From b4e8ba1919764a02d91e5b4eb7db4177bbd487fb Mon Sep 17 00:00:00 2001 From: Youssef Kashef Date: Wed, 20 May 2020 21:10:48 +0200 Subject: [PATCH 1/6] split ito files --- notes/06_fastica/0_recap_ica_whitening.tex | 100 ++ notes/06_fastica/1_ica_ambiguous.tex | 135 ++ notes/06_fastica/1_pcaica.tex | 1295 -------------------- notes/06_fastica/2_pcaica.tex | 191 +++ notes/06_fastica/3_badgaussians.tex | 162 +++ notes/06_fastica/4_kurt.tex | 488 ++++++++ notes/06_fastica/5_fastica.tex | 161 +++ notes/06_fastica/tutorial.tex | 30 +- 8 files changed, 1266 insertions(+), 1296 deletions(-) create mode 100644 notes/06_fastica/0_recap_ica_whitening.tex create mode 100644 notes/06_fastica/1_ica_ambiguous.tex delete mode 100644 notes/06_fastica/1_pcaica.tex create mode 100644 notes/06_fastica/2_pcaica.tex create mode 100644 notes/06_fastica/3_badgaussians.tex create mode 100644 notes/06_fastica/4_kurt.tex create mode 100644 notes/06_fastica/5_fastica.tex diff --git a/notes/06_fastica/0_recap_ica_whitening.tex b/notes/06_fastica/0_recap_ica_whitening.tex new file mode 100644 index 0000000..20e20a0 --- /dev/null +++ b/notes/06_fastica/0_recap_ica_whitening.tex @@ -0,0 +1,100 @@ + +\section{The ICA Problem} +\begin{frame} + +independent sources: $\vec s = (s_1, s_2,...,s_N)^\top \in \R^N$\\ +observations: $\vec x \in \R^N$ + +\begin{equation} +\label{eq:ica} +\vec x = \vec A \, \vec s +\end{equation} + +\begin{equation} +\widehat{\vec s} = \vec W \cdot \vec x +\end{equation} + +Methods for solving the ICA problem: + +\begin{itemize} +\item maximizing the \emph{mutual information} between $\vec x$ and $\vec {\hat s}$ \\ +(e.g. Infomax) +\item maximizing the \emph{nongaussianity} of $\widehat {\vec s}$ \\ +(e.g. Kurtosis-based ICA, FastICA) +\end{itemize} +\end{frame} + +\begin{frame} +\underline{Outline:} +\begin{itemize} + \item ICA on whitened data + \begin{itemize} + \item Whitening/sphering + \item Amibguities in ICA + \item PCA is \emph{half} the ICA Problem + \end{itemize} + \item the problem with gaussians + \item maximizing nongaussianity + \begin{itemize} + \item Kurtosis-based + \item negentropy + \end{itemize} +\end{itemize} +\end{frame} + +\newpage + +\section{Whitening} + +\begin{frame} + +\notesonly{ +The purpose of whitening is to decorrelate the data. +} + +Let the data $\vec X \in \R^{N \times p}$ be centered: + +\begin{equation} +\label{eq:centered} +\E \lbrack \vec x \rbrack = 0 +\end{equation} + +\notesonly{ +From this follows: +} + +\begin{equation} +\label{eq:cov} +\vec \Sigma_x = \mathrm{Cov}(\vec x) = \E \lbrack \, \vec x \, \vec x^\top \rbrack +\end{equation} + +The whitening transformation yields: + +\begin{equation} +\label{eq:whitening} +\vec v^{(\alpha)} = \vec \Lambda^{-\frac{1}{2}} \vec M^\top \vec x^{(\alpha)} +\end{equation} + +where $\vec M = (\vec e_1, \vec e_2, \ldots,\vec e_N)$ +\notesonly{ +and $\vec \Lambda^{-\frac{1}{2}}$ is a diagonal matrix containig the square roots of the corresponding eigenvalues. +} +\begin{equation} +\label{eq:covw} +\vec \Sigma_v = \mathrm{Cov}(\vec v) = \E \lbrack \, \vec v \, \vec v^\top \rbrack = \vec I_N +\end{equation} + +\notesonly{ +Uncorrelted means zero covariance. Therefore, the covariance matrix for uncorrelated data is a diagonal matrix because it only contains the variances of the individual variables. +Whitening decorrelates the variables and normalizes the variances to 1. +} +\end{frame} + +\begin{frame} +\begin{figure}[ht] +\label{fig:sphering} +\includegraphics[width=12cm]{img/cov.png} +\caption{A visual interpretation of whitening} +\end{figure} + +\end{frame} diff --git a/notes/06_fastica/1_ica_ambiguous.tex b/notes/06_fastica/1_ica_ambiguous.tex new file mode 100644 index 0000000..6f8d83c --- /dev/null +++ b/notes/06_fastica/1_ica_ambiguous.tex @@ -0,0 +1,135 @@ + +\section{Ambiguities in ICA and limitations} +\begin{frame} + +Sources can be recovered up to: +\begin{itemize} +\item sign +\item scale +\item permutation i.e. ordering +\item only one gaussian distributed source +\end{itemize} +\begin{align*} +\vec P &:= \text{arbitrary permutation matrix}\\ +\vec \Lambda &:= \text{arbitrary diagonal matrix} +\end{align*} +\begin{align} +\vec x &= \vec A \, \vec s\\ +\vec x &= \lbrack \, \vec A\, \vec P^{-1} \vec \Lambda^{-1}\, \rbrack \, \lbrack \, \vec \Lambda \, \vec P \, \vec s\, \rbrack +\end{align} + +\end{frame} + + + +ICA cannot resolve if the mixing matrix is $\vec A$ or a permuatated and/or scaled version of $\vec A$. +It can \textbf{also} not resolve if the independent sources are $\vec s$ or a permutated and/or scaled version of $\vec s$. + +Permutations and scaling are not an issue for ICA because permutation and scaling do not interfere with statistical independence. + +$$ +P_{s_1, s_2}(\widehat {\vec s}) \eqexcl P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) +$$ + +\begin{frame} +Permutations of sources +{\footnotesize +\begin{equation*} + \arraycolsep=1.4pt%\def\arraystretch{2.2} + \begin{array}{ccc} + \left( \begin{array}{ll} + \textcolor{gray}{\widehat{s}_1} \\ \widehat{s}_2 + \end{array} \right) + = + \left( \begin{array}{ll} + \textcolor{gray}{\mathrm{w}_{11}} & \textcolor{gray}{\mathrm{w}_{12}} \\ + \mathrm{w}_{21} & \mathrm{w}_{22} + \end{array} \right) + \left( \begin{array}{ll} + \mathrm{x}_1 \\ \mathrm{x}_2 + \end{array} \right) + & \corresponds & + \left( \begin{array}{ll} + \widehat{s}_2 \\ \textcolor{gray}{\widehat{s}_1} + \end{array} \right) + = + \left( \begin{array}{ll} + \mathrm{w}_{21} & \mathrm{w}_{22} \\ + \textcolor{gray}{\mathrm{w}_{11}} & \textcolor{gray}{\mathrm{w}_{12}} + \end{array} \right) + \left( \begin{array}{ll} + \mathrm{x}_1 \\ \mathrm{x}_2 + \end{array} \right) + \\\\ + P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) + && + P_{s_2} (\widehat{s}_2) \cdot P_{s_1} (\widehat{s}_1) + \end{array} +\end{equation*} +} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +Scaling of source amplitudes: + +{\footnotesize +\begin{equation*} + \begin{array}{ccc} + \arraycolsep=1.4pt + \left( \begin{array}{ll} + \widehat{s}_1 \\ \widehat{s}_2 + \end{array} \right) + = + \left( \begin{array}{ll} + \mathrm{w}_{11} & \mathrm{w}_{12} \\ + \mathrm{w}_{21} & \mathrm{w}_{22} + \end{array} \right) + \left( \begin{array}{ll} + \mathrm{x}_1 \\ \mathrm{x}_2 + \end{array} \right) + & \corresponds & + \left( \begin{array}{ll} + \textcolor{gray}{a}\,\widehat{s}_1 \\ + \textcolor{gray}{b}\,\widehat{s}_2 + \end{array} \right) + = + \left( \begin{array}{ll} + \textcolor{gray}{a}\,\mathrm{w}_{11} & \textcolor{gray}{a}\,\mathrm{w}_{12} \\ + \textcolor{gray}{b}\,\mathrm{w}_{21} & \textcolor{gray}{b}\,\mathrm{w}_{22} + \end{array} \right) + \left( \begin{array}{ll} + \mathrm{x}_1 \\ \mathrm{x}_2 + \end{array} \right) + \\\\ + P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) + && + aP_{s_1} (a\widehat{s}_1) \cdot bP_{s_2} (b\, \widehat{s}_2) + \end{array} +\end{equation*} +} +\end{frame} +\subsection{Implications of the ambiguities} + +\begin{frame} + +We can assume: +$$ +\E \lbrack \, \vec s \, \rbrack = 0 +$$ + +Removing the mean from $\vec x$ does not change $\vec A$: + +$$ +\vec x - \E \lbrack \, \vec x \, \rbrack = \vec A \left( \vec s - \E \lbrack \, \vec s \, \rbrack \right) +$$ + +\notesonly{Note that} $\E \lbrack \, \vec s \, \rbrack$ and $\E \lbrack \, \vec x \, \rbrack$ are not necessarily equal. +\pause + +We can also assume: +$$ +\mathrm{Cov}(\vec s) = \E \lbrack \, \vec s \, \vec s^\top \rbrack = \vec I_N +$$ + +Any scaling in $\mathrm{Cov}(\widehat{\vec s})$ can be assumed to come from $\vec A$ and can be undone. + +\end{frame} + diff --git a/notes/06_fastica/1_pcaica.tex b/notes/06_fastica/1_pcaica.tex deleted file mode 100644 index d431126..0000000 --- a/notes/06_fastica/1_pcaica.tex +++ /dev/null @@ -1,1295 +0,0 @@ - -\section{The ICA Problem} -\begin{frame} - -independent sources: $\vec s = (s_1, s_2,...,s_N)^\top \in \R^N$\\ -observations: $\vec x \in \R^N$ - -\begin{equation} -\label{eq:ica} -\vec x = \vec A \, \vec s -\end{equation} - -\begin{equation} -\widehat{\vec s} = \vec W \cdot \vec x -\end{equation} - -Methods for solving the ICA problem: - -\begin{itemize} -\item maximizing the \emph{mutual information} between $\vec x$ and $\vec {\hat s}$ \\ -(e.g. Infomax) -\item maximizing the \emph{nongaussianity} of $\widehat {\vec s}$ \\ -(e.g. Kurtosis-based ICA, FastICA) -\end{itemize} -\end{frame} - -\begin{frame} -\underline{Outline:} -\begin{itemize} - \item ICA on whitened data - \begin{itemize} - \item Whitening/sphering - \item Amibguities in ICA - \item PCA is \emph{half} the ICA Problem - \end{itemize} - \item the problem with gaussians - \item maximizing nongaussianity - \begin{itemize} - \item Kurtosis-based - \item negentropy - \end{itemize} -\end{itemize} -\end{frame} - -\newpage -\section{Whitening} - -\begin{frame} - -\notesonly{ -The purpose of whitening is to decorrelate the data. -} - -Let the data $\vec X \in \R^{N \times p}$ be centered: - -\begin{equation} -\label{eq:centered} -\E \lbrack \vec x \rbrack = 0 -\end{equation} - -\notesonly{ -From this follows: -} - -\begin{equation} -\label{eq:cov} -\vec \Sigma_x = \mathrm{Cov}(\vec x) = \E \lbrack \, \vec x \, \vec x^\top \rbrack -\end{equation} - -The whitening transformation yields: - -\begin{equation} -\label{eq:whitening} -\vec v^{(\alpha)} = \vec \Lambda^{-\frac{1}{2}} \vec M^\top \vec x^{(\alpha)} -\end{equation} - -where $\vec M = (\vec e_1, \vec e_2, \ldots,\vec e_N)$ -\notesonly{ -and $\vec \Lambda^{-\frac{1}{2}}$ is a diagonal matrix containig the square roots of the corresponding eigenvalues. -} -\begin{equation} -\label{eq:covw} -\vec \Sigma_v = \mathrm{Cov}(\vec v) = \E \lbrack \, \vec v \, \vec v^\top \rbrack = \vec I_N -\end{equation} - -\notesonly{ -Uncorrelted means zero covariance. Therefore, the covariance matrix for uncorrelated data is a diagonal matrix because it only contains the variances of the individual variables. -Whitening decorrelates the variables and normalizes the variances to 1. -} -\end{frame} - -\begin{frame} -\begin{figure}[ht] -\label{fig:sphering} -\includegraphics[width=12cm]{img/cov.png} -\caption{A visual interpretation of whitening} -\end{figure} - -\end{frame} -\clearpage -\section{Ambiguities in ICA and limitations} -\begin{frame} - -Sources can be recovered up to: -\begin{itemize} -\item sign -\item scale -\item permutation i.e. ordering -\item only one gaussian distributed source -\end{itemize} -\begin{align*} -\vec P &:= \text{arbitrary permutation matrix}\\ -\vec \Lambda &:= \text{arbitrary diagonal matrix} -\end{align*} -\begin{align} -\vec x &= \vec A \, \vec s\\ -\vec x &= \lbrack \, \vec A\, \vec P^{-1} \vec \Lambda^{-1}\, \rbrack \, \lbrack \, \vec \Lambda \, \vec P \, \vec s\, \rbrack -\end{align} - -\end{frame} - - - -ICA cannot resolve if the mixing matrix is $\vec A$ or a permuatated and/or scaled version of $\vec A$. -It can \textbf{also} not resolve if the independent sources are $\vec s$ or a permutated and/or scaled version of $\vec s$. - -Permutations and scaling are not an issue for ICA because permutation and scaling do not interfere with statistical independence. - -$$ -P_{s_1, s_2}(\widehat {\vec s}) \eqexcl P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) -$$ - -\begin{frame} -Permutations of sources -{\footnotesize -\begin{equation*} - \arraycolsep=1.4pt%\def\arraystretch{2.2} - \begin{array}{ccc} - \left( \begin{array}{ll} - \textcolor{gray}{\widehat{s}_1} \\ \widehat{s}_2 - \end{array} \right) - = - \left( \begin{array}{ll} - \textcolor{gray}{\mathrm{w}_{11}} & \textcolor{gray}{\mathrm{w}_{12}} \\ - \mathrm{w}_{21} & \mathrm{w}_{22} - \end{array} \right) - \left( \begin{array}{ll} - \mathrm{x}_1 \\ \mathrm{x}_2 - \end{array} \right) - & \corresponds & - \left( \begin{array}{ll} - \widehat{s}_2 \\ \textcolor{gray}{\widehat{s}_1} - \end{array} \right) - = - \left( \begin{array}{ll} - \mathrm{w}_{21} & \mathrm{w}_{22} \\ - \textcolor{gray}{\mathrm{w}_{11}} & \textcolor{gray}{\mathrm{w}_{12}} - \end{array} \right) - \left( \begin{array}{ll} - \mathrm{x}_1 \\ \mathrm{x}_2 - \end{array} \right) - \\\\ - P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) - && - P_{s_2} (\widehat{s}_2) \cdot P_{s_1} (\widehat{s}_1) - \end{array} -\end{equation*} -} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -Scaling of source amplitudes: - -{\footnotesize -\begin{equation*} - \begin{array}{ccc} - \arraycolsep=1.4pt - \left( \begin{array}{ll} - \widehat{s}_1 \\ \widehat{s}_2 - \end{array} \right) - = - \left( \begin{array}{ll} - \mathrm{w}_{11} & \mathrm{w}_{12} \\ - \mathrm{w}_{21} & \mathrm{w}_{22} - \end{array} \right) - \left( \begin{array}{ll} - \mathrm{x}_1 \\ \mathrm{x}_2 - \end{array} \right) - & \corresponds & - \left( \begin{array}{ll} - \textcolor{gray}{a}\,\widehat{s}_1 \\ - \textcolor{gray}{b}\,\widehat{s}_2 - \end{array} \right) - = - \left( \begin{array}{ll} - \textcolor{gray}{a}\,\mathrm{w}_{11} & \textcolor{gray}{a}\,\mathrm{w}_{12} \\ - \textcolor{gray}{b}\,\mathrm{w}_{21} & \textcolor{gray}{b}\,\mathrm{w}_{22} - \end{array} \right) - \left( \begin{array}{ll} - \mathrm{x}_1 \\ \mathrm{x}_2 - \end{array} \right) - \\\\ - P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) - && - aP_{s_1} (a\widehat{s}_1) \cdot bP_{s_2} (b\, \widehat{s}_2) - \end{array} -\end{equation*} -} -\end{frame} -\subsection{Implications of the ambiguities} - -\begin{frame} - -We can assume: -$$ -\E \lbrack \, \vec s \, \rbrack = 0 -$$ - -Removing the mean from $\vec x$ does not change $\vec A$: - -$$ -\vec x - \E \lbrack \, \vec x \, \rbrack = \vec A \left( \vec s - \E \lbrack \, \vec s \, \rbrack \right) -$$ - -\notesonly{Note that} $\E \lbrack \, \vec s \, \rbrack$ and $\E \lbrack \, \vec x \, \rbrack$ are not necessarily equal. -\pause - -We can also assume: -$$ -\mathrm{Cov}(\vec s) = \E \lbrack \, \vec s \, \vec s^\top \rbrack = \vec I_N -$$ - -Any scaling in $\mathrm{Cov}(\widehat{\vec s})$ can be assumed to come from $\vec A$ and can be undone. - -\end{frame} - - -\subsection{Whitening in the context of ICA} - -\begin{frame} - -\begin{align} -\label{eq:expxina} -\vec \Sigma_x = \mathrm{Cov}(\vec x) &= \E \lbrack \, \vec x \, \vec x^\top \rbrack\\ -&= \E \lbrack \, \vec A\,\vec s \, \left( \vec A\,\vec s \right)^\top \rbrack\\ -&= \E \lbrack \, \vec A\; \underbrace{\;\vec s \, \vec s^\top}_{= \vec I_N} \vec A^\top \rbrack\\ -\label{eq:sigmax} -&= \vec A\, \vec A^\top -\end{align} - -\end{frame} - -\begin{frame} - -\notesonly{ -If we were to apply \emph{Singular Value Decomposition} on the symmetric matrix $\Sigma_x$: -} - -\begin{equation} -\mathrm{SVD}(\vec \Sigma_x) = \vec U\, \vec D \, \vec U^\top -\end{equation} - -where\\ -\notesonly{ -$\vec U$ is an orthogonal matrix,} $\vec U^\top\vec U = \vec I_N$ and -\notesonly{$\vec D$ is a diagonal matrix with rank $N$, }$\vec D^\top = \vec D$.\\ - -We define the two following transformation $\vec Q$ and $\widetilde{\vec A}$: -\begin{equation} -\label{eq:defq} -\vec Q := \vec D^{-\frac{1}{2}} \vec U^\top -\end{equation} -\begin{equation} -\label{eq:defatilde} -\widetilde{\vec A} := \vec Q \, \vec A -\end{equation} - -Which yields a new ICA problem: -\begin{equation} -\vec u = \vec Q\, \vec x = \vec Q\,\vec A \, \vec s = \widetilde{\vec A} \, \vec s -\end{equation} - -\question{What is so special about the new mixing matrix $\widetilde{\vec A}$?} \\ - -\notesonly{-$\widetilde{\vec A}$ is orthogonal (i.e. $\widetilde{\vec A}\, \widetilde{\vec A}^\top = \widetilde{\vec A}^\top \widetilde{\vec A} = \widetilde{\vec A}^{-1} \widetilde{\vec A} = \vec I_N$). - We will see the implications of this in how it reduces the number of free parameters for solving the new ICA problem. - } -\end{frame} - -\begin{frame} - -\slidesonly{ -$$ -\vec Q := \vec D^{-\frac{1}{2}} \vec U^\top -$$ -$$ -\vec \Sigma_x = \E \lbrack\, \vec x \, \vec x^\top \rbrack = \vec U \, \vec D \, \vec U^{\top} -$$ -} - -\begin{align} -\E \lbrack\, \vec u \, \vec u^\top \rbrack -&\mystackrel{\vec u = \vec Q\, \vec x}{\vec u = \vec Q\, \vec x} - \E \lbrack\, \vec Q \, \vec x \left( \vec Q \, \vec x\right)^\top \rbrack\\ -&\mystackrel{\vec u = \vec Q\, \vec x}{}\E \lbrack\, \vec Q \, \vec x \; \vec x^\top \vec Q^\top \rbrack\\ -&\mystackrel{\vec u = \vec Q\, \vec x}{} \E \lbrack\, \vec Q \, \vec \Sigma_{X} \, \vec Q^\top \rbrack\\ -&\mystackrel{\vec u = \vec Q\, \vec x}{} \vec Q \, \vec \Sigma_{x} \, \vec Q^\top\\ -&\mystackrel{\vec u = \vec Q\, \vec x}{} -\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) -\left( \vec U \, \vec D \, \vec U^\top \right) -\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) -\end{align} - -\end{frame} -\begin{frame} - -\slidesonly{ -\begin{align} -\E \lbrack\, \vec u \, \vec u^\top \rbrack -&\mystackrel{\vec u = \vec Q\, \vec x}{} -\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) -\left( \vec U \, \vec D \, \vec U^\top \right) -\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) -\end{align} -} - -From $\vec U^\top\vec U = \vec I_N$ and $\vec D^\top = \vec D$ follows: - -\begin{align} -\E \lbrack\, \vec u \, \vec u^\top \rbrack -&= -\vec D^{-\frac{1}{2}} -\underbrace{\left( \vec U^\top \vec U \right)}_{= \vec I_N} -\, \vec D \, -\underbrace{\left( \vec U^\top \, \vec U \right)}_{= \vec I_N} -\, -\left(\vec D^{-\frac{1}{2}} \right)^\top \\ -&= \vec D^\top -\, -\underbrace{\vec D^{-\frac{1}{2}} \, -\left(\vec D^{-\frac{1}{2}} \right)^\top}_{=\vec D^{-1}} = \vec I_N -\end{align} - - -Recall -\notesonly{from \eqref{eq:sigmax}:} -$$ -\vec \Sigma_x = \E \lbrack \, \vec x \, \vec x^\top \rbrack -= \vec A\, \vec A^\top -$$ -Therefore: -$$ -\E \lbrack \, \vec u \, \vec u^\top \rbrack -= \widetilde {\vec A}\, \widetilde {\vec A}^\top = \vec I_N -$$ - -\notesonly{ -The new mixing matrix} -$\widetilde{\vec A}$ is orthonormal.\\ -The space of solutions is restricted to $\vec W$ that are orthogonal. -The can speed up convergence of ICA. -The number of free parameters -\notesonly{ -for solving the new ICA problem -$\vec u = \widetilde{\vec A}\, \vec s$ -is reduced from $N^2$ to $N(N-1)/2$ -} -\slidesonly{ -$N^2 \rightarrow N(N-1)/2$ -} -\end{frame} - -\begin{frame} -\slidesonly{ -\frametitle{What just happend?} - -\begin{itemize} -\setlength\itemsep{1em} -\item We've transformed $\vec x$ to get $\vec u$: -$ \qquad \qquad \qquad -\vec u := \vec D^{-\frac{1}{2}} \vec U^\top \vec x -$ - -\item Where did $\vec D$ and $\vec U$ come from? -$ \qquad \qquad -\mathrm{SVD}(\vec \Sigma_x) = \vec U\, \vec D \, \vec U^\top -$ - -\item What do $\vec D$ and $\vec U$ represent? - the eigenval. and eigenvect. of $\vec \Sigma_x$ - -\item What's so special about $\vec u$? -$ \qquad \qquad \qquad \qquad -\vec \Sigma_u = \vec I_N -$ - -\item ...So? - -- new ICA problem: $\vec u := \widetilde{\vec A}\, \vec s$\\ -- $\widetilde{\vec A}$ is orthogonal\\ -- unmixing the new problem involves only ``half'' the number of weights. - -\question{What is the transformation $\vec D^{-\frac{1}{2}} \vec U^\top$ called?} - -\end{itemize} -} - -\end{frame} - - -\section{Summary so far:} - -\slidesonly{ -\begin{frame} -\begin{enumerate} -\item Initial ICA Problem: $\vec x = \vec A\, \vec s$ -\item New ICA Problem: $\vec u := \widetilde{\vec A}\, \vec s$,\\ - -where $\vec u = \vec D^{-\frac{1}{2}} \vec U^\top \vec x$ and $\vec \Sigma_u = \vec I_N$. -\item $\vec u$ is the \emph{whitened} version of $\vec x$. -\item $\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. -\item Applying ICA on whitened data reduces the number of free parameters. -\item PCA simplifies the ICA problem. -\item ICA on whitened data is about ``rotating'' it back. -\end{enumerate} - -\end{frame} -} -\subsection{Gaussians are bad for ICA} - -\begin{frame} - -\textbf{A visual argument:} - -\notesonly{Consider the following uniformly distributed independent sources $s1$ and $s2$ and corresponding arbitrary mixtures $x_1$ and $x_2$: -} -\slidesonly{ -Mixing uniformly distributed independent sources $s1$ and $s2$: -} - -\begin{tabular}[h]{c c} -\includegraphics[width=4.5cm]{img/uniform-s_centered.png} & -\includegraphics[width=4.5cm]{img/uniform-x_whitened.png}\\ -original $\vec s$& $\vec x = \vec A \, \vec s$ -\end{tabular} - -\notesonly{The task of ICA is to find an unmixing matrix $\vec W$ that rotates $\vec x$ back to its original orientation. -If the density $\vec x$ were to follow more the shape of a rectangle or parallelogram, whitening -would decorrelate the data and normalized the variances such that the density appears as a square, which then reduces the ICA problem, of finding the right rotation. -The same applies to other densities (not necesarly shaped as rectangles or squares) in that ICA on whitened data ``rotates things back''.} -\end{frame} - -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} -} -\notesonly{ -If our independent sources were normally distributed (left), a mixing matrix will effectively rotate in some way, which yields the same circular ``shape'': - -} -\slidesonly{ -Mixing normally distributed independent sources $s1$ and $s2$: -} -\begin{tabular}[h]{c c} -\includegraphics[width=4.5cm]{img/normal-s_centered.png} & -\includegraphics[width=4.5cm]{img/normal-x_whitened.png}\\ -original $\vec s$& $\vec x = \vec A \, \vec s$ -\end{tabular} - -\question{How should we rotate $\vec x$ to get back $\vec s$?} -\pause -\notesonly{It's not possible. The gaussian distribution is \emph{rotationally symmetric}. We cannot resolve the original axes. -} -\end{frame} - -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} -} -\newpage -\textbf{A more formal argument for why Gaussians are bad for ICA}: - -Recall that the joint density of independent sources is a factorizing density: - -\begin{equation} -\label{eq:facts} -P_{\vec s}(\vec s) = \prod_{i=1}^{N} P_{s_i}(s_i) \,. -\end{equation} - -If we assumed gaussian distributed sources, the following factorization becomes possible: - -\begin{equation} - \label{eq:factsgauss} - \begin{array}{ll} - P_{\vec{s}}({\vec{s}}) - & = \frac{1}{2\pi} - \exp \left( -\frac{\lVert{\vec{s}}\rVert^2}{2} \right) - \\ - & = \underbrace{ - \Bigg[ \frac{1}{\sqrt{2\pi}} \exp \left( - - \frac{{s_1}^2}{2} \right) \Bigg] - }_{{P}_{s_1}({s}_1)} - \underbrace{\Bigg[ - \frac{1}{\sqrt{2\pi}} \exp \left( - - \frac{{s_2}^2}{2} \right) - \Bigg] - }_{{P}_{s_2}({s}_2)} - \end{array} -\end{equation} -\end{frame} - -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} -} -\slidesonly{\textbf{A more formal argument (cont'd):}} - -Now consider applying an orthognal mixing matrix that is \textbf{known}. -\slidesonly{(orthogonal from whitening the $\vec x$)\\ -Consequently: -} -\notesonly{We've seen how whitening takes an ICA problem with any valid\footnote{invertible} mixing matrix $A$ -and reformulates it into a new problem with an orthogonal mixing matrix $\widetilde{\vec A}$. -The following holds for such an orthogonal mixing matrix $\widetilde{\vec A}$: -} -$$ -\widetilde{\vec A}^\top = \widetilde{\vec A}^{-1} \Leftrightarrow \widetilde{\vec A}^{-1}\widetilde{\vec A}=\vec I_N \qquad \text{and} \qquad |\det \widetilde{\vec A}| = |\det \widetilde{\vec A}^\top| = 1 -$$ - -$$ -\vec s = \widetilde{\vec A}^{-1} \vec{x} = \widetilde{\vec A}^\top \vec x -$$ - -\notesonly{ -We are now interested in the joint density $P_x(\vec x)$ of the mixtures $x_1$ and $x_2$. -} - -Density transformation tells us that: -\begin{equation} - \label{eq:gausstransformeddt} - {P}_{\vec x}(\vec x) = - {P}_{\vec s}(\vec s) \left|\det \frac{\partial \vec s}{\partial \vec x}\right| - = {P}_{\vec s}(\vec s) \left|\det \widetilde{\vec A}^{-1}\right| - = {P}_{\vec s}(\vec s) \left|\det \widetilde{\vec A}^\top\right| -\end{equation} -Therefore, -\slidesonly{ -\begin{equation} - {P}_{\vec x}(\vec x) - = \frac{1}{2\pi} - \exp \left( - -\frac{\lVert{\widetilde{\vec A}^\top \vec x}\rVert^2}{2} - \right) - \left|\det \widetilde{\vec A}^\top\right| - = \ldots? -\end{equation} -} -\end{frame} - -\begin{frame} -\slidesonly{ -Orthogonal mixing matrix $\widetilde{\vec A}$ implies: -$$ -\widetilde{\vec A}^\top = \widetilde{\vec A}^{-1} \Leftrightarrow \widetilde{\vec A}^{-1}\widetilde{\vec A}=\vec I_N \qquad \text{and} \qquad |\det \widetilde{\vec A}| = |\det \widetilde{\vec A}^\top| = 1 -$$ -$$ -\vec s = \widetilde{\vec A}^{-1} \vec{x} = \widetilde{\vec A}^\top \vec x -$$} -\begin{align} - \label{eq:gaussx} - {P}_{\vec x}(\vec x) - &= \frac{1}{2\pi} - \exp - \Big( - -\frac{ - \overbrace{ - \lVert{\widetilde{\vec A}^\top \vec x}\rVert^2 - }^{\mathclap{ - \lVert{\widetilde{\vec A}^\top \vec x}\rVert^2 - = \left(\widetilde{\vec A}^\top \vec x\right)^\top \widetilde{\vec A}^\top \vec x - \,=\, \vec x^\top \widetilde{\vec A} \, \widetilde{\vec A}^\top \vec x - = \vec x^\top \vec x - = \lVert{\vec x}\rVert^2 - }}}{2} - \Big) - \underbrace{ - \left|\det \widetilde{\vec A}^\top\right| - }_{=\, 1} - \\ - &= \frac{1}{2\pi} - \exp \left( - -\frac{\lVert{\vec x}\rVert^2}{2} - \right)\\ - \label{eq:gaussxfact} - & = \underbrace{ - \Bigg[ \frac{1}{\sqrt{2\pi}} \exp \left( - - \frac{{x_1}^2}{2} \right) \Bigg] - }_{{P}_{x_1}({x}_1)={P}_{s_1}({s}_1)} - \underbrace{\Bigg[ - \frac{1}{\sqrt{2\pi}} \exp \left( - - \frac{{x_2}^2}{2} \right) - \Bigg] - }_{{P}_{x_2}({x}_2)={P}_{s_2}({s}_2)} - \text{no change in pdf!} -\end{align} -\end{frame} - -We see that the factorization in \eqref{eq:gaussxfact} describes the pdf for the mixtures $\vec x$ identically to the pdf of the original sources. -The original and mixed distributions are identical. The mixing matrix did not change this. Therefore, it would be impossible to find its corresponding unmixing matrix to undo the rotation. - -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} -} - -\notesonly{ -Mixing two independent Gaussians leads to a joint mixed distribution that is equal to that of the original sources. -This is actually justified by the property that \emph{ -uncorrelated jointly Gaussian variables are necessarily independent.}\footnote{ -Further details can be found in Hyv{\"a}rinen Ch. 2.5. -} -} - -A mixture of sources where at most one is Gaussian, is still fine. It only becomes a problem when we have more than one. - - -\slidesonly{ -\begin{itemize} -\item Mixing two independent Gaussians leads to a joint mixed distribution that is equal to that of the original sources. -\item No surprise: \emph{uncorrelated jointly Gaussian variables are necessarily independent.} -\item One Gaussian + other distributions is fine. -\item Two ore more Gaussians. No way. -\end{itemize} -} -\end{frame} - - -\section{Solving ICA by maximizing nongaussianity} - -\begin{frame} - -\begin{block}{Intuition from the Central Limit Theorem} -\emph{The distribution of the sum of independent random variables is ``more Gaussian'' than the original distributions of the random variables.}\\\vspace{2mm} - -Searching for the direction of maximum deviation from a Gaussian distribution may recover the original sources. -\end{block} -\end{frame} - - -\begin{frame} -\frametitle{Maximizing nongaussianity leads to independent sources} - -\textbf{The setting:}\\ - -Two statistically independent sources with $\langle s_i s_j \rangle = \delta_{ij} \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N$ -\notesonly{ -The sources are mixed using a mixing matrix $\vec A$ resulting in observations $\vec x$: -} -\begin{equation} -\label{eq:icaproblemx} -\vec x = \vec A \, \vec s -\end{equation} -\notesonly{ -One out of the $N$ independent sources can be reconstructed using a row vector from an unmixing matrix $\vec W$: -} -\begin{equation} -\label{eq:singlesource} -\widehat s_i = \vec w_i^\top \vec x -\end{equation} -\notesonly{ -By substituting \eqref{eq:icaproblemx} for $\vec x$ in \eqref{eq:singlesource}, -we describe each reconstructed source $\widehat s_i$ as a linear combination of the original sources: -} - - -\notesonly{According to the CLT we can think of the variables in $\vec x$ to be more Gaussian distributed than the original variables in $\vec s$. -Therefore, a solution to the ICA problem is finding an inverse to $\vec A$ that undoes this effect and removes the ``accumulated Gaussianity'' from $\vec x$. -The role of any $\vec w_i$ becomes to maximize the nongaussianity of $\widehat{s}_i$ when we multiply it by $\vec x$. This is the same role $w_i^\top \vec A = \vec z_i^\top$ has when applied to $\vec s$. - - -} - - -\begin{equation} -\label{eq:szs} -\widehat s_i = \vec w_i^\top \vec A \, \vec s = \vec z_i^{\top} \vec s = - \left( \begin{array}{ll} - {z}_1 \\ {z}_2 - \end{array} \right)^\top_i - \left( \begin{array}{ll} - {s}_1 \\ {s}_2 - \end{array} \right) -= z_1 s_1 + z_2 s_2 -\end{equation} - -\notesonly{ -Looking at \eqref{eq:szs} we recognize that $\vec z_i$ describes how to route the information in $\vec s$ such that $\widehat{\vec s}_i$ fully describes -one of the independent sources in $\vec s$. -This can be accomplished with a vector containing a single non-zero element: -} - -\begin{equation*} - \vec{z}_{\text{opt.}} = \left( \begin{array}{c} - 0 \\ \pm 1 - \end{array} \right) - \quad \text{ or }\quad - \vec{z}_{\text{opt.}} = \left( \begin{array}{c} - \pm 1 \\ 0 - \end{array} \right) -\end{equation*} - -\end{frame} - -\notesonly{ -Recall that ICA cannot resolve scale or permutation of the sources and thirdly it cannot resolve the sign. -This is not an issue. -The role of $\vec z_i$ is to route either $s_1$ or $s_2$ to $\widehat{\vec s}_i$. This covers the ambiguitiy in terms of permutation. -We cannot have both independent sources contribute to $\widehat{s}_i$, only one can. Therefore, we only need a single non-zero component for $\vec z_i$. -Wether $s_1$ is scaled by any factor before reaching $\widehat{s}_i$ does not make it more or less independent of $s_2$. Choosing $1$ for the non-zero component is therefore sufficient. -Finally, negating the source by multiplying it by $(-1)$ also has no consequences on the independence criterion. - -We won't actually try and find $\vec z_i$ because we don't have $\vec s$ to apply them to. We use the requirements for $\vec z_i$ by finding a $\vec w_i$ that satisfies these requirements through: -\begin{equation} -\label{eq:zfromw} -\vec z_i = \left(\vec w_i^\top \vec A\right)^\top = \vec A^\top \vec w_i -\end{equation} -} - -\begin{frame} -\question{Does maximizing nongaussianity deliver independent components?} -\slidesonly{ - -\begin{equation} -\widehat s_i = \vec z_i^{\top} \vec s = z_1 s_1 + z_2 s_2 -\end{equation} - -\begin{equation*} - \vec{z}_{\text{opt.}} = \left( \begin{array}{c} - 0 \\ \pm 1 - \end{array} \right) - \quad \text{ or }\quad - \vec{z}_{\text{opt.}} = \left( \begin{array}{c} - \pm 1 \\ 0 - \end{array} \right) -\end{equation*} - -$\vec z_i$ ensures independent $\widehat s_i$ - -\begin{equation} -\widehat s_i = \vec w_i^\top \vec x -\end{equation} - -$\vec w_i$ finds less gaussian $\widehat s_i$ - -\begin{equation} -\vec z_i = \vec A^\top \vec w_i -\end{equation} - -Maximizing nongaussianity is ensured to keep $\widehat s_i$ independent. - -} -\end{frame} -\notesonly{ -- By (1) maximizing the nongaussianity of $\vec w_i^\top \vec x$ and (2) having $\vec z_i = \vec A^\top \vec w_i$ yield independent components and (3) knowing that -$\widehat s_i = \vec w_i^\top \vec x = \vec z_i^{\top} \vec s$, we conclude that maximizing $\vec w_i^\top \vec x$ gives us one independent component. -} -\newpage - -\section{Kurtosis as a measure for nongaussianity} - -\begin{frame} - -\notesonly{ -Kurtosis represents the fourth-order cumulant\footnote{ -Cumulants allow us to express the i-th moment in terms of a cumulative sum of the moments preceeding it. -This simplifies the expression of higher-order moments such as kurtosis which is the fourth-order moment. -} of a random variable. -} - -\begin{block}{Definition} - Let $x$ be a random variable with zero-mean, i.e. $\E \lbrack\,x\,\rbrack = 0$. - \begin{equation} - \label{eq:kurt} - \kurt (x) = \langle x^4 \rangle - 3 - \left( \langle x^2 \rangle \right)^2 \quad - \stackrel{\text{sphered data}}{=} \quad \langle x^4 \rangle - 3 - \end{equation} - - \notesonly{ - By assuming zero-mean and unit-variance, we see that kurtosis is simply a normalized version of the fourth moment. - - Useful properties of kurtosis: - - } - Let $x_1$ and $x_2$ be two independent random variables, then: - \begin{eqnarray*} - \kurt(x_1 + x_2)& = & \kurt(x_1) + \kurt(x_2) \\ - \kurt(z_1 x_1) & = & z_1^4 \kurt(x_1) - \end{eqnarray*} -\end{block} -\end{frame} - -\begin{frame} - -\question{What does kurtosis measure?}\\ - -\begin{tabular}[h]{c c c c} -& -\includegraphics[width=2.7cm]{img/section2_fig20} & -\includegraphics[width=2.7cm]{img/section2_fig21} & -\includegraphics[width=2.7cm]{img/section2_fig22} \\ \\ - -& $\kurt(x) = 0$ & $\kurt(x) > 0$ & $\kurt(x) < 0$\\ \\ - -& -Gaussian PDF & -super-Gaussian PDF& -sub-Gaussian PDF\\ -&bell shaped & peaky, long tails (``outliers'')& bulky, no ``outliers'' \\\\ -e.g.&normal & Laplace & uniform -\end{tabular}\\[1cm] - -\notesonly{This implies that we can use} -$|kurt(x)| > 0$ as a measure of nongaussianity - -\textbf{Caveat:} -\slidesonly{ -sensitive to outliers -} -\end{frame} - - -\notesonly{ -Kurtosis, just like other higher-order cumulants are sensitive to outliers, in that an outlier - will register a much higher kurtosis value. This will be addressed later by FastICA. -} - -\clearpage - -\subsection{kurtosis-based ICA} - -\begin{frame} - -Two statistically independent sources with - -$\langle s_i s_j \rangle = \delta_{ij} \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N$ (any scaling can be attributed to $\vec A$) - -\begin{equation*} -\widehat{s}_i \quad -= \quad \vec{W}^\top \vec{x} \quad -= \quad \vec{W}^\top \vec{A} \cdot \vec{s} \quad -= \quad \vec{z}^\top \vec{s} \quad -= \quad z_1 s_1 + z_2 s_2 -\end{equation*} -\vspace{1mm} -We want the covariance of our reconstructions to match that of the original sources. -\begin{equation*} -\langle \widehat{\vec s} \, \widehat{\vec s}^\top \rangle \eqexcl \langle \vec s \, \vec s^\top \rangle = \vec I_N -\end{equation*} -This implies, -\begin{align*} -\var(\widehat{s}_i) - \; &= \; \langle \big( z_1 s_1 + z_2 s_2 \big)^2 \rangle_{P_{\vec s}}\\ - \; &= \; \langle z_1^2 \, s_1^2 \rangle \;+\; 2 \, \langle z_1\, s_1\, z_2 \, s_2 \rangle \;+\; \langle z_2^2 \, s_2^2 \rangle \\ - \; &= \; z_1^2 \, \langle s_1^2 \rangle \;+\; 2 \, z_1\, z_2 \, \underbrace{\langle s_1\, s_2 \rangle}_{= 0} \;+\; z_2^2 \, \langle s_2^2 \rangle \\ - \; &= \; z_1^2 \, \langle s_1^2 \rangle \;+\; z_2^2 \,\langle s_2^2 \rangle \\ - \; &= \; z_1^2 + z_2^2 \eqexcl 1 -\end{align*} -Making the constraint of unit variance for $\widehat{s}_i$ is to match the variance assumed for the orgiinal sources $s_1$ and $s_2$. This implies that solutions for $\vec z$ are constrained to lie on a unit circle. -\vspace{1mm} -\begin{align*} -\kurt(\widehat{s}) \;\; &= \;\; \kurt(z_1 s_1 + z_2 s_2) \;\; \\ &= \;\; \kurt(z_1 s_1) + \kurt(z_2 s_2) \; = \; z_1^4 \kurt(s_1) + z_2^4 \kurt(s_2) -\end{align*} - -\begin{block}{Kurtosis-based optimization problem} -\begin{equation*} - \begin{array}{rllc} - \left| \kurt(\widehat{s}) \right| & \eqexcl \max_{\vec{z}} - & \leftarrow & \substack{ \text{search for the direction} \\ - \text{with extreme kurtosis}} \\\\ - \var(\widehat{s}) = z_1^2 + z_2^2 & \eqexcl 1 - & \leftarrow & \substack{ \text{such that the data} \\ - \text{remains sphered}} - \end{array} -\end{equation*} - \end{block} -\end{frame} - -\begin{frame} -\slidesonly{ -\begin{block}{Kurtosis-based optimization problem} -\begin{equation*} - \begin{array}{rllc} - \left| \kurt(\widehat{s}) \right| & \eqexcl \max_{\vec{z}} - & \leftarrow & \substack{ \text{search for the direction} \\ - \text{with extreme kurtosis}} \\\\ - \var(\widehat{s}) = z_1^2 + z_2^2 & \eqexcl 1 - & \leftarrow & \substack{ \text{such that the data} \\ - \text{remains sphered}} - \end{array} -\end{equation*} - \end{block} -} - - -\begin{center} -$\hat{s} \,=\, \vec{z}^\top \vec{s} \,=\, \vec{w}^\top \underbrace{\vec{A} \, \vec{s}}_{\vec{x}} \,=\, \vec{b}^\top \underbrace{\tilde{\vec{A}} \, \vec{s}}_{\vec{u}}$ \hspace{4cm} $\kurt(s_i) \neq 0$ -\end{center} - -\question{What can we optimize kurtosis with?} - -\pause - \begin{enumerate} - \item $\max_{\vec{z}} | \kurt{(\vec{z}^\top \vec{s})} | \; \; \;\quad s.t. \quad |\vec{z}| = 1$ - \item $\max_{\vec{w}} | \kurt{(\vec{w}^\top \vec{x})} | \quad s.t. \quad |\vec{A}^\top \vec{w}| = 1$ - \item $ - \max_{\vec{b}} | \kurt{(\vec{b}^\top \vec{u})} | \;\quad s.t. \quad - \underbrace{|\tilde{\vec{A}}^\top \vec{b}| - }_{\substack{= |\vec{b}| \\ \text{ since } \tilde{\vec{A}} \\ \text{ is orthogonal}}} = 1 - $ - \end{enumerate} -\notesonly{ -The different maximization approaches are equivalent. We opt for maximizing -$| \kurt{(\vec{b}^\top \vec{u})} |$ -because it is the only which we can obtain. The other terms either require access to the $\vec s$ or $\vec A$ which is not possible. -Whitening $\vec x$ yields $\vec u$. We also do not know the orthogonal unmixing matrix $\widetilde{\vec A}$ but this -is not an issue because the orthogonality of $\widetilde{\vec A}$ lets the constraint -reduce to only ensuring that $\vec b$ is kept at unit length. -} - -\newpage - -\end{frame} - -\subsection{Kurtosis-based ICA: the gradient algorithm} - -\begin{frame} - -\notesonly{ -$| \kurt{(\vec{b}^\top \vec{u})} |$ can be maximized by moving $\vec b$ -in the direction of the gradient until this becomes zero, whilst keeping the length of $\vec b$ equal to 1. -} - -\begin{equation} -\label{eq:kurtgradient} - \frac{\partial |\text{kurt}(\vec b^\top \vec u)|}{\partial \vec{b}} - = 4 \operatorname{sign} \left[ \kurt{(\vec{b}^\top \vec{u})} \right] \bigg( \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle - 3 \vec{b} \, | \vec{b} |^2 \bigg) \eqexcl \vec{0} -\end{equation} -\begin{equation} -\label{eq:kurtgradientconstraint} - \text{s.t. } \lVert{\vec{b}}\rVert^2 = 1 -\end{equation} -\slidesonly{ -\small{last term changes only length of $\vec{b} \leadsto$ can be removed due to constraint $\lVert{\vec{b}}\rVert^2 = 1$} -} -\notesonly{ -We can omit the last term $3 \vec{b} \, | \vec{b} |^2$ as it only modifies the -length of $\vec b$ which we want to keep equal to 1. -} -\normalsize -\begin{align} -\label{eq:kurtgradientsimple} - \Delta \vec b &\propto 4 \operatorname{sign} \left[ \kurt{(\vec{b}^\top \vec{u})} \right] \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle = \vec{0}\\ - \vec{b} &\leftarrow \vec{b} / \lVert{\vec{b}}\rVert^2 -\end{align} - -\notesonly{ -where normalizing $\vec{b}$ ensures the constraint is satisfied. -We can now describe an implementation of the Kurtosis-based gradient algorithm in its ``batch'' as well as its ``online'' form. -} - -\end{frame} - -\begin{frame} -\slidesonly{ -\frametitle{Kurtosis-based ICA: the gradient algorithm} -} -\begin{block}{I. batch learning:} - Initialization: random vector $\vec{b}$ of unit length - \begin{eqnarray*} - \Delta \vec{b} &=& \varepsilon \operatorname{sign}\left[ \kurt{(\vec{b}^\top \vec{u})} \right] \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle \\ - \vec{b} &\leftarrow& \vec{b} / |\vec{b}| \text{ (normalization to fulfill constraint |\vec{b}| = 1)} - \end{eqnarray*} - - \small - ERM: replace expectations ($\kurt{(\cdot)}$ and $\langle \cdot \rangle$) by their respective empirical averages - \normalsize -\end{block} -\end{frame} - -\begin{frame} -\question{How to do online learning if the gradient requires computing expectations?} - -\slidesonly{ -\begin{align} - \Delta \vec b \; &\propto \; 4 \operatorname{sign} \left[ \kurt{(\vec{b}^\top \vec{u})} \right] \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle = \vec{0}\\ - \vec{b} \;&\leftarrow \;\vec{b} / |{\vec{b}}| -\end{align} - -Recall: - \begin{equation} - \kurt (\vec{b}^\top \vec{u}) = \left\langle \left(\vec{b}^\top \vec{u}\right)^4 \right\rangle - 3 - \end{equation} -} - -\notesonly{ -In order to apply the gradient algorithm in an online fashion as described by \eqref{eq:kurtgradientsimple}, we have to account for the fact that the kurtosis term inside our expression for the gradient involves an expectation operator which cannot be omitted (cf. \eqref{eq:kurt} for how kurtosis is defined). We therefore resort to estimating the kurtosis from a moving average $\gamma$ which starts at zero and is updated at each iteration using: -} -\slidesonly{ -Estimate kurtosis via moving average: -} -\begin{equation} -\label{eq:gammaupdate} -\Delta \gamma = \eta \left[ (\vec{b}^\top \vec{u})^4 -3 - \gamma \right] -\end{equation} -\slidesonly{ -where $\gamma$ is initialized with 0. -} -\end{frame} -\begin{frame} - -\begin{block}{II. online learning:} - Initialization: random vector $\vec{b}$ of unit length, $\gamma = 0$ \\\vspace{0.2cm} - choose a data point $\vec{u}$ - \vspace{-0.2cm} - \begin{eqnarray*} - \Delta \vec{b} &=& \varepsilon \operatorname{sign} (\gamma) \; \vec{u} (\vec{b}^\top \vec{u})^3 \hspace{0.25cm} \quad\;\, \substack{\text{\hspace{0.6mm}(weight update per data point)}} \\ - \Delta \gamma &=& \eta \left[ (\vec{b}^\top \vec{u})^4 -3 - \gamma \right] \hspace{0.25cm} \quad \substack{\text{(running average of the kurtosis with learning rate } \eta )} \\ - \vec{b} &\leftarrow& \vec{b} / |\vec{b}| - \end{eqnarray*} -\end{block} -\end{frame} - -\begin{frame} -\frametitle{The gradient algorithm - advantages and disadvantages} -\textbf{Advantage(s)}: -\pause -\begin{itemize} -\item online learning to adapt to non-stationary data -\end{itemize} -\textbf{Disadvantage(s)}: -\pause -\begin{itemize} -\item dependent on good choice of learning rate and its schedule (i.e. decay over time) -\end{itemize} - -\slidesonly{ -\vspace{2cm} -$\leadsto$ fixed-point iteration alternative -} -\notesonly{ -A fixed-point iteration algorithm provides an alternative to make the learning faster and more reliable without the need for deciding on a learning rate and its sequence. -} -\end{frame} - -\begin{frame} - -\notesonly{ -A stable point of the gradient algorithm is when the gradient points in the same direction of $\vec b$ which leads to not having to update $\vec b$ (i.e. change its direction) any further. This is also the case when the gradient algorithm has converged. We won't go into a rigorous justification of this \footnote{If interested, see Hyv{\"a}rinen Ch. 8.2.3 and Ex 3.9 from the same book.} - -Below is a realization of this faster fixed-point iteration alternative of the Kurtosis-based ICA (The kurtosis-based fastICA should not be confused with the fastICA algorithm we will discussed later.) -} - -\begin{block}{III. fixed-point algorithm (\textbf{kurtosis-based} fastICA)} - fixed point condition of gradient descent: $\vec{b} \propto \Delta \vec{b}$ \\ - $\leadsto \vec{b} \propto \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle - 3 |\vec{b}|^2 \vec{b}$ \\\vspace{0.2cm} - exploiting normalization $(|\vec{b}|^2 = 1)$ we have: - \vspace{-0.3cm} - \begin{eqnarray*} - \vec{b} &\leftarrow& \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle - 3 \vec{b} \\ - \vec{b} &\leftarrow& \vec{b} / |\vec{b}| - \end{eqnarray*} \\ - kurtosis-based fastICA-algorithm for whitened data $\vec{u}^{(\alpha)},\, \alpha = 1, \dots, p$: \\[5pt] - initialization: random vector $\vec{b}$ of unit length, then iterate:\\ - \vspace{-0.6cm} - \begin{eqnarray*} - \vec{b} &\leftarrow& \frac{1}{p} \sum_{\alpha=1}^{p} \vec{u}^{(\alpha)} (\vec{b}^\top \vec{u}^{(\alpha)})^3 - 3 \vec{b} \\ - \vec{b} &\leftarrow& \vec{b} / |\vec{b}| - \end{eqnarray*} \\ -\end{block} -\end{frame} - -\slidesonly{ -\begin{frame} -\frametitle{Summary so far:} -\begin{enumerate} -\item \textcolor{gray}{ -Initial ICA Problem: $\vec x = \vec A\, \vec s$ -} -\item \textcolor{gray}{ -New ICA Problem: $\vec u = \widetilde{\vec A}\, \vec s$,\\ -where $\vec u = \vec D^{-\frac{1}{2}} \vec U^\top \vec x$ and $\vec \Sigma_u = \vec I_N$. -} -\item \textcolor{gray}{ -$\vec u$ is the \emph{whitened} version of $\vec x$. -} -\item \textcolor{gray}{ -$\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. -} -\item \textcolor{gray}{ -Applying ICA on whitened data reduced the number of free parameters. -} -\item \textcolor{gray}{ -PCA simplifies the ICA problem. -} -\item Ambiguities in ICA -\item Why are Gaussians bad for ICA? -\item ICA by maximizing nongaussianity -\item Kurtosis-based ICA - -\end{enumerate} - - -\textbf{Next: Can we do better than kurtosis-based ICA?} - - -\end{frame} -} -\notesonly{ -Next, we will look for an alternative that mitigates the sensitivity to outliers which kurtosis-based ICA is prone to. -} - -\begin{frame} -Kurtosis is easy to compute but can be \emph{sensitive to outliers}. -This is a usual problem with higher-order statistics. -\begin{block}{Example} -\begin{itemize} - \item Sample of 1000 values from a distribution with mean = 0 and std=1 - \item One observation with $x=10$ after sphering: - \itl contribution to kurtosis: $ \geq 10^4/1000 -3 = 7$ -\end{itemize} -\end{block} -\end{frame} - -We therefore turn to an alternate measure for nongaussianity, namely \emph{negentropy} for brevity (not the same as negative entropy $-H(\cdot)$). Negentropy of the reconstructed source $\widehat{\vec s}$ measures the difference between the differential entropy of $\widehat{\vec s}$ and the differential entropy of a Gaussian distribution with the same variance as $\widehat{\vec s}$. - -\newpage - -Negentropy $J(\widehat{s})$ of the reconstructed sources $\widehat{\vec s}$ is defined as: - -\begin{equation} -\label{eq:negentropy} - J(\widehat{s}) \coloneqq H(\widehat{s})_\normal - H(\widehat{s}) -\end{equation} - -where - -\begin{equation} -\label{eq:diffentropyshat} -H(\widehat{s}) := - \int p(\widehat{s}) \log p(\widehat{s}) d\widehat{s} -\end{equation} - -\begin{frame} -\frametitle{Negentropy} -\slidesonly{ -$$ -H(\hat{s}) := - \int p(\hat{s}) \log p(\hat{s}) d\hat{s} \qquad \qquad \text{(differential entropy)} -$$ - -\begin{block}{Definition of negentropy} -\begin{equation*} - J(\widehat{s}) \coloneqq \underbrace{ H(\widehat{s})_\normal}_{ - \substack{ \text{entropy of a Gaussian} \\ - \text{distribution with} \\ - \text{same variance}}} - - \underbrace{ H(\widehat{s}) }_{ - \substack{ \text{entropy of the true} \\ - \text{distribution} \\ - \text{(variance } \sigma^2 \text{)} }} -\end{equation*} -\end{block} -} - -\notesonly{ -The properties that make negentropy suitable: -} - -\begin{itemize} - \itR theoretically well motivated measure. Considered in some cases the optimzal estimator for nongaussianity. - \itR non-negative - \itR scale-invariant: $J(\alpha \widehat{s}) = J(\widehat{s}), \ \ \forall \alpha \ne 0$ (cf. exercise sheet) - \itR \textbf{Problem:} requires estimation of density $p(\widehat{s})$ -\end{itemize} - -\question{Should we minimize or maximize negentropy?} - -\end{frame} - -\subsection{Approximations of negentropy} - -\begin{frame} - -\notesonly{ -Estimating negentropy using the definition in \eqref{eq:negentropy} is computationally costly. It would require estimating the density of the random variable. We therefore resort to simpler approximations for negentropy. Such as the following use of cumulants: -} - -\begin{equation} -\label{eq:negentropyapprox} -J(\hat{s}) \approx \frac{1}{12} \langle (\hat{s})^3 \rangle^2 + \frac{1}{48} (\kurt{(\hat{s}))^2} + \text{higher order terms} -\end{equation} - -\notesonly{ -For symmetric distributions the first term in the approximation in \eqref{eq:negentropyapprox} is effectivley zero, which makes the approximation equivalent to the square of the kurtosis. The approximation would therefore from the same sensitvity to outliers. -} - -\slidesonly{ -$\rightarrow$ for symmetric distributions optimizing this is equivalent to optimizing $|\kurt{(\hat{s})}|$ sharing its outlier sensitivity \\\vspace{0.4cm} -} -The approximation is modified using -``nonpolynomial moments'' contrast functions $G$: - -\begin{equation} - J(\hat{s}) \approx \left( \langle G(\hat{s}) \rangle - \langle G(u_{\text{Gauss}}) \rangle \right)^2 -\end{equation} - -\end{frame} - -\clearpage -\begin{frame}{Common contrast functions} - - \notesonly{ -\textbf{Common contrast functions} - -The contrast function can be chosen depending on the assumed shape of the source densities. - -(e.g. speech: highly super-Gaussian) - -} - \begin{figure} - \centering - \includegraphics[width=4.5cm]{./img/contrast_functions.pdf} - \vspace{-0.5cm} - \caption*{\hspace{5cm}\textit{\tiny{Source: Hyv\"arinen, 2001}}} - \end{figure} - - \begin{equation*} - \smaller - \begin{array}{lll} - G_1(\hat{s}) = \frac{1}{a} \log \cosh (a \cdot \hat{s}) - \;\;& \;\;G_2(\hat{s}) = -\exp \Big( -\frac{(\hat{s})^2}{2} \Big) - & \;\;G_3(\hat{s}) = \frac{1}{4} (\hat{s})^4 - \end{array} - \end{equation*} - Any even, non-constant and non-quadratic (contrast) function $G$ can be used for ICA - - \question{When to choose which contrast function?} - - \end{frame} - -\begin{frame} -\slidesonly{ -\frametitle{Common contrast functions:} -} -\slidesonly{ - \smaller - \begin{tabular}{ccc} - $G_1(\hat{s}) = \frac{1}{a} \log \cosh (a \cdot \hat{s})$ & $G'_1(\hat{s}) = \tanh{(a\hat{s})}$ & $G''_1(\hat{s}) = a (1 - \tanh^2{(a\hat{s})})$\\[7pt] - \multicolumn{3}{c}{general purpose} \\[25pt] - $G_2(\hat{s}) = -\exp \Big( -\frac{(\hat{s})^2}{2} \Big)$ & $G'_2(\hat{s}) = \hat{s} \exp{(-\frac{(\hat{s})^2}{2})}$ & $G''_2(\hat{s}) = (1-(\hat{s})^2) \exp{(-\frac{(\hat{s})^2}{2})}$ \\[7pt] - \multicolumn{3}{c}{good for ``super''-Gaussian sources with many ``outliers''} \\[25pt] - $G_3(\hat{s}) = \frac{1}{4} (\hat{s})^4 $ & $G'_3(\hat{s}) = (\hat{s})^3$ & $G''_3(\hat{s}) = 3(\hat{s})^2$ \\[7pt] - \multicolumn{3}{c}{kurtosis: good for ``sub''-Gaussian sources with few ``outliers''} - \end{tabular} -} -\notesonly{ -\begin{itemize} - \item general purpose: - \begin{itemize} - \item $G_1(\hat{s}) = \frac{1}{a} \log \cosh (a \cdot \hat{s})$ - \item $G'_1(\hat{s}) = \tanh{(a\hat{s})}$ - \item $G''_1(\hat{s}) = a (1 - \tanh^2{(a\hat{s})})$ - \end{itemize} - \item for ``super''-Gaussian sources with many ``outliers'': - \begin{itemize} - \item $G_2(\hat{s}) = -\exp \Big( -\frac{(\hat{s})^2}{2} \Big)$ - \item $G'_2(\hat{s}) = \hat{s} \exp{(-\frac{(\hat{s})^2}{2})}$ - \item $G''_2(\hat{s}) = (1-(\hat{s})^2) \exp{(-\frac{(\hat{s})^2}{2})}$ - \end{itemize} - \item kurtosis: good for ``sub''-Gaussian sources with few ``outliers'': - \begin{itemize} - \item $G_3(\hat{s}) = \frac{1}{4} (\hat{s})^4 $ - \item $G'_3(\hat{s}) = (\hat{s})^3$ - \item $G''_3(\hat{s}) = 3(\hat{s})^2$ - \end{itemize} -\end{itemize} -} -\end{frame} - -\begin{frame} -cf. lecture slides for optmization of negentropy using contrast functions. -\end{frame} - -\begin{frame} -\question{How do we evaluate ICA?}\\ - --cf. https://research.ics.aalto.fi/ica/icasso/ -\end{frame} - diff --git a/notes/06_fastica/2_pcaica.tex b/notes/06_fastica/2_pcaica.tex new file mode 100644 index 0000000..88c42b2 --- /dev/null +++ b/notes/06_fastica/2_pcaica.tex @@ -0,0 +1,191 @@ +\subsection{Whitening in the context of ICA} + +\begin{frame} + +\begin{align} +\label{eq:expxina} +\vec \Sigma_x = \mathrm{Cov}(\vec x) &= \E \lbrack \, \vec x \, \vec x^\top \rbrack\\ +&= \E \lbrack \, \vec A\,\vec s \, \left( \vec A\,\vec s \right)^\top \rbrack\\ +&= \E \lbrack \, \vec A\; \underbrace{\;\vec s \, \vec s^\top}_{= \vec I_N} \vec A^\top \rbrack\\ +\label{eq:sigmax} +&= \vec A\, \vec A^\top +\end{align} + +\end{frame} + +\begin{frame} + +\notesonly{ +If we were to apply \emph{Singular Value Decomposition} on the symmetric matrix $\Sigma_x$: +} + +\begin{equation} +\mathrm{SVD}(\vec \Sigma_x) = \vec U\, \vec D \, \vec U^\top +\end{equation} + +where\\ +\notesonly{ +$\vec U$ is an orthogonal matrix,} $\vec U^\top\vec U = \vec I_N$ and +\notesonly{$\vec D$ is a diagonal matrix with rank $N$, }$\vec D^\top = \vec D$.\\ + +We define the two following transformation $\vec Q$ and $\widetilde{\vec A}$: +\begin{equation} +\label{eq:defq} +\vec Q := \vec D^{-\frac{1}{2}} \vec U^\top +\end{equation} +\begin{equation} +\label{eq:defatilde} +\widetilde{\vec A} := \vec Q \, \vec A +\end{equation} + +Which yields a new ICA problem: +\begin{equation} +\vec u = \vec Q\, \vec x = \vec Q\,\vec A \, \vec s = \widetilde{\vec A} \, \vec s +\end{equation} + +\question{What is so special about the new mixing matrix $\widetilde{\vec A}$?} \\ + +\notesonly{-$\widetilde{\vec A}$ is orthogonal (i.e. $\widetilde{\vec A}\, \widetilde{\vec A}^\top = \widetilde{\vec A}^\top \widetilde{\vec A} = \widetilde{\vec A}^{-1} \widetilde{\vec A} = \vec I_N$). + We will see the implications of this in how it reduces the number of free parameters for solving the new ICA problem. + } +\end{frame} + +\begin{frame} + +\slidesonly{ +$$ +\vec Q := \vec D^{-\frac{1}{2}} \vec U^\top +$$ +$$ +\vec \Sigma_x = \E \lbrack\, \vec x \, \vec x^\top \rbrack = \vec U \, \vec D \, \vec U^{\top} +$$ +} + +\begin{align} +\E \lbrack\, \vec u \, \vec u^\top \rbrack +&\mystackrel{\vec u = \vec Q\, \vec x}{\vec u = \vec Q\, \vec x} + \E \lbrack\, \vec Q \, \vec x \left( \vec Q \, \vec x\right)^\top \rbrack\\ +&\mystackrel{\vec u = \vec Q\, \vec x}{}\E \lbrack\, \vec Q \, \vec x \; \vec x^\top \vec Q^\top \rbrack\\ +&\mystackrel{\vec u = \vec Q\, \vec x}{} \E \lbrack\, \vec Q \, \vec \Sigma_{X} \, \vec Q^\top \rbrack\\ +&\mystackrel{\vec u = \vec Q\, \vec x}{} \vec Q \, \vec \Sigma_{x} \, \vec Q^\top\\ +&\mystackrel{\vec u = \vec Q\, \vec x}{} +\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) +\left( \vec U \, \vec D \, \vec U^\top \right) +\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) +\end{align} + +\end{frame} +\begin{frame} + +\slidesonly{ +\begin{align} +\E \lbrack\, \vec u \, \vec u^\top \rbrack +&\mystackrel{\vec u = \vec Q\, \vec x}{} +\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) +\left( \vec U \, \vec D \, \vec U^\top \right) +\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) +\end{align} +} + +From $\vec U^\top\vec U = \vec I_N$ and $\vec D^\top = \vec D$ follows: + +\begin{align} +\E \lbrack\, \vec u \, \vec u^\top \rbrack +&= +\vec D^{-\frac{1}{2}} +\underbrace{\left( \vec U^\top \vec U \right)}_{= \vec I_N} +\, \vec D \, +\underbrace{\left( \vec U^\top \, \vec U \right)}_{= \vec I_N} +\, +\left(\vec D^{-\frac{1}{2}} \right)^\top \\ +&= \vec D^\top +\, +\underbrace{\vec D^{-\frac{1}{2}} \, +\left(\vec D^{-\frac{1}{2}} \right)^\top}_{=\vec D^{-1}} = \vec I_N +\end{align} + + +Recall +\notesonly{from \eqref{eq:sigmax}:} +$$ +\vec \Sigma_x = \E \lbrack \, \vec x \, \vec x^\top \rbrack += \vec A\, \vec A^\top +$$ +Therefore: +$$ +\E \lbrack \, \vec u \, \vec u^\top \rbrack += \widetilde {\vec A}\, \widetilde {\vec A}^\top = \vec I_N +$$ + +\notesonly{ +The new mixing matrix} +$\widetilde{\vec A}$ is orthonormal.\\ +The space of solutions is restricted to $\vec W$ that are orthogonal. +The can speed up convergence of ICA. +The number of free parameters +\notesonly{ +for solving the new ICA problem +$\vec u = \widetilde{\vec A}\, \vec s$ +is reduced from $N^2$ to $N(N-1)/2$ +} +\slidesonly{ +$N^2 \rightarrow N(N-1)/2$ +} +\end{frame} + +\begin{frame} +\slidesonly{ +\frametitle{What just happened?} + +\begin{itemize} +\setlength\itemsep{1em} +\item We've transformed $\vec x$ to get $\vec u$: +$ \qquad \qquad \qquad +\vec u := \vec D^{-\frac{1}{2}} \vec U^\top \vec x +$ + +\item Where did $\vec D$ and $\vec U$ come from? +$ \qquad \qquad +\mathrm{SVD}(\vec \Sigma_x) = \vec U\, \vec D \, \vec U^\top +$ + +\item What do $\vec D$ and $\vec U$ represent? - the eigenval. and eigenvect. of $\vec \Sigma_x$ + +\item What's so special about $\vec u$? +$ \qquad \qquad \qquad \qquad +\vec \Sigma_u = \vec I_N +$ + +\item ...So? + +- new ICA problem: $\vec u := \widetilde{\vec A}\, \vec s$\\ +- $\widetilde{\vec A}$ is orthogonal\\ +- unmixing the new problem involves only ``half'' the number of weights. + +\question{What is the transformation $\vec D^{-\frac{1}{2}} \vec U^\top$ called?} + +\end{itemize} +} + +\end{frame} + + +\section{Summary so far:} + +\slidesonly{ +\begin{frame} +\begin{enumerate} +\item Initial ICA Problem: $\vec x = \vec A\, \vec s$ +\item New ICA Problem: $\vec u := \widetilde{\vec A}\, \vec s$,\\ + +where $\vec u = \vec D^{-\frac{1}{2}} \vec U^\top \vec x$ and $\vec \Sigma_u = \vec I_N$. +\item $\vec u$ is the \emph{whitened} version of $\vec x$. +\item $\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. +\item Applying ICA on whitened data reduces the number of free parameters. +\item PCA simplifies the ICA problem. +\item ICA on whitened data is about ``rotating'' it back. +\end{enumerate} + +\end{frame} +} + diff --git a/notes/06_fastica/3_badgaussians.tex b/notes/06_fastica/3_badgaussians.tex new file mode 100644 index 0000000..3a7ec97 --- /dev/null +++ b/notes/06_fastica/3_badgaussians.tex @@ -0,0 +1,162 @@ + + +\begin{frame} +\slidesonly{ +\frametitle{Gaussians are bad for ICA} +} +\newpage +\textbf{A more formal argument for why Gaussians are bad for ICA}: + +Recall that the joint density of independent sources is a factorizing density: + +\begin{equation} +\label{eq:facts} +P_{\vec s}(\vec s) = \prod_{i=1}^{N} P_{s_i}(s_i) \,. +\end{equation} + +If we assumed gaussian distributed sources, the following factorization becomes possible: + +\begin{equation} + \label{eq:factsgauss} + \begin{array}{ll} + P_{\vec{s}}({\vec{s}}) + & = \frac{1}{2\pi} + \exp \left( -\frac{\lVert{\vec{s}}\rVert^2}{2} \right) + \\ + & = \underbrace{ + \Bigg[ \frac{1}{\sqrt{2\pi}} \exp \left( - + \frac{{s_1}^2}{2} \right) \Bigg] + }_{{P}_{s_1}({s}_1)} + \underbrace{\Bigg[ + \frac{1}{\sqrt{2\pi}} \exp \left( - + \frac{{s_2}^2}{2} \right) + \Bigg] + }_{{P}_{s_2}({s}_2)} + \end{array} +\end{equation} +\end{frame} + +\begin{frame} +\slidesonly{ +\frametitle{Gaussians are bad for ICA} +} +\slidesonly{\textbf{A more formal argument (cont'd):}} + +Now consider applying an orthognal mixing matrix that is \textbf{known}. +\slidesonly{(orthogonal from whitening the $\vec x$)\\ +Consequently: +} +\notesonly{We've seen how whitening takes an ICA problem with any valid\footnote{invertible} mixing matrix $A$ +and reformulates it into a new problem with an orthogonal mixing matrix $\widetilde{\vec A}$. +The following holds for such an orthogonal mixing matrix $\widetilde{\vec A}$: +} +$$ +\widetilde{\vec A}^\top = \widetilde{\vec A}^{-1} \Leftrightarrow \widetilde{\vec A}^{-1}\widetilde{\vec A}=\vec I_N \qquad \text{and} \qquad |\det \widetilde{\vec A}| = |\det \widetilde{\vec A}^\top| = 1 +$$ + +$$ +\vec s = \widetilde{\vec A}^{-1} \vec{x} = \widetilde{\vec A}^\top \vec x +$$ + +\notesonly{ +We are now interested in the joint density $P_x(\vec x)$ of the mixtures $x_1$ and $x_2$. +} + +Density transformation tells us that: +\begin{equation} + \label{eq:gausstransformeddt} + {P}_{\vec x}(\vec x) = + {P}_{\vec s}(\vec s) \left|\det \frac{\partial \vec s}{\partial \vec x}\right| + = {P}_{\vec s}(\vec s) \left|\det \widetilde{\vec A}^{-1}\right| + = {P}_{\vec s}(\vec s) \left|\det \widetilde{\vec A}^\top\right| +\end{equation} +Therefore, +\slidesonly{ +\begin{equation} + {P}_{\vec x}(\vec x) + = \frac{1}{2\pi} + \exp \left( + -\frac{\lVert{\widetilde{\vec A}^\top \vec x}\rVert^2}{2} + \right) + \left|\det \widetilde{\vec A}^\top\right| + = \ldots? +\end{equation} +} +\end{frame} + +\begin{frame} +\slidesonly{ +Orthogonal mixing matrix $\widetilde{\vec A}$ implies: +$$ +\widetilde{\vec A}^\top = \widetilde{\vec A}^{-1} \Leftrightarrow \widetilde{\vec A}^{-1}\widetilde{\vec A}=\vec I_N \qquad \text{and} \qquad |\det \widetilde{\vec A}| = |\det \widetilde{\vec A}^\top| = 1 +$$ +$$ +\vec s = \widetilde{\vec A}^{-1} \vec{x} = \widetilde{\vec A}^\top \vec x +$$} +\begin{align} + \label{eq:gaussx} + {P}_{\vec x}(\vec x) + &= \frac{1}{2\pi} + \exp + \Big( + -\frac{ + \overbrace{ + \lVert{\widetilde{\vec A}^\top \vec x}\rVert^2 + }^{\mathclap{ + \lVert{\widetilde{\vec A}^\top \vec x}\rVert^2 + = \left(\widetilde{\vec A}^\top \vec x\right)^\top \widetilde{\vec A}^\top \vec x + \,=\, \vec x^\top \widetilde{\vec A} \, \widetilde{\vec A}^\top \vec x + = \vec x^\top \vec x + = \lVert{\vec x}\rVert^2 + }}}{2} + \Big) + \underbrace{ + \left|\det \widetilde{\vec A}^\top\right| + }_{=\, 1} + \\ + &= \frac{1}{2\pi} + \exp \left( + -\frac{\lVert{\vec x}\rVert^2}{2} + \right)\\ + \label{eq:gaussxfact} + & = \underbrace{ + \Bigg[ \frac{1}{\sqrt{2\pi}} \exp \left( - + \frac{{x_1}^2}{2} \right) \Bigg] + }_{{P}_{x_1}({x}_1)={P}_{s_1}({s}_1)} + \underbrace{\Bigg[ + \frac{1}{\sqrt{2\pi}} \exp \left( - + \frac{{x_2}^2}{2} \right) + \Bigg] + }_{{P}_{x_2}({x}_2)={P}_{s_2}({s}_2)} + \text{no change in pdf!} +\end{align} +\end{frame} + +We see that the factorization in \eqref{eq:gaussxfact} describes the pdf for the mixtures $\vec x$ identically to the pdf of the original sources. +The original and mixed distributions are identical. The mixing matrix did not change this. Therefore, it would be impossible to find its corresponding unmixing matrix to undo the rotation. + +\begin{frame} +\slidesonly{ +\frametitle{Gaussians are bad for ICA} +} + +\notesonly{ +Mixing two independent Gaussians leads to a joint mixed distribution that is equal to that of the original sources. +This is actually justified by the property that \emph{ +uncorrelated jointly Gaussian variables are necessarily independent.}\footnote{ +Further details can be found in Hyv{\"a}rinen Ch. 2.5. +} +} + +A mixture of sources where at most one is Gaussian, is still fine. It only becomes a problem when we have more than one. + + +\slidesonly{ +\begin{itemize} +\item Mixing two independent Gaussians leads to a joint mixed distribution that is equal to that of the original sources. +\item No surprise: \emph{uncorrelated jointly Gaussian variables are necessarily independent.} +\item One Gaussian + other distributions is fine. +\item Two ore more Gaussians. No way. +\end{itemize} +} +\end{frame} diff --git a/notes/06_fastica/4_kurt.tex b/notes/06_fastica/4_kurt.tex new file mode 100644 index 0000000..67effb2 --- /dev/null +++ b/notes/06_fastica/4_kurt.tex @@ -0,0 +1,488 @@ +\begin{frame} +\frametitle{Maximizing nongaussianity leads to independent sources} + +\textbf{The setting:}\\ + +Two statistically independent sources with $\langle s_i s_j \rangle = \delta_{ij} \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N$ +\notesonly{ +The sources are mixed using a mixing matrix $\vec A$ resulting in observations $\vec x$: +} +\begin{equation} +\label{eq:icaproblemx} +\vec x = \vec A \, \vec s +\end{equation} +\notesonly{ +One out of the $N$ independent sources can be reconstructed using a row vector from an unmixing matrix $\vec W$: +} +\begin{equation} +\label{eq:singlesource} +\widehat s_i = \vec w_i^\top \vec x +\end{equation} +\notesonly{ +By substituting \eqref{eq:icaproblemx} for $\vec x$ in \eqref{eq:singlesource}, +we describe each reconstructed source $\widehat s_i$ as a linear combination of the original sources: +} + + +\notesonly{According to the CLT we can think of the variables in $\vec x$ to be more Gaussian distributed than the original variables in $\vec s$. +Therefore, a solution to the ICA problem is finding an inverse to $\vec A$ that undoes this effect and removes the ``accumulated Gaussianity'' from $\vec x$. +The role of any $\vec w_i$ becomes to maximize the nongaussianity of $\widehat{s}_i$ when we multiply it by $\vec x$. This is the same role $w_i^\top \vec A = \vec z_i^\top$ has when applied to $\vec s$. + + +} + + +\begin{equation} +\label{eq:szs} +\widehat s_i = \vec w_i^\top \vec A \, \vec s = \vec z_i^{\top} \vec s = + \left( \begin{array}{ll} + {z}_1 \\ {z}_2 + \end{array} \right)^\top_i + \left( \begin{array}{ll} + {s}_1 \\ {s}_2 + \end{array} \right) += z_1 s_1 + z_2 s_2 +\end{equation} + +\notesonly{ +Looking at \eqref{eq:szs} we recognize that $\vec z_i$ describes how to route the information in $\vec s$ such that $\widehat{\vec s}_i$ fully describes +one of the independent sources in $\vec s$. +This can be accomplished with a vector containing a single non-zero element: +} + +\begin{equation*} + \vec{z}_{\text{opt.}} = \left( \begin{array}{c} + 0 \\ \pm 1 + \end{array} \right) + \quad \text{ or }\quad + \vec{z}_{\text{opt.}} = \left( \begin{array}{c} + \pm 1 \\ 0 + \end{array} \right) +\end{equation*} + +\end{frame} + +\notesonly{ +Recall that ICA cannot resolve scale or permutation of the sources and thirdly it cannot resolve the sign. +This is not an issue. +The role of $\vec z_i$ is to route either $s_1$ or $s_2$ to $\widehat{\vec s}_i$. This covers the ambiguitiy in terms of permutation. +We cannot have both independent sources contribute to $\widehat{s}_i$, only one can. Therefore, we only need a single non-zero component for $\vec z_i$. +Wether $s_1$ is scaled by any factor before reaching $\widehat{s}_i$ does not make it more or less independent of $s_2$. Choosing $1$ for the non-zero component is therefore sufficient. +Finally, negating the source by multiplying it by $(-1)$ also has no consequences on the independence criterion. + +We won't actually try and find $\vec z_i$ because we don't have $\vec s$ to apply them to. We use the requirements for $\vec z_i$ by finding a $\vec w_i$ that satisfies these requirements through: +\begin{equation} +\label{eq:zfromw} +\vec z_i = \left(\vec w_i^\top \vec A\right)^\top = \vec A^\top \vec w_i +\end{equation} +} + +\begin{frame} +\question{Does maximizing nongaussianity deliver independent components?} +\slidesonly{ + +\begin{equation} +\widehat s_i = \vec z_i^{\top} \vec s = z_1 s_1 + z_2 s_2 +\end{equation} + +\begin{equation*} + \vec{z}_{\text{opt.}} = \left( \begin{array}{c} + 0 \\ \pm 1 + \end{array} \right) + \quad \text{ or }\quad + \vec{z}_{\text{opt.}} = \left( \begin{array}{c} + \pm 1 \\ 0 + \end{array} \right) +\end{equation*} + +$\vec z_i$ ensures independent $\widehat s_i$ + +\begin{equation} +\widehat s_i = \vec w_i^\top \vec x +\end{equation} + +$\vec w_i$ finds less gaussian $\widehat s_i$ + +\begin{equation} +\vec z_i = \vec A^\top \vec w_i +\end{equation} + +Maximizing nongaussianity is ensured to keep $\widehat s_i$ independent. + +} +\end{frame} +\notesonly{ +- By (1) maximizing the nongaussianity of $\vec w_i^\top \vec x$ and (2) having $\vec z_i = \vec A^\top \vec w_i$ yield independent components and (3) knowing that +$\widehat s_i = \vec w_i^\top \vec x = \vec z_i^{\top} \vec s$, we conclude that maximizing $\vec w_i^\top \vec x$ gives us one independent component. +} +\newpage + +\section{Kurtosis as a measure for nongaussianity} + +\begin{frame} + +\notesonly{ +Kurtosis represents the fourth-order cumulant\footnote{ +Cumulants allow us to express the i-th moment in terms of a cumulative sum of the moments preceeding it. +This simplifies the expression of higher-order moments such as kurtosis which is the fourth-order moment. +} of a random variable. +} + +\begin{block}{Definition} + Let $x$ be a random variable with zero-mean, i.e. $\E \lbrack\,x\,\rbrack = 0$. + \begin{equation} + \label{eq:kurt} + \kurt (x) = \langle x^4 \rangle - 3 + \left( \langle x^2 \rangle \right)^2 \quad + \stackrel{\text{sphered data}}{=} \quad \langle x^4 \rangle - 3 + \end{equation} + + \notesonly{ + By assuming zero-mean and unit-variance, we see that kurtosis is simply a normalized version of the fourth moment. + + Useful properties of kurtosis: + + } + Let $x_1$ and $x_2$ be two independent random variables, then: + \begin{eqnarray*} + \kurt(x_1 + x_2)& = & \kurt(x_1) + \kurt(x_2) \\ + \kurt(z_1 x_1) & = & z_1^4 \kurt(x_1) + \end{eqnarray*} +\end{block} +\end{frame} + +\begin{frame} + +\question{What does kurtosis measure?}\\ + +\begin{tabular}[h]{c c c c} +& +\includegraphics[width=2.7cm]{img/section2_fig20} & +\includegraphics[width=2.7cm]{img/section2_fig21} & +\includegraphics[width=2.7cm]{img/section2_fig22} \\ \\ + +& $\kurt(x) = 0$ & $\kurt(x) > 0$ & $\kurt(x) < 0$\\ \\ + +& +Gaussian PDF & +super-Gaussian PDF& +sub-Gaussian PDF\\ +&bell shaped & peaky, long tails (``outliers'')& bulky, no ``outliers'' \\\\ +e.g.&normal & Laplace & uniform +\end{tabular}\\[1cm] + +\notesonly{This implies that we can use} +$|kurt(x)| > 0$ as a measure of nongaussianity + +\textbf{Caveat:} +\slidesonly{ +sensitive to outliers +} +\end{frame} + + +\notesonly{ +Kurtosis, just like other higher-order cumulants are sensitive to outliers, in that an outlier + will register a much higher kurtosis value. This will be addressed later by FastICA. +} + +\clearpage + +\subsection{kurtosis-based ICA} + +\begin{frame} + +Two statistically independent sources with + +$\langle s_i s_j \rangle = \delta_{ij} \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N$ (any scaling can be attributed to $\vec A$) + +\begin{equation*} +\widehat{s}_i \quad += \quad \vec{W}^\top \vec{x} \quad += \quad \vec{W}^\top \vec{A} \cdot \vec{s} \quad += \quad \vec{z}^\top \vec{s} \quad += \quad z_1 s_1 + z_2 s_2 +\end{equation*} +\vspace{1mm} +We want the covariance of our reconstructions to match that of the original sources. +\begin{equation*} +\langle \widehat{\vec s} \, \widehat{\vec s}^\top \rangle \eqexcl \langle \vec s \, \vec s^\top \rangle = \vec I_N +\end{equation*} +This implies, +\begin{align*} +\var(\widehat{s}_i) + \; &= \; \langle \big( z_1 s_1 + z_2 s_2 \big)^2 \rangle_{P_{\vec s}}\\ + \; &= \; \langle z_1^2 \, s_1^2 \rangle \;+\; 2 \, \langle z_1\, s_1\, z_2 \, s_2 \rangle \;+\; \langle z_2^2 \, s_2^2 \rangle \\ + \; &= \; z_1^2 \, \langle s_1^2 \rangle \;+\; 2 \, z_1\, z_2 \, \underbrace{\langle s_1\, s_2 \rangle}_{= 0} \;+\; z_2^2 \, \langle s_2^2 \rangle \\ + \; &= \; z_1^2 \, \langle s_1^2 \rangle \;+\; z_2^2 \,\langle s_2^2 \rangle \\ + \; &= \; z_1^2 + z_2^2 \eqexcl 1 +\end{align*} +Making the constraint of unit variance for $\widehat{s}_i$ is to match the variance assumed for the orgiinal sources $s_1$ and $s_2$. This implies that solutions for $\vec z$ are constrained to lie on a unit circle. +\vspace{1mm} +\begin{align*} +\kurt(\widehat{s}) \;\; &= \;\; \kurt(z_1 s_1 + z_2 s_2) \;\; \\ &= \;\; \kurt(z_1 s_1) + \kurt(z_2 s_2) \; = \; z_1^4 \kurt(s_1) + z_2^4 \kurt(s_2) +\end{align*} + +\begin{block}{Kurtosis-based optimization problem} +\begin{equation*} + \begin{array}{rllc} + \left| \kurt(\widehat{s}) \right| & \eqexcl \max_{\vec{z}} + & \leftarrow & \substack{ \text{search for the direction} \\ + \text{with extreme kurtosis}} \\\\ + \var(\widehat{s}) = z_1^2 + z_2^2 & \eqexcl 1 + & \leftarrow & \substack{ \text{such that the data} \\ + \text{remains sphered}} + \end{array} +\end{equation*} + \end{block} +\end{frame} + +\begin{frame} +\slidesonly{ +\begin{block}{Kurtosis-based optimization problem} +\begin{equation*} + \begin{array}{rllc} + \left| \kurt(\widehat{s}) \right| & \eqexcl \max_{\vec{z}} + & \leftarrow & \substack{ \text{search for the direction} \\ + \text{with extreme kurtosis}} \\\\ + \var(\widehat{s}) = z_1^2 + z_2^2 & \eqexcl 1 + & \leftarrow & \substack{ \text{such that the data} \\ + \text{remains sphered}} + \end{array} +\end{equation*} + \end{block} +} + + +\begin{center} +$\hat{s} \,=\, \vec{z}^\top \vec{s} \,=\, \vec{w}^\top \underbrace{\vec{A} \, \vec{s}}_{\vec{x}} \,=\, \vec{b}^\top \underbrace{\tilde{\vec{A}} \, \vec{s}}_{\vec{u}}$ \hspace{4cm} $\kurt(s_i) \neq 0$ +\end{center} + +\question{What can we optimize kurtosis with?} + +\pause + \begin{enumerate} + \item $\max_{\vec{z}} | \kurt{(\vec{z}^\top \vec{s})} | \; \; \;\quad s.t. \quad |\vec{z}| = 1$ + \item $\max_{\vec{w}} | \kurt{(\vec{w}^\top \vec{x})} | \quad s.t. \quad |\vec{A}^\top \vec{w}| = 1$ + \item $ + \max_{\vec{b}} | \kurt{(\vec{b}^\top \vec{u})} | \;\quad s.t. \quad + \underbrace{|\tilde{\vec{A}}^\top \vec{b}| + }_{\substack{= |\vec{b}| \\ \text{ since } \tilde{\vec{A}} \\ \text{ is orthogonal}}} = 1 + $ + \end{enumerate} +\notesonly{ +The different maximization approaches are equivalent. We opt for maximizing +$| \kurt{(\vec{b}^\top \vec{u})} |$ +because it is the only which we can obtain. The other terms either require access to the $\vec s$ or $\vec A$ which is not possible. +Whitening $\vec x$ yields $\vec u$. We also do not know the orthogonal unmixing matrix $\widetilde{\vec A}$ but this +is not an issue because the orthogonality of $\widetilde{\vec A}$ lets the constraint +reduce to only ensuring that $\vec b$ is kept at unit length. +} + +\newpage + +\end{frame} + +\subsection{Kurtosis-based ICA: the gradient algorithm} + +\begin{frame} + +\notesonly{ +$| \kurt{(\vec{b}^\top \vec{u})} |$ can be maximized by moving $\vec b$ +in the direction of the gradient until this becomes zero, whilst keeping the length of $\vec b$ equal to 1. +} + +\begin{equation} +\label{eq:kurtgradient} + \frac{\partial |\text{kurt}(\vec b^\top \vec u)|}{\partial \vec{b}} + = 4 \operatorname{sign} \left[ \kurt{(\vec{b}^\top \vec{u})} \right] \bigg( \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle - 3 \vec{b} \, | \vec{b} |^2 \bigg) \eqexcl \vec{0} +\end{equation} +\begin{equation} +\label{eq:kurtgradientconstraint} + \text{s.t. } \lVert{\vec{b}}\rVert^2 = 1 +\end{equation} +\slidesonly{ +\small{last term changes only length of $\vec{b} \leadsto$ can be removed due to constraint $\lVert{\vec{b}}\rVert^2 = 1$} +} +\notesonly{ +We can omit the last term $3 \vec{b} \, | \vec{b} |^2$ as it only modifies the +length of $\vec b$ which we want to keep equal to 1. +} +\normalsize +\begin{align} +\label{eq:kurtgradientsimple} + \Delta \vec b &\propto 4 \operatorname{sign} \left[ \kurt{(\vec{b}^\top \vec{u})} \right] \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle = \vec{0}\\ + \vec{b} &\leftarrow \vec{b} / \lVert{\vec{b}}\rVert^2 +\end{align} + +\notesonly{ +where normalizing $\vec{b}$ ensures the constraint is satisfied. +We can now describe an implementation of the Kurtosis-based gradient algorithm in its ``batch'' as well as its ``online'' form. +} + +\end{frame} + +\begin{frame} +\slidesonly{ +\frametitle{Kurtosis-based ICA: the gradient algorithm} +} +\begin{block}{I. batch learning:} + Initialization: random vector $\vec{b}$ of unit length + \begin{eqnarray*} + \Delta \vec{b} &=& \varepsilon \operatorname{sign}\left[ \kurt{(\vec{b}^\top \vec{u})} \right] \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle \\ + \vec{b} &\leftarrow& \vec{b} / |\vec{b}| \text{ (normalization to fulfill constraint |\vec{b}| = 1)} + \end{eqnarray*} + + \small + ERM: replace expectations ($\kurt{(\cdot)}$ and $\langle \cdot \rangle$) by their respective empirical averages + \normalsize +\end{block} +\end{frame} + +\begin{frame} +\question{How to do online learning if the gradient requires computing expectations?} + +\slidesonly{ +\begin{align} + \Delta \vec b \; &\propto \; 4 \operatorname{sign} \left[ \kurt{(\vec{b}^\top \vec{u})} \right] \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle = \vec{0}\\ + \vec{b} \;&\leftarrow \;\vec{b} / |{\vec{b}}| +\end{align} + +Recall: + \begin{equation} + \kurt (\vec{b}^\top \vec{u}) = \left\langle \left(\vec{b}^\top \vec{u}\right)^4 \right\rangle - 3 + \end{equation} +} + +\notesonly{ +In order to apply the gradient algorithm in an online fashion as described by \eqref{eq:kurtgradientsimple}, we have to account for the fact that the kurtosis term inside our expression for the gradient involves an expectation operator which cannot be omitted (cf. \eqref{eq:kurt} for how kurtosis is defined). We therefore resort to estimating the kurtosis from a moving average $\gamma$ which starts at zero and is updated at each iteration using: +} +\slidesonly{ +Estimate kurtosis via moving average: +} +\begin{equation} +\label{eq:gammaupdate} +\Delta \gamma = \eta \left[ (\vec{b}^\top \vec{u})^4 -3 - \gamma \right] +\end{equation} +\slidesonly{ +where $\gamma$ is initialized with 0. +} +\end{frame} +\begin{frame} + +\begin{block}{II. online learning:} + Initialization: random vector $\vec{b}$ of unit length, $\gamma = 0$ \\\vspace{0.2cm} + choose a data point $\vec{u}$ + \vspace{-0.2cm} + \begin{eqnarray*} + \Delta \vec{b} &=& \varepsilon \operatorname{sign} (\gamma) \; \vec{u} (\vec{b}^\top \vec{u})^3 \hspace{0.25cm} \quad\;\, \substack{\text{\hspace{0.6mm}(weight update per data point)}} \\ + \Delta \gamma &=& \eta \left[ (\vec{b}^\top \vec{u})^4 -3 - \gamma \right] \hspace{0.25cm} \quad \substack{\text{(running average of the kurtosis with learning rate } \eta )} \\ + \vec{b} &\leftarrow& \vec{b} / |\vec{b}| + \end{eqnarray*} +\end{block} +\end{frame} + +\begin{frame} +\frametitle{The gradient algorithm - advantages and disadvantages} +\textbf{Advantage(s)}: +\pause +\begin{itemize} +\item online learning to adapt to non-stationary data +\end{itemize} +\textbf{Disadvantage(s)}: +\pause +\begin{itemize} +\item dependent on good choice of learning rate and its schedule (i.e. decay over time) +\end{itemize} + +\slidesonly{ +\vspace{2cm} +$\leadsto$ fixed-point iteration alternative +} +\notesonly{ +A fixed-point iteration algorithm provides an alternative to make the learning faster and more reliable without the need for deciding on a learning rate and its sequence. +} +\end{frame} + +\begin{frame} + +\notesonly{ +A stable point of the gradient algorithm is when the gradient points in the same direction of $\vec b$ which leads to not having to update $\vec b$ (i.e. change its direction) any further. This is also the case when the gradient algorithm has converged. We won't go into a rigorous justification of this \footnote{If interested, see Hyv{\"a}rinen Ch. 8.2.3 and Ex 3.9 from the same book.} + +Below is a realization of this faster fixed-point iteration alternative of the Kurtosis-based ICA (The kurtosis-based fastICA should not be confused with the fastICA algorithm we will discussed later.) +} + +\begin{block}{III. fixed-point algorithm (\textbf{kurtosis-based} fastICA)} + fixed point condition of gradient descent: $\vec{b} \propto \Delta \vec{b}$ \\ + $\leadsto \vec{b} \propto \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle - 3 |\vec{b}|^2 \vec{b}$ \\\vspace{0.2cm} + exploiting normalization $(|\vec{b}|^2 = 1)$ we have: + \vspace{-0.3cm} + \begin{eqnarray*} + \vec{b} &\leftarrow& \langle \vec{u} (\vec{b}^\top \vec{u})^3 \rangle - 3 \vec{b} \\ + \vec{b} &\leftarrow& \vec{b} / |\vec{b}| + \end{eqnarray*} \\ + kurtosis-based fastICA-algorithm for whitened data $\vec{u}^{(\alpha)},\, \alpha = 1, \dots, p$: \\[5pt] + initialization: random vector $\vec{b}$ of unit length, then iterate:\\ + \vspace{-0.6cm} + \begin{eqnarray*} + \vec{b} &\leftarrow& \frac{1}{p} \sum_{\alpha=1}^{p} \vec{u}^{(\alpha)} (\vec{b}^\top \vec{u}^{(\alpha)})^3 - 3 \vec{b} \\ + \vec{b} &\leftarrow& \vec{b} / |\vec{b}| + \end{eqnarray*} \\ +\end{block} +\end{frame} + +\slidesonly{ +\begin{frame} +\frametitle{Summary so far:} +\begin{enumerate} +\item \textcolor{gray}{ +Initial ICA Problem: $\vec x = \vec A\, \vec s$ +} +\item \textcolor{gray}{ +New ICA Problem: $\vec u = \widetilde{\vec A}\, \vec s$,\\ +where $\vec u = \vec D^{-\frac{1}{2}} \vec U^\top \vec x$ and $\vec \Sigma_u = \vec I_N$. +} +\item \textcolor{gray}{ +$\vec u$ is the \emph{whitened} version of $\vec x$. +} +\item \textcolor{gray}{ +$\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. +} +\item \textcolor{gray}{ +Applying ICA on whitened data reduced the number of free parameters. +} +\item \textcolor{gray}{ +PCA simplifies the ICA problem. +} +\item Ambiguities in ICA +\item Why are Gaussians bad for ICA? +\item ICA by maximizing nongaussianity +\item Kurtosis-based ICA + +\end{enumerate} + + +\textbf{Next: Can we do better than kurtosis-based ICA?} + + +\end{frame} +} +\notesonly{ +Next, we will look for an alternative that mitigates the sensitivity to outliers which kurtosis-based ICA is prone to. +} + +\begin{frame} +Kurtosis is easy to compute but can be \emph{sensitive to outliers}. +This is a usual problem with higher-order statistics. +\begin{block}{Example} +\begin{itemize} + \item Sample of 1000 values from a distribution with mean = 0 and std=1 + \item One observation with $x=10$ after sphering: + \itl contribution to kurtosis: $ \geq 10^4/1000 -3 = 7$ +\end{itemize} +\end{block} +\end{frame} + +We therefore turn to an alternate measure for nongaussianity, namely \emph{negentropy} for brevity (not the same as negative entropy $-H(\cdot)$). Negentropy of the reconstructed source $\widehat{\vec s}$ measures the difference between the differential entropy of $\widehat{\vec s}$ and the differential entropy of a Gaussian distribution with the same variance as $\widehat{\vec s}$. + + diff --git a/notes/06_fastica/5_fastica.tex b/notes/06_fastica/5_fastica.tex new file mode 100644 index 0000000..73a720e --- /dev/null +++ b/notes/06_fastica/5_fastica.tex @@ -0,0 +1,161 @@ + +Negentropy $J(\widehat{s})$ of the reconstructed sources $\widehat{\vec s}$ is defined as: + +\begin{equation} +\label{eq:negentropy} + J(\widehat{s}) \coloneqq H(\widehat{s})_\normal - H(\widehat{s}) +\end{equation} + +where + +\begin{equation} +\label{eq:diffentropyshat} +H(\widehat{s}) := - \int p(\widehat{s}) \log p(\widehat{s}) d\widehat{s} +\end{equation} + +\begin{frame} +\frametitle{Negentropy} +\slidesonly{ +$$ +H(\hat{s}) := - \int p(\hat{s}) \log p(\hat{s}) d\hat{s} \qquad \qquad \text{(differential entropy)} +$$ + +\begin{block}{Definition of negentropy} +\begin{equation*} + J(\widehat{s}) \coloneqq \underbrace{ H(\widehat{s})_\normal}_{ + \substack{ \text{entropy of a Gaussian} \\ + \text{distribution with} \\ + \text{same variance}}} + - \underbrace{ H(\widehat{s}) }_{ + \substack{ \text{entropy of the true} \\ + \text{distribution} \\ + \text{(variance } \sigma^2 \text{)} }} +\end{equation*} +\end{block} +} + +\notesonly{ +The properties that make negentropy suitable: +} + +\begin{itemize} + \itR theoretically well motivated measure. Considered in some cases the optimzal estimator for nongaussianity. + \itR non-negative + \itR scale-invariant: $J(\alpha \widehat{s}) = J(\widehat{s}), \ \ \forall \alpha \ne 0$ (cf. exercise sheet) + \itR \textbf{Problem:} requires estimation of density $p(\widehat{s})$ +\end{itemize} + +\question{Should we minimize or maximize negentropy?} + +\end{frame} + +\subsection{Approximations of negentropy} + +\begin{frame} + +\notesonly{ +Estimating negentropy using the definition in \eqref{eq:negentropy} is computationally costly. It would require estimating the density of the random variable. We therefore resort to simpler approximations for negentropy. Such as the following use of cumulants: +} + +\begin{equation} +\label{eq:negentropyapprox} +J(\hat{s}) \approx \frac{1}{12} \langle (\hat{s})^3 \rangle^2 + \frac{1}{48} (\kurt{(\hat{s}))^2} + \text{higher order terms} +\end{equation} + +\notesonly{ +For symmetric distributions the first term in the approximation in \eqref{eq:negentropyapprox} is effectivley zero, which makes the approximation equivalent to the square of the kurtosis. The approximation would therefore from the same sensitvity to outliers. +} + +\slidesonly{ +$\rightarrow$ for symmetric distributions optimizing this is equivalent to optimizing $|\kurt{(\hat{s})}|$ sharing its outlier sensitivity \\\vspace{0.4cm} +} +The approximation is modified using +``nonpolynomial moments'' contrast functions $G$: + +\begin{equation} + J(\hat{s}) \approx \left( \langle G(\hat{s}) \rangle - \langle G(u_{\text{Gauss}}) \rangle \right)^2 +\end{equation} + +\end{frame} + +\clearpage +\begin{frame}{Common contrast functions} + + \notesonly{ +\textbf{Common contrast functions} + +The contrast function can be chosen depending on the assumed shape of the source densities. + +(e.g. speech: highly super-Gaussian) + +} + \begin{figure} + \centering + \includegraphics[width=4.5cm]{./img/contrast_functions.pdf} + \vspace{-0.5cm} + \caption*{\hspace{5cm}\textit{\tiny{Source: Hyv\"arinen, 2001}}} + \end{figure} + + \begin{equation*} + \smaller + \begin{array}{lll} + G_1(\hat{s}) = \frac{1}{a} \log \cosh (a \cdot \hat{s}) + \;\;& \;\;G_2(\hat{s}) = -\exp \Big( -\frac{(\hat{s})^2}{2} \Big) + & \;\;G_3(\hat{s}) = \frac{1}{4} (\hat{s})^4 + \end{array} + \end{equation*} + Any even, non-constant and non-quadratic (contrast) function $G$ can be used for ICA + + \question{When to choose which contrast function?} + + \end{frame} + +\begin{frame} +\slidesonly{ +\frametitle{Common contrast functions:} +} +\slidesonly{ + \smaller + \begin{tabular}{ccc} + $G_1(\hat{s}) = \frac{1}{a} \log \cosh (a \cdot \hat{s})$ & $G'_1(\hat{s}) = \tanh{(a\hat{s})}$ & $G''_1(\hat{s}) = a (1 - \tanh^2{(a\hat{s})})$\\[7pt] + \multicolumn{3}{c}{general purpose} \\[25pt] + $G_2(\hat{s}) = -\exp \Big( -\frac{(\hat{s})^2}{2} \Big)$ & $G'_2(\hat{s}) = \hat{s} \exp{(-\frac{(\hat{s})^2}{2})}$ & $G''_2(\hat{s}) = (1-(\hat{s})^2) \exp{(-\frac{(\hat{s})^2}{2})}$ \\[7pt] + \multicolumn{3}{c}{good for ``super''-Gaussian sources with many ``outliers''} \\[25pt] + $G_3(\hat{s}) = \frac{1}{4} (\hat{s})^4 $ & $G'_3(\hat{s}) = (\hat{s})^3$ & $G''_3(\hat{s}) = 3(\hat{s})^2$ \\[7pt] + \multicolumn{3}{c}{kurtosis: good for ``sub''-Gaussian sources with few ``outliers''} + \end{tabular} +} +\notesonly{ +\begin{itemize} + \item general purpose: + \begin{itemize} + \item $G_1(\hat{s}) = \frac{1}{a} \log \cosh (a \cdot \hat{s})$ + \item $G'_1(\hat{s}) = \tanh{(a\hat{s})}$ + \item $G''_1(\hat{s}) = a (1 - \tanh^2{(a\hat{s})})$ + \end{itemize} + \item for ``super''-Gaussian sources with many ``outliers'': + \begin{itemize} + \item $G_2(\hat{s}) = -\exp \Big( -\frac{(\hat{s})^2}{2} \Big)$ + \item $G'_2(\hat{s}) = \hat{s} \exp{(-\frac{(\hat{s})^2}{2})}$ + \item $G''_2(\hat{s}) = (1-(\hat{s})^2) \exp{(-\frac{(\hat{s})^2}{2})}$ + \end{itemize} + \item kurtosis: good for ``sub''-Gaussian sources with few ``outliers'': + \begin{itemize} + \item $G_3(\hat{s}) = \frac{1}{4} (\hat{s})^4 $ + \item $G'_3(\hat{s}) = (\hat{s})^3$ + \item $G''_3(\hat{s}) = 3(\hat{s})^2$ + \end{itemize} +\end{itemize} +} +\end{frame} + +\begin{frame} +cf. lecture slides for optmization of negentropy using contrast functions. +\end{frame} + +\begin{frame} +\question{How do we evaluate ICA?}\\ + +-cf. https://research.ics.aalto.fi/ica/icasso/ +\end{frame} + diff --git a/notes/06_fastica/tutorial.tex b/notes/06_fastica/tutorial.tex index 4304f0b..ede48b8 100644 --- a/notes/06_fastica/tutorial.tex +++ b/notes/06_fastica/tutorial.tex @@ -62,11 +62,39 @@ \newpage \mode -\input{./1_pcaica} +\input{./0_recap_ica_whitening} \mode* \clearpage +\mode +\input{./1_ica_ambiguous} +\mode* + +\clearpage + +\mode +\input{./2_pcaica} +\mode* + +\clearpage + +\mode +\input{./3_badgaussians} +\mode* + +\clearpage + +\mode +\input{./4_kurt} +\mode* + +\clearpage + +\mode +\input{./5_fastica} +\mode* + %\section{References} %\begin{frame}[allowframebreaks] \frametitle{References} %\scriptsize From fe490922693e26a861ab2a4202b068a5a21220b8 Mon Sep 17 00:00:00 2001 From: Youssef Kashef Date: Sun, 24 May 2020 22:50:02 +0200 Subject: [PATCH 2/6] ambiguity and whitening --- notes/06_fastica/0_recap_ica_whitening.tex | 29 ++-- notes/06_fastica/1_ica_ambiguous.tex | 69 +++++--- notes/06_fastica/2_pcaica.tex | 153 ++++++++++++------ notes/06_fastica/3_badgaussians.tex | 57 ++++--- notes/06_fastica/4_kurt.tex | 16 +- notes/06_fastica/Makefile | 2 + .../06_fastica/img/meme_doesnotinterfere.jpg | Bin 0 -> 54234 bytes notes/06_fastica/img/meme_icagaussian.jpg | Bin 0 -> 70157 bytes notes/06_fastica/img/meme_newalwaysbetter.jpg | Bin 0 -> 73511 bytes 9 files changed, 222 insertions(+), 104 deletions(-) create mode 100644 notes/06_fastica/img/meme_doesnotinterfere.jpg create mode 100644 notes/06_fastica/img/meme_icagaussian.jpg create mode 100644 notes/06_fastica/img/meme_newalwaysbetter.jpg diff --git a/notes/06_fastica/0_recap_ica_whitening.tex b/notes/06_fastica/0_recap_ica_whitening.tex index 20e20a0..0db31fc 100644 --- a/notes/06_fastica/0_recap_ica_whitening.tex +++ b/notes/06_fastica/0_recap_ica_whitening.tex @@ -1,6 +1,6 @@ \section{The ICA Problem} -\begin{frame} +\begin{frame}{\secname} independent sources: $\vec s = (s_1, s_2,...,s_N)^\top \in \R^N$\\ observations: $\vec x \in \R^N$ @@ -11,7 +11,7 @@ \section{The ICA Problem} \end{equation} \begin{equation} -\widehat{\vec s} = \vec W \cdot \vec x +\widehat{\vec s} = \vec W \vec x \end{equation} Methods for solving the ICA problem: @@ -44,9 +44,9 @@ \section{The ICA Problem} \newpage -\section{Whitening} +\section{Whitening revisited} -\begin{frame} +\begin{frame}{\secname} \notesonly{ The purpose of whitening is to decorrelate the data. @@ -75,22 +75,31 @@ \section{Whitening} \vec v^{(\alpha)} = \vec \Lambda^{-\frac{1}{2}} \vec M^\top \vec x^{(\alpha)} \end{equation} -where $\vec M = (\vec e_1, \vec e_2, \ldots,\vec e_N)$ -\notesonly{ -and $\vec \Lambda^{-\frac{1}{2}}$ is a diagonal matrix containig the square roots of the corresponding eigenvalues. -} +where +\begin{itemize} +\item[] $\vec M = (\vec e_1, \vec e_2, \ldots,\vec e_N)$ is the orthonormal eigenbasis of $\Sigma_x$ +\item[] and $\vec\Lambda$ is diagonal matrix with the corresponding eigenvalues. +\end{itemize} + +\pause + +\question{What do we know about the variables in $\vec v$?} + +\pause + \begin{equation} \label{eq:covw} \vec \Sigma_v = \mathrm{Cov}(\vec v) = \E \lbrack \, \vec v \, \vec v^\top \rbrack = \vec I_N \end{equation} \notesonly{ -Uncorrelted means zero covariance. Therefore, the covariance matrix for uncorrelated data is a diagonal matrix because it only contains the variances of the individual variables. +Uncorrelated means zero covariance. Therefore, the covariance matrix for uncorrelated data is a diagonal matrix because it only contains the variances of the individual variables. Whitening decorrelates the variables and normalizes the variances to 1. } \end{frame} -\begin{frame} +\begin{frame}{\secname} + \begin{figure}[ht] \label{fig:sphering} \includegraphics[width=12cm]{img/cov.png} diff --git a/notes/06_fastica/1_ica_ambiguous.tex b/notes/06_fastica/1_ica_ambiguous.tex index 6f8d83c..e340fdf 100644 --- a/notes/06_fastica/1_ica_ambiguous.tex +++ b/notes/06_fastica/1_ica_ambiguous.tex @@ -1,6 +1,6 @@ \section{Ambiguities in ICA and limitations} -\begin{frame} +\begin{frame}{\secname} Sources can be recovered up to: \begin{itemize} @@ -20,21 +20,35 @@ \section{Ambiguities in ICA and limitations} \end{frame} - - +\notesonly{ ICA cannot resolve if the mixing matrix is $\vec A$ or a permuatated and/or scaled version of $\vec A$. It can \textbf{also} not resolve if the independent sources are $\vec s$ or a permutated and/or scaled version of $\vec s$. +} + +\begin{frame}{\secname} Permutations and scaling are not an issue for ICA because permutation and scaling do not interfere with statistical independence. -$$ +\begin{equation} P_{s_1, s_2}(\widehat {\vec s}) \eqexcl P_{s_1} (\widehat{s}_1) \cdot P_{s_2} (\widehat{s}_2) -$$ +\end{equation} + +\slidesonly{ +\begin{center} + \includegraphics[width=0.4\textwidth]{img/meme_doesnotinterfere}% +\end{center} +} + +\end{frame} + +\notesonly{ +We can verify that ambiuities to scale and permutation do not interfere with statistical independence. +} -\begin{frame} +\begin{frame}{Verification} Permutations of sources {\footnotesize -\begin{equation*} +\begin{equation} \arraycolsep=1.4pt%\def\arraystretch{2.2} \begin{array}{ccc} \left( \begin{array}{ll} @@ -65,13 +79,13 @@ \section{Ambiguities in ICA and limitations} && P_{s_2} (\widehat{s}_2) \cdot P_{s_1} (\widehat{s}_1) \end{array} -\end{equation*} +\end{equation} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Scaling of source amplitudes: {\footnotesize -\begin{equation*} +\begin{equation} \begin{array}{ccc} \arraycolsep=1.4pt \left( \begin{array}{ll} @@ -103,33 +117,48 @@ \section{Ambiguities in ICA and limitations} && aP_{s_1} (a\widehat{s}_1) \cdot bP_{s_2} (b\, \widehat{s}_2) \end{array} -\end{equation*} +\end{equation} } \end{frame} + \subsection{Implications of the ambiguities} -\begin{frame} +\begin{frame}{\subsecname} We can assume: -$$ -\E \lbrack \, \vec s \, \rbrack = 0 -$$ +\begin{equation} +\E \lbrack \, \vec s \, \rbrack = \vec 0 +\end{equation} -Removing the mean from $\vec x$ does not change $\vec A$: +Substracting the mean from $\vec x$ does not change $\vec A$: -$$ +\begin{equation} \vec x - \E \lbrack \, \vec x \, \rbrack = \vec A \left( \vec s - \E \lbrack \, \vec s \, \rbrack \right) -$$ +\end{equation} \notesonly{Note that} $\E \lbrack \, \vec s \, \rbrack$ and $\E \lbrack \, \vec x \, \rbrack$ are not necessarily equal. -\pause + +\end{frame} + +\begin{frame}{\subsecname} We can also assume: -$$ +\begin{equation} \mathrm{Cov}(\vec s) = \E \lbrack \, \vec s \, \vec s^\top \rbrack = \vec I_N -$$ +\end{equation} Any scaling in $\mathrm{Cov}(\widehat{\vec s})$ can be assumed to come from $\vec A$ and can be undone. +\pause + +\begin{align} +\label{eq:expxina} +\vec \Sigma_x = \mathrm{Cov}(\vec x) &= \E \lbrack \, \vec x \, \vec x^\top \rbrack\\ +&= \E \lbrack \, \vec A\,\vec s \, \left( \vec A\,\vec s \right)^\top \rbrack\\ +&= \E \lbrack \, \vec A\; \underbrace{\;\vec s \, \vec s^\top}_{= \vec I_N} \vec A^\top \rbrack\\ +\label{eq:sigmax} +&= \vec A\, \vec A^\top +\end{align} + \end{frame} diff --git a/notes/06_fastica/2_pcaica.tex b/notes/06_fastica/2_pcaica.tex index 88c42b2..748def2 100644 --- a/notes/06_fastica/2_pcaica.tex +++ b/notes/06_fastica/2_pcaica.tex @@ -1,19 +1,27 @@ \subsection{Whitening in the context of ICA} -\begin{frame} +\mode{ +\begin{frame} + \begin{center} \huge + \subsecname + \end{center} + \begin{center} + Bringing PCA into the mix. + How PCA solves half of the ICA problem. + \end{center} +\end{frame} +} -\begin{align} -\label{eq:expxina} -\vec \Sigma_x = \mathrm{Cov}(\vec x) &= \E \lbrack \, \vec x \, \vec x^\top \rbrack\\ -&= \E \lbrack \, \vec A\,\vec s \, \left( \vec A\,\vec s \right)^\top \rbrack\\ -&= \E \lbrack \, \vec A\; \underbrace{\;\vec s \, \vec s^\top}_{= \vec I_N} \vec A^\top \rbrack\\ -\label{eq:sigmax} -&= \vec A\, \vec A^\top -\end{align} +\begin{frame}{\subsecname} -\end{frame} +\slidesonly{ +We've established that +\begin{equation} +\vec \Sigma_x = \vec A\, \vec A^\top +\end{equation} +} -\begin{frame} +\pause \notesonly{ If we were to apply \emph{Singular Value Decomposition} on the symmetric matrix $\Sigma_x$: @@ -38,40 +46,73 @@ \subsection{Whitening in the context of ICA} \widetilde{\vec A} := \vec Q \, \vec A \end{equation} -Which yields a new ICA problem: +Which yields\only<3>{ a new ICA problem}: \begin{equation} \vec u = \vec Q\, \vec x = \vec Q\,\vec A \, \vec s = \widetilde{\vec A} \, \vec s \end{equation} +\end{frame} + +\begin{frame}{A new ICA problem} + +\slidesonly{ +\begin{equation} +\vec u := \vec Q\, \vec x = \vec Q\,\vec A \, \vec s = \widetilde{\vec A} \, \vec s +\end{equation} +} + \question{What is so special about the new mixing matrix $\widetilde{\vec A}$?} \\ \notesonly{-$\widetilde{\vec A}$ is orthogonal (i.e. $\widetilde{\vec A}\, \widetilde{\vec A}^\top = \widetilde{\vec A}^\top \widetilde{\vec A} = \widetilde{\vec A}^{-1} \widetilde{\vec A} = \vec I_N$). We will see the implications of this in how it reduces the number of free parameters for solving the new ICA problem. - } +} + +\pause + +\slidesonly{ +\begin{center} + \includegraphics[width=0.4\textwidth]{img/meme_newalwaysbetter}% +\end{center} +} + \end{frame} -\begin{frame} +\begin{frame}{A new ICA problem} \slidesonly{ -$$ -\vec Q := \vec D^{-\frac{1}{2}} \vec U^\top -$$ -$$ -\vec \Sigma_x = \E \lbrack\, \vec x \, \vec x^\top \rbrack = \vec U \, \vec D \, \vec U^{\top} -$$ +\vspace{-14mm} +\hspace{8.0cm} +\StickyNote[1.7cm]{ + \begingroup + \scriptsize + %\begin{equation} + %\widehat{P}_{u_i} (\widehat{u}_i) = + %\frac{1}{\big| \widehat{f}_i^{'} (\widehat{s}_i) \big|} + %\widehat{P}_{s_i}(\widehat{s}_i) + %\end{equation} + \begin{equation} + \vec Q := \vec D^{-\frac{1}{2}} \vec U^\top + \end{equation} +\vspace{-2mm} + \begin{equation} + \vec \Sigma_x = \E \lbrack\, \vec x \, \vec x^\top \rbrack = \vec U \, \vec D \, \vec U^{\top} + \end{equation} + \endgroup +}[3.3cm] % width +\vspace{-22mm} } \begin{align} \E \lbrack\, \vec u \, \vec u^\top \rbrack -&\mystackrel{\vec u = \vec Q\, \vec x}{\vec u = \vec Q\, \vec x} +&\mystackrel{\vec u := \vec Q\, \vec x}{\vec u = \vec Q\, \vec x} \E \lbrack\, \vec Q \, \vec x \left( \vec Q \, \vec x\right)^\top \rbrack\\ &\mystackrel{\vec u = \vec Q\, \vec x}{}\E \lbrack\, \vec Q \, \vec x \; \vec x^\top \vec Q^\top \rbrack\\ -&\mystackrel{\vec u = \vec Q\, \vec x}{} \E \lbrack\, \vec Q \, \vec \Sigma_{X} \, \vec Q^\top \rbrack\\ +&\mystackrel{\vec u = \vec Q\, \vec x}{} \E \lbrack\, \vec Q \, \vec \Sigma_{x} \, \vec Q^\top \rbrack\\ &\mystackrel{\vec u = \vec Q\, \vec x}{} \vec Q \, \vec \Sigma_{x} \, \vec Q^\top\\ &\mystackrel{\vec u = \vec Q\, \vec x}{} \left( \vec D^{-\frac{1}{2}} \vec U^\top \right) \left( \vec U \, \vec D \, \vec U^\top \right) -\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) +\left( \vec D^{-\frac{1}{2}} \vec U^\top \right)^\top \end{align} \end{frame} @@ -83,7 +124,7 @@ \subsection{Whitening in the context of ICA} &\mystackrel{\vec u = \vec Q\, \vec x}{} \left( \vec D^{-\frac{1}{2}} \vec U^\top \right) \left( \vec U \, \vec D \, \vec U^\top \right) -\left( \vec D^{-\frac{1}{2}} \vec U^\top \right) +\left( \vec D^{-\frac{1}{2}} \vec U^\top \right)^\top \end{align} } @@ -98,38 +139,49 @@ \subsection{Whitening in the context of ICA} \underbrace{\left( \vec U^\top \, \vec U \right)}_{= \vec I_N} \, \left(\vec D^{-\frac{1}{2}} \right)^\top \\ -&= \vec D^\top +&= \vec D^{\top} \, \underbrace{\vec D^{-\frac{1}{2}} \, -\left(\vec D^{-\frac{1}{2}} \right)^\top}_{=\vec D^{-1}} = \vec I_N +\left(\vec D^{-\frac{1}{2}} \right)^{\cancel{\top}}}_{=\vec D^{-1}} = {\color{blue}\vec I_N} \end{align} Recall \notesonly{from \eqref{eq:sigmax}:} -$$ +\begin{equation} \vec \Sigma_x = \E \lbrack \, \vec x \, \vec x^\top \rbrack = \vec A\, \vec A^\top -$$ +\end{equation} Therefore: -$$ +\begin{equation} \E \lbrack \, \vec u \, \vec u^\top \rbrack -= \widetilde {\vec A}\, \widetilde {\vec A}^\top = \vec I_N -$$ += \widetilde {\vec A}\, \widetilde {\vec A}^\top = {\color{blue}\vec I_N} +\end{equation} \notesonly{ The new mixing matrix} $\widetilde{\vec A}$ is orthonormal.\\ -The space of solutions is restricted to $\vec W$ that are orthogonal. -The can speed up convergence of ICA. -The number of free parameters -\notesonly{ -for solving the new ICA problem -$\vec u = \widetilde{\vec A}\, \vec s$ -is reduced from $N^2$ to $N(N-1)/2$ -} + +\end{frame} + +\begin{frame}{\subsecname} + \slidesonly{ -$N^2 \rightarrow N(N-1)/2$ + +What makes $\widetilde{\vec A}$ so special is that it is orthonormal. +} + +\pause + +\svspace{5mm} + +The space of solutions restricted to finding unmixing matrices that are also orthogonal. +This restriction can speed up convergence of ICA. +The number of free parameters for solving the new ICA problem +$\vec u = \widetilde{\vec A}\, \vec s$ +is reduced\notesonly{ from $N^2$ to $N(N-1)/2$ +}\slidesonly{:\\ +$$N^2 \rightarrow N(N-1)/2$$ } \end{frame} @@ -153,27 +205,32 @@ \subsection{Whitening in the context of ICA} \item What's so special about $\vec u$? $ \qquad \qquad \qquad \qquad -\vec \Sigma_u = \vec I_N +\E \lbrack \, \vec u \, \vec u^\top \rbrack = \vec I_N $ +\pause + \item ...So? -- new ICA problem: $\vec u := \widetilde{\vec A}\, \vec s$\\ -- $\widetilde{\vec A}$ is orthogonal\\ -- unmixing the new problem involves only ``half'' the number of weights. +\pause + +\begin{itemize} +\item[-] new ICA problem: $\vec u := \widetilde{\vec A}\, \vec s$\\ +\item[-] $\widetilde{\vec A}$ is orthogonal\\ +\item[-] unmixing the new problem involves only ``half'' the number of weights. +\end{itemize} \question{What is the transformation $\vec D^{-\frac{1}{2}} \vec U^\top$ called?} +\notesonly{- It's a whitening transformation. The same we know from PCA.} + \end{itemize} } \end{frame} - -\section{Summary so far:} - \slidesonly{ -\begin{frame} +\begin{frame}{Summary so far} \begin{enumerate} \item Initial ICA Problem: $\vec x = \vec A\, \vec s$ \item New ICA Problem: $\vec u := \widetilde{\vec A}\, \vec s$,\\ diff --git a/notes/06_fastica/3_badgaussians.tex b/notes/06_fastica/3_badgaussians.tex index 3a7ec97..49d38cc 100644 --- a/notes/06_fastica/3_badgaussians.tex +++ b/notes/06_fastica/3_badgaussians.tex @@ -1,11 +1,8 @@ +\subsection{Gaussians are bad for ICA} +\subsubsection{A more formal argument for why Gaussians are bad for ICA} -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} -} -\newpage -\textbf{A more formal argument for why Gaussians are bad for ICA}: +\begin{frame}{\subsecname} Recall that the joint density of independent sources is a factorizing density: @@ -14,7 +11,7 @@ P_{\vec s}(\vec s) = \prod_{i=1}^{N} P_{s_i}(s_i) \,. \end{equation} -If we assumed gaussian distributed sources, the following factorization becomes possible: +If we assumed gaussian distributed sources, the following factorization becomes possible (e.g. $N=2$): \begin{equation} \label{eq:factsgauss} @@ -36,27 +33,27 @@ \end{equation} \end{frame} -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} -} -\slidesonly{\textbf{A more formal argument (cont'd):}} +\begin{frame}{\subsecname (cont'd)} + +%\slidesonly{\textbf{A more formal argument (cont'd):}} -Now consider applying an orthognal mixing matrix that is \textbf{known}. -\slidesonly{(orthogonal from whitening the $\vec x$)\\ +Now consider applying an orthognal mixing matrix $\widetilde{\vec A}$ that is \textbf{known}. +\slidesonly{(orthogonal because we whitened the data $\vec x$)\\ Consequently: } \notesonly{We've seen how whitening takes an ICA problem with any valid\footnote{invertible} mixing matrix $A$ and reformulates it into a new problem with an orthogonal mixing matrix $\widetilde{\vec A}$. The following holds for such an orthogonal mixing matrix $\widetilde{\vec A}$: } -$$ +\begin{equation} \widetilde{\vec A}^\top = \widetilde{\vec A}^{-1} \Leftrightarrow \widetilde{\vec A}^{-1}\widetilde{\vec A}=\vec I_N \qquad \text{and} \qquad |\det \widetilde{\vec A}| = |\det \widetilde{\vec A}^\top| = 1 -$$ +\end{equation} -$$ +\begin{equation} \vec s = \widetilde{\vec A}^{-1} \vec{x} = \widetilde{\vec A}^\top \vec x -$$ +\end{equation} + +\pause \notesonly{ We are now interested in the joint density $P_x(\vec x)$ of the mixtures $x_1$ and $x_2$. @@ -84,7 +81,7 @@ } \end{frame} -\begin{frame} +\begin{frame}{\subsecname~(cont'd)} \slidesonly{ Orthogonal mixing matrix $\widetilde{\vec A}$ implies: $$ @@ -132,14 +129,13 @@ \end{align} \end{frame} +\notesonly{ We see that the factorization in \eqref{eq:gaussxfact} describes the pdf for the mixtures $\vec x$ identically to the pdf of the original sources. The original and mixed distributions are identical. The mixing matrix did not change this. Therefore, it would be impossible to find its corresponding unmixing matrix to undo the rotation. - -\begin{frame} -\slidesonly{ -\frametitle{Gaussians are bad for ICA} } +\begin{frame}{\subsecname~(cont'd)} + \notesonly{ Mixing two independent Gaussians leads to a joint mixed distribution that is equal to that of the original sources. This is actually justified by the property that \emph{ @@ -152,11 +148,24 @@ \slidesonly{ +\begin{minipage}{0.65\textwidth} + +\svspace{5mm} + \begin{itemize} -\item Mixing two independent Gaussians leads to a joint mixed distribution that is equal to that of the original sources. +\item Mixing 2 independent Gaussians leads to a joint mixed distribution ${P}_{\vec x}(\vec x)$ that is effectively equal to the joint distribution of the original sources ${P}_{\vec s}(\vec s)$. +\pause \item No surprise: \emph{uncorrelated jointly Gaussian variables are necessarily independent.} +\pause \item One Gaussian + other distributions is fine. \item Two ore more Gaussians. No way. \end{itemize} +\end{minipage} +} + +\slidesonly{ +\only<4>{ + \placeimage{10.5}{6}{img/meme_icagaussian}{width=4cm}% +} } \end{frame} diff --git a/notes/06_fastica/4_kurt.tex b/notes/06_fastica/4_kurt.tex index 67effb2..b4d916b 100644 --- a/notes/06_fastica/4_kurt.tex +++ b/notes/06_fastica/4_kurt.tex @@ -1,5 +1,17 @@ -\begin{frame} -\frametitle{Maximizing nongaussianity leads to independent sources} +\section{ICA by maximizing nongaussianity} + +\mode{ +\begin{frame} + \begin{center} \huge + \subsecname + \end{center} + \begin{center} + Maximizing nongaussianity leads to independent sources + \end{center} +\end{frame} +} + +\begin{frame}{Maximizing nongaussianity leads to independent sources} \textbf{The setting:}\\ diff --git a/notes/06_fastica/Makefile b/notes/06_fastica/Makefile index dfe51f7..86bfda4 100644 --- a/notes/06_fastica/Makefile +++ b/notes/06_fastica/Makefile @@ -10,6 +10,7 @@ projnameA = $(projname).notes slides: $(projname).slides.tex $(projname).tex 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z*ABWY9A(MOVutMuUlhK!JR4!i%B}C^)EpCNRK2mLfu_K?8AE|b#Cd=43tw{mrqRH+ zOxX^7KD@pF literal 0 HcmV?d00001 From 3264dc27f93e9494e138a00412a781be66637685 Mon Sep 17 00:00:00 2001 From: Youssef Kashef Date: Mon, 25 May 2020 01:07:56 +0200 Subject: [PATCH 3/6] clt [wip] --- notes/06_fastica/1_ica_ambiguous.tex | 2 +- notes/06_fastica/3_badgaussians.tex | 4 ++-- notes/06_fastica/4_kurt.tex | 14 ++++++++++++-- notes/06_fastica/img/clt_uniform_0.pdf | Bin 0 -> 15023 bytes notes/06_fastica/img/clt_uniform_1.pdf | Bin 0 -> 15942 bytes notes/06_fastica/img/clt_uniform_2.pdf | Bin 0 -> 16284 bytes notes/06_fastica/img/clt_uniform_3.pdf | Bin 0 -> 17668 bytes notes/06_fastica/img/clt_uniform_4.pdf | Bin 0 -> 18318 bytes 8 files changed, 15 insertions(+), 5 deletions(-) create mode 100644 notes/06_fastica/img/clt_uniform_0.pdf create mode 100644 notes/06_fastica/img/clt_uniform_1.pdf create mode 100644 notes/06_fastica/img/clt_uniform_2.pdf create mode 100644 notes/06_fastica/img/clt_uniform_3.pdf create mode 100644 notes/06_fastica/img/clt_uniform_4.pdf diff --git a/notes/06_fastica/1_ica_ambiguous.tex b/notes/06_fastica/1_ica_ambiguous.tex index e340fdf..a74c15e 100644 --- a/notes/06_fastica/1_ica_ambiguous.tex +++ b/notes/06_fastica/1_ica_ambiguous.tex @@ -42,7 +42,7 @@ \section{Ambiguities in ICA and limitations} \end{frame} \notesonly{ -We can verify that ambiuities to scale and permutation do not interfere with statistical independence. +We can verify that ambiguities to scale and permutation do not interfere with statistical independence. } \begin{frame}{Verification} diff --git a/notes/06_fastica/3_badgaussians.tex b/notes/06_fastica/3_badgaussians.tex index 49d38cc..2843efc 100644 --- a/notes/06_fastica/3_badgaussians.tex +++ b/notes/06_fastica/3_badgaussians.tex @@ -144,7 +144,7 @@ \subsubsection{A more formal argument for why Gaussians are bad for ICA} } } -A mixture of sources where at most one is Gaussian, is still fine. It only becomes a problem when we have more than one. +A mixture of sources where, at most, one is Gaussian, can still be separated by ICA. It only becomes a problem when we have more than one Gaussian. \slidesonly{ @@ -153,7 +153,7 @@ \subsubsection{A more formal argument for why Gaussians are bad for ICA} \svspace{5mm} \begin{itemize} -\item Mixing 2 independent Gaussians leads to a joint mixed distribution ${P}_{\vec x}(\vec x)$ that is effectively equal to the joint distribution of the original sources ${P}_{\vec s}(\vec s)$. +\item Mixing two independent Gaussians leads to a joint mixed distribution ${P}_{\vec x}(\vec x)$ that is effectively equal to the joint distribution of the original sources ${P}_{\vec s}(\vec s)$. \pause \item No surprise: \emph{uncorrelated jointly Gaussian variables are necessarily independent.} \pause diff --git a/notes/06_fastica/4_kurt.tex b/notes/06_fastica/4_kurt.tex index b4d916b..8d54a32 100644 --- a/notes/06_fastica/4_kurt.tex +++ b/notes/06_fastica/4_kurt.tex @@ -3,7 +3,7 @@ \section{ICA by maximizing nongaussianity} \mode{ \begin{frame} \begin{center} \huge - \subsecname + \secname \end{center} \begin{center} Maximizing nongaussianity leads to independent sources @@ -11,7 +11,17 @@ \section{ICA by maximizing nongaussianity} \end{frame} } -\begin{frame}{Maximizing nongaussianity leads to independent sources} +\begin{frame}{Maximizing nongaussianity} + +\begin{block}{Intuition from the Central Limit Theorem} +\emph{The distribution of the sum of independent random variables is ''more Gaussian'' than the original distributions of the random variables.}\\\vspace{2mm} +Searching for the direction of maximum deviation from a Gaussian distribution may recover the original sources. +\end{block} +\end{frame} + + + +\begin{frame}{The setting} \textbf{The setting:}\\ diff --git a/notes/06_fastica/img/clt_uniform_0.pdf b/notes/06_fastica/img/clt_uniform_0.pdf new file mode 100644 index 0000000000000000000000000000000000000000..ef6e69e4af0e37a8efdb9412f4aa2232e7eca546 GIT binary patch literal 15023 zcmeHuc|29$*T150sYFPrE<%JmyXMR?3rV5O^X!^3loBZ#%w$Z+n0byO^Hf4KXUsfi 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zh6X&L|G-Y)XfS>#`U&`%#wP#+=J_j)4-nihG(q6p;x`)bkpDshqycEqPi=tcca>jg0_aWTXPO}F z4>VrnZ}Sxt{C&-#aP*e*Q-3fZ=>IFvfJ4puJMDM*!JvWw4)vG*V85>i0KSI)y1p|FlK9nb|s6xgGy*j)Idlu)_lkpv{)1 nvom^6J1%S4`dR^-0(zrxN13^yj<*RIKQI9hGqbFk9O!=mGKs;y literal 0 HcmV?d00001 From 4afa0d8b814f8136d0f4637fe3ed7224a9ab6327 Mon Sep 17 00:00:00 2001 From: Youssef Kashef Date: Wed, 27 May 2020 20:30:44 +0200 Subject: [PATCH 4/6] whitening,gaussians, clt --- notes/06_fastica/2_pcaica.tex | 92 ++++++++++++++- notes/06_fastica/3_badgaussians.tex | 32 +++++- notes/06_fastica/4_kurt.tex | 105 +++++++++++------- .../img/meme_breakingicabadgaussian.jpg | Bin 0 -> 63397 bytes notes/06_fastica/img/meme_nomoregaussians.jpg | Bin 0 -> 53448 bytes .../06_fastica/img/meme_thismuchgaussian.jpg | Bin 0 -> 68025 bytes .../img/uniform_mixtures_centered.pdf | Bin 0 -> 18508 bytes .../img/uniform_mixtures_decorrelated.pdf | Bin 0 -> 19422 bytes .../img/uniform_mixtures_whitened.pdf | Bin 0 -> 19471 bytes .../img/uniform_original_sources.pdf | Bin 0 -> 18861 bytes 10 files changed, 182 insertions(+), 47 deletions(-) create mode 100644 notes/06_fastica/img/meme_breakingicabadgaussian.jpg create mode 100644 notes/06_fastica/img/meme_nomoregaussians.jpg create mode 100644 notes/06_fastica/img/meme_thismuchgaussian.jpg create mode 100644 notes/06_fastica/img/uniform_mixtures_centered.pdf create mode 100644 notes/06_fastica/img/uniform_mixtures_decorrelated.pdf create mode 100644 notes/06_fastica/img/uniform_mixtures_whitened.pdf create mode 100644 notes/06_fastica/img/uniform_original_sources.pdf diff --git a/notes/06_fastica/2_pcaica.tex b/notes/06_fastica/2_pcaica.tex index 748def2..537088f 100644 --- a/notes/06_fastica/2_pcaica.tex +++ b/notes/06_fastica/2_pcaica.tex @@ -73,6 +73,8 @@ \subsection{Whitening in the context of ICA} \begin{center} \includegraphics[width=0.4\textwidth]{img/meme_newalwaysbetter}% \end{center} + +- $\widetilde{\vec A}$ is orthogonal. Not necessarily the case for the original mixing matrix $\vec A$. } \end{frame} @@ -183,6 +185,7 @@ \subsection{Whitening in the context of ICA} }\slidesonly{:\\ $$N^2 \rightarrow N(N-1)/2$$ } + \end{frame} \begin{frame} @@ -217,7 +220,7 @@ \subsection{Whitening in the context of ICA} \begin{itemize} \item[-] new ICA problem: $\vec u := \widetilde{\vec A}\, \vec s$\\ \item[-] $\widetilde{\vec A}$ is orthogonal\\ -\item[-] unmixing the new problem involves only ``half'' the number of weights. +\item[-] unmixing the new problem involves only ``half'' the number of free parameters. \end{itemize} \question{What is the transformation $\vec D^{-\frac{1}{2}} \vec U^\top$ called?} @@ -229,6 +232,91 @@ \subsection{Whitening in the context of ICA} \end{frame} +\subsubsection{Whitening solves half of the ICA problem} + +\begin{frame}{\subsubsecname} + +\slidesonly{ + +Unmixing the new problem involves only ``half'' the number of free parameters. +$$N^2 \rightarrow N(N-1)/2$$ +} + +\pause + +\svspace{-5mm} + +\question{Does this mean that $\vec W$ is no longer $N \times N$?} + +-No $\vec W$ is still $N \times N$, but we no longer have to search for each component individually. + +\end{frame} + +\begin{frame}{\subsubsecname} + +Example with $N=2$ uniformly distributed sources $s_1, s_2$: + + \begin{center} + \begin{minipage}{0.29\textwidth} +\includegraphics[width=0.99\textwidth]{./img/uniform_original_sources} + \end{minipage} + \hspace{5mm} + \begin{minipage}{0.29\textwidth} +\includegraphics[width=0.99\textwidth]{./img/uniform_mixtures_centered} + \end{minipage}\\ + + \begin{minipage}{0.29\textwidth} +\includegraphics[width=0.99\textwidth]{./img/uniform_mixtures_decorrelated} + \end{minipage} + \hspace{5mm} + \begin{minipage}{0.29\textwidth} +\includegraphics[width=0.99\textwidth]{./img/uniform_mixtures_whitened} + \end{minipage} + \notesonly{ + \captionof{figure}{Example with $N=2$ uniformly distributed sources $s_1, s_2$} + } +\end{center} + +\end{frame} + +\begin{frame}{\subsubsecname} + +\slidesonly{ +Example with $N=2$ uniformly distributed sources $s_1, s_2$: + + \begin{center} + \begin{minipage}{0.25\textwidth} +\includegraphics[width=0.99\textwidth]{./img/uniform_original_sources} + \end{minipage} + \hspace{5mm} + \begin{minipage}{0.25\textwidth} +\includegraphics[width=0.99\textwidth]{./img/uniform_mixtures_whitened} + \end{minipage} +\end{center} +} + +ICA on whitened mixtures is about ``rotating it back''. + +2D data can be rotated by an angle $\theta$ using: + +\begin{align} + \vec{x}_{\theta} & = + \begin{pmatrix} + \cos(\theta) & -\sin(\theta) \\ + \sin(\theta) & \cos(\theta) + \end{pmatrix} + \vec{x} \qquad \text{with} \quad \theta = 0, \, \ldots\, , 2\pi +\end{align} + +\pause + +Find the right rotation for \textcolor{blue}{2}D data involves a \textcolor{red}{single} free parameter. +\begin{equation} +\# \text{free parameters} = \textcolor{blue}{N} \cdot (\textcolor{blue}{N}-1)/2 = \textcolor{blue}{2} \cdot (\textcolor{blue}{2}-1)/2 = \textcolor{red}{1} +\end{equation} + +\end{frame} + \slidesonly{ \begin{frame}{Summary so far} \begin{enumerate} @@ -240,7 +328,7 @@ \subsection{Whitening in the context of ICA} \item $\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. \item Applying ICA on whitened data reduces the number of free parameters. \item PCA simplifies the ICA problem. -\item ICA on whitened data is about ``rotating'' it back. +\item ICA on whitened data is about ``rotating it back''. \end{enumerate} \end{frame} diff --git a/notes/06_fastica/3_badgaussians.tex b/notes/06_fastica/3_badgaussians.tex index 2843efc..9b586da 100644 --- a/notes/06_fastica/3_badgaussians.tex +++ b/notes/06_fastica/3_badgaussians.tex @@ -1,8 +1,20 @@ \subsection{Gaussians are bad for ICA} -\subsubsection{A more formal argument for why Gaussians are bad for ICA} +\mode{ +\begin{frame} + \begin{center} \huge + \subsecname + \end{center} + \begin{center} + \includegraphics[width=0.4\textwidth]{./img/meme_breakingicabadgaussian}\\ + And \textbf{white}ning won't help either. + \end{center} +\end{frame} +} + +\subsubsection{A formal argument for why Gaussians are bad for ICA} -\begin{frame}{\subsecname} +\begin{frame}{\subsubsecname} Recall that the joint density of independent sources is a factorizing density: @@ -33,7 +45,7 @@ \subsubsection{A more formal argument for why Gaussians are bad for ICA} \end{equation} \end{frame} -\begin{frame}{\subsecname (cont'd)} +\begin{frame}{\subsecname~(cont'd)} %\slidesonly{\textbf{A more formal argument (cont'd):}} @@ -153,19 +165,27 @@ \subsubsection{A more formal argument for why Gaussians are bad for ICA} \svspace{5mm} \begin{itemize} -\item Mixing two independent Gaussians leads to a joint mixed distribution ${P}_{\vec x}(\vec x)$ that is effectively equal to the joint distribution of the original sources ${P}_{\vec s}(\vec s)$. +\item Mixing two independent Gaussians leads to a joint mixed distribution ${P}_{\vec x}(\vec x)$ that is effectively equal to the joint distribution of the original sources ${P}_{\vec s}(\vec s)$.\\ + \pause + \item No surprise: \emph{uncorrelated jointly Gaussian variables are necessarily independent.} \pause -\item One Gaussian + other distributions is fine. + +\pause + +\item One Gaussian + other distribution(s) is fine. \item Two ore more Gaussians. No way. \end{itemize} \end{minipage} } \slidesonly{ -\only<4>{ +\only<2>{ \placeimage{10.5}{6}{img/meme_icagaussian}{width=4cm}% } +\only<4>{ + \placeimage{10.5}{6}{img/meme_nomoregaussians}{width=4cm}% +} } \end{frame} diff --git a/notes/06_fastica/4_kurt.tex b/notes/06_fastica/4_kurt.tex index 8d54a32..7f4cf82 100644 --- a/notes/06_fastica/4_kurt.tex +++ b/notes/06_fastica/4_kurt.tex @@ -14,18 +14,30 @@ \section{ICA by maximizing nongaussianity} \begin{frame}{Maximizing nongaussianity} \begin{block}{Intuition from the Central Limit Theorem} -\emph{The distribution of the sum of independent random variables is ''more Gaussian'' than the original distributions of the random variables.}\\\vspace{2mm} +\emph{The distribution of the sum of independent random variables is ``more Gaussian'' than the original distributions of the random variables.}\\\vspace{2mm} Searching for the direction of maximum deviation from a Gaussian distribution may recover the original sources. \end{block} -\end{frame} - +\begin{center} + \includegraphics<2>[width=0.8\textwidth]{./img/clt_uniform_0} + \includegraphics<3>[width=0.8\textwidth]{./img/clt_uniform_1} + \includegraphics<4>[width=0.8\textwidth]{./img/clt_uniform_2} + \includegraphics<5>[width=0.8\textwidth]{./img/clt_uniform_3} + \includegraphics<6>[width=0.8\textwidth]{./img/clt_uniform_4} + \notesonly{ + \captionof{figure}{The sum of independent random variables is ``more Gaussian'' than the original distributions.} + } +\end{center} + +\end{frame} \begin{frame}{The setting} -\textbf{The setting:}\\ +\notesonly{\textbf{The setting:}\\} -Two statistically independent sources with $\langle s_i s_j \rangle = \delta_{ij} \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N$ +Two statistically independent sources with +\begin{equation}\langle s_i s_j \rangle = \delta_{ij} \quad \forall i,j=1,\ldots,N \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N +\end{equation} \notesonly{ The sources are mixed using a mixing matrix $\vec A$ resulting in observations $\vec x$: } @@ -72,7 +84,7 @@ \section{ICA by maximizing nongaussianity} This can be accomplished with a vector containing a single non-zero element: } -\begin{equation*} +\begin{equation} \vec{z}_{\text{opt.}} = \left( \begin{array}{c} 0 \\ \pm 1 \end{array} \right) @@ -80,7 +92,7 @@ \section{ICA by maximizing nongaussianity} \vec{z}_{\text{opt.}} = \left( \begin{array}{c} \pm 1 \\ 0 \end{array} \right) -\end{equation*} +\end{equation} \end{frame} @@ -91,57 +103,72 @@ \section{ICA by maximizing nongaussianity} We cannot have both independent sources contribute to $\widehat{s}_i$, only one can. Therefore, we only need a single non-zero component for $\vec z_i$. Wether $s_1$ is scaled by any factor before reaching $\widehat{s}_i$ does not make it more or less independent of $s_2$. Choosing $1$ for the non-zero component is therefore sufficient. Finally, negating the source by multiplying it by $(-1)$ also has no consequences on the independence criterion. - -We won't actually try and find $\vec z_i$ because we don't have $\vec s$ to apply them to. We use the requirements for $\vec z_i$ by finding a $\vec w_i$ that satisfies these requirements through: -\begin{equation} -\label{eq:zfromw} -\vec z_i = \left(\vec w_i^\top \vec A\right)^\top = \vec A^\top \vec w_i -\end{equation} } -\begin{frame} -\question{Does maximizing nongaussianity deliver independent components?} -\slidesonly{ -\begin{equation} -\widehat s_i = \vec z_i^{\top} \vec s = z_1 s_1 + z_2 s_2 -\end{equation} +%\begin{frame} +%\question{Does maximizing nongaussianity deliver independent components?} +%\slidesonly{ -\begin{equation*} - \vec{z}_{\text{opt.}} = \left( \begin{array}{c} - 0 \\ \pm 1 - \end{array} \right) - \quad \text{ or }\quad - \vec{z}_{\text{opt.}} = \left( \begin{array}{c} - \pm 1 \\ 0 - \end{array} \right) -\end{equation*} +%\begin{equation} +%\widehat s_i = \vec z_i^{\top} \vec s = z_1 s_1 + z_2 s_2 +%\end{equation} -$\vec z_i$ ensures independent $\widehat s_i$ +%\begin{equation*} + %\vec{z}_{\text{opt.}} = \left( \begin{array}{c} + %0 \\ \pm 1 + %\end{array} \right) + %\quad \text{ or }\quad + %\vec{z}_{\text{opt.}} = \left( \begin{array}{c} + %\pm 1 \\ 0 + %\end{array} \right) +%\end{equation*} + +%$\vec z_i$ ensures independent $\widehat s_i$ -\begin{equation} -\widehat s_i = \vec w_i^\top \vec x -\end{equation} -$\vec w_i$ finds less gaussian $\widehat s_i$ +%\begin{equation} +%\widehat s_i = \vec w_i^\top \vec x +%\end{equation} + +%$\vec w_i$ finds less gaussian $\widehat s_i$ + +\notesonly{ + +We won't actually try and find $\vec z_i$ because we don't have $\vec s$ to apply them to. We use the requirements for $\vec z_i$ by finding a $\vec w_i$ that satisfies these requirements through: \begin{equation} -\vec z_i = \vec A^\top \vec w_i +\label{eq:zfromw} +\vec z_i = \left(\vec w_i^\top \vec A\right)^\top = \vec A^\top \vec w_i \end{equation} +} -Maximizing nongaussianity is ensured to keep $\widehat s_i$ independent. +%Maximizing nongaussianity is ensured to keep $\widehat s_i$ independent. -} -\end{frame} +%} +%\end{frame} \notesonly{ - By (1) maximizing the nongaussianity of $\vec w_i^\top \vec x$ and (2) having $\vec z_i = \vec A^\top \vec w_i$ yield independent components and (3) knowing that $\widehat s_i = \vec w_i^\top \vec x = \vec z_i^{\top} \vec s$, we conclude that maximizing $\vec w_i^\top \vec x$ gives us one independent component. } \newpage +\mode{ +\begin{frame} + + \begin{center} + Deciding on a measure for nongaussianity + \end{center} + \begin{center} + \includegraphics[width=0.4\textwidth]{./img/meme_thismuchgaussian} + \end{center} + +\end{frame} +} + \section{Kurtosis as a measure for nongaussianity} -\begin{frame} +\begin{frame}{\secname} \notesonly{ Kurtosis represents the fourth-order cumulant\footnote{ @@ -175,7 +202,7 @@ \section{Kurtosis as a measure for nongaussianity} \begin{frame} -\question{What does kurtosis measure?}\\ +\question{How do we interpret the kurtosis measure?}\\ \begin{tabular}[h]{c c c c} & diff --git a/notes/06_fastica/img/meme_breakingicabadgaussian.jpg b/notes/06_fastica/img/meme_breakingicabadgaussian.jpg new file mode 100644 index 0000000000000000000000000000000000000000..ab432daa5cc236fcdf1fa52f07cf6e40a4d4cf15 GIT binary patch literal 63397 zcmbq)Wl$VX^z9NPxC9IC?(XjHy12W$CBa>TTd;-2eUSjc-50mu?iM8E@%w*x_1?!< z@66Ox_r2Xy(>*hFs!!khZ~5PE0H&g>f-C?A1^|G0KLG#M08#+h5C7r6E9^Vq5a9j; zA_4+D0umxJG7=&Z5;7_VDl!T>3K9|;78*JRCMFgpGU_MnPng*6Wz7F>0`p%_SUAM@ zj+iJ&DDT$)NBrvpU?IYEeQ<(>p#*%uf`P??`8NO{0RR9WVg7sE{|%TAuyF7Q9}$t> z-9#{e53v8Y`vc7Xxc{vJP+{M_n6Q}dL!Aj#8yERTDaFjbmb6^qKFs9dj#432s=*2e 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zJ70gR3jsrc9nW9$z?{F919Ni%aNaNFATY#O|E)bRHw=I)|DFfo0GR$YkDco`d2E35 z#4mL@ffj$wgTR2@+F!~+IN5&X83F|ZGyN~+07r&jd4@naflcUN%W?m6EMZ{8q>1?b zIUW!g6#UOTwtw~whWG%_uXVZpNtX=*{s%9BSb$0P?{op@f`7<`jT^C){Jk7I?2YWp;OC$S>&nih<8v|>Vy1B?uSb Owt`JVBd#cc{l5S|tbWA+ literal 0 HcmV?d00001 From 99431c6d95f5463e5d92428645ff43f07488f5e7 Mon Sep 17 00:00:00 2001 From: Youssef Kashef Date: Wed, 27 May 2020 22:32:25 +0200 Subject: [PATCH 5/6] kurtosis based ica --- notes/06_fastica/1_ica_ambiguous.tex | 6 +- notes/06_fastica/3_badgaussians.tex | 2 +- notes/06_fastica/4_kurt.tex | 128 +++++++++++++++++---------- notes/06_fastica/5_fastica.tex | 32 +++++-- 4 files changed, 106 insertions(+), 62 deletions(-) diff --git a/notes/06_fastica/1_ica_ambiguous.tex b/notes/06_fastica/1_ica_ambiguous.tex index a74c15e..60a306f 100644 --- a/notes/06_fastica/1_ica_ambiguous.tex +++ b/notes/06_fastica/1_ica_ambiguous.tex @@ -21,8 +21,8 @@ \section{Ambiguities in ICA and limitations} \end{frame} \notesonly{ -ICA cannot resolve if the mixing matrix is $\vec A$ or a permuatated and/or scaled version of $\vec A$. -It can \textbf{also} not resolve if the independent sources are $\vec s$ or a permutated and/or scaled version of $\vec s$. +ICA cannot resolve if the mixing matrix is $\vec A$ or a permuted and/or scaled version of $\vec A$. +It can \textbf{also} not resolve if the independent sources are $\vec s$ or a permuted and/or scaled version of $\vec s$. } \begin{frame}{\secname} @@ -130,7 +130,7 @@ \subsection{Implications of the ambiguities} \E \lbrack \, \vec s \, \rbrack = \vec 0 \end{equation} -Substracting the mean from $\vec x$ does not change $\vec A$: +Subtracting the mean from $\vec x$ does not change $\vec A$: \begin{equation} \vec x - \E \lbrack \, \vec x \, \rbrack = \vec A \left( \vec s - \E \lbrack \, \vec s \, \rbrack \right) diff --git a/notes/06_fastica/3_badgaussians.tex b/notes/06_fastica/3_badgaussians.tex index 9b586da..5e59bbb 100644 --- a/notes/06_fastica/3_badgaussians.tex +++ b/notes/06_fastica/3_badgaussians.tex @@ -49,7 +49,7 @@ \subsubsection{A formal argument for why Gaussians are bad for ICA} %\slidesonly{\textbf{A more formal argument (cont'd):}} -Now consider applying an orthognal mixing matrix $\widetilde{\vec A}$ that is \textbf{known}. +Now consider applying an orthogonal mixing matrix $\widetilde{\vec A}$ that is \textbf{known}. \slidesonly{(orthogonal because we whitened the data $\vec x$)\\ Consequently: } diff --git a/notes/06_fastica/4_kurt.tex b/notes/06_fastica/4_kurt.tex index 7f4cf82..2bf96ac 100644 --- a/notes/06_fastica/4_kurt.tex +++ b/notes/06_fastica/4_kurt.tex @@ -99,7 +99,7 @@ \section{ICA by maximizing nongaussianity} \notesonly{ Recall that ICA cannot resolve scale or permutation of the sources and thirdly it cannot resolve the sign. This is not an issue. -The role of $\vec z_i$ is to route either $s_1$ or $s_2$ to $\widehat{\vec s}_i$. This covers the ambiguitiy in terms of permutation. +The role of $\vec z_i$ is to route either $s_1$ or $s_2$ to $\widehat{\vec s}_i$. This covers the ambiguity in terms of permutation. We cannot have both independent sources contribute to $\widehat{s}_i$, only one can. Therefore, we only need a single non-zero component for $\vec z_i$. Wether $s_1$ is scaled by any factor before reaching $\widehat{s}_i$ does not make it more or less independent of $s_2$. Choosing $1$ for the non-zero component is therefore sufficient. Finally, negating the source by multiplying it by $(-1)$ also has no consequences on the independence criterion. @@ -166,13 +166,13 @@ \section{ICA by maximizing nongaussianity} \end{frame} } -\section{Kurtosis as a measure for nongaussianity} +\subsection{Kurtosis as a measure for nongaussianity} -\begin{frame}{\secname} +\begin{frame}{\subsecname} \notesonly{ Kurtosis represents the fourth-order cumulant\footnote{ -Cumulants allow us to express the i-th moment in terms of a cumulative sum of the moments preceeding it. +Cumulants allow us to express the i-th moment in terms of a cumulative sum of the moments preceding it. This simplifies the expression of higher-order moments such as kurtosis which is the fourth-order moment. } of a random variable. } @@ -239,11 +239,13 @@ \section{Kurtosis as a measure for nongaussianity} \subsection{kurtosis-based ICA} -\begin{frame} +\begin{frame}{\subsecname} +\notesonly{ Two statistically independent sources with $\langle s_i s_j \rangle = \delta_{ij} \quad \Leftrightarrow \quad \langle \vec s \, \vec s^\top \rangle = \vec I_N$ (any scaling can be attributed to $\vec A$) +} \begin{equation*} \widehat{s}_i \quad @@ -252,21 +254,34 @@ \subsection{kurtosis-based ICA} = \quad \vec{z}^\top \vec{s} \quad = \quad z_1 s_1 + z_2 s_2 \end{equation*} + \vspace{1mm} -We want the covariance of our reconstructions to match that of the original sources. +We want the covariance of our reconstructions $\widehat{\vec s}$ to match that of the original sources $\vec s$. \begin{equation*} \langle \widehat{\vec s} \, \widehat{\vec s}^\top \rangle \eqexcl \langle \vec s \, \vec s^\top \rangle = \vec I_N \end{equation*} This implies, -\begin{align*} +\begin{align} \var(\widehat{s}_i) \; &= \; \langle \big( z_1 s_1 + z_2 s_2 \big)^2 \rangle_{P_{\vec s}}\\ \; &= \; \langle z_1^2 \, s_1^2 \rangle \;+\; 2 \, \langle z_1\, s_1\, z_2 \, s_2 \rangle \;+\; \langle z_2^2 \, s_2^2 \rangle \\ \; &= \; z_1^2 \, \langle s_1^2 \rangle \;+\; 2 \, z_1\, z_2 \, \underbrace{\langle s_1\, s_2 \rangle}_{= 0} \;+\; z_2^2 \, \langle s_2^2 \rangle \\ \; &= \; z_1^2 \, \langle s_1^2 \rangle \;+\; z_2^2 \,\langle s_2^2 \rangle \\ \; &= \; z_1^2 + z_2^2 \eqexcl 1 -\end{align*} -Making the constraint of unit variance for $\widehat{s}_i$ is to match the variance assumed for the orgiinal sources $s_1$ and $s_2$. This implies that solutions for $\vec z$ are constrained to lie on a unit circle. +\end{align} + +\end{frame} + +\begin{frame}{\subsecname} + +\slidesonly{ +$$ +\var(\widehat{s}_i) + \; = \; z_1^2 + z_2^2 \eqexcl 1 + $$ +} + +Making the constraint of unit variance for $\widehat{s}_i$ is to match the variance assumed for the original sources $s_1$ and $s_2$. This implies that solutions for $\vec z$ are constrained to lie on a unit circle. \vspace{1mm} \begin{align*} \kurt(\widehat{s}) \;\; &= \;\; \kurt(z_1 s_1 + z_2 s_2) \;\; \\ &= \;\; \kurt(z_1 s_1) + \kurt(z_2 s_2) \; = \; z_1^4 \kurt(s_1) + z_2^4 \kurt(s_2) @@ -332,9 +347,9 @@ \subsection{kurtosis-based ICA} \end{frame} -\subsection{Kurtosis-based ICA: the gradient algorithm} +\subsubsection{Kurtosis-based ICA: the gradient algorithm} -\begin{frame} +\begin{frame}{\subsubsecname} \notesonly{ $| \kurt{(\vec{b}^\top \vec{u})} |$ can be maximized by moving $\vec b$ @@ -371,10 +386,8 @@ \subsection{Kurtosis-based ICA: the gradient algorithm} \end{frame} -\begin{frame} -\slidesonly{ -\frametitle{Kurtosis-based ICA: the gradient algorithm} -} +\begin{frame}{\subsubsecname} + \begin{block}{I. batch learning:} Initialization: random vector $\vec{b}$ of unit length \begin{eqnarray*} @@ -480,47 +493,55 @@ \subsection{Kurtosis-based ICA: the gradient algorithm} \end{block} \end{frame} -\slidesonly{ -\begin{frame} -\frametitle{Summary so far:} -\begin{enumerate} -\item \textcolor{gray}{ -Initial ICA Problem: $\vec x = \vec A\, \vec s$ -} -\item \textcolor{gray}{ -New ICA Problem: $\vec u = \widetilde{\vec A}\, \vec s$,\\ -where $\vec u = \vec D^{-\frac{1}{2}} \vec U^\top \vec x$ and $\vec \Sigma_u = \vec I_N$. -} -\item \textcolor{gray}{ -$\vec u$ is the \emph{whitened} version of $\vec x$. -} -\item \textcolor{gray}{ -$\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. -} -\item \textcolor{gray}{ -Applying ICA on whitened data reduced the number of free parameters. -} -\item \textcolor{gray}{ -PCA simplifies the ICA problem. -} -\item Ambiguities in ICA -\item Why are Gaussians bad for ICA? -\item ICA by maximizing nongaussianity -\item Kurtosis-based ICA - -\end{enumerate} +%\slidesonly{ +%\begin{frame} +%\frametitle{Summary so far:} +%\begin{enumerate} +%\item \textcolor{gray}{ +%Initial ICA Problem: $\vec x = \vec A\, \vec s$ +%} +%\item \textcolor{gray}{ +%New ICA Problem: $\vec u = \widetilde{\vec A}\, \vec s$,\\ +%where $\vec u = \vec D^{-\frac{1}{2}} \vec U^\top \vec x$ and $\vec \Sigma_u = \vec I_N$. +%} +%\item \textcolor{gray}{ +%$\vec u$ is the \emph{whitened} version of $\vec x$. +%} +%\item \textcolor{gray}{ +%$\vec D$ and $\vec U$ can be obtained via PCA on $\vec x$. +%} +%\item \textcolor{gray}{ +%Applying ICA on whitened data reduced the number of free parameters. +%} +%\item \textcolor{gray}{ +%PCA simplifies the ICA problem. +%} +%\item Ambiguities in ICA +%\item Why are Gaussians bad for ICA? +%\item ICA by maximizing nongaussianity +%\item Kurtosis-based ICA +%\end{enumerate} -\textbf{Next: Can we do better than kurtosis-based ICA?} -\end{frame} -} +%\end{frame} +%} \notesonly{ Next, we will look for an alternative that mitigates the sensitivity to outliers which kurtosis-based ICA is prone to. } \begin{frame} + +\slidesonly{ + +\textbf{Next: Can we do better than kurtosis-based ICA?} + +\vspace{5mm} + +\pause +} + Kurtosis is easy to compute but can be \emph{sensitive to outliers}. This is a usual problem with higher-order statistics. \begin{block}{Example} @@ -530,8 +551,17 @@ \subsection{Kurtosis-based ICA: the gradient algorithm} \itl contribution to kurtosis: $ \geq 10^4/1000 -3 = 7$ \end{itemize} \end{block} -\end{frame} -We therefore turn to an alternate measure for nongaussianity, namely \emph{negentropy} for brevity (not the same as negative entropy $-H(\cdot)$). Negentropy of the reconstructed source $\widehat{\vec s}$ measures the difference between the differential entropy of $\widehat{\vec s}$ and the differential entropy of a Gaussian distribution with the same variance as $\widehat{\vec s}$. +\pause +\slidesonly{ +$\Rightarrow\;\;$ a more robust measure for nongaussianity\\ +} +\notesonly{We therefore turn to an alternate measure for nongaussianity, namely }\emph{negentropy} \notesonly{for brevity }(not the same as negative entropy $-H(\cdot)$).\\ +%\svspace{5mm} +\notesonly{ +Negentropy of the reconstructed source $\widehat{\vec s}$ measures the difference between the differential entropy of $\widehat{\vec s}$ and the differential entropy of a Gaussian distribution with the same variance as $\widehat{\vec s}$. +} + +\end{frame} diff --git a/notes/06_fastica/5_fastica.tex b/notes/06_fastica/5_fastica.tex index 73a720e..01a29c9 100644 --- a/notes/06_fastica/5_fastica.tex +++ b/notes/06_fastica/5_fastica.tex @@ -1,3 +1,15 @@ +\subsection{Negentropy} + +\mode{ +\begin{frame} + \begin{center} \huge + \subsecname + \end{center} + \begin{center} + A more robust alternative to Kurtosis-based ICA + \end{center} +\end{frame} +} Negentropy $J(\widehat{s})$ of the reconstructed sources $\widehat{\vec s}$ is defined as: @@ -39,7 +51,7 @@ } \begin{itemize} - \itR theoretically well motivated measure. Considered in some cases the optimzal estimator for nongaussianity. + \itR theoretically well motivated measure. Considered in some cases the optimal estimator for nongaussianity. \itR non-negative \itR scale-invariant: $J(\alpha \widehat{s}) = J(\widehat{s}), \ \ \forall \alpha \ne 0$ (cf. exercise sheet) \itR \textbf{Problem:} requires estimation of density $p(\widehat{s})$ @@ -49,9 +61,9 @@ \end{frame} -\subsection{Approximations of negentropy} +\subsubsection{Approximations of negentropy} -\begin{frame} +\begin{frame}{\subsubsecname} \notesonly{ Estimating negentropy using the definition in \eqref{eq:negentropy} is computationally costly. It would require estimating the density of the random variable. We therefore resort to simpler approximations for negentropy. Such as the following use of cumulants: @@ -63,7 +75,7 @@ \subsection{Approximations of negentropy} \end{equation} \notesonly{ -For symmetric distributions the first term in the approximation in \eqref{eq:negentropyapprox} is effectivley zero, which makes the approximation equivalent to the square of the kurtosis. The approximation would therefore from the same sensitvity to outliers. +For symmetric distributions the first term in the approximation in \eqref{eq:negentropyapprox} is effectivley zero, which makes the approximation equivalent to the square of the kurtosis. The approximation would therefore from the same sensitivity to outliers. } \slidesonly{ @@ -79,10 +91,12 @@ \subsection{Approximations of negentropy} \end{frame} \clearpage -\begin{frame}{Common contrast functions} + +\subsubsection{Contrast functions} + +\begin{frame}{\subsubsecname} \notesonly{ -\textbf{Common contrast functions} The contrast function can be chosen depending on the assumed shape of the source densities. @@ -112,7 +126,7 @@ \subsection{Approximations of negentropy} \begin{frame} \slidesonly{ -\frametitle{Common contrast functions:} +\frametitle{Common contrast functions} } \slidesonly{ \smaller @@ -150,12 +164,12 @@ \subsection{Approximations of negentropy} \end{frame} \begin{frame} -cf. lecture slides for optmization of negentropy using contrast functions. +cf. lecture slides for optimization of negentropy using contrast functions. \end{frame} \begin{frame} \question{How do we evaluate ICA?}\\ --cf. https://research.ics.aalto.fi/ica/icasso/ +- Visualization methods\footnote{If interested cf. https://research.ics.aalto.fi/ica/icasso/} \end{frame} From ab8bcc0ecbdd652acc43775eb3ee3aaa7a7b39f2 Mon Sep 17 00:00:00 2001 From: Youssef Kashef Date: Mon, 1 Jun 2020 20:55:48 +0200 Subject: [PATCH 6/6] fixes and fixed point iteration --- notes/06_fastica/0_recap_ica_whitening.tex | 49 +++++++++++------- notes/06_fastica/2_pcaica.tex | 3 +- notes/06_fastica/4_kurt.tex | 31 +++++++++++ notes/06_fastica/5_fastica.tex | 45 ++++++++++++++-- notes/06_fastica/img/fixed_point_iter_cos.pdf | Bin 0 -> 12613 bytes notes/06_fastica/img/meme_maxormin.jpg | Bin 0 -> 79487 bytes notes/06_fastica/tutorial.tex | 5 ++ 7 files changed, 108 insertions(+), 25 deletions(-) create mode 100644 notes/06_fastica/img/fixed_point_iter_cos.pdf create mode 100644 notes/06_fastica/img/meme_maxormin.jpg diff --git a/notes/06_fastica/0_recap_ica_whitening.tex b/notes/06_fastica/0_recap_ica_whitening.tex index 0db31fc..d8b8ca0 100644 --- a/notes/06_fastica/0_recap_ica_whitening.tex +++ b/notes/06_fastica/0_recap_ica_whitening.tex @@ -24,28 +24,36 @@ \section{The ICA Problem} \end{itemize} \end{frame} -\begin{frame} -\underline{Outline:} -\begin{itemize} - \item ICA on whitened data - \begin{itemize} - \item Whitening/sphering - \item Amibguities in ICA - \item PCA is \emph{half} the ICA Problem - \end{itemize} - \item the problem with gaussians - \item maximizing nongaussianity - \begin{itemize} - \item Kurtosis-based - \item negentropy - \end{itemize} -\end{itemize} -\end{frame} - -\newpage +%\begin{frame} +%\underline{Outline:} +%\begin{itemize} + %\item ICA on whitened data + %\begin{itemize} + %\item Whitening/sphering + %\item Amibguities in ICA + %\item PCA is \emph{half} the ICA Problem + %\end{itemize} + %\item the problem with gaussians + %\item maximizing nongaussianity + %\begin{itemize} + %\item Kurtosis-based + %\item negentropy + %\end{itemize} +%\end{itemize} +%\end{frame} + +%\newpage \section{Whitening revisited} +\mode{ +\begin{frame} + \begin{center} \huge + \secname + \end{center} +\end{frame} +} + \begin{frame}{\secname} \notesonly{ @@ -87,6 +95,9 @@ \section{Whitening revisited} \pause + +\svspace{-5mm} + \begin{equation} \label{eq:covw} \vec \Sigma_v = \mathrm{Cov}(\vec v) = \E \lbrack \, \vec v \, \vec v^\top \rbrack = \vec I_N diff --git a/notes/06_fastica/2_pcaica.tex b/notes/06_fastica/2_pcaica.tex index 537088f..9c8b8c6 100644 --- a/notes/06_fastica/2_pcaica.tex +++ b/notes/06_fastica/2_pcaica.tex @@ -46,10 +46,11 @@ \subsection{Whitening in the context of ICA} \widetilde{\vec A} := \vec Q \, \vec A \end{equation} -Which yields\only<3>{ a new ICA problem}: +Which yields\only<3>{ a new ICA problem: \begin{equation} \vec u = \vec Q\, \vec x = \vec Q\,\vec A \, \vec s = \widetilde{\vec A} \, \vec s \end{equation} +} \end{frame} diff --git a/notes/06_fastica/4_kurt.tex b/notes/06_fastica/4_kurt.tex index 2bf96ac..8b339bc 100644 --- a/notes/06_fastica/4_kurt.tex +++ b/notes/06_fastica/4_kurt.tex @@ -466,6 +466,37 @@ \subsubsection{Kurtosis-based ICA: the gradient algorithm} } \end{frame} +\subsubsection{Fixed-point iteration} + +\begin{frame}{\subsubsecname} + +\begin{itemize} +\item A fixed point: $x=g(x)=g(g(\ldots g(x)\ldots))$\\ + +Example: +\begin{align} + g(x) &= x^{2} - 3x + 4\\ + g(2) &= 2^{2} - 3\cdot 2 + 4 = 4 - 6 + 4 = 2 +\end{align} + +\item Fixed point iteration: +\begin{equation} +x^{(t+1)} = g(x^{(t)}), t=0,1,2,\ldots +\end{equation} +\end{itemize} + +\begin{center} + \includegraphics[height=3.5cm]{img/fixed_point_iter_cos} + \mode

{ + \captionof{figure}{ + Fixed point iteration example + } + \label{fig:fixedpointcos} + } +\end{center} + +\end{frame} + \begin{frame} \notesonly{ diff --git a/notes/06_fastica/5_fastica.tex b/notes/06_fastica/5_fastica.tex index 01a29c9..b83e7db 100644 --- a/notes/06_fastica/5_fastica.tex +++ b/notes/06_fastica/5_fastica.tex @@ -10,12 +10,19 @@ \subsection{Negentropy} \end{center} \end{frame} } +\notesonly{ Negentropy $J(\widehat{s})$ of the reconstructed sources $\widehat{\vec s}$ is defined as: \begin{equation} -\label{eq:negentropy} - J(\widehat{s}) \coloneqq H(\widehat{s})_\normal - H(\widehat{s}) + J(\widehat{s}) \coloneqq \underbrace{ H(\widehat{s})_\normal}_{ + \substack{ \text{entropy of a Gaussian} \\ + \text{distribution with} \\ + \text{same variance}}} + - \underbrace{ H(\widehat{s}) }_{ + \substack{ \text{entropy of the true} \\ + \text{distribution} \\ + \text{(variance } \sigma^2 \text{)} }} \end{equation} where @@ -24,6 +31,7 @@ \subsection{Negentropy} \label{eq:diffentropyshat} H(\widehat{s}) := - \int p(\widehat{s}) \log p(\widehat{s}) d\widehat{s} \end{equation} +} \begin{frame} \frametitle{Negentropy} @@ -33,7 +41,7 @@ \subsection{Negentropy} $$ \begin{block}{Definition of negentropy} -\begin{equation*} +\begin{equation} J(\widehat{s}) \coloneqq \underbrace{ H(\widehat{s})_\normal}_{ \substack{ \text{entropy of a Gaussian} \\ \text{distribution with} \\ @@ -42,7 +50,7 @@ \subsection{Negentropy} \substack{ \text{entropy of the true} \\ \text{distribution} \\ \text{(variance } \sigma^2 \text{)} }} -\end{equation*} +\end{equation} \end{block} } @@ -57,8 +65,32 @@ \subsection{Negentropy} \itR \textbf{Problem:} requires estimation of density $p(\widehat{s})$ \end{itemize} +\end{frame} + +\begin{frame} +\frametitle{Optimizing negentropy} + +\slidesonly{ +\begin{equation} + J(\widehat{s}) \coloneqq \underbrace{ H(\widehat{s})_\normal}_{ + \substack{ \text{entropy of a Gaussian} \\ + \text{distribution with} \\ + \text{same variance}}} + - \underbrace{ H(\widehat{s}) }_{ + \substack{ \text{entropy of the true} \\ + \text{distribution} \\ + \text{(variance } \sigma^2 \text{)} }} +\end{equation} +} + \question{Should we minimize or maximize negentropy?} +\slidesonly{ +\begin{center} + \includegraphics[width=0.25\textwidth]{img/meme_maxormin}% +\end{center} +} + \end{frame} \subsubsection{Approximations of negentropy} @@ -164,7 +196,10 @@ \subsubsection{Contrast functions} \end{frame} \begin{frame} -cf. lecture slides for optimization of negentropy using contrast functions. +\begin{itemize} +\item[] cf. lecture slides for optimization of negentropy using contrast functions. +\item[] Hyv\"arinen, Karhunen \& Oja, Independent Component Analysis, Wiley, 2001 +\end{itemize} \end{frame} \begin{frame} diff --git a/notes/06_fastica/img/fixed_point_iter_cos.pdf b/notes/06_fastica/img/fixed_point_iter_cos.pdf new file mode 100644 index 0000000000000000000000000000000000000000..ba53cae77b0c723b70bc0febc740b4b782916147 GIT binary 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+\mode
{ \tableofcontents +} \end{frame} \newpage