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completed lecture 6
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Chapter 1.
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\begin{document}
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\chapter{Lecture 6 - 07-04-2020}
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$(X, Y)$ We random variables drawn iid from $D$ on $X \cdot Y$ $\longrightarrow$ where $D$ is fixed but unknown\\\\
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Independence does not hold. We do not collect datapoints to an independent
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process.\\
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Example: identify new article and i want to put categories. The feed is highly
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depend on what is happening in the world and there are some news highly
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correlated. Why do we make an assumption that follows reality?
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Is very convenient in mathematical term.
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If you assume Independence you can make a lot of process in mathematical
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term in making the algorithm.\\
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If you have enough data they look independent enough. Statistical learning is
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not the only way of analyse algorithms —> we will see in linear ML algorithm
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and at the end you can use both statistical model s
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\section{Bayes Optimal Predictor}
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$$ f^* : X \rightarrow Y$$
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$$ f^*(x) = argmin \, \barra{E}\left[ \, \ell(y,\hat{y})| X=x \, \right] \qquad \hat{y} \in Y$$
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\\
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In general $Y$ given $X$ has distribution $D_y|X=x$
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\\
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Clearly $\forall$ $h$ \quad $X\rightarrow Y$
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\\
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$$
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\barra{E} \left[ \, \ell(y, f^*(x)) | X=x \, \right] \leq \barra{E}\left[ \, \ell(y,h(x)| X = x \, \right]
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$$
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$$
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X,Y \qquad \barra{E} \left[ \, Y|X = x \, \right] = F(x) \quad \longrightarrow \red{Conditional Expectation}
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$$
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$$
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\barra{E} \left[ \, \barra{E} \left[ \, Y|X \, \right] \, \right] = \barra{E}(Y)
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$$
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\\
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Now take Expectation for distribution
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$$
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\barra{E} \left[ \, \ell(y, f^*(x))\, \right] \leq \left[ \, \barra{E} (\ell(y, h(x)) \, \right]
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$$
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\\ where \red{risk is smaller in $f^*$}
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\\
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I can look at the quantity before\\
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$l_d$ Bayes risk $\longrightarrow$ Smallest possible risk given a learning problm
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\\\\
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$$
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l_d(f^*) > 0 \qquad \textit{because y are still stochastic given X}
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$$
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\\
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Learning problem can be complem $\rightarrow$ large risk
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\\\\
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\subsection{Square Loss}
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$$\ell(y,\hat{y} = (y - \hat{y})^2$$
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I want to compute bayes optimal predictor\\
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$\hat{y}, y \in \barra{R}$
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\\
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$$
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f^*(x) = argmin \, \barra{E} \left[ \, (y-\hat{y})^2 | X = x \, \right] = \qquad \hat{y} \in \barra{R}
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$$\
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$$
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\textit{we use }\qquad \barra{E}\left[\,X+Y\,\right] = \barra{E}[X] + \barra{E}[Y] = argmin \, \barra{E}\left[\,\red{y^2} + \hat{y}^2- 2\cdot y \cdot \hat{y}^2 | X = x \, \right] =
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$$
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\\
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Dropping $\red{y^2}$ i remove something that is not important for $\hat{y}$
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\\
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$$
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= argmin ( \barra{E} \left[\, y^2 | X = x\, \right] + \hat{y}^2 - 2 \cdot \hat{y} \cdot \barra{E} \left[ \, y | X = x \, \right] ) =
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$$
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$$
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= argmin (\hat{y}^2 - 2 \cdot \hat{y} \cdot \barra{E} \left[ \, y | X = x \, \right] ) =
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$$
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\\ Expectation is a number, so it's a \red{constant}
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\\
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Assume $ \boxdot = y^2 $
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$$
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argmin \, \left[\, \boxdot + \hat{y}^2 + 2 \cdot \hat{y} \cdot \barra{E} \left[\, Y|X =x\,\right] \right]
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$$
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where red{$G(\hat{y})$ is equal to the part between $\left[...\right]$}
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$$
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\frac{d G(\hat{y})}{d\hat{y}} = 2 \cdot \hat{y}- 2 \cdot \barra{E} \left[ \, y | X= x \, \right] = 0 \quad \longrightarrow \quad \red{\textit{So setting derivative to 0}}
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$$
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\\ --- DISEGNO OPT CURVE ---\\\\
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$G' (\hat{y}) = \hat{y}^2 - 2\cdot b \cdot \hat{y}$
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\\
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$$
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\hat{y} = \barra{E} \left[ \, y| X= x \, \right] \qquad f^*(x) = \barra{E} \left[ \, y | X = x \, \right]
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$$
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\\
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Square loss is nice because expected prediction is ...\\
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In order to predict the best possibile we have to estimate the value given data
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point.
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\\
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$$
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\barra{E} \left[ \, (y- f^*(x))^2 | X = x \, \right] =
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$$
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$$
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= \barra{E} \left[ \, (y- \barra{E} \left[ \, y | X = x \,\right] )^2 | X = x \, \right] = Var \left[ \, Y | X = x \, \right]
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$$
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\\
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\subsection{Zero-one loss for binary classification}
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$ Y = \{-1,1\}
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$
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$$
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\ell(y,\hat{y}) = I \{ \hat{y} \neq y \}
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\qquad I_A (x) =
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\begin{cases}
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1 \quad x \in A
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\\
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0 \quad x \not\in A
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\end{cases}
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$$
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\\
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\red{If $\hat{y} \neq y$ true, indicator function will give us 1, otherwise it will give 0}
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\\
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$$
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D \quad on \qquad X \cdot Y \qquad D_x^* \quad D_{y|x} = D
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$$\
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$$
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D_x \qquad \eta: X \longrightarrow \left[ \, 0,1 \, \right] \qquad \eta = \barra{P} \,(y = 1 | X = x )
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$$\
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$$
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D \leadsto (D_x, \eta) \quad \longrightarrow \quad \red{\textit{Distribution 0-1 loss}}
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$$\
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$$
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X \backsim D_x \quad \longrightarrow \quad \red{ \textit{Where $\backsim$ mean "draw from" and $D_x$ is marginal distribution} }
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$$
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$$
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Y = 1 \qquad \textit{ with probability } \eta(x)
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$$\
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$$
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D_{y|x} = \{ \eta(x), 1- \eta(x) \}
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$$
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\\
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Suppose we have a learning domain\\
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--- DISEGNO --
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\\
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where $\eta$ is a function of $x$, so i can plot it\\
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$\eta$ will te me $Prob (x) = $
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\\
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$\eta$ tells me a lot how hard is learning problem in the domain
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\\
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$\eta(x)$ is not necessary continous
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\\
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--- DISEGNO ---
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\\\\
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$\eta(x) \in \{0,1\} $ \qquad $y$ is always determined by $x$
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\\
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How to get $f^*$ from the graph?
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\\
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$$
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f^+ : X \rightarrow \{-1,1\}
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$$
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$$
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Y = \{-1, +1 \}
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$$
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--- DISEGNO ---\\
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===============================\\
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MANCA ROBAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\\
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==============================
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$$
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f^*(x) = argmin \, \barra{E} \left[ \, \ell(y, \hat{y}) | X= x\, \right] = \qquad \longrightarrow \hat{y} \in \{-1,+1 \}
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$$
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$$
|
||||
= argmin \, \barra{E} \left[ \, I\{\hat{y} = 1\} \cdot I\{Y=-1\} + I\{\hat{y}=-1\} \cdot I\{y=1\} \, | \, X = x \, \right] =
|
||||
$$
|
||||
\\
|
||||
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}
|
||||
|
||||
@ -1,54 +1,58 @@
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This is pdfTeX, Version 3.14159265-2.6-1.40.21 (MiKTeX 2.9.7300 64-bit) (preloaded format=pdflatex 2020.4.13) 13 APR 2020 12:51
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||||
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||||
(main.tex
|
||||
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|
||||
|
||||
[1
|
||||
|
||||
]
|
||||
] [2]
|
||||
(lectures/lecture1.tex
|
||||
Chapter 1.
|
||||
|
||||
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]
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[]
|
||||
|
||||
[3]) [4] (lectures/lecture2.tex
|
||||
[4]) [5] (lectures/lecture2.tex
|
||||
Chapter 2.
|
||||
|
||||
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[6]
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[7]
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[9]
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[]
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) [11] (lectures/lecture3.tex
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||||
Chapter 3.
|
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[11
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[12
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[13]
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[]
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) [16] (lectures/lecture4.tex
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) [17] (lectures/lecture4.tex
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Chapter 4.
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[17
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[18
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[19]
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[19] [20] (lectures/lecture5.tex
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[20] [21] (lectures/lecture5.tex
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||||
Chapter 5.
|
||||
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Loading…
Reference in New Issue
Block a user