Factor Analysis Derivation
Author
Daniel Kurniadi
Last Updated
5年前
License
LaTeX Project Public License 1.3c
Abstract
Just another Factor Analysis derivation. Refer to Stanford Lecture Notes CS229.
Just another Factor Analysis derivation. Refer to Stanford Lecture Notes CS229.
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\Large Factor Analysis
\end{center}
\section{Review of Unsupervised Learning}
Recall that in unsupervised learning, we're given $m$ unlabelled training set $x^{(i)}\in$ $\mathbb{R}^n$ ($0<i<m$) that comes from mixture of Gaussians, and we would like to model the density of $P(x)=\sum_{z} P(x | z) . P(z)$ where $z^{(i)}\in$ $\mathbb{R}^k$ is a latent variable that denote which distribution does $x^{(i)}$ belongs to and thus, we assume there are $k$ distributions.
Here, $z^{(i)}$ ~ $Multinomial(\phi)$, where $\phi_{j}$ > 0 and $\sum_{j} = 1$ and ($X^{(i)}|z^{(i)}=j)$ ~ $\mathcal{N}(\mu_{j},
\Sigma_{j})$
$\begin{aligned}
\ l(\phi, \mu, \sigma) &= \sum_{i=1}^{m}\log p(x^{(i)}, \phi, \mu, \Sigma) \ \\
&= \sum_{i=1}^{m}\log \sum_{z^{(i)}=1}^{k} p(x^{(i)}|z^{(i)}; \mu, \Sigma) . p(z^{(i)}; \phi) \ \textrm{} \\
\end{aligned}$
The EM algorithm can be applied to fit a mixture model.
1. E- step:
$\begin{aligned}
\ w_{j} &= P(z^{(i)}|x^{(i)}, \phi, \mu, \Sigma) \ \\
\ &= \frac{P(x^{(i)}|z^{(i)}=j) . P(z^{(i)}=j)}{\sum_{l=1}^{k}P(x^{(i)})|z^{(i)}} . P(z^{(i)}=l)\ \\
\ &= \frac{\frac{1}{{\sigma \sqrt {2\pi } }} exp[(x^{(i)}-\mu_{j}^{(i)})^T . \sum_{j}^-1 . (x^{(i)} - \mu_{j})] . \phi_{j}}
{\sum_{l=1}^{k}(\frac{1}{{\sigma \sqrt {2\pi } }} exp[(x^{(i)}-\mu_{j}^{(i)})^T . \sum_{j}^-1 . (x^{(i)} - \mu_{j})]) } \ \\
\end{aligned}$
2. M- step:
$\begin{aligned}
\phi_{j}&=\dfrac{1}{m} \sum_{i=1}^{m} w_{j}^{(i)}\\
\end{aligned}$; $\begin{aligned}
\Sigma_{j}&=\dfrac{\sum_{i=1}^{m} w_{j}^{(i)}.(x^{(i)}-\mu_{j}).(x^{(i)}-\mu_{j})^T}{\sum_{i=1}^{m} w_{j}^{(i)}}\\
\end{aligned}$
$\begin{aligned}
\mu_{j}&=\dfrac{\sum_{i=1}^{m} w_{j}^{(i)} . x^{(i)}}{\sum_{i=1}^{m} w_{j}^{(i)}};
\end{aligned}$
\subsection{Problem Statement}
Now suppose that we have $n>>m$, we will find that $\Sigma$ is singular matrix. This means $\Sigma^{-1}$ is doesn't not exist and we find ${1}/{|\Sigma|^{\frac{1}{2}}}$ = $1/0$. Those terms are needed to compute EM algorithm (refer to E-step).
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