We cannot recover the "real" scale of S. Any scale of the
original signals can be normalized by the mixing matrix W. If S’ is a set of
signals with non-unit variances, then let M be a diagonal scaling matrix such
that with that a_{ii}S_{i}’ results in a column whose elements
have variance of one.

X = A’S’ = (A’M^{-1})(MS’) = AS

Since the scale of the original signals cannot be recovered, we find signals with a variance of one by convention.

The scaling matrix above could also be positive or negative. For this reason, ICA produces two answers for each independent component. It does not matter which of these are selected.

There is no order to the independent signals either. Again, a permutation matrix can show that permuted columns have equivalent answers:

X = A’S’ = (A’P^{-1})(PS’) = AS

The main constraint that allows a solution to be found is
that the columns of S are statistically independent. Statistical independence
is described in the next section. We can think of each signal as a random
variable and we want to find a matrix W=A^{-1} such that WX=S such that
the signals in S are maximally independent. WX=S says that the each independent
signal S_{i} can be expressed as a linear combination of the signals in X.

Note that A must be invertible for WX=S to be valid._{}

Here is a review on introductory statistics for the reader who needs it.

The following graphs show readings from two gaussian signals, two uniform signals, and the S and X example signals from above.

**
These graphs show the joint distribution between the
two measured signals (X _{1} and X_{2}) and the two measured
signals (S_{1} and S_{2}). These graphs were created with the
following MATLAB commands:**

*
plot(s1,s2,’o’);
plot(x1,x2,’o’);
*

Up

Prev Next