[eng] Indepent component analysis. Linear discriminant analysis.tex

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\textbf{ICA}

$ X: n \times d \\ $

$ X = UDV^{\top} \\ $

$ U^{\top} \sqrt{n} = S \\ $

$\frac{1}{\sqrt{n}}SV^{\top} = A^{\top} \\$

X = AS \\

$X = ARR^{\top}S \\$

$ X = A^{*}S^{*} \\ $

$\bar{x} = 0, \omega v(x) = I \\$

S - независ.$ \rightarrow \omega v(S) = I \\$
X = AS \\

Задача: \\
$S = A^{\top}X\\$

$H(y) = \int g(y) log(g) dy \\$

$I(y) = \sum_{j=1}^d H(y_i) - H(x) - log(A) \\$

$\min_{y} I(y)$ \\



 \textbf{Linear discriminant analysis} \\

1) $k=1, K, \pi_k, F_k(x) \\ $
$ \max_{k} \Pr(C=k|X=x) = \frac{\pi_k f_k(x)}{\sum_{l} \pi_l f_l(x)} \\$
 
 $\mu$ - mean
 $\Sigma$ - dispersion

 $f_k=\frac{1}{(2\pi)^{\frac{d}{2}} \mid \Sigma_k \mid ^{\frac{1}{2}}} e^{-(x - \mu_k)^{\top} \Sigma_k^{-1}(x- \mu_k)} \\$

 $log (\frac{\Pr(C=k|X=x)}{\Pr(C=l|X=x)}) = log({\frac{\pi_k}{\pi_l}}) - \frac{1}{2}(\mu_k + \mu_l)\Sigma^{-1}(\mu_k + \mu_l) + x^{\top}\Sigma^{-1}(\mu_k - \mu_l) \\$

 2) $\hat{\pi_k} = \frac{\sum C=k}{\pi} \\ $
 
 
  $\bar{\mu_k} = \frac{\sum I(C=k)x}{\sum I(C=k)x} \\ $
  
  
  $\Sigma=\frac{1}{n} \sum_{k} \sum_{i=1}^n I(C=k) \\$

 $W = \sum_{k} \sum I(C=k)(x-\mu_k)^{\top}(x-\mu_k) \\
  M = 
  \begin{vmatrix}
  \mu_1 \\
  \ldots \\
  \mu_n
  \end{vmatrix}$ \\

$ \bar{M} = \frac{1}{k}\sum \hat{\mu_k} \\$


$B = (M-\bar{M})^{\top}(M-\bar{M}) \\$

$\max_{a} \frac{a^{\top}Ba}{a^{\top}wa}$ - Fisher distribution \\

$(x - \hat{\mu_k})^{\top}(x - \hat{\mu_k})$ \\

\textbf{GSVD} \\

$B = U_B D_B V^{\top} \\ $

$ W = U_w D_w V^{\top} \\$

$a^{\top}V D_B U_B^{\top} U_B D_B V^{\top} a = b^{\top} D_B^{\top}b$ \\


\textbf{Common Spatial Patterns} \\

$X: n \times c \times t ; 
y = n \times 1 $ \\
n - experiments; c - channels; t - time \\

K = 2 \\
$P^{(k)} = \frac{1}{n_k} \sum_{l \in (y=k)} X(i, \cdot, \cdot) \\
P \in c \times t $ \\

$max_\omega = \frac{\omega P^{(l)}P^{(1)^{\top}}\omega^{\top}}{\omega(P^{(1)}P^{(1)^{\top}} + P^{(2)}P^{(2)^{\top}})\omega^{\top}} $ \\
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