Steps for estimating alpha in 1/f^alpha noise using OLS

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I have a signal (the vector "pink"), which I know is pink noise, 1/f^alfa, where alpha = 1. But let's assume alpha is unknown, and I need to estimate alpha. In other words, Let's assume I just have the signal.
I use the following procedure to get into a position to estimate alpha using OLS:
F = abs(fft(pink)); x = log(1:(length(pink)/2)); y = log(F(1:(length(pink)/2)));
Now I run OLS regression:
[r,m,b] = regression(x,y)
where I get
r = -0.5948, m = -0.4915, b = 6.8223.
I would have thought my estimate of alpha was m, but m is not equal to 1. So, what am I doing wrong? Is there an additional step I am forgetting, or is my procedure plain wrong?
  3 个评论
Ulrik William Nash
Ulrik William Nash 2017-5-28
编辑:John D'Errico 2017-5-28
Hi John,
I didn't expect that twist.
Here is the code for the particular function:
function [r,m,b] = regression(targets,outputs,flag)
%REGRESSION Linear regression.
%
% <a href="matlab:doc regression">regression</a> calculates the linear regression between each element
% of the network response and the corresponding target.
%
% [R,M,B] = <a href="matlab:doc regression">regression</a>(T,Y) takes cell array or matrix targets T and
% output Y, each with total matrix rows of N, and returns the linear
% regression for each of the N rows: the regression values R, slopes M,
% and y-intercepts B.
%
% <a href="matlab:doc regression">regression</a>(T,Y,'one') returns scalar R, M and B values across all
% rows of targets and outputs.
%
% Here a feedforward network is trained and regression performed on its
% targets and outputs.
%
% [x,t] = <a href="matlab:doc simplefit_dataset">simplefit_dataset</a>;
% net = <a href="matlab:doc feedforwardnet">feedforwardnet</a>(10);
% net = <a href="matlab:doc train">train</a>(net,x,t);
% y = net(x);
% [r,m,b] = <a href="matlab:doc regression">regression</a>(t,y)
%
% See also PLOTREGRESSION
% Copyright 2010 The MathWorks, Inc.
if nargin < 2, error(message('nnet:Args:NotEnough')); end
if iscell(targets), targets = cell2mat(targets); end
if iscell(outputs), outputs = cell2mat(outputs); end
if all(size(targets) ~= size(outputs))
error(message('nnet:NNData:TYMismatch'))
end
if (nargin >= 3) && ischar(flag) && strcmp(flag,'one')
targets = targets(:)';
outputs = outputs(:)';
end
[N,Q] = size(targets);
m = zeros(N,1);
b = zeros(N,1);
r = zeros(N,1);
for i=1:N
t = targets(i,:);
y = outputs(i,:);
ignore = find(isnan(t) | isnan(y));
t(ignore) = [];
y(ignore) = [];
Quse = Q - length(ignore);
h = [t' ones(size(t'))];
yt = y';
rankStatus = warning('off','MATLAB:rankDeficientMatrix');
rankRestore = onCleanup(@() warning(rankStatus));
theta = h\yt;
m(i) = theta(1);
b(i) = theta(2);
yn = y - mean(y);
tn = t - mean(t);
sty = std(yn);
stt = std(tn);
r(i) = yn*tn'/(Quse - 1);
if (sty~=0)&&(stt~=0)
r(i) = r(i)/(sty*stt);
end
end
Ulrik William Nash
Ulrik William Nash 2017-5-28
I just double-checked that the "regression" and "regress" functions compute the same output, which they do, except the first also calculates r.

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回答(1 个)

Nachiket Katakkar
The regression function is part of the Neural Network Toolbox:
The coefficient "m" is the slope of the curve between the inputs "x" and "y". The example on the documentation page shows a value of m equal to 1, however, this would not necessarily be the case for any regression problem.
Consider this example with random data:
>> pink = rand(1,100);
F = abs(fft(pink));
x = log(1:(length(pink)/2));
y = log(F(1:(length(pink)/2)));
[r,m,b] = regression(x,y)
r =
-0.3448
m =
-0.3469
b =
1.7326
You can visualize the relationship between "x" and 'y" using:
>> plotregression(x,y)
Also notice, that in your case the value of "b", the offset of regression is also quite high. The regression function seems to be working as expected, so I would recommend confirming that the algorithm you are using is correct.

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