standard deviation:
condition = @(x) (x-mean(x))./std(x); %Function to subtract mean AND normalize standard deviation
Then the correlation appears to be (A_1 * A_2)/(A_1^2), like this:
(condition(A_1)' * condition(A_2)) / sum(condition(A_1).^2); %Returns 0.956766573975185
By symmetry, this should also work
(condition(A_1)' * condition(A_2)) / sum(condition(A_2).^2); %Returns 0.956766573975185
And it does.
I believe, but don't have the energy to confirm right now, that the same math can be used to compute correlation and cross correlation terms when dealing with multi-dimensiotnal inputs, so long as care is taken when handling the dimensions and orientations of the input arrays.
Try xcorr
, it's a built-in function in MATLAB for cross-correlation:
c = xcorr(A_1, A_2);
However, note that it requires the Signal Processing Toolbox installed. If not, you can look into the corrcoef command instead.
For correlations you can just use the corr function (statistics toolbox)
corr(A_1(:), A_2(:))
Note that you can also just use
corr(A_1, A_2)
But the linear indexing guarantees that your vectors don't need to be transposed.
To perform a linear regression between two vectors x
and y
follow these steps:
[p,err] = polyfit(x,y,1); % First order polynomial
y_fit = polyval(p,x,err); % Values on a line
y_dif = y - y_fit; % y value difference (residuals)
SSdif = sum(y_dif.^2); % Sum square of difference
SStot = (length(y)-1)*var(y); % Sum square of y taken from variance
rsq = 1-SSdif/SStot; % Correlation 'r' value. If 1.0 the correlelation is perfect
For x=[10;200;7;150]
and y=[0.001;0.45;0.0007;0.2]
I get rsq = 0.9181
.
Reference URL: http://www.mathworks.com/help/matlab/data_analysis/linear-regression.html