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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture12.tex

@@ -5,6 +5,213 @@
 
 \section{Non parametrics algorithms}
 
+We talk about \bred{consistency}: as the training size grows unbounded the expected risk of algorithms converge to Bayes Risk.
+\\\\
+Now we talk about \bred{non parametric algorithm}: the structure of the model is determined by the data.\\
+Structure of the model is fixed, like the structure of a Neural Network but in non parametric algorithm will change structure of the model as the data grows ($\knn$ and tree predictor).\\
+If I live the tree grow unboundenly then we get a non parametric tree, but if we bound the grows then we get a parametric one.
+\\\\
+The converve rate of Bayes Risk (in this case doubled) was small.
+Converge of $1$-$NN$ to $2 \, \ell_D(f^*)$ is $m^{-\frac{1}{d+1}}$ so we need an esponential  in the dimension. And we need this is under Lips assumption of $\eta$.
+\\ It's possible to converge to Bayes Risk and it's called \bred{No free lunch}.
+\subsection{Theorem: No free lunch}
+Let a sequenece of number 
+$a_1, a_2$ ... $\in \barra{R} $ 
+such that they converge to 0. 
+\\Also $\frac{1}{22222222} \geq a_1 \geq a_2 \geq ...$ $\forall A $ 
+for binary classification $\exists D$ s. t. 
+\\$\ell_D(f^*) = 0 $ (zero-one loss) so Bayes risk is zero and 
+\expt{\ell_D\left(A(S_M)\right)} $\geq a_m \quad \forall m \geq 1$
+\\
+Any Bayes Optimal you should be prepared to do so on long period of time. This means that:
+\begin{itemize}
+\item For specific data distribution $D$, then $A$ may converge fast to Bayes Risk.
+\item If $\eta$ is Lipschitz then it is continous. This mean that we perturb the input by the output doesno't change too much.
+\item If Bayes Risk is 0 ($\ell_D(f^*) = 0$) function will be discontinous
+\end{itemize}
+This result typically people think twice for using consistent algorithm because
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.6\linewidth]{../img/lez12-img1.JPG}
+    \caption{Tree building}
+    %\label{fig:}
+\end{figure}\\
+\\
+I have Bayes risk and some non conssitent algorithm that will converge to some value ($\ell_D(\hat{h}^*)$).
+Maybe i have Bayes risk and the convergence takes a lot on increasing data points. Before converging was better non parametric (?..)
+\\\\
+Picture for binary classification, (similar for other  losses)
+\begin{itemize}
+\item Under no assumption on $\eta$, the typicall "parametric" converge rate to risk of best model in $H$ (including ERM) is $m^{-\frac{1}{2}}$. (Bias error may be high)
+\item Under no assumption on $\eta$ there is no guaranteed convergence to Bayes Risk (in general) and this is \bred{no-free-lunch} that guaranteed me no convergence rate.
+\item Under Lipshtiz assunption on $\eta$ the typical non parametric convergence to Bayes Risk is $m^{-\frac{1}{d+1}}$. This is exponentially worst than the parametric convergency rate.
+\end{itemize}
+The exponential depencendece on $d$ is called \bred{Curse of dimnsionality}.
+\\ But if I assume small number of dimension $\longrightarrow$ $\knn$ is ok if $d$ is small (and $\eta$ is "easy")
+\\
+If you have a non parametric algorithm (no Bayes error but may have exponentially infinity training error).
+I want them to be balanced and avoid bias and variance. We need to introduce a bit of bias in controlled way.
+\\
+Inserting bias to reducing variance error. So we sacrify a bit to get a better variance error.
+\\\\
+It could be good to inject bias in order to reduce the variance error. In practise instead of having 0 training error i want to have a larger training error and hope to reduce overfitting sacrifing a bit in the training error.
+\\
+I can increase bias in different technics: one is the unsamble methods.
+
+
+\section{Highly Parametric Learning Algorithm}
+
+\subsection{Linear Predictors}
+Our domain is Euclidean space (so we have points of numbers).
+\\
+$$
+X \ is \ \barra{R}^d \qquad x = (x_1,..,x_d)
+$$
+A linear predictor will be a linear function of the data points.
+$$
+h: \barra{R}^d \longrightarrow Y \qquad h\left(x\right) = f(w^T \, x) \quad w \in \barra{R}^d
+$$
+$$
+f: \barra{R} \longrightarrow Y
+$$
+And this is the dot product that is
+$$
+w^T \, x = \sum_{t = 1}^{d} w_i x_i = \| w \| \, \| x \| \cos \Theta
+$$
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.2\linewidth]{../img/lez12-img2.JPG}
+    \caption{Dot product}
+    %\label{fig:}
+\end{figure}\\
+Suppose we look a regression with square loss.\\
+$$ Y = \barra{R} \qquad h(x) = w^T\, x \quad w \in \barra{R}^d
+$$
+$
+f^*(x) = $\expt{ Y| X=x }
+\\
+Binary classification with zero-one loss 
+$ Y = \{ -1,1\}$
+We cannot use this since is not a real number but i can do:
+$$
+h(x) = sgn\left(w^T\, x\right) \qquad sgn(x) = \begin{cases}
++1 \ if \ z > 0\\
+-1 \ if \ z \leq 0 
+\end{cases}
+$$
+where sgn is a sign function.
+Linear classifier.
+\\
+$\| X \| \cos \Theta $ is the length of the projection of $x$ onto $w$
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.2\linewidth]{../img/lez12-img3.JPG}
+    \caption{Dot product}
+    %\label{fig:}
+\end{figure}\\
+Now let's look at this set:
+$$
+\{ x \in \barra{R}^d : w^Tx = c\}
+$$
+This is a hyperplane.
+$$ 
+\|w \| \| x \| \cos \Theta = c
+\qquad
+\|x \| \cos \Theta = \frac{c}{\| w\|}
+$$
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.4\linewidth]{../img/lez12-img4.JPG}
+    \caption{Hyperplane}
+    %\label{fig:}
+\end{figure}\\
+So $ (w,c)$ describe an hyperplane.
+\\
+We can do binary classification using the hyperplane. Any points that lives in the positive half space and the negative. So the hyperplane is splitting in halfs.
+$ H \equiv \{ x \in \barra{R}^d : w^T x = c \}$\
+$$
+H^+ \equiv \{x \in \barra{R}^d : w^Tx > c \} \qquad \textbf{positive $h_s$}
+$$
+$$
+H^- \equiv \{x \in \barra{R}^d : w^Tx \leq \} \qquad \textbf{negative $h_s$}
+$$\
+$$ h(x) = 
+\begin{cases}
++1 \ if \ x \in H^+\\
+-1 \ if \ x \not\in H^-
+\end{cases} \qquad
+h(x) = sgn (w^T -c)
+$$
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.6\linewidth]{../img/lez12-img5.JPG}
+    \caption{Hyperplane}
+    %\label{fig:}
+\end{figure}
+\newpage
+$h_1$ is non-homogenous linear classifier.\\
+$h_2$ is homogenous linear classifier.
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.3\linewidth]{../img/lez12-img6.JPG}
+    \caption{Hyperplane}
+    %\label{fig:}
+\end{figure}
+Any homogenous classifier is equivalent to this:
+$$
+\{x \in \barra{R}^d : X = c \} \textbf{ is equivalent to \quad} \{x: \barra{R}^{d+1} : \nu^T x = 0 \}
+$$
+$$
+\nu = (w_1,..,w_d, -c) \qquad x' = (x_1,..., x_d, 1)
+$$
+So we added a dimension.
+$$
+w^T x = c \ \Leftrightarrow \ \nu^T x' = 0 
+$$
+$$
+\sum_{i} w_1 x_1 = c \ \Leftrightarrow \ \sum_{i} w_1 x_1 -c = 0
+$$\\\\
+\bred{Rule}:\\
+\textbf{When you learn predictor just add an extra feature to your data points, set it ot 1 and forget about non- homogenous stuff.}
+\newpage
+One dimensional example
+\begin{figure}[h]
+    \centering
+    \includegraphics[width=0.6\linewidth]{../img/lez12-img7.JPG}
+    \caption{Example of one dimensional hyperplane}
+    %\label{fig:}
+\end{figure}\\
+I have negative (left of $(x,1)$ and positive point (left of $(z,1$) classified 
+\\\\
+Now i want to learn linear classifier. How can i do it?
+$$
+H_d = \{ \ h : \exists w \in \barra{R}^d \ h(x) = sgn(w^T x) \ \}
+$$
+Parametric!
+\\
+We expect high bias a low variance.
+\\
+$$
+ERM  \qquad \hat{h}_S = arg \min_{h \in H_d} \ \frac{1}{m} \cdot \sum_{t = 1}^{m} I \{h(x_t) \neq y_t \}  = 
+$$
+$$
+= \  arg \min_{w \in \barra{R}^d} \ \frac{1}{m} \cdot \sum_{t = 1}^{m} I \, \{ \, y_t \, w^T x_t \leq 0 \, \} 
+$$
+A bad optimisation problem!
+\\\\
+\bred{FACT}:\\
+It is unlikely to find an algorithm that solves ERM for $H_d$ and zero-one loss efficiently. 
+\\
+\bred{NP completeness problems!}
+\\ It's very unlikely to solve this problem.
+\\
+This problem is called \textbf{MinDisagreement}
+\\
+\subsection{MinDisagreement}
+Instance: $(x_1, y_1) ... (x_m, y_m) \in \{ 0,1 \}^d \, x \, \{-1, 1 \}, \quad k \in \barra{N}$\\
+Question: Is there $w \in x \, D^d $ \\ s.t $y_t \, w^T   x_t \leq 0$ for at most $k$ indices $t \in \{1,...m\}$
+\\
+This is NP-complete!
 %\expt{\ell(\hat{\ell}) + 2}
 
 

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 es/MiKTeX 2.9/fonts/type1/public/amsfonts/symbols/msbm10.pfb>
-Output written on main.pdf (77 pages, 2082324 bytes).
+Output written on main.pdf (79 pages, 2109982 bytes).
 PDF statistics:
- 806 PDF objects out of 1000 (max. 8388607)
+ 816 PDF objects out of 1000 (max. 8388607)
  0 named destinations out of 1000 (max. 500000)
- 246 words of extra memory for PDF output out of 10000 (max. 10000000)
+ 251 words of extra memory for PDF output out of 10000 (max. 10000000)
 

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

@@ -40,7 +40,7 @@
 \titlespacing*{\chapter}{0pt}{-80pt}{40pt}
 \chapterfont{\color{Blue}}
 \sectionfont{\color{DarkGreen}}
-\subsectionfont{\color{BrickRed}}
+\subsectionfont{\color{red}}
 
 
   

1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/main.toc

@@ -52,3 +52,5 @@
 \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}%
+\contentsline {subsection}{\numberline {12.1.1}Theorem: No free lunch}{75}%
+\contentsline {section}{\numberline {12.2}Highly Parametric Learning Algorithm}{77}%