Identifiability of a particular Independent Component Analysis model

Question

I am considering the model :

$$ \mathbf{x} = \mathbf{A}\mathbf{s} $$

where $\mathbf A \in \mathcal{M}_{n,p}(\mathbb{R})$ and $\mathbf s \in \mathbb{R}^{p}$ such that the entries of $\mathbf s$ are i.i.d. with $\text{Laplace}(1/2)$ distribution. In other words:

$$ p(\mathbf{s}) = \prod \limits_{i=1}^{p} \frac{1}{2} \, \exp\left( -\frac{1}{2}\vert s_{i} \vert \right) = \frac{1}{2^{p}} \, \exp \left( -\frac{1}{2} \Vert \mathbf{s} \Vert_{1} \right). $$

I am wondering whether this is identifiable. Because if I replace $\mathbf{A}$ with $-\mathbf{A}$ and $\mathbf{s}$ with $-\mathbf{s}$, I will obtain the same observations $\mathbf{x}$ and the components of $-\mathbf{s}$ are still i.i.d. with $\text{Laplace}(1/2)$ distribution. Am I right? How can I make this model identifiable? Shall I choose another distribution instead of the Laplace distribution?

Answer 1

No, it's not identifiable, in the sense that you can choose some matrix $\boldsymbol J$ which is diagonal with every nonzero being $\pm 1$ and do:

$\boldsymbol x = \boldsymbol A \boldsymbol J \boldsymbol J \boldsymbol s$

So that $\boldsymbol A'=\boldsymbol A \boldsymbol J$ and $\boldsymbol s'=\boldsymbol J \boldsymbol s$ but $p(\boldsymbol s)=p(\boldsymbol s')$.

So yes you are right.

One option is to sample a $\boldsymbol J$ at each iteration (guessing this is frequentist though). Another is to constrain $\boldsymbol s$ or $\boldsymbol A$ to be positive on the understanding that any appropriate (column, entry) pair could be flipped in sign.