Estimatenoise

Some curve fitting or smoothing tools can benefit from knowledge of the noise variance to expect on your data. Kalman filters use this information, also some spline fitting tools. So I wrote a function to extract the noise variance from a signal vector. It also works on any specified dimension of an array.

A few examples of this code in use:

Simple linear data, with purely additive N(0,1) gaussian noise:

t = 0:10000;
x = t + randn(size(t));
mv = estimatenoise(x)
mv =
1.0166

Gaussian noise added to a sine wave (Nominal variance = 0.01)
t = linspace(0,1,1000)';
x = sin(t*50) + randn(size(t))/10;
mv = estimatenoise(x)
mv =
0.0096887

Pure gaussian noise, with a nominal variance of 9. (Note that var would have been a better estimator for this particular case...)

mv = estimatenoise(3*randn(2,3,1000),3)
mv =
9.6584 8.2696 8.632
9.2404 8.5346 9.7725

A piecewise constant function with multiple discontinuities. The true noise variance should be 0.01.

t = linspace(0,1,1000);
X = round(cos(t*6*pi)) + randn(size(t))/10;
plot(t,X)
var(X) % var will be wildly in error
ans =
0.68256

estimatenoise(X)
ans =
0.010882

Test if estimatenoise is able to recover the variance of a normally distributed random sample with unit variance. (Yes, it will be much slower than var.)

mean(estimatenoise(randn(1000,1000)))
ans =
1.0002

Estimatenoise can now handle non-uniformly spaced series (by request.) In the next example,
the actual noise variance was 1.0 here. Perform the operation 1000 times, then look at the median variance estimate. How well did we do?

t = sort([randn(1,100) , randn(1,100)+5]);
X = repmat(sin(t*5)*100,1000,1) + ...
randn(1000,length(t));

Estimatenoise is clearly wrong when the spacing is ignored.

median(estimatenoise(X,2))
ans =
16.438

Supplying the sampling "times", we get quite a reasonable result.
median(estimatenoise(X,2,t))
ans =
1.1307

Estimatenoise also works on data with replicates. In this example, each point will be replicated up to 3 times. The actual noise variance was again 1.0 here. I'll also compare the increase in times required for estimatenoise when t is supplied.

t = sort([0:1:100, 1:2:100, 1:4:100]);
X = repmat(sin(t/10)*100,1000,1)+ ...
randn(1000,length(t));

Again, estimatenoise is clearly wrong when the non-uniform spacing is ignored.

tic,median(estimatenoise(X,2)),toc
ans =
4.2056
Elapsed time is 2.690135 seconds.

Supplying the sampling "times", we again get quite a reasonable result. The time penalty is not quite 2x.

tic,median(estimatenoise(X,2,t)),toc
ans =
1.0116
Elapsed time is 4.864486 seconds.