ICA for Noise Reduction over a Dataset Suppose my dataset consists of $N$ example vectors $\mathbf{x}_{1}, \ldots, \mathbf{x}_{N}$ where $\mathbf{x}_{n} \in \mathbb{R}^{p}$ $\forall n$. I assume that each vector  $\mathbf{x}_{n}$ is comprised of an underlying true datapoint $\mathbf{s}_{n}$ that is corrupted by additive Gaussian noise $\mathbf{x}_{n} = \mathbf{s}_{n} + \mathbf{w}_{n}$.

What I want to do is estimate the $\mathbf{s}_{n}$. Can I do this with ICA?



So the ICA model is of the form:

$$ \mathbf{x}_{n} = a_{1,n} \mathbf{s}_{1} + a_{2,n} \mathbf{s}_{2} + \ldots + a_{N,n} \mathbf{s}_{N} $$ 

Does this mean I need to fit my problem to the model by assuming that $ \mathbf{s}_{n} = a_{1,n} \mathbf{s}_{1} $ and $\mathbf{w}_{n} = a_{2,n} \mathbf{s}_{2}$ ? If so, how do I recover the vectors $\mathbf{s}_{n}$ $\forall n$?

I ask because I see the ICA problem formulated as:

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

Would this then imply that: $\mathbf{s} = \mathbf{A}^{+} \mathbf{x}$, where $\mathbf{A}^{+}$ is the psuedo-inverse of $\mathbf{A}$? But how would I get the $\mathbf{s}_{n}$ out of this formulation?

Note: I cross-posted this on stats.stackexchange to get the statisticians' point of view, but I'd like to hear what the signal people have to say.



Edits: 

So I do not know $\mathbf{A}$ but I do know that each $\mathbf{s}_{n} \sim f_{s}(\cdot | \theta)$ for some parameter vector $\theta$. I have specified $f_{s}(\cdot | \cdot)$ but $\theta$ is unknown. I also know that the noise vectors are not sampled from that same distribution, but that they are independent of the signal vectors.
This appears similar to a classic least squared solution of an overdetermined equation that proceeds as follows:

Starting with:

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

$\mathbf{A}$ is not a square matrix if overdetermined (more equations than unknownns) so therefore an inverse does not exist. What you do then is multiply both sides by the transpose of $\mathbf{A}$ since $\mathbf{A^TA}$ is a square matrix. From that, assuming and inverse exists for$\mathbf{A^TA}$, you can solve for s:

$$ \mathbf{A^Tx} = \mathbf{A^TA} \mathbf{s}  $$

$$ \mathbf{s} =  (\mathbf{A^TA})^{-1}\mathbf{A^Tx} $$