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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture2.tex

@@ -30,7 +30,14 @@ One example of loss is the absolute loss: absolute difference between numbers\\
 \section{Loss}
 \subsection{Absolute Loss}
 $$\ell(y,\hat{y} = | y - \hat{y} |  \Rightarrow absolute \quad loss\\ $$
---- DISEGNO ---\\\\
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez2-img1.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
+\\
 Some inconvenient properties:
 
 \begin{itemize}
@@ -40,8 +47,14 @@ Some inconvenient properties:
 
 \subsection{Square Loss}
 $$ \ell(y,\hat{y} = ( y - \hat{y} )^2  \Rightarrow \textit{square loss}\\$$
--- DISEGNO ---\\
-Derivative :
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez2-img2.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
+\\Derivative :
 \begin{itemize}
 \item more informative
 \item and differentible 

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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture3.tex

@@ -19,8 +19,14 @@ Example $(x,y)$ \qquad y is the label associated with x\\
 Learning with example $(x_1,y_1)...(x_m,y_m)  \quad \textit{training set} $\\\\
 Training set is a set of examples with every algorithm can learn.......\\\\
 Learning algorithm take training set as input and produces a predictor as output.\\\\
- ......DISEGNO \\\\
-With image recognition we use as measurement pixels.\\
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.8\linewidth]{../img/lez3-img1.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
+\\With image recognition we use as measurement pixels.\\
 How do we measure the power of a predictor?\\
 A learning algorithm will look at training set, algorithm and generate the predictor. Now the problem is verify the score. \\
 Now we can consider a test set collection of example
@@ -194,20 +200,29 @@ $
 S is the traning set $(x_1,y_1)...(x_m,y_m) \\ x_t \in \barra{R}^d \qquad y_t \in \{-1,1\} \\\\
 d = 2 \rightarrow \textit{2-dimensional vector}\\
 $\\
-....-- DISEGNO --...
-\\
 where + and - are labels
-\\\\
+\newpage
 \textbf{Point of test set}
 \\
 If i want to predict this point?
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez3-img2.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
 Maybe if point is close to point with label i know then. Maybe they have the same label.
 \\
 $\hat{y} = + \quad or \quad  \hat{y} = - $
 \\\\
-.....-- DISEGNO -- ...
-\\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez3-img3.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
+\\
 I can came up with some sort of classifier.
 \\\\
 Given $S$ training set, i can define $\hnn$  $X \rightarrow \{-1,1\}\\

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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture4.tex

@@ -49,8 +49,12 @@ You can vary this algorithm as you want.\\\\
 Let’s go back to Binary classification.\\
 The $k$ parameter is the effect of making the structure of classifier more
 complex and less complex for small value of $k$.\\\\
---.. DISEGNO ..--
-\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez4-img1.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
 Fix training set and test set\\
 Accury as oppose to the error
 \\\\
@@ -83,6 +87,12 @@ I want to avoid comparing $x_i$ with $x_j$,  $i\neq j $\\
 so comparing different feature and we want to compare each feature with
 each self. I don’t want to mix them up.\\
 We can use a tree!
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.5\linewidth]{../img/lez4-img2.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
 \\
 I have 3 features:
 \begin{itemize}
@@ -90,17 +100,21 @@ I have 3 features:
 \item humidity $= \{[0,100]\}$
 \item windy $ = \{yes,no\}$
 \end{itemize}
-... -- DISEGNO -- ...\\\\
 Tree is a natural way of doing decision and abstraction of decision process of
 one person. It is a good way to deal with categorical variables.\\
 What kind of tree we are talking about?\\
 Tree has inner node and leaves. Leaves are associated with labels $(Y)$ and
 inner nodes are associated with test.
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.3\linewidth]{../img/lez4-img3.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
 \begin{itemize}
 \item Inner node $\rightarrow$ test
 \item Leaves $\rightarrow$ label in Y
 \end{itemize}
-%... -- DISEGNO -- ... 
 Test if a function $f$ (NOT A PREDICTOR!) \\
 Test $ \qquad f_i \, X_i \rightarrow \{1,...,k\}$
 \\ where $k$ is the number of children (inner node) to which test is assigned
@@ -146,19 +160,37 @@ $ Y = \{-1, +1 \}$
 \\\\
 What's the simplest way?\\
 Initial tree and correspond to a costant classifier
-\\\\
--- DISEGNO --
-\\\\
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.2\linewidth]{../img/lez4-img4.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
 \textbf{Majority of all example}
-\\\\
--- DISEGNO --
-\\\\
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.2\linewidth]{../img/lez4-img5.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
 $(x_1, y_1) ... (x_m, y_m)$ \\
 $ x_t \in X$ \qquad $ y_t \in \{-1,+1\}$\\
 Training set $S = \{ (x,y) \in S$, x is routed to $\ell\}$\\
 $S_{\ell}^+$ 
-\\\\
--- DISEGNO --
-\\\\
-$ S_{\ell}$ and $ S’_{\ell}$ are given by the result of the test, not the labels  and $\ell$ and $\ell'$.
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.8\linewidth]{../img/lez4-img6.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}\\
+$ S_{\ell}$ and $ S’_{\ell}$ are given by the result of the test, not the labels  and $\ell$ and $\ell'$.\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.7\linewidth]{../img/lez4-img7.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \end{document}

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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture5.tex

@@ -9,9 +9,14 @@ 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
-\\\\
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez5-img1.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 $$
 \sll \equiv \{(x_1,y_1) \, x_t \quad \textit{is router to } \ell \}
 $$
@@ -57,22 +62,39 @@ where $\psi(a) = min \{a, 1-a \} \qquad a \in [0,1] $
 \\
 I want to replace inner node with other leaves.
 \\
--- DISEGNO --
-\\\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.8\linewidth]{../img/lez5-img2.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 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).\\
-
+testing error goes to 0).
+\newpage
 \section{Jensen’s inequality}
-If $\psi$ is a concave function $\longrightarrow $ (like $log$ or $\sqrt[2]{..}$ )\\
+If $\psi$ is a concave function $\longrightarrow $ (like $log$ or $\sqrt[2]{..}$ )
+\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez5-img3.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 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 --
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.5\linewidth]{../img/lez5-img4.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
 $$ \hat{\ell}(h_T) = \frac{1}{m} \cdot \sum_{\ell}{} \psi  (\frac{\nl^+}{\nl}) \cdot \nl
 $$
@@ -89,8 +111,13 @@ $
 \\\\
 I want to check function $min$ concave between 0 and 1.\\
 $$min (0,1) = 0  \qquad \psi(a) = min(\alpha, 1- \alpha) $$
-\\ -- DISEGNO --
-\\\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez5-img5.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 \red{This is a concave function and now I can apply Jensen's inquality}
 \\\\
 $$
@@ -139,8 +166,13 @@ the approximately since we are not every time sure.
 \\\\
 --- MANCA PARTE ---
 \\
---- IMMAGINE ---
-\\\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.8\linewidth]{../img/lez5-img6.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 $ p = 0.8 \qquad q = 1 \qquad r = 1 \qquad \alpha = 60\%$
 \\
 Net Change in number of mistakes\\
@@ -155,9 +187,19 @@ Fraction of example miss classified $\ell -$ error $\ell' +$  error $\ell"$ \\
 $$
 = 0.2 - ( \frac{1}{2} \cdot 0.4 + \frac{1}{2} \cdot 0 ) = 0  
 $$
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.6\linewidth]{../img/lez5-img7.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez5-img8.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
---- DISEGNO ---
-\\\\
 Idea is to replace minimum function with convex combination.
 $$
 \psi(\alpha) = min\ \{\alpha, 1-\alpha\} \qquad \psi(a) \geq \psi(\alpha)
@@ -170,11 +212,15 @@ $$
 )}
 \end{cases}
 $$
-All this functions has this shape (concave???)\\
--- DISEGNO --
+All this functions has this shape (concave???)
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez5-img9.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
 In practise Machine Learning algorithm use GNI or entropy to control the split
-\\\\
 \section{Tree Predictor}
 \begin{itemize}
 \item Multi class classification $|Y| > 2$ $\longrightarrow$ \red{take majority}
@@ -189,7 +235,12 @@ 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 ----
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.7\linewidth]{../img/lez5-img10.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
 $$ if (x_1 = \alpha) \wedge (x_2 = \geq \alpha) \vee (x_1 = b) \vee (x_1 = c) \wedge (x_3= y) \qquad 
 $$

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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture6.tex

@@ -79,7 +79,14 @@ where red{$G(\hat{y})$ is equal to the part between $\left[...\right]$}
 $$
 \frac{d G(\hat{y})}{d\hat{y}} = 2 \cdot \hat{y}- 2 \cdot \barra{E} \left[ \, y | X= x \, \right] = 0  \quad \longrightarrow \quad \red{\textit{So setting derivative to 0}}
 $$
-\\ --- DISEGNO OPT CURVE ---\\\\
+Suppose we have a learning domain\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.1\linewidth]{../img/lez6-img0.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 $G' (\hat{y}) = \hat{y}^2 - 2\cdot b \cdot \hat{y}$
 \\
 $$
@@ -130,9 +137,14 @@ $$\
 $$
 D_{y|x} = \{ \eta(x), 1- \eta(x) \}
 $$
-\\
+\newpage
 Suppose we have a learning domain\\
---- DISEGNO --
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.6\linewidth]{../img/lez6-img1.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
 where $\eta$ is a function of $x$, so i can plot it\\
 $\eta$ will te me $Prob (x) = $
@@ -141,10 +153,15 @@ $\eta$ tells me a lot how hard is learning problem in the domain
 \\
 $\eta(x)$ is not necessary continous
 \\
---- DISEGNO ---
-\\\\
-$\eta(x) \in \{0,1\} $ \qquad $y$ is always determined by $x$
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.6\linewidth]{../img/lez6-img2.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
 \\
+$\eta(x) \in \{0,1\} $ \qquad $y$ is always determined by $x$
+\newpage
 How to get $f^*$ from the graph?
 \\
 $$
@@ -153,7 +170,13 @@ $$
 $$
 Y = \{-1, +1 \}
 $$
---- DISEGNO ---\\
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.7\linewidth]{../img/lez6-img3.JPG}
+    \caption{Example of domain of $\knn$}
+    %\label{fig:}
+\end{figure}
+\\
 ===============================\\
 MANCA ROBAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\\
 ==============================
@@ -217,10 +240,11 @@ $$\
 $$
 \barra{E} \left[ \, \ell , f^*(x) \, \right] = \barra{E} \left[ \, min \, \{ \eta(x) , 1- \eta(x) \} \, \right] 
 $$
+\\
 \begin{figure}[h]
     \centering
-    \includegraphics[width=1\textwidth]{bayesrisk.jpg}
-    \caption{Example of Bayes Risk}
+    \includegraphics[width=0.7\linewidth]{../img/lez6-img4.JPG}
+    \caption{Example of domain of $\knn$}
     %\label{fig:}
 \end{figure}
 \\

1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture7.log

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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture8.log

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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture9.log

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+\contentsline {section}{\numberline {8.2}Cross-validation}{55}%
+\contentsline {section}{\numberline {8.3}Nested cross validation}{57}%
+\contentsline {chapter}{\numberline {9}Lecture 9 - 07-04-2020}{58}%
+\contentsline {section}{\numberline {9.1}Tree predictors}{58}%
+\contentsline {subsection}{\numberline {9.1.1}Catalan Number}{60}%
+\contentsline {chapter}{\numberline {10}Lecture 10 - 07-04-2020}{64}%
+\contentsline {section}{\numberline {10.1}TO BE DEFINE}{64}%
+\contentsline {section}{\numberline {10.2}MANCANO 20 MINUTI DI LEZIONE}{64}%
+\contentsline {section}{\numberline {10.3}Compare risk for zero-one loss}{66}%
+\contentsline {chapter}{\numberline {11}Lecture 11 - 20-04-2020}{68}%
+\contentsline {section}{\numberline {11.1}Analysis of $K_{NN}$}{68}%
+\contentsline {subsection}{\numberline {11.1.1}Study of $K_{NN}$}{71}%
+\contentsline {subsection}{\numberline {11.1.2}study of trees}{72}%
+\contentsline {section}{\numberline {11.2}Non-parametric Algorithms}{73}%
+\contentsline {subsection}{\numberline {11.2.1}Example of parametric algorithms}{74}%
+\contentsline {chapter}{\numberline {12}Lecture 12 - 21-04-2020}{75}%
+\contentsline {section}{\numberline {12.1}Non parametrics algorithms}{75}%