MATLAB: How to estimate the auto-correlation function value r(k)=E[X(n)X(n-k)]

matlab; xcorr

If X(n) is an available time series, I want to estimate the auto-correlation function value of r(k). The definition of r(k) is r(k)=E[X(n)X(n-k)]. Should I use the command "xcorr(x)"? But I will obtain a sequence, how could I get a value? Is this way right? I define y(n)=x(n-k), and use the xcorr(x,y)? But still I need a value and how to do the time delay process y(n)=x(n-k)?

Best Answer

XCORR will return the autocorrelation values and the associated lags at which these values were calculated. The examples section has one:

ww = randn(1000,1);
[c_ww,lags] = xcorr(ww,10,'coeff');
stem(lags,c_ww)

What do you mean by just one value representing ACF? The sequence of values represent correlation between values at different lag indices (time) and a single value does not make sense.

Related Solutions

MATLAB: How to find cells corresponding to zero lag in an autocorrelation function

The zero lag produces the highest value in the autocorrelation output. Hence simply finding the maximum value of the autocorrelation matrix and finding its coordinates gives the pixel location.

In the below example xcorr2 functions compute the 2-D autocorrelation.

I = imread('cameraman.tif');       
acf = xcorr2(I);
maximum = max(max(acf));
[r_0,c_0]=find(acf==maximum);

Also, the following code will work since the maximum value of the autocorrelation output will be at the middle position.

I = imread('cameraman.tif');       
[m,n] = size(I);
r_0 = ceil((m+n-1)/2);
c_0 = ceil((m+n-1)/2);

Refer the following link for further information on xcorr2.

https://www.mathworks.com/help/signal/ref/xcorr2.html

MATLAB: Autocorrelation Matrix from a vector

No, that is not correct. The way you are doing it is not giving the autocorrelation matrix. For one thing your matrix is not going to be Toeplitz.

For example:

    x = randn(10,1);
    rxx = x*x';

Note the above, rxx, is not Toeplitz.

But

[xc,lags] = xcorr(x,x,9,'biased');
r = xc(10:end);
rxx = toeplitz(r,conj(r)); % the conj() of course here is not needed

Now, rxx is Toeplitz and not that the autocorrelation matrix has full column rank.

rank(rxx)

Or you could do:

    X = fft(x,2^nextpow2(2*size(x,1)-1));
    R = ifft(abs(X).^2);
    m = length(x);
    R = R./m; % Biased autocorrelation estimate
    rxx = toeplitz(R(1:length(x)),conj(R(1:length(x))));

Best Answer

Related Solutions

MATLAB: How to find cells corresponding to zero lag in an autocorrelation function

MATLAB: Autocorrelation Matrix from a vector

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