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In the above wiki page, how do I calculate the expectation values in the second step of the algorithm for "Single Component Extraction"?

My interpretation is as follows:

$$ X = \begin{bmatrix}a & b & c\\d & e & f\end{bmatrix} \\ E\{X\} = \frac{1}{3}\begin{bmatrix}a + b + c\\d + e + f\end{bmatrix} \\ $$

Am I right? (I assumed this way because the page says "average over the column-vectors")

My second question is, how do I calulate $x g(w^{T}x)$ in the first expectation value of the same step? (The matrix orders don't agree; $x$ is $n \times m$ and $w$ is $n \times 1$)

EDIT: The definition of $X$ can be found on this page http://en.wikipedia.org/wiki/Independent_component_analysis#Mathematical_definitions

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can you give more information about X? What you do is not the same as the link you posted –  Stef Feb 28 at 20:55
@Stefanos, Why is my approach not the same as the one that has been done there? –  Priyatham Feb 28 at 21:08
i checked it again. i see what you mean –  Stef Feb 28 at 21:10

1 Answer 1

up vote 1 down vote accepted

Your interpretation is right about the $E\{\}$ operation.

Let us focus on the update algorithm.

The vector $x$ is a single column of the matrix $X$ which is $n\times m$, so $x$ is of size $n\times 1$.

Weight vector $w$ is $n\times 1$, so its transpose $w^T$ is $1\times n$.

Thus $w^Tx$ is $1\times 1$, i.e. it is a scalar value, leading to $g(w^Tx)$ being a scalar value resulting from the nonlinear operator $g(\cdot)$.

Suppose you have a matrix $X$ of size $n\times m$.

You would calculate $x_ig(w^Tx_i)$ for each column vector $x_i$ in $X$, where $i\in \{1,2,..,m\}$.

Basically you scale each column $x_i$ by the scalar $g(w^Tx_i)$, and average over these scaled columns.

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