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97 lines
2.9 KiB
TeX
97 lines
2.9 KiB
TeX
\documentclass[../main.tex]{subfiles}
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\begin{document}
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\chapter{Lecture 10 - 07-04-2020}
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\section{TO BE DEFINE}
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$|E[z] = |E[|E[z|x]]$
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\\\\
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$|E[X] = \sum_{t = 1}^{m} |E[x \Pi(A\begin{small}
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t \end{small} ) ]$
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\\\\
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$x \in \mathbb{R}^d
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$
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\\
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$\mathbb{P}(Y_{\Pi(s,x)} = 1) = \\\\ \mathbb{E}[\Pi { Y_{\Pi(s,x)} = 1 } ] = \\\\
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= \sum_{t = 1}^{m} \mathbb{E}[\Pi\{Y_t = 1\} \cdot \Pi { Pi(s,x) = t}] = \\\\
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= \sum_{t = 1}^{m} \mathbb{E}[\mathbb{E}[\Pi\{Y_t = 1\} \cdot \Pi\{\Pi(s,x) = t\} | X_t]] = \\\\
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given the fact that Y_t \sim \eta(X_t) \Rightarrow give me probability \\
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Y_t = 1 and \Pi(s,x) = t are independent given X_Y (e. g. \mathbb{E}[Zx] = \mathbb{E}[x] \ast \cdot \mathbb{E}[z]\\\\
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= \sum_{t = 1}^{m} \barra{E}[\barra{E}[\Pi\{Y_t = 1\}|X_t] \cdot \barra{E} [ \Pi(s,x) = t | Xt]] = \\\\
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= \sum_{t = 1}^{m} \barra{E}[\eta(X_t) \cdot \Pi \cdot \{\Pi (s,x) = t \}] = \\\\
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= \barra{E} [ \eta(X_{\Pi(s,x)}]
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$
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\[ \barra{P} (Y_{\Pi(s,x)}| X=x = \barra{E}[\eta(X_\Pi (s,x))] \]
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\\\\
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$
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\barra{P} (Y_{\Pi(s,x)} = 1, y = -1 ) = \\\\
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= \barra{E}[\Pi\{Y_{\Pi(s,x) }= 1\} \dot \Pi\{Y= -1|X\} ]] = \\\\
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= \barra{E}[\Pi \{ Y_{\Pi(s,x)} = 1\} \cdot \Pi \{ y = -1 \} ] = \\\\
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= \barra{E}[\barra{E}[\Pi \{ Y_{\Pi(s,x)} = 1\} \cdot \Pi \{ y = -1 | X \} ]] = \\\\
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$
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\[ Y_{\Pi(s,x)} = 1 \quad \quad y = -1 (1- \eta(x)) \quad when \quad X = x\]
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$
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\\\\ = \barra{E}[\barra{E}[\Pi \{Y_\Pi(s,x)\} = 1 | X] \cdot \barra{E}[\Pi \{y = -1\} |X ]] = \\\\
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= \barra {E}[\eta_{\Pi(s,x)} \cdot (1-\eta(x))] = \\\\
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similarly: \quad \barra{P}(Y_{\Pi(s,x)} = -1 , y = 1) = \\
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\barra{E} [(1- \eta_{\Pi(s,x)}) \cdot \eta(x)]
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\\\\
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\barra{E} [ \ell_D (\hat{h}_s)] = \barra{P}(Y_{\Pi(s,x)} \neq y ) =
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\\\\
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= \barra{P}(Y_{\Pi(s,x)} = 1, y = -1) + \barra{P}(Y_{Pi(s,x)} = -1, y = 1) =
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\\\\
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= \barra{E} [\eta_{\Pi(s,x)} \cdot (1-eta(x))] + \barra{E}[( 1- \eta_{\Pi(s,x)})\cdot \eta(x)]$
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\\\\
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Make assumptions on $D_x \quad and \quad \eta$: \\
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MANCAAAAAAA ROBAAA
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\\\\
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$
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\eta(x') <= \eta(x) + c || X-x'|| --> euclidean distance
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\\\\
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1-\eta(x') <= 1- \eta(x) + c||X-x'||
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\\\\
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$
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$
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X' = X_{Pi(s,x)}
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\\\\
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\eta(X) \cdot (1-\eta(x')) + (1-\eta(x))\cdot \eta(x') <=
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\\\\
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<= \eta(x) \cdot((1-\eta(x))+\eta(x)\cdot c||X-x'|| + (1-\eta(x))\cdot c||X-x'|| =
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\\\\
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= 2 \cdot \eta(x) \cdot (1- \eta(x)) + c||X-x'|| \\\\
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\barra{E}[\ell_d \cdot (\hat{h}_s)] <= 2 \cdot \barra{E} [\eta(x) - (1-\eta(x))] + c \cdot \barra(E)[||X-x_{\Pi(s,x)}||]
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$
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\\ where $<=$ mean at most
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\\\\
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Compare risk for zero-one loss
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\\
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$
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\barra{E}[min\{\eta(x),1-\eta(x)\}] = \ell_D (f*)
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\\\\
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\eta(x) \cdot( 1- \eta(X)) <= min\{\eta(x), 1-eta(x) \} \quad \forall x
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\\\\
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\barra{E}[\eta(x)\cdot(1-\eta(x)] <= \ell_D(f*)
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\\\\
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\barra{E}[\ell_d(\hat{l}_s)] <= 2 \cdot \ell_D(f*) + c \cdot \barra{E}[||X-X_{\Pi(s,x)}||]
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\\\\
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\eta(x) \in \{0,1\}
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$
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\\\\
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Depends on dimension: curse of dimensionality
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\\\\--DISEGNO--
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\\\\
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$
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\ell_d(f*) = 0 \iff min\{ \eta(x), 1-\eta(x)\} =0 \quad$ with probability = 1
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\\
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to be true $\eta(x) \in \{0,1\}$
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\end{document} |