From d010cdae55a7e19d78a24e46c1db3058d69347fc Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Wed, 15 Aug 2012 18:27:30 -0700
Subject: Repository restructuration (has to be clean now that it is shared ;)

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+\documentclass{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsmath,amsthm,amsfonts}
+\usepackage{comment}
+\newtheorem{lemma}{Lemma}
+\newtheorem{prop}{Proposition}
+\newcommand{\var}{\mathop{\mathrm{Var}}}
+\newcommand{\condexp}[2]{\mathop{\mathbb{E}}\left[#1|#2\right]}
+\newcommand{\expt}[1]{\mathop{\mathbb{E}}\left[#1\right]}
+\newcommand{\norm}[1]{\lVert#1\rVert}
+\newcommand{\tr}[1]{#1^*}
+\newcommand{\ip}[2]{\langle #1, #2 \rangle}
+\newcommand{\mse}{\mathop{\mathrm{MSE}}}
+\newcommand{\trace}{\mathop{\mathrm{tr}}}
+\begin{document}
+
+
+\section{pomme}
+
+In this section, we will consider that we are given a \emph{universe}
+set $U$ such that all the sets we consider are subsets or $U$ and all
+functions defined on sets are defined on the power set of $U$,
+$\mathfrak{P}(U)$.
+
+A function $f$ defined on $\mathfrak{P}(U)$ will be said
+\emph{increasing} if it is increasing with regards to inclusion, that
+is:
+\begin{displaymath}
+  \forall\,S\subseteq T,\quad f(S)\leq f(T)
+\end{displaymath}
+A \emph{decreasing} function on $\mathfrak{P}(U)$ is defined similarly.
+
+\begin{prop}
+  Let $R:\mathbf{R}\rightarrow \mathbf{R}$ be a decreasing concave
+  function and $f:\mathfrak{P}(U)\rightarrow\mathbf{R}$ be a
+  decreasing submodular function, then the composed function $R\circ
+  f$ is increasing and supermodular.
+\end{prop}
+
+\begin{proof}
+  The increasingness of $R\circ f$ follows immediately from the
+  decreasingness of $R$ and $f$.
+  
+  For the supermodularity, let $S$ and $T$ be two sets such that
+  $S\subseteq T$. By decreasingness of $f$, we have:
+  \begin{displaymath}
+    \forall\,V,\quad f(T)\leq f(S)\quad\mathrm{and}\quad f(T\cup V)\leq f(S\cup V)
+  \end{displaymath}
+  
+  Thus, by concavity of $R$:
+  \begin{displaymath}\label{eq:base}
+    \forall\,V,\quad\frac{R\big(f(S)\big)-R\big(f(S\cup V)\big)}{f(S)-f(S\cup V)}
+    \leq\frac{R\big(f(T)\big)-R\big(f(T\cup V)\big)}{f(T)-f(T\cup V)}
+  \end{displaymath}
+  
+  $f$ is decreasing, so multiplying this last inequality by
+  $f(S)-f(S\cup V)$ and $f(T)-f(T\cup V)$ yields:
+  \begin{multline}
+    \forall V,\quad\Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(T)-f(T\cup V)\big)\\
+    \leq \Big(R\big(f(T)\big)-R\big(f(T\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)
+  \end{multline}
+
+  $f$ is submodular, so:
+  \begin{displaymath}
+    f(T\cup V)-f(T)\leq f(S\cup V) - f(S)
+  \end{displaymath}
+  
+  $R\circ f$ is increasing, so:
+  \begin{displaymath}
+    R\big(f(S)\big)-R\big(f(S\cup V)\big)\leq 0
+  \end{displaymath}
+
+  By combining the two previous inequalities, we get:
+    \begin{multline*}
+    \forall V,\quad\Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)\\
+    \leq \Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(T)-f(T\cup V)\big)
+  \end{multline*}
+
+  Injecting this last inequality into \eqref{eq:base} gives:
+  \begin{multline*}
+    \forall V,\quad\Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)\\
+    \leq \Big(R\big(f(T)\big)-R\big(f(T\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)
+  \end{multline*}
+
+  Dividing left and right by $f(S)-f(S\cup V)$ yields:
+  \begin{displaymath}
+    \forall V,\quad R\big(f(S)\big)-R\big(f(S\cup V)\big)
+    \leq R\big(f(T)\big)-R\big(f(T\cup V)\big)
+  \end{displaymath}
+  which is exactly the supermodularity of $R\circ f$.
+\end{proof}
+
+
+
+\section{Understanding the recommender system}
+
+\subsection{General problem}
+
+We already have a database $D_n$ of $n$ users. For each user $i$ we
+have a set of $k$ explanatory variables (features), this is a vector
+$x_i$.
+
+The problem is the following: we are about to start an experiment where for
+some of the users in the database, a variable of interest $y_i$ will
+be revealed (it could be for example a medical survey, a website which
+buys some data from another website). From the pairs $(x_i, y_i)$ that
+we will acquire through the experiment, we are going to compute a
+regression function $f_n$ which will allow us to predict for a new $x$
+its associated $y$. The accuracy of this prediction will be measured
+by the Mean Squared Error :
+\begin{displaymath}
+  \mse(f_n) = \expt{\big(f_n(x)-y\big)^2}
+\end{displaymath}
+
+We would like to understand the impact of the number of users who take
+part in the experiment. Especially, how much does adding a user to the
+experiment impacts the MSE.
+
+\subsection{From the bivariate normal case to linear regression}
+If $(X,Y)$ is drawn from a bivariate normal distribution with mean
+vector $\mu$ and covariance matrix $\Sigma$. Then, one can
+write:
+\begin{displaymath}
+  Y = \condexp{Y}{X} + \big(Y-\condexp{Y}{X}\big)
+\end{displaymath}
+In this particular case, $\condexp{Y}{X}$ is a linear function of $X$: 
+\begin{displaymath}
+\condexp{Y}{X} = \alpha X + \beta  
+\end{displaymath}
+where $\alpha$ and $\beta$ can be expressed as a function of $\mu$ and
+$\Sigma$. Writing $\varepsilon = Y-\condexp{Y}{X}$, it is easy to see
+that $\expt{X\varepsilon}=0$. Furthermore $\varepsilon$ is also normally
+distributed. Under these assumptions, it can be proven that the least
+square estimator for $(\alpha,\beta)$ is optimal (it reaches the
+Cramér-Rao bound).
+
+\subsection{Linear regression}
+
+We assume a linear model:
+\begin{displaymath}
+  y_i = \beta\cdot x_i + \varepsilon_i
+\end{displaymath}
+
+Where $\varepsilon_i$ is a normal random variable uncorrelated to
+$x_i$. We also assume that $\var(\varepsilon_i)$ is constant
+(homoscedasticity), $\sigma^2$ will denote the common value.
+
+From the database we compute the least-squared estimator of $\beta$:
+\begin{displaymath}
+  \hat{\beta}_n = (\tr XX)^{-1}\tr XY
+\end{displaymath}
+where $X$ is a $n\times k$ matrix ($k$ is the number of explanatory
+variables)  whose $i$-th row $\tr x_i$ and $Y$ is $(y_1,\ldots,y_n)$.
+
+The regression function is simply the inner product of $\hat{\beta}_n$
+and the new vector of explanatory variables $x$.
+
+From there we can compute the MSE:
+\begin{displaymath}
+  \mse(D_n)
+  =\expt{\left(\beta\cdot x + \varepsilon - \hat\beta_n\cdot x\right)^2}
+  = \tr x\expt{(\hat\beta_n-\beta)\cdot (\hat\beta_n-\beta)} x + \expt{\varepsilon^2}
+\end{displaymath}
+where we used that
+$\expt{x\varepsilon}=0$. The variance-covariance matrix of
+$\hat\beta_n$ is equal to $\sigma^2(\tr XX)^{-1}$. Finally we get:
+\begin{displaymath}
+  \mse(D_n)
+  = \sigma^2\tr x(\tr XX)^{-1}x + \sigma^2
+\end{displaymath}
+
+\subsubsection*{Monotonicity}
+
+We first want to study the impact of adding one observation to the
+database. First, notice that:
+\begin{displaymath}
+  \tr XX = \sum_{i=1}^n x_i \tr x_i
+\end{displaymath}
+Let write $A= \tr X X$. Then, adding $x_0$ to the database will change
+$A$ to $A+x_0\tr x_0$.
+
+The following derivation will make an extensive use of the
+Sherman-Morrisson formula \cite{sm} (which can be proven by direct
+verification). For any invertible matrix $A$:
+\begin{equation}\label{eq:inverse}
+(A+x\tr y)^{-1} = A^{-1} - \frac{A^{-1}x\tr yA^{-1}}{1+\tr x A^{-1}y}
+\end{equation}
+
+$A^{-1}$ is the inverse of a positive semidefinite matrix. Hence it is
+also positive semidefinite. We will denote by $\ip{x}{y}$ the scalar product defined by $A^{-1}$,
+that is:
+\begin{displaymath}
+  \ip{x}{y} = \tr x A^{-1} y = \tr y A^{-1} x
+\end{displaymath}
+
+Using \eqref{eq:inverse} we get:
+\begin{displaymath}
+\begin{split}
+  \tr x (A + x_0\tr x_0)^{-1} x & = \tr x A^{-1} x - \frac{\tr x
+  A^{-1}x_0\tr x_0A^{-1} x}{1+\tr x_0 A^{-1}x_0 }\\
+& = \tr x A^{-1} x - \frac{\ip{x}{x_0}^2}{1+\norm{x_0}^2} 
+\end{split}
+\end{displaymath}
+
+Thus:
+\begin{equation}\label{eq:decrease}
+\mse(D_n\cup\{x_0\}) = \mse(D_n) - \frac{\sigma^2\ip{x}{x_0}^2}{1+\norm{x_0}^2}
+\end{equation}
+
+\emph{Adding one observation to the database decreases the MSE.}
+
+\subsubsection*{Submodularity}
+
+Let $D_m$ a database of size $m$ containing $D_n$: $D_m$ is obtained
+from $D_n$ by adding some observations. We would like to show that
+adding one observation to $D_m$ yields a smaller decrease in MSE than
+adding the same observation to $D_n$:
+\begin{displaymath}
+  \mse(D_m)-\mse(D_m\cup\{x_0\})\leq \mse(D_n)-\mse(D_n\cup\{x_0\})
+\end{displaymath}
+
+First, remark that it is necessary and sufficient to prove this property when $D_n$ and
+$D_m$ differ by only one observation. Indeed, if the property is true
+in general, then it will also be true when the two databases differ by
+only one observation. Conversely, if the property is true when the two
+databases differ by only one observation, then applying the property
+repeatedly yields the general property. 
+
+Using \eqref{eq:decrease}, the decrease of MSE when adding $x_0$ to
+the database is:
+\begin{displaymath}
+  \frac{\sigma^2(\tr x A^{-1} x_0)^2}{1+\tr x_0 A^{-1} x_0}
+\end{displaymath}
+
+If we denote by $z$ the additional observation present in $D_m$ and
+not in $D_n$, then we would like to prove that:
+\begin{displaymath}
+  \frac{\sigma^2(\tr x A^{-1} x_0)^2}{1+\tr x_0 A^{-1} x_0}
+\geq   \frac{\sigma^2\left(\tr x (A+z\tr z)^{-1} x_0\right)^2}{1+\tr x_0 (A+z\tr z)^{-1} x_0}
+\end{displaymath}
+
+Using the same notations as before, this is equivalent
+to:
+\begin{displaymath}
+  \frac{\ip{x}{x_0}^2}{1+\norm{x_0}^2}
+\geq \frac{\left(\left(1+\norm{z}^2\right)\ip{x}{x_0}-\ip{x}{z}\ip{z}{x_0}\right)^2}
+{\left(1+\norm{z}^2\right)\big((1+\norm{z}^2)(1+\norm{x_0}^2)-\ip{x_0}{z}^2\big)}
+\end{displaymath}
+
+By the Cauchy-Schwarz inequality:
+\begin{displaymath}
+ (1+\norm{z}^2)(1+\norm{x_0}^2)-\ip{x_0}{z}^2 > 0 
+\end{displaymath}
+
+Thus the previous inequality is consequently equivalent to:
+\begin{multline*}
+  \left(1+\norm{z}^2\right)^2\left(1+\norm{x_0}^2\right)\ip{x}{x_0}^2
+-\left(1+\norm{z}^2\right)\ip{x_0}{z}^2\ip{x}{x_0}^2\\
+\geq \left(1+\norm{x_0}^2\right)\left(1+\norm{z}^2\right)^2\ip{x}{x_0}^2
++ \left(1+\norm{x_0}^2\right)\ip{x}{z}^2\ip{z}{x_0}^2\\
+-2\left(1+\norm{x_0}^2\right)\left(1+\norm{z}^2\right)\ip{x}{x_0}\ip{x}{z}\ip{z}{x_0}
+\end{multline*}
+
+\begin{multline*}
+2\left(1+\norm{x_0}^2\right)\left(1+\norm{z}^2\right)\ip{x}{x_0}\ip{x}{z}\ip{z}{x_0}\\
+\geq \left(1+\norm{x_0}^2\right)\ip{x}{z}^2\ip{z}{x_0}^2
++ \left(1+\norm{z}^2\right)\ip{x_0}{z}^2\ip{x}{x_0}^2
+\end{multline*}
+
+This last inequality is not true in general. As soon as $x$, $x_0$ and
+$z$ span a 2-dimensional space, it is possible that for example
+$\ip{x}{x_0}$ and $\ip{x}{z}$ are positive and $\ip{z}{x_0}$
+negative. Then the left term of the inequality will be negative and
+cannot be greater than the right term which is always positive.
+
+In the one-dimensional case, the inner product $\ip{x}{z}$ can be
+written as $\lambda xz$ for some positive $\lambda$. Then the last
+inequality becomes:
+\begin{displaymath}
+  2\geq \frac{\lambda z^2}{1+\lambda z^2} 
++ \frac{\lambda x_0^2}{1+\lambda x_0^2}
+\end{displaymath}
+which is trivially true (a more direct proof for the one-dimensional
+case is of course possible).
+
+In order to understand more precisely under which assumptions the
+above inequality could become true, it is convenient to look at it
+from the quadratic form perspective. Indeed this inequality can be
+rewritten as:
+
+\begin{equation}\label{eq-inequality}
+\tr x B x \geq 0 
+\end{equation}
+
+with:
+\begin{align*}
+ B = &\, \left(1+\norm{x_0}^2\right)\left(1+\norm{z}^2\right)\ip{x_0}{z}
+ (x_0\tr z+z\tr x_0)\\
+& -\ip{x_0}{z}^2\Big( \left(1+\norm{x_0}^2\right)z\tr z +  \left(1+\norm{z}^2\right)x_0\tr z\big)
+\end{align*}
+
+This quadratic form is degenerate, its kernel is $x_0^{\bot}\cap
+z^\bot$ which is of dimension $k-2$.
+
+\paragraph{Case when $\norm{x_0}=\norm{z}=1$} In this case, it suffices to study the quadratic form given by matrix
+$B'$ with:
+\begin{displaymath}
+  B' = 2\ip{x_0}{z}(x_0\tr z+z\tr x_0) -\ip{x_0}{z}^2(z\tr z + x_0\tr x_0)
+\end{displaymath}
+
+Writing $a = \ip{x_0}{z}$, the two non-zero eigenvalues are:
+\begin{align*}
+  \lambda_1 & = -2a^3 + 2a^2 + 4a = -2a(a+1)(a-2)\\
+  \lambda_2 & = 2a^3 + 2a^2 - 4a = 2a(a-1)(a+2)
+\end{align*}
+
+which are respectively associated with the eigenvectors:
+\begin{align*}
+  x_1 & = x_0+z\\
+  x_2 & = x_0 - z
+\end{align*}
+
+By the Cauchy-Schwarz inequality, $a\in]-1,1[$, and the two
+eigenvalues are of opposite sign on this interval. Thus inequality
+\eqref{eq-inequality} does not hold for all $x$.
+
+\paragraph{In expectation?} If we assume a prior knowledge on the
+distribution of $x$, writing $\Sigma$ the variance-covariance matrix
+of $x$ and $\mu$ its mean vector, then taking the expectation of
+\eqref{eq-inequality} we get:
+\begin{displaymath}
+  \expt{\tr x B' x} = \trace(B'\Sigma) + \tr\mu B'\mu
+\end{displaymath}
+
+\nocite{shapley,inverse,recommendation,cook,shapleyor,subsetselection11,lse}
+\bibliographystyle{plain}
+\bibliography{notes.bib}
+
+\section*{Appendix}
+
+\paragraph{Previous attempt at taming the submodularity}
+
+The inequality only depends on the projection of $x$ on the plane
+spanned by $x_0$ and $z$. Writing
+\begin{displaymath}
+  x = \lambda x_0 + \mu z + v,\quad\mathrm{with}\quad v \,\bot\,
+  \mathrm{span}\{x_0, v\}
+\end{displaymath}
+the previous inequality becomes:
+\begin{multline*}
+2\left(1+\norm{x_0}^2\right)\left(1+\norm{z}^2\right)\ip{x_0}{z}
+\left(\lambda\norm{x_0}^2+\mu\ip{x_0}{z}\right)
+\left(\lambda\ip{x_0}{z}+\mu\norm{z}^2\right)\\
+- \left(1+\norm{x_0}^2\right)\ip{x_0}{z}^2\left(\lambda\ip{x_0}{z}+\mu\norm{z}^2\right)^2\\
+-
+\left(1+\norm{z}^2\right)\ip{x_0}{z}^2\left(\lambda\norm{x_0}^2+\mu\ip{x_0}{z}\right)^2\geq 0
+\end{multline*}
+
+By expanding and reordering the terms, we obtain a quadratic function
+of $\lambda$ and $\mu$:
+\begin{multline*}
+  \lambda^2\ip{x_0}{z}^2\left[2\norm{x_0}^2+\norm{x_0}^4+\norm{x_0}^2\norm{z}^2
+  +(1+\norm{x_0})^2\big(\norm{x_0}^2\norm{z}^2-\ip{x_0}{z}^2\big)\right]\\
++\mu^2\ip{x_0}{z}^2\left[2\norm{z}^2+\norm{z}^4+\norm{x_0}^2\norm{z}^2
+  +(1+\norm{z})^2\big(\norm{x_0}^2\norm{z}^2-\ip{x_0}{z}^2\big)\right]\\
++2\lambda\mu\ip{x_0}{z}\Big[\norm{x_0}^2\norm{z}^2
+\big(\norm{x_0}^2\norm{z}^2-\ip{x_0}{z}^2\big)\\
++\ip{x_0}{z}^2+\norm{x_0}^2\norm{z}^2
+\big(1+\norm{x_0}^2+\norm{z}^2\big)\Big]\geq 0
+\end{multline*}
+
+This inequality will be true for all $\lambda$ and $\mu$ if and only
+if the quadratic form is positive semidefinite. As its trace is
+positive, this is equivalent to the positiveness of its determinant.
+
+
+\end{document}
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