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lecture 5 completed
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\section{Lecture 10 - 07-04-2020}
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\subsection{TO BE DEFINE}
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@ -89,4 +92,6 @@ Depends on dimension: curse of dimensionality
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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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to be true $\eta(x) \in \{0,1\}$
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Output written on lecture4.pdf (4 pages, 125496 bytes).
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Output written on lecture4.pdf (4 pages, 125307 bytes).
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@ -6,7 +6,7 @@
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We spoke about Knn classifier with voronoi diagram
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$$
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\hat{\ell}(\hnn) = 0 \qquad \forall Traning set
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\hat{\ell}(\hnn) = 0 \qquad \forall \, \textit{Traning set}
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$$
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\\
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$\hnn$ predictor needs to store entire dataset.
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@ -16,7 +16,7 @@ Computing $\hnn(x)$ requires computing distances between x and points in the tra
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\\
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$$
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\Theta(d) \quad \textit{time for each distance}
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$$\\
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$$
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NN $\rightarrow$ 1-NN\\
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We can generalise NN in K-NN with $k = 1,3,5,7$ so odd $K$ \\
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@ -36,7 +36,7 @@ $K = 1$ error is 0.\\
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In the second line we switch to $k =3$. Second point doesn’t switch and third will
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be classify to positive and we have training mistake.\\
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Switches corresponds to border of voronoi partition.
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$$\knn \qquad \textit{For multiclass classification}$$\\
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$$\knn \qquad \textit{For multiclass classification}$$
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$$
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(|Y| > 2 ) \qquad \textit{for regression } Y\equiv \barra{R}
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$$
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\documentclass[../main.tex]{subfiles}
|
||||
\begin{document}
|
||||
\section{Lecture 5 - 07-04-2020}
|
||||
|
||||
\subsection{Tree Classifier}
|
||||
|
||||
Supposed we groped a tree up to this point and we are wandering how to
|
||||
grow it.
|
||||
\\
|
||||
$S$ Training set $(x_1,y_1)...(x_m,y_m)$, $x_1 \in X$
|
||||
\\\\
|
||||
-- DISEGNO
|
||||
\\\\
|
||||
$$
|
||||
\sll \equiv \{(x_1,y_1) \, x_t \quad \textit{is router to } \ell \}
|
||||
$$
|
||||
\\
|
||||
$y_1 \in \{-1,1\}$
|
||||
\\
|
||||
$$
|
||||
\sll ^+ \equiv \{(x_1,y_1) \in \sll : \quad y_t = +1 \}
|
||||
$$
|
||||
$$
|
||||
\sll^- \equiv \{(x_1,y_1) \in \sll : \quad y_t = -1 \}
|
||||
\qquad
|
||||
\sll^+ \cap \sll^- \equiv 0 \qquad \sll \equiv \sll^+ \cup \sll^-
|
||||
$$
|
||||
|
||||
$$
|
||||
\qquad \nl = | \sll|
|
||||
\qquad \nl^+ = |\sll^+|
|
||||
\qquad \nl^- = |\sll^-| $$
|
||||
$$
|
||||
\nl = \nl^- + \nl^+
|
||||
$$
|
||||
leaf $\ell$ classifies all traning example ($\sll$)
|
||||
\\
|
||||
$$
|
||||
Y_{\ell} =
|
||||
\begin{cases}
|
||||
+1, & \mbox{If } \nl^+ \geq \nl^-
|
||||
\\
|
||||
-1, & \mbox{If } otherwise
|
||||
\end{cases}
|
||||
$$
|
||||
\\
|
||||
$\ell$ makes a mistake on $min \{\nl^+, \nl^- \}$ example in $\sll$
|
||||
$$
|
||||
\hat{\ell}(h_T) = \frac{1}{m}
|
||||
\cdot \sum_{\ell}{} min \{ \frac{\nl^+}{\nl} , \frac{\nl^-}{\nl} \} \cdot \nl =
|
||||
$$
|
||||
$$
|
||||
= \frac{1}{m} \cdot \sum_{\ell}{}\psi \cdot (\frac{\nl+}{\nl}) \cdot \nl \quad \longrightarrow \qquad \frac{\nl^+}{\nl} = 1 - \frac{\nl}{\nl ??}
|
||||
$$
|
||||
where $\psi(a) = min \{a, 1-a \} \qquad a \in [0,1] $
|
||||
\\
|
||||
I want to replace inner node with other leaves.
|
||||
\\
|
||||
-- DISEGNO --
|
||||
\\\\
|
||||
How is traning error going to change?
|
||||
(when i replace inner nodes with other leaves)
|
||||
\\
|
||||
I’m hoping my algorithm is not going to overfit (if training error goes to 0 also
|
||||
testing error goes to 0).\\
|
||||
|
||||
\subsection{Jensen’s inequality}
|
||||
If $\psi$ is a concave function $\longrightarrow $ (like $log$ or $\sqrt[2]{..}$ )\\
|
||||
Also $\psi$ is a function that map $0$ to $1$, \quad $\longrightarrow$ \quad $\psi\:[0,1]\rightarrow \barra{R}$\\
|
||||
$$
|
||||
\psi(\alpha \cdot a + (1-\alpha) \cdot b ) \geq \alpha \cdot \psi(a) + (1-\alpha) \cdot \psi(b)
|
||||
\qquad \textit{Also 2° derivative is negative}$$
|
||||
\\
|
||||
-- DISEGNO --
|
||||
\\
|
||||
$$ \hat{\ell}(h_T) = \frac{1}{m} \cdot \sum_{\ell}{} \psi (\frac{\nl^+}{\nl}) \cdot \nl
|
||||
$$
|
||||
\\
|
||||
Look a single contribution fo a leaf $\ell$ to training error
|
||||
\\
|
||||
$$
|
||||
\psi(\frac{\nl^+}{\nl}) \cdot \nl = \psi(\frac{\nl'^+}{\nl'} \cdot \red{\frac{\nl'}{\nl}} + \frac{\nl"^+}{\nl"} \cdot \red{\frac{\nl"}{\nl}}) \cdot \nl
|
||||
$$
|
||||
where $\red{\frac{\nl'}{\nl}} = \alpha$ and $\red{\frac{\nl"}{\nl}} = 1-\alpha$ \qquad so \quad $\red{\frac{\nl'}{\nl}} + \red{\frac{\nl"}{\nl}} = 1$ \qquad $\longrightarrow$ $\alpha + 1 -\alpha = 1$
|
||||
\\\\
|
||||
$N_{\ell'}^+ + N_{\ell"}^+ = \nl
|
||||
$
|
||||
\\\\
|
||||
I want to check function $min$ concave between 0 and 1.\\
|
||||
$$min (0,1) = 0 \qquad \psi(a) = min(\alpha, 1- \alpha) $$
|
||||
\\ -- DISEGNO --
|
||||
\\\\
|
||||
\red{This is a concave function and now I can apply Jensen's inquality}
|
||||
\\\\
|
||||
$$
|
||||
\psi(\frac{\nl^+}{\nl}) \cdot \nl
|
||||
\geq
|
||||
(\frac{\nl'}{\nl} \cdot \psi (\frac{\nl'^+}{\nl'})
|
||||
+
|
||||
\frac{\nl"}{\nl}\cdot \psi(\frac{\nl"^+}{\nl"})) \cdot \nl =
|
||||
$$
|
||||
|
||||
$\qquad \qquad\qquad \qquad \quad= $
|
||||
\framebox[1.1\width]
|
||||
{
|
||||
$
|
||||
\psi(\frac{\nl'^+}{\nl'})\cdot \nl'
|
||||
+
|
||||
\psi(\frac{\nl"^+}{\nl"})\cdot \nl"
|
||||
$
|
||||
}
|
||||
\\\\
|
||||
\red{This are the contribuion of $\ell'$ and $\ell"$ to the training error}
|
||||
\\\\
|
||||
Every time i split my tree my training error is never going to increase since we
|
||||
have a concave function.\\
|
||||
Whenever I’m growing my tree training error is going to be smaller.\\\\
|
||||
\textbf{Every time a leaf is expanded the training error never goes up.
|
||||
(Hopelly will go down)}
|
||||
\\
|
||||
I’ll should always grow the tree by expanding leave that decrease the training
|
||||
error as much as possible.\\
|
||||
If i take the effort of growing the tree i should get benefits. I can imaging that if
|
||||
i grow the tree at random my training error is going to drop down error (but
|
||||
maybe will derive overfitting).\\
|
||||
For now is just an intuition since we will introduced statistical learning model.\\\\
|
||||
Could be complicated and tree big may have 100 leave and there could be
|
||||
many way of associating a test with that leaves.\\
|
||||
I can spent a lot of time to select which leave is the best promising to split.\\
|
||||
\begin{itemize}
|
||||
\item Grow the tree by expanding leave that decrease the training error as much
|
||||
as possible
|
||||
\item In general we can assume:\\greedy algorithm at each step pick the pair leaf and test that cause
|
||||
(approximative) the largest decrease in training error\\
|
||||
\end{itemize}
|
||||
In practise we want optimise this all the way since it’s time expensive. That’s
|
||||
the approximately since we are not every time sure.
|
||||
\\\\
|
||||
--- MANCA PARTE ---
|
||||
\\
|
||||
--- IMMAGINE ---
|
||||
\\\\
|
||||
$ p = 0.8 \qquad q = 1 \qquad r = 1 \qquad \alpha = 60\%$
|
||||
\\
|
||||
Net Change in number of mistakes\\
|
||||
$$
|
||||
\psi(p) - (\alpha \cdot \psi(q) + (1- \alpha ) \cdot \psi (r) ) =
|
||||
$$
|
||||
$$
|
||||
\red{
|
||||
\qquad \qquad \ell \quad- \qquad \ell' \quad+ \qquad \ell" \qquad \qquad \qquad \qquad
|
||||
}$$
|
||||
Fraction of example miss classified $\ell -$ error $\ell' +$ error $\ell"$ \\
|
||||
$$
|
||||
= 0.2 - ( \frac{1}{2} \cdot 0.4 + \frac{1}{2} \cdot 0 ) = 0
|
||||
$$
|
||||
\\
|
||||
--- DISEGNO ---
|
||||
\\\\
|
||||
Idea is to replace minimum function with convex combination.
|
||||
$$
|
||||
\psi(\alpha) = min\ \{\alpha, 1-\alpha\} \qquad \psi(a) \geq \psi(\alpha)
|
||||
$$
|
||||
$$
|
||||
\begin{cases}
|
||||
\psi_1(\alpha) = 2\cdot\alpha \cdot (1-\alpha) \longrightarrow \red{GNI} \\
|
||||
\psi_2(\alpha)= -\frac{\alpha}{2}\cdot \ln \alpha - \frac{1-\alpha}{2} \cdot \ln (1-\alpha)\longrightarrow \red{ENTROPY} \\
|
||||
\psi_3(\alpha) = \sqrt[]{\alpha \cdot (1-\alpha
|
||||
)}
|
||||
\end{cases}
|
||||
$$
|
||||
All this functions has this shape (concave???)\\
|
||||
-- DISEGNO --
|
||||
\\
|
||||
In practise Machine Learning algorithm use GNI or entropy to control the split
|
||||
\\\\
|
||||
\subsection{Tree Predictor}
|
||||
\begin{itemize}
|
||||
\item Multi class classification $|Y| > 2$ $\longrightarrow$ \red{take majority}
|
||||
\item Regression $Y = \barra{R} $ $\longrightarrow$ \red{take average of labels in $\sll$}
|
||||
\end{itemize}
|
||||
I still take majority among different classes.\\
|
||||
Take average of labels in $\sll$
|
||||
\\
|
||||
Unless $\frac{\nl^+}{\nl} \in {0,1} $ \qquad $\forall$ leaves $\ell$, $\hat{\ell}(h_T) > 0$
|
||||
\\
|
||||
Unless leaves are \textit{"pured"}, the training error will be bigger than 0.
|
||||
\\\\
|
||||
In general, i can always write $\hat{\ell}(h_t)$ to 0 by growing enough the tree unless there are $x_1$ in the Time Series such that $(x_t, y_t)(x_t,y_t’)$ with $y_t \neq y_t’$ both occur.
|
||||
\\
|
||||
--- DISEGNO ----
|
||||
\\
|
||||
$$ if (x_1 = \alpha) \wedge (x_2 = \geq \alpha) \vee (x_1 = b) \vee (x_1 = c) \wedge (x_3= y) \qquad
|
||||
$$
|
||||
$$
|
||||
\textit{then predict 1} \qquad \qquad
|
||||
$$
|
||||
$
|
||||
\qquad \quad \,\,else
|
||||
$
|
||||
$$
|
||||
\textit{then predict -1} \qquad \qquad
|
||||
$$
|
||||
\\
|
||||
--- Picture of tree classifier of iris dataset. ---\\
|
||||
I’m using due attribute at the time.\\
|
||||
Each data point is a flower and i can measure how petal and sepal are long.
|
||||
I can use two attribute and i test this two. I can see the plot of the tree
|
||||
classifier (second one) making test splitting data space into region that has
|
||||
this sort of “blackish” shape ( like boxes: blue box, red box, yellow box)\\
|
||||
A good exercise in which I want to reconstruct the tree given this picture.
|
||||
\\\\
|
||||
\subsection{Statistical model for Machine Learning}
|
||||
|
||||
To understand Tree classifier, nearest neighbour and other algorithm...\\
|
||||
It’s important to understand that the only way to have a guideline in which
|
||||
model to choose.\\\\
|
||||
\textbf{This mathematical model are developed to learning and choose learning
|
||||
algorithm.}\\\\
|
||||
Now let start with theoretical model.
|
||||
\begin{itemize}
|
||||
\item How example $(x,y)$ are generated to create test set and training set?\\
|
||||
We get the dataset but we need to have a mathematical model for this
|
||||
process.
|
||||
$(x,y)$ are drawn from a fixed but unknown probability distribution on the pairs $X$
|
||||
and $Y$ ($X$ data space, $Y$ label set o label space)
|
||||
\item Why $X$ should be random? \\
|
||||
In general we assumed that not all the $x$ in $X$ are equally likely to be observed.
|
||||
I have some distribution over my data point and this said that I’m most like to
|
||||
get a datapoint to another.
|
||||
\item How much label?\\
|
||||
Often labels are not determined uniquely by their datapoints because labels
|
||||
are given by human that have their subjective thoughts and also natural
|
||||
phenomena. Labels are stochastic phenomena given a datapoint: i will have a
|
||||
distribution.
|
||||
\end{itemize}
|
||||
We’re going to write (in capital) $(X, Y)$ since they are random variable drawn
|
||||
from $D$ on $X \cdot Y$
|
||||
A dataset $(X_1, Y_1) ... (X_m, Y_m)$ they are drawn independently from $D$
|
||||
(distribution on examples)\\
|
||||
When I get a training the abstraction of process collecting a training set\\
|
||||
$D$ is a joint probability distribution over $X\cdot Y$\\
|
||||
where $D_x$ is the marginal over $X \rightarrow D_y|x$ (conditional of $Y$ given $X$).\\
|
||||
I can divided my draw in two part.
|
||||
I draw sample and label from conditional.??\\
|
||||
Any dataset ( training or test ) is a random sample (campione casuale) in the
|
||||
statistical sense $\longrightarrow$ so we can use all stastical tools to make inference.
|
||||
\end{document}
|
||||
@ -1 +1,7 @@
|
||||
\documentclass[../main.tex]{subfiles}
|
||||
\begin{document}
|
||||
\section{Lecture 6 - 07-04-2020}
|
||||
|
||||
|
||||
|
||||
\end{document}
|
||||
|
||||
@ -1 +1,6 @@
|
||||
\documentclass[../main.tex]{subfiles}
|
||||
\begin{document}
|
||||
\section{Lecture 7 - 07-04-2020}
|
||||
|
||||
|
||||
\end{document}
|
||||
@ -1 +1,5 @@
|
||||
\documentclass[../main.tex]{subfiles}
|
||||
\begin{document}
|
||||
|
||||
\section{Lecture 8 - 07-04-2020}
|
||||
\end{document}
|
||||
@ -1 +1,6 @@
|
||||
\documentclass[../main.tex]{subfiles}
|
||||
\begin{document}
|
||||
\section{Lecture 9 - 07-04-2020}
|
||||
|
||||
|
||||
\end{document}
|
||||
@ -1,26 +1,34 @@
|
||||
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|
||||
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||||
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||||
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||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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||||
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||||
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[27]) [28] (lectures/lecture6.tex) [29] (lectures/lecture7.tex) [30]
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\usepackage{systeme}
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\usepackage[english]{babel}
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\usepackage[dvipsnames]{xcolor}
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\usepackage{framed}
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\newcommand\barra[1]{\mathbb{#1}}
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\newcommand\hnn{h_{NN}}
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\newcommand\hknn{h_{k-NN}}
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\newcommand\nl{N_{\ell}}
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\newcommand\red[1]{\textcolor{BrickRed}{#1}}
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\begin{document}
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\maketitle
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@ -1,23 +1,30 @@
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\contentsline {section}{\numberline {1}Lecture 1 - 09-03-2020}{3}%
|
||||
\contentsline {subsection}{\numberline {1.1}Introduction}{3}%
|
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\contentsline {section}{\numberline {2}Lecture 2 - 07-04-2020}{6}%
|
||||
\contentsline {subsection}{\numberline {2.1}Argomento}{6}%
|
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\contentsline {subsection}{\numberline {2.2}Loss}{6}%
|
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|
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\contentsline {subsubsection}{\numberline {2.2.2}Square Loss}{7}%
|
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\contentsline {subsubsection}{\numberline {2.2.3}Example of information of square loss}{7}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.4}labels and losses}{9}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.5}Example TF(idf) documents encoding}{10}%
|
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\contentsline {section}{\numberline {3}Lecture 3 - 07-04-2020}{12}%
|
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|
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\contentsline {subsection}{\numberline {3.3}Nearest neighbour}{16}%
|
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\contentsline {section}{\numberline {4}Lecture 4 - 07-04-2020}{18}%
|
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\contentsline {section}{\numberline {5}Lecture 5 - 07-04-2020}{19}%
|
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\contentsline {section}{\numberline {6}Lecture 6 - 07-04-2020}{20}%
|
||||
\contentsline {section}{\numberline {7}Lecture 7 - 07-04-2020}{21}%
|
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\contentsline {section}{\numberline {8}Lecture 8 - 07-04-2020}{22}%
|
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\contentsline {section}{\numberline {9}Lecture 9 - 07-04-2020}{23}%
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\contentsline {section}{\numberline {10}Lecture 10 - 07-04-2020}{24}%
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||||
\contentsline {subsection}{\numberline {10.1}TO BE DEFINE}{24}%
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\babel@toc {english}{}
|
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\contentsline {section}{\numberline {1}Lecture 1 - 09-03-2020}{4}%
|
||||
\contentsline {subsection}{\numberline {1.1}Introduction}{4}%
|
||||
\contentsline {section}{\numberline {2}Lecture 2 - 07-04-2020}{7}%
|
||||
\contentsline {subsection}{\numberline {2.1}Argomento}{7}%
|
||||
\contentsline {subsection}{\numberline {2.2}Loss}{7}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.1}Absolute Loss}{7}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.2}Square Loss}{8}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.3}Example of information of square loss}{8}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.4}labels and losses}{10}%
|
||||
\contentsline {subsubsection}{\numberline {2.2.5}Example TF(idf) documents encoding}{11}%
|
||||
\contentsline {section}{\numberline {3}Lecture 3 - 07-04-2020}{13}%
|
||||
\contentsline {subsection}{\numberline {3.1}Overfitting}{15}%
|
||||
\contentsline {subsubsection}{\numberline {3.1.1}Noise in the data}{15}%
|
||||
\contentsline {subsection}{\numberline {3.2}Underfitting}{17}%
|
||||
\contentsline {subsection}{\numberline {3.3}Nearest neighbour}{17}%
|
||||
\contentsline {section}{\numberline {4}Lecture 4 - 07-04-2020}{19}%
|
||||
\contentsline {subsection}{\numberline {4.1}Computing $h_{NN}$}{19}%
|
||||
\contentsline {subsection}{\numberline {4.2}Tree Predictor}{20}%
|
||||
\contentsline {section}{\numberline {5}Lecture 5 - 07-04-2020}{23}%
|
||||
\contentsline {subsection}{\numberline {5.1}Tree Classifier}{23}%
|
||||
\contentsline {subsection}{\numberline {5.2}Jensen’s inequality}{24}%
|
||||
\contentsline {subsection}{\numberline {5.3}Tree Predictor}{26}%
|
||||
\contentsline {subsection}{\numberline {5.4}Statistical model for Machine Learning}{27}%
|
||||
\contentsline {section}{\numberline {6}Lecture 6 - 07-04-2020}{29}%
|
||||
\contentsline {section}{\numberline {7}Lecture 7 - 07-04-2020}{30}%
|
||||
\contentsline {section}{\numberline {8}Lecture 8 - 07-04-2020}{31}%
|
||||
\contentsline {section}{\numberline {9}Lecture 9 - 07-04-2020}{32}%
|
||||
\contentsline {section}{\numberline {10}Lecture 10 - 07-04-2020}{33}%
|
||||
\contentsline {subsection}{\numberline {10.1}TO BE DEFINE}{33}%
|
||||
|
||||
Loading…
Reference in New Issue
Block a user