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1year/3trimester/Machine Learning, Statistical Learning, Deep Learning and Artificial Intelligence/Machine Learning/lectures/lecture11.aux
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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) 20 APR 2020 09:36
This is pdfTeX, Version 3.14159265-2.6-1.40.21 (MiKTeX 2.9.7300 64-bit) (preloaded format=pdflatex 2020.4.13) 20 APR 2020 11:33
entering extended mode
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(lecture11.tex
@ -318,700 +318,259 @@ LaTeX Font Info: Trying to load font information for U+msb on input line 6.
("C:\Program Files\MiKTeX 2.9\tex/latex/amsfonts\umsb.fd"
File: umsb.fd 2013/01/14 v3.01 AMS symbols B
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! Missing delimiter (. inserted).
<to be read again>
{
l.8 \barra{E} \left{
\ell_d} (\hat{\ell}_s ) \right] \leq 2 \cdot \ell_D \lef...
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`\}' here. If you typed, e.g., `{' instead of `\{', you
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Acceptable delimiters are characters whose \delcode is
nonnegative, or you can use `\delimiter <delimiter code>'.
<../img/lez11-img1.JPG, id=1, 214.55156pt x 150.5625pt>
File: ../img/lez11-img1.JPG Graphic file (type jpg)
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! Missing } inserted.
<inserted text>
}
l.8 ...ra{E}\left[ \, \| X = x_{\Pi(s,x) \| \right
]
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Underfull \hbox (badness 10000) in paragraph at lines 9--34
Underfull \hbox (badness 10000) in paragraph at lines 9--39
[]
Underfull \hbox (badness 10000) in paragraph at lines 9--34
Underfull \hbox (badness 10000) in paragraph at lines 9--39
[]
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Underfull \hbox (badness 10000) in paragraph at lines 9--39
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<to be read again>
\varepsilon
l.56 ...| \right] \leq \barra{E} \left \varepsilon
\sqrt[]{d} \sum_{i = 1}^{...
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! Extra \right.
l.56 ...i \} \cdot I\{C_i \cap S \neq 0 \} \right]
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MS/cmsy/m/n/12  [] [] \OML/cmm/m/it/12 I\OMS/cmsy/m/n/12 f\OML/cmm/m/it/12 X \
OMS/cmsy/m/n/12 2 \OML/cmm/m/it/12 C[]\OMS/cmsy/m/n/12 g  \OML/cmm/m/it/12 I\O
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[]
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}
l.59 ..._i \} I \{c_1 \cap S \neq 0 \} \right] }
+ 2 \sqrt[]{d} \sum_{i = ...
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\@textcolor ...otect \leavevmode {\color #1{#2}#3}
l.59 ..._i \} I \{c_1 \cap S \neq 0 \} \right] }
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l.61 ...i \} I \{c_1 \cap S \neq 0 \} \right] $}
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File: ../img/lez11-img2.JPG Graphic file (type jpg)
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<to be read again>
$
l.68 $
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A number should have been here; I inserted `0'.
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l.74 \barra{E} \left \I
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The control sequence at the end of the top line
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misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
! Missing \right. inserted.
<inserted text>
\right .
l.75 $
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With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
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Underfull \hbox (badness 10000) in paragraph at lines 45--65
[]
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l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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\textdef@ ...th {#1}\let \f@size #2\selectfont #3}
}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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deleted material, e.g., by typing `I$}'.
Underfull \hbox (badness 10000) in paragraph at lines 45--65
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\textdef@ ...h {#1}\let \f@size #2\selectfont #3}}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\text@ ...e {\textdef@ \displaystyle \f@size {#1}}
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l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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deleted material, e.g., by typing `I$}'.
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l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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you left one out. Proceed, with fingers crossed.
! Extra }, or forgotten $.
\textdef@ ...th {#1}\let \f@size #2\selectfont #3}
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l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\textdef@ ...h {#1}\let \f@size #2\selectfont #3}}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\text@ ...xtstyle \f@size {\firstchoice@false #1}}
{\textdef@ \textstyle \sf@...
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
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<inserted text>
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l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Extra }, or forgotten $.
\textdef@ ...th {#1}\let \f@size #2\selectfont #3}
}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\textdef@ ...h {#1}\let \f@size #2\selectfont #3}}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\text@ ...tstyle \sf@size {\firstchoice@false #1}}
{\textdef@ \textstyle \ssf...
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Missing $ inserted.
<inserted text>
$
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
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you left one out. Proceed, with fingers crossed.
! Extra }, or forgotten $.
\textdef@ ...th {#1}\let \f@size #2\selectfont #3}
}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\textdef@ ...h {#1}\let \f@size #2\selectfont #3}}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\text@ ...style \ssf@size {\firstchoice@false #1}}
\check@mathfonts }
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\text@ ...firstchoice@false #1}}\check@mathfonts }
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
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spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\@textcolor ...otect \leavevmode {\color #1{#2}#3}
l.93 ...0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}}
=
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
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<inserted text>
$
l.95 where $\bred{p \, e ^{-m\,p}}
$ is $F(p)$
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Extra }, or forgotten $.
<recently read> \egroup
l.95 where $\bred{p \, e ^{-m\,p}}
$ is $F(p)$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Extra }, or forgotten $.
\@textcolor ...otect \leavevmode {\color #1{#2}#3}
l.95 where $\bred{p \, e ^{-m\,p}}
$ is $F(p)$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
! Missing $ inserted.
<inserted text>
$
l.99 F^
(p) = 0 \Leftrightarrow p = \frac{1}{m} \quad check!
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Missing $ inserted.
<inserted text>
$
l.108 \end{document}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
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\newpage ...everypar {}\fi \par \ifdim \prevdepth
>\z@ \vskip -\ifdim \prevd...
l.108 \end{document}
You can refer to \spacefactor only in horizontal mode;
you can refer to \prevdepth only in vertical mode; and
neither of these is meaningful inside \write. So
I'm forgetting what you said and using zero instead.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing $ inserted.
<inserted text>
$
l.108 \end{document}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing $ inserted.
<inserted text>
$
l.108 \end{document}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing $ inserted.
<inserted text>
$
l.108 \end{document}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
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<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
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<inserted text>
$
l.108 \end{document}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing { inserted.
<to be read again>
$
l.108 \end{document}
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing { inserted.
<to be read again>
$
l.108 \end{document}
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing { inserted.
<to be read again>
$
l.108 \end{document}
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Missing } inserted.
<inserted text>
}
l.108 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
! Display math should end with $$.
<to be read again>
\vfil
l.108 \end{document}
The `$' that I just saw supposedly matches a previous `$$'.
So I shall assume that you typed `$$' both times.
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@ -5,7 +5,7 @@
\section{Analysis of $\knn$}
$$
\barra{E} \left{\ell_d} (\hat{\ell}_s ) \right] \leq 2 \cdot \ell_D \left( f^* \right) + c \cdot \barra{E}\left[ \, \| X = x_{\Pi(s,x) \| \right]
\barra{E} \left[{\ell_D} (\hat{\ell}_s ) \right] \leq 2 \cdot \ell_D \left( f^* \right) + c \cdot \barra{E}\left[ \, \| X = x_{\Pi(s,x)} \| \, \right]
$$
At which rate this thing goes down? If number of dimension goes up then a lot of point are far away from $X$. \\
So this quantity must depend on the space in which X live.
@ -16,7 +16,12 @@ We are going to use the assumption that:
\\
$| X_t | \leq 1 \qquad \forall $ cordinates $i = 1, ..., d$
\\
--- DISEGNO ---
\begin{figure}[h]
\centering
\includegraphics[width=0.6\linewidth]{../img/lez11-img1.JPG}
\caption{Example of domain of $\knn$}
%\label{fig:}
\end{figure}\\
\\
Hyper box in bydimension. All point live in this box and we exploit that.
Look at the little suare in which is divided and we assume that we are dividing the box in small boxes of size $\varepsilon$. Now the training points will be a strincle of point distributed in the big square. \\
@ -32,74 +37,128 @@ How big is this when we have this two cases?
(We lookjing at specific choices of x and s)
\\
$$
\| X - X_{s,x} \| leq
\| X - X_{s,x} \| \leq \
\begin{cases}
\varepsilon \sqrt[]{d} \qquad c_i \cup S \neq 0 \\
\sqrt[]{d} \qquad c_i \cup S = 0
\varepsilon \ \sqrt[]{d} \qquad C_i \cup S \neq 0 \\
\sqrt[]{d} \qquad C_i \cup S = 0
\end{cases}
$$
were $X \in C_i$
\\
We have to multiply by the lenght of the cube.
Will be $\varepsilon \sqrt[]{d}$
Will be $\varepsilon \ \, \sqrt[]{d}$
\\
-- DISEGNO ---
\begin{figure}[h]
\centering
\includegraphics[width=0.1\linewidth]{../img/lez11-img2.JPG}
\caption{Diagonal length}
%\label{fig:}
\end{figure}\\
\\
If things go badly can be very far away like the length of the domain.
Lenght is $2$ and diagonal is $ \sqrt[]{d}$
Lenght is $2$ and diagonal is $ \ \sqrt[]{d}$
\\
if close they are going to be $\varepsilon close$ or far as domain.
\\
We can split that the expression inside the expectation according to the two cases.
\\
$$
\barra{E} \left[ \| X - X_{\Pi(s,x)} \| \right] \leq \barra{E} \left \varepsilon \sqrt[]{d} \sum_{i = 1}^{r} I \{ X \in C_i \} \cdot I \{ C_1\cap S \neq 0 \} \right] + 2 \cdot \sqrt[]{d} \sum_{i=1}^{r} I \{ X \in C_i \} \cdot I\{C_i \cap S \neq 0 \} \right] =
\barra{E} \left[ \, \| X - X_{\Pi(s,x)} \| \, \right]
\leq
$$
$$
= \varepsilon \sqrt[]{d} \barra{E} \left[ \red{\sum_{t = 1} ^ {r} I \{ X \in C_i \} I \{c_1 \cap S \neq 0 \} \right] } + 2 \sqrt[]{d} \sum_{i = 1}^{r} \barra{E} \left[ I \{ X \in C_1 \} \cdot I \{ C_1 \cap S \neq 0 \} \right]
\barra{E} \left[
\varepsilon \cdot \sqrt[]{d} \cdot \sum_{i = 1}^{r} I \{ X \in C_i \}
\cdot
I \{ C_i\cap S \neq 0 \}
\right]
+ \ 2 \, \cdot \
\sqrt[]{d} \, \cdot \,
\sum_{i=1}^{r} I \{ X \in C_i \}
\cdot I\{C_i \cap S \neq 0 \}
=
$$
I don't care about this one \bred{$\sum_{t = 1} ^ {r} I \{ X \in C_i \} I \{c_1 \cap S \neq 0 \} \right] $}
$$
= \varepsilon \cdot \ \sqrt[]{d} \cdot \ \barra{E} \left[ \,
\bred{ $
\sum_{t = 1}^{r} I \{ X \in C_i \} I \cdot \{C_i \cap S \neq 0 \} $}\, \right]
+ 2 \cdot \ \sqrt[]{d}\ \cdot \ \sum_{i = 1}^{r}
\barra{E} \left[ \,
I \{ X \in C_1 \}
\cdot
I \{ C_1 \cap S \neq 0 \} \,
\right] \leq
$$
I don't care about this one \red{$\sum_{t = 1}^{r} I \{ X \in C_i \}\cdot I
\{ C_i \cap S \neq 0 \} $}
\\
Can be either $0$ or $1$ (if for some $i$, $X$ belong to some $C_i$
\\
So at most 1 the expectation
$$
\leq \ \varepsilon \sqrt[]{d} + \box
\leq \ \varepsilon \cdot \ \sqrt[]{d} + \square
$$
We can bound this square. Are the event I in the summation of the term after +. If they are indepednt the product will be the product of the two expectation. If I fix the cube.
We can bound this square. Are the event I in the summation of the term after $+$. If they are indepednt the product will be the product of the two expectation. If I fix the cube.
$X$ and $S$ are independent.
\\
Now the two events are independent \\ \bred{ $X \in C_1$ is inepdend of $C_i \cap S \neq $}
Now the two events are independent \\ \bred{ $X \in C_i$ is independent of $C_i \cap S \neq $}
$$
\barra{E} \left \I\{X \in C_i \} \cdot I \{ C_1 \cap S \neq 0 \} \right] = \barra{E} \left[ I \{ X \in C_i \} \right] \cdot \barra{E} \left[ I \{ C_i
\barra{E}
\left[
\ I \{ X \in C_i \}
\cdot
I \{ C_i \cap S \neq 0 \}
\right]
= \barra{E}
\left[
I \{ X \in C_i \}
\right]
\cdot
\barra{E}
\left[ I \{ C_i \right \} ]
$$
MANCAAAAAAA 9.26
\\
$$
\barra{P} \left( C_i \cap S \right) = \left( 1- \barra{P} \left( X \in C_1 \right) \right)^m \leq \exp (- m \barra{O} (x \in C_1 ))
\barra{P} \left( C_i \cap S \right) = \left( 1- \barra{P} \left( X \in C_1 \right) \right)^m \leq \exp (- m \cdot \barra{P} (x \in C_1 ))
$$
The probability of the point fall there and will be the probability of falling in the cube.
\\
Probability of Xs to fall in the cube with a m (samples?)
\\
Now use inequality $ (1 - p)^m \in e^{-pm}$ $\longrightarrow$ $1 + x \leq e^x$
-- IMMAGINE --
\begin{figure}[h]
\centering
\includegraphics[width=0.2\linewidth]{../img/lez11-img3.JPG}
\caption{Shape of the function}
%\label{fig:}
\end{figure}\\
$$
\sum_{t = 1}^{r} \barra{E} \left[ \barra{P} (X \in C_1 ) \barra{P} (C_1 \cap S \neq ) \right] \leq sum_{i = 1}^{r} p_i \, e^{-m \, p_i} \leq
\sum_{t = 1}^{r} \ \barra{E} \left[ \, \barra{P} \left(X \in C_1 \right) \cdot \barra{P} \left(C_1 \cap S \neq \right)\, \right] \leq \sum_{i = 1}^{r} p_i \cdot e^{-m \, p_i} \leq
$$
given that $p_i = \barra{P} (X \in C_i)$
I can upper bound this
$$
\leq \sum_{t=1}^{r} \left( \max_{0 \leq p \leq 1} \, p \, e^{- m \, p} \right) \leq r \, \max_{0 \leq p \leq 1} \, \bred{p \, e ^{-m\,p}} =
\leq
\sum_{t=1}^{r}
\left(
\max_{0 \leq p \leq 1}
\, p \, e^{- m \, p}
\right)
\leq
r \, \max_{0 \leq p \leq 1} \, \red{p \, e ^{-m\,p}} \, =
$$
where $\bred{p \, e ^{-m\,p}}$ is $F(p)$
where \bred{$p \, e ^{-m\,p}$} is $F(p)$
it is concave function so i'm going to take first order derivative to maximise it.
\\
$$
F^(p) = 0 \Leftrightarrow p = \frac{1}{m} \quad check!
F'(p) = 0 \Leftrightarrow p = \frac{1}{m} \quad check!
$$
$$
F''(p) \leq 0
F''(p) \leq 0 \qquad \qquad \qquad \qquad
$$
Check this two condition!
$$
@ -109,56 +168,76 @@ $$
Now get expectation
\\
$$
\barra{E} \left[ \| X - X_{\Pi(s,x)} \| \right] \leq \varepsilon \sqrt[]{d} + \left( 2 \cdot \sqrt[]{d}\right) \frac{r}{e \, m} =
\barra{E} \left[ \, \| X - X_{\Pi(s,x)} \| \, \right]
\leq
\varepsilon \cdot \ \sqrt[]{d} + \left( 2 \cdot \sqrt[]{d}\right) \frac{r}{e \, m} =
$$
I have $(2/epsilon)^2$ squares.
This bring $\variepsilon$ in the game
I have $(\frac{2}{\varepsilon})^2$ squares.
This bring $\varepsilon$ in the game
$$
\varepsilon \sqrt[]{d} + \left( 2 \cdot \sqrt[]{d}\right) \frac{1}{e \, m} \cdot \left( \frac{2}{\varepsilon}\right) ^d =
\varepsilon \cdot \ \sqrt[]{d} + \left( 2 \cdot \sqrt[]{d}\right) \frac{1}{e \, m} \cdot \left( \frac{2}{\varepsilon}\right)^d =
$$
$$
= \sqrt[]{d} \left( \varepsilon + \frac{2}{e \, m } \cdot \left( \frac{2}{\varepsilon}^d \right) \right)
= \sqrt[]{d} \left( \varepsilon + \frac{2}{e \, m } \cdot \left( \frac{2}{\varepsilon} \right)^d \right)
$$
\blue{HE MISS THE "c" costant from the start}
we can choose $\varepsilon$ to take them balanced
\\set $\varepsilon = 2 \, m^{\frac{-1}{(d+1)}} $
\bred{
$$
\bred{ \left( \varepsilon + \frac{2}{e \, m } \cdot \left( \frac{2}{\varepsilon}^d \right) \leq 4 \, m ^{\frac{-1}{(d+1)}} = }
\left( \varepsilon + \frac{2}{e \, m } \right)
\cdot
\left( \frac{2}{\varepsilon}^d \right)
\leq 4 \, m ^{\frac{-1}{(d+1)}} =
$$
}
\\
$$
\barra{E} \left[ \ell_d (\hat{h}_s) \leq 2 \ell_d (f^*) + 4 \cdot c \cdot \sqrt[]{d} \cdot m^{-\frac{1}{d+1}}
\barra{E}
\left[
\ell_d (\hat{h}_s) \right] \ \
\leq \ \
2
\ell_d (f^*)
+
4 \cdot c \cdot \ \sqrt[]{d}
\cdot m^{- \frac{1}{d+1}}
$$
We have that:\\
if $m \longrightarrow \infty$ \quad $\ell_D (f^*) \leq \barra{E} \left[ \ell_D (\hat{h}_s \right} \leq 2 \1ell_D(f^*)$
if $m$ $\longrightarrow$ $\infty$ \quad
$\ell_D (f^*) \
\leq \
\barra{E} \left[ \ell_D (\hat{h}_s \right] ) \ \
\leq \ \
2 \ell_D(f^*)$
\\
I want this smaller than twice risk + some small quantity
$$
\barra{E} \left[ \ell_d(\hat{h}_S) \right] \leq 2 \ell_D(f^*) + \varepsilon
\barra{E} \left[ \ell_d(\hat{h}_S) \right] \ \leq \ 2 \ell_D(f^*) + \varepsilon
$$
How big $m$ ?\\
Ignore this part since very small $(4 c \sqrt[]{•d})$
Ignore this part since very small $(4 \cdot c \cdot \ \sqrt[]{•d})$
\\
$$
m ^ {-\frac{1}{d+1}} \leq \varepsilon \Leftrightarrow m \beq (\frac{1}{\varepsilon}^d+1
m ^ {-\frac{1}{d+1}} \ \leq \ \varepsilon \Leftrightarrow m \ \geq \ \left(\frac{1}{\varepsilon}\right)^d+1
$$
So 1-NN require a training set size exponential "accuracy" \ $1-\varepsilon$
\\\\
We show that $1-NN$ can approach twice based risk $2 \ell_D(f^*)$\\
We show that $1-NN$ can approach twice based risk \ $2 \cdot \ell_D(f^*)$\\
but it takes a training set exponential in $d$.
\\
\subsection{Study of $\knn$}
Maybe we can use the $\knn$.
$$
\barra{E}\left[ \ell_D(\hat{h}_s)\right] \leq \left( 1+ \sqrt[]{\frac{8}{k}}\right) \ell_D(f^*) + 0 \, \left((k \, m^{-\frac{1}{d+1}}\right)
\barra{E}\left[ \ell_D(\hat{h}_s)\right] \leq \left( 1+ \sqrt[]{\frac{8}{k}}\right) \ell_D(f^*) + 0 \, \left(k \, m^{-\frac{1}{d+1}}\right)
$$
So is not exponential here.
\\
\bred{Learning algorithm $A$ is consistent for a certain loss $\ell$}
\\
If $\forall \, D$(distribution) of data we have that $A(S_m)$ predictor output by $A$\\.
If $\forall \, D$(distribution) of data we have that $A(S_m)$ predictor output by $A$\\
Now have the risk of that in $\ell_D(A(S_m))$ and we look at the expectation $\barra{E}\left[\ell_D(A(S_m))\right]$
If we give a training set size large ($\lim_{m \rightarrow \infty} \, \barra{E}\left[\ell_D(A(S_m))\right] = \ell_D (f^*)$ risk will converge in based risk.
\\If we give a training set size large ($ \ \lim_{m \rightarrow \infty} \, \barra{E}\left[\ell_D(A(S_m))\right] = \ell_D (f^*)\, $) risk will converge in based risk.
\\\\
$\knn$ where $K = K_m$ (is a function of training set size).
$K_, \rightarrow \infty $ as $m \rightarrow \infty$.
@ -170,7 +249,7 @@ For instance $K_m = \sqrt[]{m}$
\\
Then:
$$
\lim_{m \rightarrow \infty} \barra{E} \left[ \ell_D\left(A'\left(S_m\right)\right) \right] = \ell_D(f^*) \qquad \textbf{where $A'$ is $K_m-NN$ }
\lim_{m \rightarrow \infty} \barra{E} \left[ \ell_D\left(A'\left(S_m\right)\right) \right] = \ell_D(f^*) \qquad \textbf{where $A'$ is $K_m$-$NN$ }
$$
Increasing the size we will converge to this base risk for any distribution and that's nice.
\\\\
@ -183,7 +262,7 @@ Algorithm that grow tree classifiers can also be made consistent provided two co
Tree has to keep growing but not so fast.\\
Second point is: suppose you have a certain number of leaves and you can look at the fraction.
Each leaf $\ell$ gets $N_\ell$ examples. You want that this fraction at any point of time is not going to 0. The fraction of point every leaf receive a split we are reducing the smallest number of examples.
\\Example keep growing and leaves too and we want that $\frac{N_\ell$}{$.$} this not going to 0. $.$ since not showed the formula.
\\Example keep growing and leaves too and we want that $\frac{N_\ell}{manca}$ this not going to 0. $.$ since not showed the formula.
\\\\
Given $A$, how do I know wheter $A$ could be consistent?
$$
@ -192,7 +271,7 @@ $$
$S$ can be any size.
If $A$ is $ERM$ then $H_A = H$, so where ERM minimise it.
\\
If $\exists f^* : X \longrightarrow $ such that $f^* \not{\in} H_A$ and $\exists D$ such that $f^*$ is Bayes optimal for some distribution $D$.
If $\exists f^* : X \longrightarrow $ such that $f^* \not\in H_A$ and $\exists D$ such that $f^*$ is Bayes optimal for some distribution $D$.\\
This cannot be consistent because distribution will not be able to generate the Bayes optimal predictor.
Maybe is there another predictor $f$ which is not equal to $f^*$ risk.
\\\\
@ -227,7 +306,12 @@ More training set for tree, then will grow more, even more larger will be ever g
\\
Any algorithm as a give training points is no parametric, while growing with parametric will stop a some point.
\\
--- IMMAGINE ---
\begin{figure}[h]
\centering
\includegraphics[width=0.5\linewidth]{../img/lez11-img4.JPG}
\caption{Parametric and non parametric growing as training set getting larger}
%\label{fig:}
\end{figure}\\
\\
If algorithm is more parametric as i give training points\\
If a certain point stop growing, $f^*$ will be out and i will grow more.

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Chapter 11.
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\contentsline {section}{\numberline {10.1}TO BE DEFINE}{54}%
\contentsline {section}{\numberline {10.2}MANCANO 20 MINUTI DI LEZIONE}{54}%
\contentsline {section}{\numberline {10.3}Compare risk for zero-one loss}{56}%
\contentsline {chapter}{\numberline {11}Lecture 11 - 20-04-2020}{58}%
\contentsline {section}{\numberline {11.1}Analysis of $K_{NN}$}{58}%
\contentsline {subsection}{\numberline {11.1.1}Study of $K_{NN}$}{61}%
\contentsline {subsection}{\numberline {11.1.2}study of trees}{62}%
\contentsline {section}{\numberline {11.2}Non-parametric Algorithms}{63}%
\contentsline {subsection}{\numberline {11.2.1}Example of parametric algorithms}{64}%
\contentsline {chapter}{\numberline {12}Lecture 12 - 21-04-2020}{65}%
\contentsline {section}{\numberline {12.1}Non parametrics algorithms}{65}%