0
$\begingroup$

I have read a lot about ICA. I think I could not find the answer that why non-Gaussian Variables are independent. What I understand, Central Limit Theorem states that

distribution of sum of independent variable tends toward more Gaussian than its original random variables.

$ s_i $ are the number of original independent sources in ICA; whereas ICA model is $x=As$. So we can define $y=w^Ts$. The main goal is to find the unmixing matrix $w$ that maximize the non-Gausaanity. So my question is what is non-Gaussanity here and why its necessary to maximize it to extract the original sources.

Any Enlightenment Please!

$\endgroup$
3
  • 2
    $\begingroup$ Welcome to DSP.SE! Any random variable can be independent from any other one. It does not depend on whether they are Gaussian distributed or non-Gaussian distributed. Why do you think they should be? $\endgroup$
    – Peter K.
    Commented Oct 29, 2015 at 8:58
  • $\begingroup$ Well, Thats the confusion, how non-gaussian variables are independent. Central Limit Theorem Says that distribution of sum of independent variable tends toward more Gaussian than its original random variables. Also maximizing non-Gaussanity yields one of independent component. So What is the connection of non-Gaussanity with independence of components? I think non-gaussanity gives local maxima for that particular component and based on local maxima we can tell all uncorrelated components $\endgroup$ Commented Oct 29, 2015 at 10:55
  • $\begingroup$ Please reword your question using the definitions in the comment on your deleted "answer". Please edit the question rather that post another non-answer. $\endgroup$
    – Peter K.
    Commented Oct 29, 2015 at 12:16

2 Answers 2

4
$\begingroup$

The model ICA uses says that there exist some unknown, statistically independent sources, $s_i$ that are non-normally distributed (their distributions are something other than Gaussian):

$$ s_i \sim S(\mu_{s_i}, \sigma^2_{s_i}) $$

where $S$ is some (possibly) known but non-Gaussian distribution with mean $\mu_{s_i}$ and variance $\sigma^2_{s_i}$.

Then it is assumed that what you can actually measure is the addition of these: $$ \mathbf{x} = \mathbf{A}\mathbf{s} $$ where $\mathbf{s}$ is the vector of the $s_i, i=1,\ldots,N$, $\mathbf{A}$ is an $N\times N$ matrix (usually assumed to be invertible) and $\mathbf{x}$ is the vector of the actual measurements.

Because the $x_i$ will just be weighted sums of the $s_i$: $$ x_i = a_{i1} s_1 + a_{i2} s_2 + \cdots + a_{iN} s_N $$ the central limit theorem says that the distribution of $x_i$ will be closer to Gaussian than the distribution of the $s_i$ (provided certain non-restrictive conditions on the true distribution of the $s_i$ are met).

Then all ICA really tries to do is to find a $\mathbf{W} = \mathbf{A}^{-1}$ the inverse of $\mathbf{A}$.

There are any number of measures for "non-Gaussianity". One of the simplest (?) is to use kurtosis as the measure of how far the sample of a random variable is from Gaussianity.

So, to attempt to answer:

So my question is what is non-Gaussianity here and why its necessary to maximize it to extract the original sources.

There are several different ways one could measure non-Gaussianity. For example, kurtosis can be used as it is known that the kurtosis of a Gaussian is 3. Any distribution with a kurtosis different from 3 is therefore "non-Gaussian" to some extent.

The reason we want non-Gaussianity is because we assumed that the original sources are non-Gaussian.

$\endgroup$
1
  • $\begingroup$ Concise and enlightening as OP requested. $\endgroup$ Commented May 22, 2016 at 14:08
2
$\begingroup$

In the explanation above, the last statement is not the actual reason why in ICA you assume non-Gaussian sources. After all, you could as well assume Gaussian sources and after the mixing process, the observed data will look Gaussian, which is not a problem at all.

The reason for assuming non-Gaussian sources was pointed out by Comon (1994), and it is because in order to learn a unique factorization $\mathbf{x=As}$, at most one component of s can be Gaussian. Which it has to do with the invariance under rotation of the Gaussian distribution. For simplicity, we usually assume non-Gaussian priors, where for instance $p_i(s_{i})\propto \dfrac{1}{\cosh s_{i}}$ yields the non-linear contrast function, $\tanh$, used in the ICA algorithm.

In another note related to the title of the post, "Why non-Gaussian variables are independent". Independence and non-Gaussianity are two different issues. Informally, independence means that a joint distribution has a factorial form on its components such that $p(s_1,\ldots,s_n)=\prod_i^n p_i(s_i)$, where the distributions $p_i(s_i)$ may or may not be Gaussian.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.