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\documentclass[../main.tex]{subfiles}
|
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\begin{document}
|
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|
||||
\chapter{Lecture 19 - 18-05-2020}
|
||||
|
||||
$$
|
||||
k(x,x') = < \phi(x), \phi(x')>
|
||||
\qquad \phi: X \rightarrow H
|
||||
$$
|
||||
where $X \rightarrow \barra{R}^2$ and $H \rightarrow barra{R}^N $
|
||||
$$
|
||||
H_\delta = \{ \sum_{i =1}^N \alpha_i \, k_\delta (x_i, \cdot), x_1,..., x_N \in \barra{R}^d, \alpha_1, ... \alpha_N, N \in \barra{N } \}
|
||||
$$
|
||||
Inner product measures "similarities" between data points.
|
||||
\\
|
||||
$$
|
||||
x^T \, x' = \|x\| \, \|x'\| \, \cos \Theta \qquad x \in X \quad k(x,x')
|
||||
$$
|
||||
$k$ sais how much similar are the structure (tree, documents etc).
|
||||
\\
|
||||
I would like to learn a predictor based on the notion of similarity.
|
||||
\\
|
||||
$$
|
||||
k(x,x') = < \phi(x), \phi(x')>
|
||||
$$
|
||||
where $<>$ is the inner product.
|
||||
\\
|
||||
So we have Data $\rightarrow$ Kernel $\rightarrow$ Kernel learning Algortithm
|
||||
\\
|
||||
Kernels offer a uniform interface to data in such way they algoriithm can learn from data.
|
||||
\\
|
||||
Given $K$ on $X$, I need to find $\exists H_k \quad \phi_k \ X \rightarrow H_k$
|
||||
\\
|
||||
$\exists <...>_k$ s.t $k(x,x') = <\phi_k(x), \phi_k(x')>_k $
|
||||
\\\\
|
||||
\bred{Theorem}
|
||||
\\
|
||||
Given $K: X \times X \rightarrow \barra{R}$, symmetric
|
||||
\\
|
||||
Then $K$ is a Kernel iif $\forall m \in \barra{N}$ $\forall x_1,...,x_m \in X$
|
||||
\\
|
||||
The $m \times m$ matrix $K$ \quad $K_{ij} = k (x_i,x_j)$ is positive semidefinite\\
|
||||
$
|
||||
\forall \alpha \in \barra{R}^m \qquad \alpha^T \, K \, \alpha \geq 0
|
||||
$
|
||||
\\
|
||||
In general, given a Kernel $K$ there is not unique representation for $\phi_k$ and $<...>_k$ (inner product).
|
||||
\\
|
||||
However, there is a "canonical" representation:
|
||||
$
|
||||
\phi_k(x) = K(x, \cdot)
|
||||
$
|
||||
$$
|
||||
\phi_k : X \rightarrow H \qquad H_k = \{ \sum_{i=1}^N \alpha_i \, k (x_i, \cdot ), \alpha_1,..., \alpha_N \in \barra{R}, x_1,...,x_N \in X, N \in \barra{N} \}
|
||||
$$
|
||||
We have to define an inner product like:
|
||||
$$
|
||||
<\phi_k(x), \phi_k(x')>_k \ = \ k(x,x')
|
||||
$$
|
||||
This is the canonical representation that helps mapping.
|
||||
\\\\
|
||||
What happen to use this mechanism to perform predictions?
|
||||
\\
|
||||
$
|
||||
x \in \barra{R}^d \ w \in \barra{R}^d \ w^T \, x \qquad \textit{\ where } g = \sum_{i=1}^N \alpha_i \, k (x_i, \cdot)
|
||||
$
|
||||
$$
|
||||
\phi_k(x) \qquad g \in H_k \qquad <g, \phi_k(x)>_k \ = \ <\sum_i \alpha_i k(x_i, \cdot), \phi_k(x)> \ = $$
|
||||
We have to satisfy allinearity
|
||||
$$
|
||||
= \ \sum_i \alpha_i <k(x_i, \cdot), k(x, \cdot) >_k \ = \ \sum_i \alpha_i <\phi(x_i), \phi_k(x)>_k \ = \ \sum_i \alpha_i k(x_i, x) = g(x)
|
||||
$$
|
||||
At the end we have:
|
||||
$$
|
||||
<g, \phi_k(x)>_k \ = \ g(x)
|
||||
$$
|
||||
\\\\
|
||||
Now, if i have two functions:
|
||||
$$
|
||||
f = \sum_{i=1}^N \alpha_i \, k(x_i, \cdot) \qquad g = \sum_{j=1}^M \beta_j \, k (x'_j, \cdot) \qquad f,g \in H_k
|
||||
$$
|
||||
$$
|
||||
<f,g>_k = <\sum_i \alpha_i \, k(x_i,\cdot) , \sum_j \beta_j \, k(x'_j, \cdot) >_k \ =
|
||||
\ \sum_i \sum_j \alpha_i \, \beta_j <k(x_i, \cdot), k(x'_j, \cdot>_k \ =
|
||||
$$
|
||||
$$
|
||||
= \ \sum_i \sum_j \alpha_i \, \beta_j \, k(x_i, x_j)
|
||||
$$
|
||||
$$
|
||||
\|f\|^2 = <f,f>_k = \sum_{ij} \alpha_i \, \alpha_j \, k(x_i, x_j)
|
||||
$$
|
||||
Perceptron convergence theorem in kernel space:
|
||||
$$
|
||||
M \leq \|U\|^2 ( \max_t \|x_t\| ^2) \qquad \forall u \in \barra{R}^d \quad y_t \, u^T \, x_t \geq 1 quad \forall g \in H_k \quad y_t \, g(x_t) \geq 1
|
||||
$$
|
||||
we know that:
|
||||
$$
|
||||
\|x_t \|^2 \rightsquigarrow \| \phi_k(x_t)\|^2_k \ = \ <\phi_k(x_t), \phi_k(\alpha_t) >_k \ = \ k (x_t,x_t)
|
||||
$$
|
||||
so
|
||||
\\
|
||||
.... MANCA ULTIMA FORUMA
|
||||
\\\\
|
||||
Ridge regression:
|
||||
$$
|
||||
w = \left( \alpha \, I + S^T \, S \right)^{-1} \, S^T \, y
|
||||
$$
|
||||
$S$ is $m \times d$ matrix whose rows are the training points $x_1,..., x_m \in \barra{R}^d$
|
||||
\\
|
||||
$y = (y_1,...,y_m) \quad y_t \in \barra{R}^d$ training labels $\alpha >0$
|
||||
$$
|
||||
\left( \alpha \, I + S^T \, S \right)^{-1} \, S^T \ = \ S^T \left( \alpha \, I_m + S\, S^T\right)^{-1}
|
||||
$$
|
||||
where $d \times d$ and $d \times m $ = $ d \times
|
||||
m$ and $m \times m$
|
||||
$$
|
||||
\left( S \, S^T\right)_{ij} = x_i^T x_j \qquad \rightsquigarrow \ <\phi(x_i),\phi(x_j)>_k = k(x_i, x_j) = K_{ij}
|
||||
$$
|
||||
$$
|
||||
S^T = \left[ x_1,...,x_m \right] \ \rightsquigarrow\ \left[ \ \phi_k(x_i),..., \phi_k(x_m) \ \right] = \left[ \ k(x_1, \cdot), ..., k(k_m, \cdot) \ \right] \ = \ k(\cdot)
|
||||
$$
|
||||
$$
|
||||
k (\cdot)^T \, \left( \alpha \, I_m + K \right)^{-1} \, y \ = \ g
|
||||
$$
|
||||
where $1 \times
|
||||
m$ and $m \times
|
||||
m$ and $m \times
|
||||
1$\\\\
|
||||
How to compute prediction?
|
||||
$$
|
||||
g(x) = y^T \left( \alpha \, I_m + K \right)^{-1} \, k(x)
|
||||
$$
|
||||
\qquad $1 \times m$ and $m \times m$ and $m \times 1$
|
||||
\\
|
||||
In fact, is the evaltuation of $g$ in any point $x$.
|
||||
\\
|
||||
The drawback is that we pass from $d \times d$ matrix to a $m \times m$ matrix that can be huge. So it is not really efficient in this way, we need to use addictional "tricks" having a more compact representation of the last matrix prediction.
|
||||
\newpage
|
||||
\section{Support Vector Machine (SVM)}
|
||||
It is a linear predictor and is a very popular one because has better performance than perceptron and we will see it for classification but there are also version for regression.
|
||||
\\\\
|
||||
The idea here is that you want to come up with an hyperplane that is defined as a solution of an optimisation problem.
|
||||
\\
|
||||
We have a classification dataset $(x_1,y_1)...(x_m,y_m) \qquad x_t \in \barra{R}^d \quad y_t \in \{ -1,1 \}$ and it is linearly separable.
|
||||
\\
|
||||
Sum as the solution $w^*$ (optimisation problem) to this problem:
|
||||
$$
|
||||
\min_{w \in \barra{R}^d} \frac{1}{2} \| w \|^2 \qquad s.t \quad y_t \, w^T \, x_t \geq 1 \quad t = 1,2,...,m
|
||||
$$
|
||||
Geometrically $w^*$ corresponds to the maximum marging separating hyperplane like:
|
||||
$$
|
||||
\gamma^* = \max_{u: \|u\|=1} y_t \, u^t \, x_t \qquad t=1,...,m
|
||||
$$\
|
||||
\textbf{$u^*$ is achieving $\gamma^*$ is the maximal margin separator.}\\
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.4\linewidth]{../img/lez19-img1.JPG}
|
||||
\caption{Draw of SVG}
|
||||
%\label{fig:}
|
||||
\end{figure}\\
|
||||
So I want to maximise this distance.
|
||||
$$
|
||||
\max_{\gamma > 0} \, \gamma^2 \qquad s.t \quad \| u \|^2 = 1 \qquad y_t \, u^t \, x_t \geq \gamma \quad t=1,...,m
|
||||
$$
|
||||
So we can maximise instead of minimising.
|
||||
\\
|
||||
What is the theorem? The equivalent between this two.
|
||||
\\\\
|
||||
\bred{Theorem}:\\
|
||||
$\forall$ linear separator $(x_1,y_1)...(x_m,y_m)$ \\
|
||||
The max margin separator $u^*$ satisfies $u^* = \gamma^* \, w^*$ where $w^*$ is the SVM solution and $\gamma^*$ is the maximum margin.
|
||||
\end{document}
|
@ -0,0 +1,6 @@
|
||||
\documentclass[../main.tex]{subfiles}
|
||||
\begin{document}
|
||||
|
||||
\chapter{Lecture 20 - 19-05-2020}
|
||||
|
||||
\end{document}
|
@ -196,5 +196,13 @@
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[]
|
||||
|
||||
[116] (main.bbl
|
||||
[116] (lectures/lecture19.tex
|
||||
Chapter 19.
|
||||
|
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|
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1/cmr/m/n/12 = [] \OML/cmm/m/it/12 [] < \OT1/cmr/m/n/12 (\OML/cmm/m/it/12 x[
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|
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OT1/cmr/m/n/12 (\OML/cmm/m/it/12 x[]; x\OT1/cmr/m/n/12 ) = \OML/cmm/m/it/12 g\O
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[118]
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[120 <./lectures/../img/lez19-img1.JPG>] (lectures/lecture20.tex
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Chapter 20.
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\contentsline {subsection}{\numberline {18.1.1}Feature expansion}{112}%
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\contentsline {subsection}{\numberline {18.1.2}Kernels implements feature expansion (Efficiently}{113}%
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\contentsline {chapter}{\numberline {19}Lecture 19 - 18-05-2020}{117}%
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\contentsline {chapter}{\numberline {20}Lecture 20 - 19-05-2020}{121}%
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Loading…
Reference in New Issue
Block a user