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We do not collect datapoints to an independent +process.\\ +Example: identify new article and i want to put categories. The feed is highly +depend on what is happening in the world and there are some news highly +correlated. Why do we make an assumption that follows reality? +Is very convenient in mathematical term. +If you assume Independence you can make a lot of process in mathematical +term in making the algorithm.\\ +If you have enough data they look independent enough. Statistical learning is +not the only way of analyse algorithms —> we will see in linear ML algorithm +and at the end you can use both statistical model s +\section{Bayes Optimal Predictor} +$$ f^* : X \rightarrow Y$$ +$$ f^*(x) = argmin \, \barra{E}\left[ \, \ell(y,\hat{y})| X=x \, \right] \qquad \hat{y} \in Y$$ +\\ +In general $Y$ given $X$ has distribution $D_y|X=x$ +\\ +Clearly $\forall$ $h$ \quad $X\rightarrow Y$ +\\ +$$ +\barra{E} \left[ \, \ell(y, f^*(x)) | X=x \, \right] \leq \barra{E}\left[ \, \ell(y,h(x)| X = x \, \right] +$$ +$$ +X,Y \qquad \barra{E} \left[ \, Y|X = x \, \right] = F(x) \quad \longrightarrow \red{Conditional Expectation} +$$ +$$ +\barra{E} \left[ \, \barra{E} \left[ \, Y|X \, \right] \, \right] = \barra{E}(Y) +$$ +\\ +Now take Expectation for distribution +$$ +\barra{E} \left[ \, \ell(y, f^*(x))\, \right] \leq \left[ \, \barra{E} (\ell(y, h(x)) \, \right] +$$ +\\ where \red{risk is smaller in $f^*$} +\\ +I can look at the quantity before\\ +$l_d$ Bayes risk $\longrightarrow$ Smallest possible risk given a learning problm +\\\\ +$$ +l_d(f^*) > 0 \qquad \textit{because y are still stochastic given X} +$$ +\\ +Learning problem can be complem $\rightarrow$ large risk +\\\\ +\subsection{Square Loss} +$$\ell(y,\hat{y} = (y - \hat{y})^2$$ +I want to compute bayes optimal predictor\\ +$\hat{y}, y \in \barra{R}$ +\\ +$$ +f^*(x) = argmin \, \barra{E} \left[ \, (y-\hat{y})^2 | X = x \, \right] = \qquad \hat{y} \in \barra{R} +$$\ +$$ +\textit{we use }\qquad \barra{E}\left[\,X+Y\,\right] = \barra{E}[X] + \barra{E}[Y] = argmin \, \barra{E}\left[\,\red{y^2} + \hat{y}^2- 2\cdot y \cdot \hat{y}^2 | X = x \, \right] = +$$ +\\ +Dropping $\red{y^2}$ i remove something that is not important for $\hat{y}$ +\\ +$$ += argmin ( \barra{E} \left[\, y^2 | X = x\, \right] + \hat{y}^2 - 2 \cdot \hat{y} \cdot \barra{E} \left[ \, y | X = x \, \right] ) = +$$ +$$ += argmin (\hat{y}^2 - 2 \cdot \hat{y} \cdot \barra{E} \left[ \, y | X = x \, \right] ) = +$$ +\\ Expectation is a number, so it's a \red{constant} +\\ +Assume $ \boxdot = y^2 $ +$$ +argmin \, \left[\, \boxdot + \hat{y}^2 + 2 \cdot \hat{y} \cdot \barra{E} \left[\, Y|X =x\,\right] \right] +$$ +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 ---\\\\ +$G' (\hat{y}) = \hat{y}^2 - 2\cdot b \cdot \hat{y}$ +\\ +$$ +\hat{y} = \barra{E} \left[ \, y| X= x \, \right] \qquad f^*(x) = \barra{E} \left[ \, y | X = x \, \right] +$$ +\\ +Square loss is nice because expected prediction is ...\\ +In order to predict the best possibile we have to estimate the value given data +point. +\\ +$$ +\barra{E} \left[ \, (y- f^*(x))^2 | X = x \, \right] = +$$ +$$ += \barra{E} \left[ \, (y- \barra{E} \left[ \, y | X = x \,\right] )^2 | X = x \, \right] = Var \left[ \, Y | X = x \, \right] +$$ +\\ +\subsection{Zero-one loss for binary classification} +$ Y = \{-1,1\} +$ +$$ +\ell(y,\hat{y}) = I \{ \hat{y} \neq y \} +\qquad I_A (x) = +\begin{cases} +1 \quad x \in A +\\ +0 \quad x \not\in A +\end{cases} +$$ +\\ +\red{If $\hat{y} \neq y$ true, indicator function will give us 1, otherwise it will give 0} +\\ +$$ +D \quad on \qquad X \cdot Y \qquad D_x^* \quad D_{y|x} = D +$$\ +$$ +D_x \qquad \eta: X \longrightarrow \left[ \, 0,1 \, \right] \qquad \eta = \barra{P} \,(y = 1 | X = x ) +$$\ +$$ +D \leadsto (D_x, \eta) \quad \longrightarrow \quad \red{\textit{Distribution 0-1 loss}} +$$\ +$$ +X \backsim D_x \quad \longrightarrow \quad \red{ \textit{Where $\backsim$ mean "draw from" and $D_x$ is marginal distribution} } +$$ +$$ +Y = 1 \qquad \textit{ with probability } \eta(x) +$$\ +$$ +D_{y|x} = \{ \eta(x), 1- \eta(x) \} +$$ +\\ +Suppose we have a learning domain\\ +--- DISEGNO -- +\\ +where $\eta$ is a function of $x$, so i can plot it\\ +$\eta$ will te me $Prob (x) = $ +\\ +$\eta$ 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 $x$ +\\ +How to get $f^*$ from the graph? +\\ +$$ +f^+ : X \rightarrow \{-1,1\} +$$ +$$ +Y = \{-1, +1 \} +$$ +--- DISEGNO ---\\ +===============================\\ +MANCA ROBAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\\ +============================== +$$ +f^*(x) = argmin \, \barra{E} \left[ \, \ell(y, \hat{y}) | X= x\, \right] = \qquad \longrightarrow \hat{y} \in \{-1,+1 \} +$$ +$$ += argmin \, \barra{E} \left[ \, I\{\hat{y} = 1\} \cdot I\{Y=-1\} + I\{\hat{y}=-1\} \cdot I\{y=1\} \, | \, X = x \, \right] = +$$ +\\ +we are splitting wrong cases +\\ +$$ += argmin \, ( \, I\{\hat{y} = 1\} \cdot \barra{E} \left[ \, I\{Y=-1\} |\, X = x\, \right] + I\{\hat{y}=-1\} \cdot \barra{E} \left[ \, I\{y=1\} \, | \, X = x \, \right] \, ) = \quad \red{\divideontimes} +$$\\ +We know that: $$ \barra{E} \left[ \, I \{y = -1 \} \, | \, X = x \, \right] = 1 \cdot \barra{P} \\ (\hat{y} = -1 | X = x ) + 0 \cdot \barra{P} (y = 1 | X= x) = +$$ +$$ +\barra{P} (x = -1 | X=x ) = \, \red{ 1- \eta(x) } +$$\\ +$$ + \red{\divideontimes} = argmin \, ( \, \col{I\{\hat{y} = 1\} \cdot (1 - \eta(x))}{Blue} + \col{I \{ \hat{y} = -1\} \cdot (\eta(x)}{Orange} \, ) +$$ +where \col{Blue}{Blue} colored $I \{...\} = 1$° and \col{Orange}{Orange} $I \{...\} = 2$° +\\\\ +I have to choose \red{-1 or +1 } so we will \textbf{remove one of the two (1° or 2°) } +\\ +It depend on $\eta(x)$: +\begin{itemize} +\item If $\eta(x) < \frac{1}{2}$ \quad $\longrightarrow$ \quad kill 1° +\item Else $\eta(x) \geq \frac{1}{2}$ \quad $\longrightarrow$ \quad kill 2° +\end{itemize} +$$ +f^*(x) = +\begin{cases} ++1 \qquad if \, \eta(x) \geq \frac{1}{2}\\ +-1 \qquad if \, \eta(x) < \frac{1}{2} +\end{cases} +$$ +\section{Bayes Risk} + +$$ +\barra{E} \left[ \, I \{ y \neq f^*(x) \}\, | \, X = x \, \right] = \barra{P}(y \neq f^*(x)|X= x) +$$\ +$$ +\eta(x) \geq \frac{1}{2} \quad \Rightarrow \quad \hat{y} = 1 \quad \Rightarrow \quad \barra{P} (y \neq 1 | X= x) = 1-\eta(x) +$$\ +$$ +\eta(x) < \frac{1}{2} \quad \Rightarrow \quad \hat{y} = -1 \quad \Rightarrow \quad \barra{P} (y \neq 1 | X= x) = \eta(x) \quad +$$ +\\ +Conditiona risk for 0-1 loss is: +\\ +$$ +\barra{E} \left[ \, \ell (y, f^*(x)) \, | \, X = x \, \right] +\quad = \quad I \{ \eta(x) \geq \frac{1}{2}\} \cdot(1-\eta(x)) + I \{ \eta(x) <\frac{1}{2}\} \cdot \eta(x) = +$$ +$$ + = min \, \{ \eta(x), 1- \eta(x) \} +$$\ +$$ +\barra{E} \left[ \, \ell , f^*(x) \, \right] = \barra{E} \left[ \, min \, \{ \eta(x) , 1- \eta(x) \} \, \right] +$$ +\includegraphics{bayesrisk.jpg} +\\ +Conditional risk will be high aroun the half so min between the two is around +the half since the labels are random i will get an error near $50\%$.\\ +My condition risk will be 0 in the region in the bottom since label are going to +be deterministic. \end{document} diff --git a/1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/main.aux b/1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/main.aux index 0773b8ac4..7b7c46461 100644 --- a/1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/main.aux +++ b/1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/main.aux @@ -1,54 +1,58 @@ \relax \@nameuse{bbl@beforestart} \babel@aux{english}{} -\@writefile{toc}{\contentsline {chapter}{\numberline {1}Lecture 1 - 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