I do not really understand the source of sign ambiguity in ICA. First, my understanding that If I apply ICA on a signal $X$ and I got 3 ICs which are represented by a set $IC^1$. Then, applying ICA on the same signal, I will get 3 ICs but with different orders and signs, let us call them $IC^2$. So the correlation of the two sets might look like this: $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & -1 \\ 0 & -1 & 0 \\ \end{bmatrix}$$

Where the permutation is due to the different orders and the negative sign is because of the sign ambiguity. My question is why does that (the sign ambiguity) happen?


1 Answer 1


We can see the ambiguity of sign as a special case of the ambiguity of scale, as explained in the Independent Component Analysis book of Hyvärinen et al. (Section 7.2.3). If we use their notation, $\mathbf x=(x_1, x_2,\dots, x_K)$ is our $K$-dimensional observed signal and $\mathbf s=(s_1, s_2, \dots, s_K)$ is the unknown source signal${}^\dagger$, and they are mixed by a (non-random) and unknown matrix $\mathbf A$ $$\mathbf{x=As}.$$ Another way to express the same relationship is to write $\mathbf x$ as the sum of the columns of the matrix $\mathbf A$, namely the vectors $\mathbf a_i$: $$\mathbf x = \sum_i \mathbf a_i s_i.$$ But the latter can be re-written as $$\mathbf{x}=\sum_i\left(\frac{1}{\alpha_i}\mathbf a_i\right)(s_i\alpha_i),$$ where the coefficients $\alpha_i$ are any (non-zero) scalars. Here we can see why sign ambiguity occurs: we could put $\alpha_i=-1$, and the resulting $\mathbf x$ wouldn't change, thus the model is not capable of distinguishing whether the source signals are $s_i$ or $-s_i$.

${}^\dagger$For simplicity, we are assuming that the matrix $\mathbf A$ is square (as is in the question), hence the sizes of $\mathbf x$ and $\mathbf s$ are the same.


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