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<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no">
<title>reveal.js</title>
<link rel="stylesheet" href="dist/reset.css">
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</head>
<body>
<div class="reveal">
<div class="slides">
<section id="title">
<h1>
PLS Fonctionnelle
</h1>
<i class="fa fa-cogs"></i> Ou comment estimer $\mathbb E[ \, Y \, \vert \, X = x \,]$ de façon smart en prenant en compte le lien entre $X$ et $Y$ <i class="fa fa-cogs"></i>
<aside class="notes">
</aside>
</section>
<section style="font-size: 0.7em;" id="avant-propos">
ce que je ne vais <strong>pas</strong> traîter :
<br>
<ul>
<li> l'algorithme numériquement plus stable pour l'arithmétique en précision finie </li>
$\rightarrow$ utile pour les informaticiens, mais pas pour mon exposé
<br>
<br>
<li>les détails des propriétés asymptotiques</li>
$\rightarrow$ sinon j'allais péter un plomb :
<br>
<br>
je vous épargnerai de ces formules :
<br>
<img src="img/meme_pas_je_traite_ca.jpeg" alt="horreur" style="height: 10em;">
<img src="img/oui_oui.jpeg" alt="aled" style="height: 10em;">
</ul>
</section>
<section id="Contexte">
<section id="contexte-title" data-background-video="img/obama-confused.mp4" data-background-video-loop data-background-video-muted data-background-opacity="0.6">
<h1>un peu de contexte</h1>
</section>
<section id="contexte-model">
<h2>modèle</h2>
$Y = a + \displaystyle\int_K b(u)X(u)du + \varepsilon$
<br />
<br />
<small>dans le cas de cet exposé (et du papier) : </small>
<br />
<span style="font-size: 0.6em;">
$$\begin{array}{rrcll}
K & & & \textsf{compact de } \mathbb R & \equiv \textsf{intervalle fermé}
\\
a& &\in& \mathbb R & \equiv scalaire
\\
b& : K \rightarrow \mathbb R &\in& \mathbb L^2(K) & \equiv fcn
\\
\varepsilon & & \in & m(\Omega , \mathbb R ) & \equiv \operatorname{VA}( \mathbb R )
\\
X & & \in & m\left[ \Omega, \mathcal C^0( \mathbb R, \mathbb R ) \right] & \equiv \operatorname{VA}\left[ C^0( \mathbb R, \mathbb R ) \right]
\\
Y && \in & m(\Omega , \mathbb R) & \equiv \operatorname{VA}( \mathbb R )
\end{array}$$
</span>
</section>
<section>
comment estimer $m(x) = \mathbb E[ Y | X = x ] ?$
</section>
<section id="contexte-model-question">
<i class="fa fa-question"></i> Comment trouver $a \in \mathbb R$ et $b \in \mathbb L^2$ <i class="fa fa-question"></i>
<br /><br />
<span class="fragment fade-down"> <i class="fa fa-warning"></i> $\mathbb L^2$ est de dimension $\infty$ <i class="fa fa-warning"></i> </span>
<span class="fragment fade-down"> $\displaystyle\sum_1^\infty \beta_i \psi_i = b \approx \sum_{1}^p \beta_i \psi_i$ </span>
</section>
<section id="contexte-b-et-sa-projection">
<h3> <i class="fa fa-pencil"></i> Notations <i class="fa fa-pencil"></i></h3>
<div class="flex-container" style="display: flex;">
<div style="flex: 1; margin-right: 10px; border: 1.5px solid black;" class="fragment fade-in">
<span style="font-size: 0.6em;">
base $( \psi_i : K \rightarrow \mathbb R )$ de $\mathbb L^2( K )$
$$F_n(\psi) = \operatorname{vect}( \psi_i )_{1,n}$$
$$F_\infty(\psi) = \operatorname{vect}( \psi_i )_{\mathbb N}=\mathbb L^2(K)$$
</span>
</div>
<div style="flex: 1;margin-left: 10px; border: 1.5px solid black;" class="fragment fade-in">
<span style="font-size: 0.6em;">
$$b = \sum\limits_{\mathbb N} \underbrace{\langle b \vert \psi_i \rangle}_{\beta_i} \, \psi_i$$
$$\hat b_p^{[\psi]} = P^\perp_{F_p(\psi)}( \, b \, ) = \sum_{i=1}^p \beta_i \psi_i$$
</span>
</div>
</div>
<br />
<p class="fragment fade-down">
<i class="fa fa-question"></i> quelle base choisir <i class="fa fa-question"></i>
</p>
</section>
<section>
<!-- <i class="fa fa-mouse-pointer"></i>
<i class="fa fa-sticky-note"></i>
<i class="fa fa-wikipedia-w"></i> -->
<p><i class="fa fa-lightbulb-o"></i> projeter sur une base adaptée aux données <i class="fa fa-lightbulb-o"></i></p>
<p class="fragment fade-down">nous appellerons cette base $\phi = (\phi_i)_{i \in \mathbb N}$</p>
</section>
<!-- —————————————————————————————————————— ACP Théorème de Mercer et rpz KL ——————————————————————————————————————————————— -->
<section id="contexte-ACP-1">
<h3> <i class="fa fa-flag"></i> ACP Fonctionnelle <i class="fa fa-flag"></i></h3>
<span style="font-size: 0.7em;">[Partie 1]</span>
<p style="font-size: 0.75em;" class="fragment fade-in"><i class="fa fa-lightbulb-o"></i> regarder la base qui est la plus adaptée à la variance de $X$ <i class="fa fa-lightbulb-o"></i></p>
<br />
<p class="fragment fade-in" style="font-size: 0.7em;" >
"théorème spectral" (Mercer $\Rightarrow$) :
<span style="font-size: 0.9em;">
$$c(s,t) = \operatorname{cov}\left[ X(s), X(t) \right] \underset {\textsf{Mercer}}{=} \sum_{\mathbb N} c_i \phi_i(s) \phi_i(t)$$
$$\int \phi_k(u) c(u,t) du =c[ \phi_k ] = c_k \phi_k \implies \sum_{\mathbb N} c_k = \int_{K^2} c(s,t) dudt$$
</span>
</p>
<br />
<p class="fragment fade-in" style="font-size: 0.7em;">
Théorème de Karhunen-Loève : <span style="font-size: 0.7em;">$\boxed{X = \mu + \sum_{\mathbb N} \langle X-\mu \vert \phi_i \rangle \phi_i}$</span>
</p>
</section>
<!-- ———————————————————————————————————————— ACP en pratique ————————————————————————————————————————————— -->
<section id="contexte-ACP-2">
<h3> <i class="fa fa-flag"></i> ACP Fonctionnelle <i class="fa fa-flag"></i></h3>
<span style="font-size: 0.7em;">[Partie 2]</span>
<br />
<p class="fragment fade-in" style="font-size: 0.7em;" >
<span style="font-size: 0.9em;">
$$\mathbb E \Vert X \Vert^2_{\mathbb L^2} = \int_K \mathbb E[X^2](u)du < \infty \implies \sum c_i < \infty \implies \sum able$$
<p style="border: 1px solid black; font-size : 0.5em; margin-left : 15%; margin-right : 15%; padding : 3px" class="fragment fade-in">
<i class="fa fa-check"></i>on peut donc réordonner les termes par valeur propre décroissante, afin de sélectionner les composantes les plus informatives
</p>
</span>
</p>
<p class="fragment fade-down" style="font-size: 0.7em;">on peut maintenant approximer $b$ par $b_p = \displaystyle\sum_{i=1}^{p} \beta_i(\phi) \phi_i$</p>
</section>
<section id="contexte-ACP-3">
<h3> <i class="fa fa-flag"></i> ACP Fonctionnelle <i class="fa fa-flag"></i></h3>
<span style="font-size: 0.7em;">[Partie 3]</span>
<p style="font-size: 0.75em;"><i class="fa fa-cogs"></i> la pratique <i class="fa fa-cogs"></i></p>
<p class="fragment fade-in" style="font-size: 0.7em;">
en se rappelant le <a href="#contexte-model">modèle</a> : $Y = \mathbb E[Y] + \sum_{i \geq 1} \beta_i \langle X - \mathbb E [X] \vert \phi_i \rangle_{\mathbb L^2} + \varepsilon$
</p>
<p class="fragment fade-in" style="font-size: 0.7em;"> on observe $D = \left(Y_i[m], T_i[m], X_i[m] = X_i(T_i[m])\right)_{\begin{array}{l}i \in ⟦ 1,n ⟧ \\ m \in ⟦ 1,M_i ⟧ \end{array}}$ <br />
$$
D
\overset{①}{\rightarrow} \hat X
\overset{②}{\rightarrow} \hat \mu
\overset{③}{\rightarrow} \hat c
\overset{④}{\rightarrow} \hat \phi
\overset{⑤}{\rightarrow} \langle \, \hat x - \hat \mu \, \vert \, \hat \phi \, \rangle
\overset{⑥}{\rightarrow} \hat \beta
$$
</p>
<p class="fragment fade-in" style="font-size: 0.7em;">
<br />
<i class="fa fa-cogs"></i>
<span style="font-size: 0.7em;"> $\boxed{\hat b_p = \displaystyle\sum_{i=1}^{p} \beta_i(\hat \phi) \hat \phi_i = P^\perp_{F_p(\hat \phi)}( \, b \, )}$</span>
<i class="fa fa-cogs"></i>
<span style="font-size: 0.7em;">
$\underset {\textsf{Moindres Carrés}}\longrightarrow \hat \beta(\hat \phi) = \begin{bmatrix} \vdots \\ \hat \beta^{LS}_i(\hat \phi) \\ \vdots \end{bmatrix}$
</span>
</p>
<aside class="notes">
<a href="#contexte-model">retour au modele</a>
</aside>
</section>
<!-- ————————————————————————————————————— ACP Questions ———————————————————————————————————— -->
<section id="contexte-ACP-4">
<i class="fa fa-check"></i>
<strong>$\phi$ représente bien X
</strong>
<p class="fragment fade-in">
<strong>
<i class="fa fa-warning"></i>
<!-- <i class="fa fa-arrow-right"></i> -->
et $b$ ?
</strong>
</p>
<p class="fragment fade-in">
<strong>
<i class="fa fa-warning"></i>
<!-- <i class="fa fa-arrow-right"></i> -->
interactions entre $b$ (et donc $Y$) et $X$ ?
<br />($Y = a + \int bX + \varepsilon$)
</strong>
</p>
</section>
</section>
<!-- ——————————————————————————————————— PLS ———————————————————————————————————————— -->
<section id="PLS">
<section id="contexte-PLS-gigachad" data-background-image="img/gigachad_by_haesch_dfbedkv.gif" data-background-opacity="0.4">
<h1>LA PLS FONCTIONNELLE</h1>
une vraie technique de GigaChad
</section>
<section id="PLS-non-fonctionnelle">
<h2>mini rappel : PLS (non fonctionnelle)</h2>
<ul style="font-size: 0.7em;" class="fragment fade-in">
<li>trouver le sous espace de l'espace engendré par $X$ qui explique le mieux la variance de l'espace engendré par $Y$</li>
</ul>
<ul style="font-size: 0.7em;" class="fragment fade-in">
<li>on projette $X$ et $Y$ dans un nouvel espace</li>
$\longrightarrow$ "latent space" [variable latente | latin : lateo - caché]
</ul>
</section>
<!-- —————————————————————————————————————————— PLS Y chapeau ——————————————————————————————————————————— -->
<section id="PLS-Y-hat" style="font-size: 0.7em;">
<h2>comme vu précédemment</h2>
<p>
$$Y =
\underset{
\mathbb E [Y \vert X = x]
\underset{\textsf{notation}}{\equiv}
m(x)
}
{
\boxed{
\mathbb E[Y] + \langle X - \mathbb E [X] \vert \, b \, \rangle_{\mathbb L^2}
}
}
+ \varepsilon
$$
note : sur la base $\underset{\Vert \cdot \Vert}{\perp}$ $\psi$ de $\mathbb L^2$ : $b = \sum\limits_{i\geq 1} \langle b \vert \psi_i \rangle \psi_i$
</p>
<p class="fragment fade-in">
on approxime en projetant sur $F_p(\psi)$
$$\hat Y_p^{[\psi]}
=
\underset{
\widehat {\mathbb{E}}_{p}^{[\psi] }[Y | X=x]
\underset{\textsf{notation}}{\equiv}
\hat m_{p}^{[\psi]}(x)
}
{\boxed{\mathbb E[Y] + \sum_{i = 1}^{\color{green}{p}} b_i \langle X - \mathbb E [X] \vert \psi_i \rangle_{\mathbb L^2}}} + \varepsilon$$
</p>
</section>
<!-- ——————————————————————————————————————— PLS Problème de minimisation —————————————————————————————————————————————— -->
<section id="PLS-pb-min" style="font-size: 0.6em;">
<h2>Problème de minimisation</h2>
<span class="fragment fade-down">
<h4>notation</h4>
$$\langle f \vert g \rangle_{c( \cdot , \cdot )} \underset{\textsf{notation}}{\equiv} \int_{K^2}f(s)c(s,t)g(t) \, ds \, dt$$
$$\Vert \cdot \Vert^2_{c(\cdot, \cdot)} : \begin{array}{ccc} \mathcal C^0(K) & \longrightarrow & \mathbb R_+ \\ f & \longmapsto & \int_{K^2} f(s) c(s,t) f(t) \, ds \, dt \end{array}$$
</span>
<span class="fragment fade-down">
<h4>expression du problème de minimisation</h4>
<p>
$$
\begin{align}
(\hat \psi_i^*)_{1,p}
&=
\underset{ \begin{array}{c}\psi_p \\ \langle \psi_i \vert \psi_p \rangle_{c(\cdot , \cdot)} = 0 \\ \Vert \psi_p \Vert = 1 \end{array} } {\operatorname{argmin}}
\left[
\underset{(\psi_i)_{1,p-1}} {\operatorname{argmin}} \operatorname{COV}\left(
\underset{\hat \varepsilon_{p-1}^{[\psi]}}{\boxed{Y - \hat m_{p-1}^{[\psi]}(x)}} \,
,
\, \langle X \vert \psi_p \rangle \right)
\right]
\end{align}
$$
</p>
</span>
<p class="fragment fade-down"> <i class="fa fa-lightbulb-o"></i> $\approx$ comme en projection-pursuit <i class="fa fa-lightbulb-o"></i> </p>
</section>
<!-- ————————————————————————————————————————————————————————————————————————————————————— -->
<!-- ————————————————————————————————————————————————————————————————————————————————————— -->
<section id="PLS-theorie-probleme" style="font-size: 0.8em;">
<p> la PLS n'est pas un nouveau concept </p>
<p> elle est était déjà utilisée et même en FDA</p>
<p
class="fragment fade-down"
> <i class="fa fa-warning"></i> <strong>MAIS</strong> sans preuve rigoureuse <br> $\rightarrow$ consistance <br> $\rightarrow$ vitesse de convergence</p>
</section>
<section id="PLS-graph-non-annote">
<img src="img/graph1_non_annote.jpeg" style="height: 8em;" alt="">
<img src="img/graph2_non_annote.jpeg" style="height: 7em;" alt="">
</section>
<!-- ————————————————————————————————————————————————————————————————————————————————————— -->
<section id="PLS-apport-du-papier">
apport de ce papier $\longrightarrow$ le développement théorique de la PLS Fonctionnelle
<br>
<br>
<p class="fragment fade-down">
<i class="fa fa-question-circle"></i> comment <i class="fa fa-question-circle"></i>
</p>
<br>
<br>
<p class="fragment fade-down">
en travaillant sur un objet plus simple et aux propriétés de convergence équivalentes
</p>
</section>
</section>
<section id="APLS-theorie">
<section id="APLS-title" data-background-image="img/think-math.gif" data-background-opacity="0.2">
<h1>Alternative-PLS</h1>
<p>la réponse à un problème de développement théorique</p>
</section>
<section id="APLS-theorie-overview" style="font-size: 0.6em;">
<h2>overview</h2>
$$
\begin{array}{rccll}
& & c(s,t) = \sum c_i \overset{\textsf{base ACP } \left(✅ \underset{\Vert \cdot \Vert}{\perp} \right)}{\boxed{\phi_i(s) \phi_i(t)}}
\\
& & \downarrow
\\
& \rightarrow&
\underset{\textsf{développement } \operatorname{vect}\left(c^k[b]\right)_{1,p}}{
\boxed{
\tilde{m}_p^{[c]}
}
}
& \leftarrow & \textsf{pour dvp la théorie}
& \left[
\begin{array}{c}
\textsf{base non } {\perp}
\\
\textsf{explicite}
\end{array}
\right]
\\
m(x) & & \Updownarrow &
\\
&\rightarrow&
\underset{\textsf{base PLS }}
{
\boxed{
\tilde{m}_p^{[\psi]}
}
}
& \leftarrow & \textsf{pour la pratique}
& \left[
\begin{array}{c}
\textsf{base } \underset{\Vert \cdot \Vert}{\perp}
\\
\textsf{non explicite}
\end{array}
\right]
\\
\end{array}
$$
<p class="fragment fade-down">
puis on va mettre des chapeaux car on est statisticiens 🤓
</p>
<p class="fragment fade-down">
et on va pleurer à regarder $\Vert \hat m_p - m \Vert$ car on est matheux ☠️
</p>
</section>
<section id="APLS-theorie-psi-dans-C", style="font-size: 0.7em;">
<h1>① équivalence ①</h1>
<p class="fragment fade-in-then-semi-out"> on aimerait montrer que $\forall p \quad \psi_p = \sum a_i c^i[b]$
<br>
ie : $\psi_p \in \operatorname{vect} \left( c^i[b] \right)_{1,p}$
</p>
<p class="fragment fade-down">$\psi_p \in \frac{ \operatorname{vect} \left( c^i[b] \right)_{1,p} }{\operatorname{sign}} = \mathcal C_{1,p}[b]$</p>
<br>
<p class="fragment fade-in-then-out">$P_{F_p(\psi)}^\perp \left[ f \in \mathcal C^0(K, \mathbb R) \right] \in \operatorname{vect}(\psi_i)_{1,p} \subset \mathcal C_{1,p}[f]$</p>
<p class="fragment fade-in-then-out">$P_{F_p(\psi)}^\perp \left[ f \in \mathcal C^0(K, \mathbb R) \right] \in \operatorname{vect}(\psi_i)_{1,p} = \mathcal C_{1,p}[f]$</p>
</section>
<section id="PLS-determination-de-la-base" style="font-size: 0.7em;">
<h3>construction de la base PLS $\; \psi$</h3>
<strong>théorème [3.1]</strong>
$$
\boxed{
\int _K \mathbb E[X^2](u)du < \infty
}
\implies
\boxed{
\begin{array}{l}
\exists ! \gamma \in \mathbb R^p \\
\rightarrow \psi_p = \gamma_0 \left( c \left[ \, b - P_{F_{p-1}(\psi)}(b) \right] + \sum_{k=1}^{p-1} \gamma_k \psi_k \, \right)
\\
\rightarrow \langle \psi_k \vert \psi_p \rangle_{c(\cdot, \cdot)} = 0 \quad k < p
\\
\rightarrow \Vert \psi_p \Vert = 1
\end{array}
}
$$
<p>
on a :
$\psi_p = \Gamma_p( c[\cdot], b , \psi_{1:p-1} )\quad$ et donc $\quad \psi_p = \sum\limits_{i=1}^p u_i c^i[b]$
</p>
<p>
$\implies \operatorname{dim} < \infty \quad \tilde m_p^{[\psi]} \iff \tilde m_p^{[c]}$
</p>
</section>
<section>
<p class="fragment semi-fade-out">
$\psi_1 = \frac{c[b]}{\Vert c[b] \Vert}$ (3.1)
</p>
<p class="fragment fade-in-then-semi-out">
$\psi_2 = \alpha_0( c[b - \langle b \vert \psi_1 \rangle \psi_1] + \alpha_1 \frac{c[b]}{\Vert c[b] \Vert} )$ (3.1)
</p>
<p class="fragment fade-in-then-semi-out">
$c[ c[b] ] = c^2[b]$ (base ACP : mercer)
</p>
<p class="fragment fade-in-then-semi-out">$\implies \psi_p = \sum_i^p a_i c^i[b]$</p>
<p class="fragment fade-in"> ⚠️ ne pas oublier que c'est au signe près car :
<br>
$\alpha_0$ déterminé par $\Vert \psi_p \Vert = 1$
</p>
</section>
<section id="APLS-theorie-estimation-de-b" style="font-size: 0.7em;">
<p>maintenant on aimerait bien les estimer, $k = \begin{bmatrix} \vdots \\ k_i \\ \vdots \end{bmatrix}_{1,p}$ qui vérifient </p>
<p>
$$
\begin{align}
m(x) &= \mathbb E[Y] + \langle \, X - \mu \, \vert \, b \, \rangle
\\
\tilde m_p^{[c]}(x) &= \mathbb E[Y] + \langle \, X - \mu \, \vert \, \sum\limits_{i=1}^p k_i c^i [b] \, \rangle
\end{align}
$$
</p>
<p class="fragment fade-down">
$$\left( k_i \right)_{1,p} = \underset{
(a_i)_{1,p}}
{\operatorname{argmin}}
\mathbb E \left[
\left(
\underbrace{
\langle \, X - \mu \, \vert \, b \, \rangle - \langle \, X - \mu \, \vert \, \sum\limits_{i=1}^p a_i c^i [b] \, \rangle
}_{
m(X) - \tilde m_p^{[c] } (X)
}
\right)^2
\right]$$
</p>
</section>
<section id="APLS-theorie-convergence-de-bp" style="font-size: 0.7em;">
<h1>② convergence ②</h1>
<strong>théorème [3.2]</strong>
<br>
<p>$$\begin{array}{c} \mathbb E \Vert X \Vert^2 < \infty \\ \operatorname{sp} c[ \, \cdot \, ] \subset \mathbb R_+^* \end{array} \implies b_p \xrightarrow[p \rightarrow \infty]{\mathbb L^2(K)} b$$</p>
<p>et donc $\boxed{b = \sum\limits_i^\infty k_i c^i[b]}$</p>
<p class="fragment fade-down">⚠️ pareil ne pas oublier que les $k_i$ ne sont pas uniques ⚠️</p>
</section>
<section id="APLS-theorie-lien-ACP" style="font-size: 0.7em;">
<h2>expression de l'APLS dans la base ACP</h2>
<br>
<p>
$$\tilde b_p = \sum_i^p k_i c^i[b] = \sum_k^\infty \beta_k \left( \sum_i^p k_i c_k^i \right) \phi_k$$
</p>
<br>
<br>
rappel : $c(s,t) = \sum_i c_i \phi_i(s) \phi_i(t)$
</section>
</section>
<section>
<section id="APLS-empirique">
<h1>L'Empirique</h1>
</section>
<section>
$$
\begin{array}
\tilde{m}_p(X) &=& \mathbb E [Y] &+ \langle X - \mu | \tilde b_p \rangle
\\
\downarrow & & \downarrow & \quad \quad \downarrow
\\
\hat m_p(x) &=& \overline Y &+ \langle X - \hat \mu | \hat b_p \rangle
\end{array}
$$
</section>
<section>
<strong>théorème [3.5]</strong>
$$\min\limits_k \mathbb E [ (m(X) - \tilde m_p^{[c]}(x))^2 ] \longrightarrow k = H^{-1} \alpha$$
$H = \begin{bmatrix} \langle c^{i+1}[b] | c^j[b] \rangle \end{bmatrix}_{ij} \quad$ et $\quad \alpha = \begin{bmatrix} \vdots \\ c^i[b] \\ \vdots \end{bmatrix}_i$
</section>
<section>
$$\hat k = \hat H^{-1} \hat \alpha$$
</section>
<section style="font-size: 0.6em;">
<div class="flex-container" style="display: flex;">
<div style="flex: 1; margin-right: 10px; border: 1.5px solid black;">
ce qu'on approxime :
$$
\begin{array}
\hat c(s,t) &
\\
\hat c[b] &
\\
&\longrightarrow \hat k
\\
\hat c^i[b] &
\\
\hat H = [\hat h_{ij}]_{ij} &
\end{array}
$$
</div>
<div style="flex: 1; margin-right: 10px; border: 1.5px solid black;" class="fragment fade-in">
$$\mathbb E \rightarrow \frac 1 n \sum_{i=1}^n$$
$$\hat k = \underset{k}{\operatorname{argmin}} \frac 1 n \sum_i(Y_i - \overline Y) - \sum^p_j k_j \int(X - \overline X) \hat c ^j[b]$$
$$\hat c(s,t) \frac 1 n \sum_i (X_i(s) - \overline X(s))(X_i(t) - \overline X(t))$$
$$\hat c[b] = \frac 1 n \sum_i (X_i - \overline X)(Y_i - \overline Y)$$
$$\hat c^{i+1}[b] = \int \hat c^i[b](u) \hat c(u,t) du$$
</div>
</div>
</section>
<section>
$\longrightarrow$ contrôler en $p$
<br> <br> <br>
$\longrightarrow$ contrôler en $n$
<br>
<p class="fragment fade-down">
$\rightarrow$ prendre $p(n)$ approprié
</p>
</section>
</section>
<section id="APLS-asymptotique">
<section id="APLS-asymptotique-">
<h1>Résultats Asymptotiques</h1>
</section>
<section>
$$\hat c = c + \frac 1 {\sqrt n} \overset{\color{blue}{[X_i \, \mu]}}{\underset{\mathbb P}{\mathcal O}(1)} + \frac 1 n \overset{\color{blue}{[\overline X \, \mu]}}{\underset{\mathbb P}{\mathcal O}(1)}$$
$$\hat c[b] = c[b] = \frac 1 {\sqrt n} \overset{\color{blue}{[X_i \, Y_i \, \mu]}}{\underset{\mathbb P}{\mathcal O}(1)} + \frac 1 n \overset{\color{blue}{[\overline X \, \overline Y \, \mu]}}{\underset{\mathbb P}{\mathcal O}(1)}$$
$$\hat c^i[b] = c^i[b] + \underset{\mathbb P}{\mathcal O}( \frac 1 {\sqrt n})$$
$$$$
</section>
<section>
<p class="fragment semi-fade-out">
$$\sqrt n (\hat h_{ij} - h_{ij}) \sim \mathcal N(0, \sigma_{ij}^2)$$
</p>
<p class="fragment fade-in-then-semi-out">
$$\Vert \hat c [b] - (c[b] + \frac \varepsilon {\sqrt n}) \Vert_\infty = \underset{\mathbb P}{\mathcal O}(n^{-1})$$
</p>
<p class="fragment fade-in">
$$\Vert \hat k - k \Vert = \underset{\mathbb P}{\mathcal O}( \frac{1 + \Vert k \Vert}{n^{1/2} \lambda_{min}(H_p)} + \frac{1}{n \lambda_{min}(H_p)^3} )$$
</p>
</section>
<section style="font-size: 0.7em;">
<p>
$$\Vert \hat k - k \Vert = \underset{\mathbb P}{\mathcal O}( \frac{1 + \Vert k \Vert}{n^{1/2} \lambda_{min}(H_p)} + \frac{1}{n \lambda_{min}^3(H_p)} )$$
</p>
<p>
$$\boxed{
\begin{array}{ccc}
\operatorname{dim} < \infty & & \operatorname{dim} = \infty
\\
\textsf{non sing} & & \textsf{sing}
\\
\lambda_{min}(H_p) & \xrightarrow[p \rightarrow \infty]{} & \lambda_{min}(H) = 0
\end{array}
}
$$
</p>
<p class="fragment fade-down">
$\implies$ on a besoin que $n^{1/2} \lambda_{min}(H_p) \xrightarrow[n \rightarrow \infty]{} \infty$
<br><br>
ie $p(n) \rightarrow \infty$ à une vitesse $\leq \sqrt n$
</p>
</section>
<section style="font-size: 0.7em;">
$$n^{-1/2} \lambda_{min}(H_p) \Vert k \Vert + \frac{1}{n \lambda_{min}^3(H_p)} \xrightarrow[n \rightarrow \infty]{} 0$$
$$\Downarrow$$
$$\mathbb E \left[ \left\{\hat m_p(X) - m(X) \right\}^2 | X_{1:n} \right] = \underset{\mathbb P}{\mathcal O}( \frac{1 + \Vert k \Vert}{n^{1/2} \lambda_{min}(H_p)} + \frac{1}{n \lambda_{min}^3(H_p)} )$$
</section>
</section>
<section id="PLS-resultats-algo">
<section>
<img src="img/graph1.jpeg" style="height: 8em;" alt="">
<img src="img/graph2.jpeg" style="height: 7em;" alt="">
</section>
</section>
<section style="font-size: 0.6em;">
<h1>💬 autres trucs à dire 💬</h1>
</section>
</div>
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