Factor Analysis Derivation
Forfatter
Daniel Kurniadi
Sidst opdateret
5 år siden
Licens
LaTeX Project Public License 1.3c
Resumé
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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