Correlation between two random signals
5 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
rihab
el 1 de Oct. de 2015
Editada: John D'Errico
el 2 de Oct. de 2015
I have two signals in MATLAB, say
a = randn(1,1e6) + randn(1,1e6)*exp(-1i * 2*pi * 1.1);
b = randn(1,1e6) + randn(1,1e6)*exp(-1i * 2*pi * 1.4);
I am finding the correlation between them as follows:
R=corrcoef(a,b);
r = R(2,1);
Now each time I run my code, the correlation coefficient is different. I even tried to increase the number of samples (from 1e6 to higher values) but that didn't work. Is there some other way to find the correlation coefficient between such signals?
0 comentarios
Respuesta aceptada
John D'Errico
el 1 de Oct. de 2015
Editada: John D'Errico
el 1 de Oct. de 2015
When you generate random data, you will NEVER be able to know the EXACT correlation coefficient. The data was composed of random numbers. Perhaps a related example will help.
What is the mean of a uniform random variable, sampled from the interval [0,1]? We know this of course to be 0.5. But try it?
n = 1e6;
x = rand(1,n);
mean(x)
ans =
0.50032
x = rand(1,n);
mean(x)
ans =
0.50006
x = rand(1,n);
mean(x)
ans =
0.4997
Hmm. I got a different number each time, and none of them were the expected value of 0.5. Close, but not exactly so.
In fact, statistics 101 will teach us that as the sample size increases, the mean will in fact approach the known expected value, but we will never expect an exact result. The sample correlation coefficient taken from data has the same property. It will be an estimate of that parameter, but it will generally never be the true value, that one could compute using theory. As well, the value we get will vary, due to the randomness of the sample. This is all exactly as we expect.
2 comentarios
John D'Errico
el 2 de Oct. de 2015
Editada: John D'Errico
el 2 de Oct. de 2015
It looks like you misunderstand that there is a difference between a parameter estimated from a SAMPLE of some random variable, compared to an expected value computed for the entire population.
This is exactly why I gave the example of the mean of a random sample. The EXPECTED value of that mean is 0.5. However, the computed mean of those random samples will essentially never be exactly the expected value. The same thing applies to a correlation coefficient. The computation that you show is no more exactly true than the computation I show for the mean.
So, again, the point is there is a difference between a sample mean and a population mean. Your problem seems to arise from the use of similar names for the parameters.
Más respuestas (1)
Jan
el 1 de Oct. de 2015
It is expected that the correlation between two random signals has a (low) random value. So what do you mean by "that didn't work"? I do not see a reason to look for another way, because corrcoeff works fine. Only your expectations do not agree with the mathematical definition.
2 comentarios
Jan
el 1 de Oct. de 2015
Please read the documentation and source code of corrcoeff.m to find out, what's going on. Because you create random inputs for this function each time, the output must be different also. Only the magnitude of the output should be "small", because the inputs are random. You cannot expect the correlation to have a fixed value as e.g. 0.0 even for uncorrelated signals.
Ver también
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!