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-\documentclass{acm_proc_article-sp}
-\usepackage[utf8]{inputenc}
-\usepackage{amsmath,amsfonts}
-\usepackage{algorithm}
-\usepackage{algpseudocode}
-\newtheorem{lemma}{Lemma}
-\newtheorem{fact}{Fact}
-\newtheorem{example}{Example}
-\newtheorem{prop}{Proposition}
-\newtheorem{theorem}{Theorem}
-\newcommand*{\defeq}{\stackrel{\text{def}}{=}}
-\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}}}
-\DeclareMathOperator{\trace}{tr}
-\DeclareMathOperator*{\argmax}{arg\,max}
-\begin{document}
-
-\section{Budget feasible mechanism}
-
-\subsection{Notations}
-
-\begin{itemize}
-    \item If $x\in\mathbf{R}^d$, $x^*$ will denote the transpose of vector $x$.
-        If $(x, y)\in\mathbf{R}^d\times\mathbf{R}^d$, $x^*y$ is the standard
-        inner product of $\mathbf{R}^d$.
-    \item We will often use the following order over symmetric matrices: if $A$
-        and $B$ are two $d\times d$ real symmetric matrices, we write that
-        $A\leq B$ iff:
-        \begin{displaymath}
-            \forall x\in\mathbf{R}^d,\quad x^*Ax \leq x^*Bx 
-        \end{displaymath}
-        That is, iff $B-A$ is symmetric semi-definite positive.
-    \item Let us recall this property of the ordered defined above: if $A$ and
-        $B$ are two symmetric definite positive matrices such that $A\leq B$,
-        then $A^{-1}\leq B^{-1}$.
-\end{itemize}
-
-\subsection{Data model}
-\begin{itemize}
-    \item set of $n$ users $\mathcal{N} = \{1,\ldots, n\}$
-    \item each user has a public vector of features $x_i\in\mathbf{R}^d$ and some
-        undisclosed variable $y_i\in\mathbf{R}$
-    \item we assume that the data has been normalized: $\forall
-        i\in\mathcal{N}$, $\norm{x_i}\leq 1$
-    \item \textbf{Ridge regression model:}
-        \begin{itemize}
-            \item $y_i = \beta^*x_i + \varepsilon_i$
-            \item $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$, 
-                $(\varepsilon_i)_{i\in \mathcal{N}}$ are mutually independent.
-            \item prior knowledge of $\beta$: $\beta\sim\mathcal{N}(0,\kappa I_d)$
-            \item $\mu = \frac{\kappa}{\sigma^2}$ is the regularization factor.
-            \item after the variables $y_i$ are disclosed, the experimenter
-                does \emph{maximum a posteriori estimation} which is equivalent
-                under Gaussian prior to solving the ridge regression
-                maximisation problem:
-                \begin{displaymath}
-                    \beta_{\text{max}} = \argmax \sum_i (y_i - \beta^*x_i)^2
-                    + \frac{1}{\mu}\sum_i \norm{\beta}_2^2
-                \end{displaymath}
-        \end{itemize}
-\end{itemize}
-
-\subsection{Economics}
-
-Value function: following the experiment design theory, the value of
-data is the decrease of uncertainty about the model. Common measure of
-uncertainty, the entropy. Thus, the value of a set $S$ of users is:
-\begin{displaymath}
-    V(S) = \mathbb{H}(\beta) - \mathbb{H}(\beta|Y_S)
-\end{displaymath}
-where $Y_S = \{y_i,\,i\in S\}$ is the set of observations.
-
-In our case (under Gaussian prior):
-\begin{align*}
-    \forall S\subset\mathcal{N},\; V(S)
-    & = \frac{1}{2}\log\det\left(I_d
-        + \mu\sum_{i\in S} x_ix_i^*\right)\\
-    & \defeq \frac{1}{2}\log\det A(S)
-\end{align*}
-
-\begin{lemma}[Marginal contribution]
-    \begin{displaymath}
-        \Delta_i V(S)\defeq V(S\cup\{i\}) - V(S) 
-        = \frac{1}{2}\log\left(1 + \mu x_i^*A(S)^{-1}x_i\right)
-    \end{displaymath}
-\end{lemma}
-
-\begin{proof}
-    We have:
-    \begin{align*}
-        V(S\cup\{i\}) & = \frac{1}{2}\log\det A(S\cup\{i\})\\
-        & = \frac{1}{2}\log\det\left(A(S) + \mu x_i x_i^*\right)\\
-        & = V(S) + \frac{1}{2}\log\det\left(I_d + \mu A(S)^{-1}x_i
-    x_i^*\right)\\
-        & = V(S) + \frac{1}{2}\log\left(1 + \mu x_i^* A(S)^{-1}x_i\right)
-    \end{align*}
-    where the last equality comes from Sylvester's determinant formula.
-\end{proof}
-\begin{itemize}
-    \item each user $i$ has a cost $c_i$
-    \item the auctioneer has a budget constraint $B$
-    \item optimisation problem:
-        \begin{displaymath}
-            OPT(V,\mathcal{N}, B) = \max_{S\subset\mathcal{N}} \left\{ V(S)\,|\,
-            \sum_{i\in S}c_i\leq B\right\}
-        \end{displaymath}
-\end{itemize}
-
-\subsection{Relaxations of the value function}
-
-We say that $R_\mathcal{N}:[0,1]^n\rightarrow\mathbf{R}$ is a relaxation of the
-value function $V$ over $\mathcal{N}$ if it coincides with $V$ at binary
-points. Formally, for any $S\subset\mathcal{N}$, let $\mathbf{1}_S$ denote the
-indicator vector of $S$. $R_\mathcal{N}$ is a relaxation of $V$ over
-$\mathcal{N}$ iff:
-\begin{displaymath}
-    \forall S\subset\mathcal{N},\; R_\mathcal{N}(\mathbf{1}_S) = V(S)
-\end{displaymath}
-
-We can extend the optimisation problem defined above to a relaxation by
-extending the cost function:
-\begin{displaymath}
-    \forall \lambda\in[0,1]^n,\; c(\lambda)
-    = \sum_{i\in\mathcal{N}}\lambda_ic_i
-\end{displaymath}
-The optimisation problem becomes:
-\begin{displaymath}
-    OPT(R_\mathcal{N}, B) =
-    \max_{\lambda\in[0,1]^n}\left\{R_\mathcal{N}(\lambda)\,|\, c(\lambda)\leq B\right\}
-\end{displaymath}
-The relaxations we will consider here rely on defining a probability
-distribution over subsets of $\mathcal{N}$.
-
-Let $\lambda\in[0,1]^n$, let us define:
-\begin{displaymath}
-    P_\mathcal{N}^\lambda(S) = \prod_{i\in S}\lambda_i
-    \prod_{i\in\mathcal{N}\setminus S}(1-\lambda_i)
-\end{displaymath}
-$P_\mathcal{N}^\lambda(S)$ is the probability of picking the set $S$ if we select
-a subset of $\mathcal{N}$ at random by deciding independently for each point to
-include it in the set with probability $\lambda_i$ (and to exclude it with
-probability $1-\lambda_i$).
-
-We will consider two relaxations of the value function $V$ over $\mathcal{N}$:
-\begin{itemize}
-    \item the \emph{multi-linear extension} of $V$:
-        \begin{align*}
-            F_\mathcal{N}(\lambda) 
-            & = \mathbb{E}_{S\sim P_\mathcal{N}^\lambda}\big[\log\det A(S)\big]\\
-            & = \sum_{S\subset\mathcal{N}} P_\mathcal{N}^\lambda(S) V(S)\\
-            & = \sum_{S\subset\mathcal{N}} P_\mathcal{N}^\lambda(S) \log\det A(S)\\
-        \end{align*}
-    \item the \emph{concave relaxation} of $V$:
-        \begin{align*}
-            L_{\mathcal{N}}(\lambda) 
-            & = \log\det \mathbb{E}_{S\sim P_\mathcal{N}^\lambda}\big[A(S)\big]\\
-            & = \log\det\left(\sum_{S\subset N}
-                P_\mathcal{N}^\lambda(S)A(S)\right)\\
-            & = \log\det\left(I_d + \mu\sum_{i\in\mathcal{N}}
-                \lambda_ix_ix_i^*\right)\\
-            & \defeq \log\det \tilde{A}(\lambda)
-        \end{align*}
-\end{itemize}
-
-\begin{lemma}
-    The \emph{concave relaxation} $L_\mathcal{N}$ is concave\footnote{Hence
-    this relaxation is well-named!}.
-\end{lemma}
-
-\begin{proof}
-    This follows from the concavity of the $\log\det$ function over symmetric
-    positive semi-definite matrices. More precisely, if $A$ and $B$ are two
-    symmetric positive semi-definite matrices, then:
-    \begin{multline*}
-        \forall\alpha\in [0, 1],\; \log\det\big(\alpha A + (1-\alpha) B\big)\\
-        \geq \alpha\log\det A + (1-\alpha)\log\det B
-    \end{multline*}
-\end{proof}
-
-\begin{lemma}[Rounding]\label{lemma:rounding}
-    For any feasible $\lambda\in[0,1]^n$, there exists a feasible
-    $\bar{\lambda}\in[0,1]^n$ such that at most one of its component is
-    fractional, that is, lies in $(0,1)$ and:
-    \begin{displaymath}
-        F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})
-    \end{displaymath}
-\end{lemma}
-
-\begin{proof}
-    We give a rounding procedure which given a feasible $\lambda$ with at least
-    two fractional components, returns some $\lambda'$ with one less fractional
-    component, feasible such that:
-    \begin{displaymath}
-        F_\mathcal{N}(\lambda) \leq F_\mathcal{N}(\lambda')
-    \end{displaymath}
-    Applying this procedure recursively yields the lemma's result.
-
-    Let us consider such a feasible $\lambda$. Let $i$ and $j$ be two
-    fractional components of $\lambda$ and let us define the following
-    function:
-    \begin{displaymath}
-        F_\lambda(\varepsilon) = F(\lambda_\varepsilon)
-        \quad\textrm{where} \quad
-        \lambda_\varepsilon = \lambda + \varepsilon\left(e_i-\frac{c_i}{c_j}e_j\right)
-    \end{displaymath}
-
-    It is easy to see that if $\lambda$ is feasible, then:
-    \begin{multline}\label{eq:convex-interval}
-        \forall\varepsilon\in\Big[\max\Big(-\lambda_i,(\lambda_j-1)\frac{c_j}{c_i}\Big), \min\Big(1-\lambda_i, \lambda_j
-        \frac{c_j}{c_i}\Big)\Big],\;\\
-            \lambda_\varepsilon\;\;\textrm{is feasible}
-    \end{multline}
-
-    Furthermore, the function $F_\lambda$ is convex, indeed:
-    \begin{align*}
-        F_\lambda(\varepsilon)
-        & = \mathbb{E}_{S'\sim P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[
-        (\lambda_i+\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i,j\})\\
-        & + (\lambda_i+\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i\})\\
-        & + (1-\lambda_i-\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{j\})\\
-        & + (1-\lambda_i-\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S')\Big]\\
-    \end{align*}
-    Thus, $F_\lambda$ is a degree 2 polynomial whose dominant coefficient is:
-    \begin{multline*}
-        \frac{c_i}{c_j}\mathbb{E}_{S'\sim
-        P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[
-            V(S'\cup\{i\})+V(S'\cup\{i\})\\
-        -V(S'\cup\{i,j\})-V(S')\Big]
-    \end{multline*}
-    which is positive by submodularity of $V$. Hence, the maximum of
-    $F_\lambda$ over the interval given in \eqref{eq:convex-interval} is
-    attained at one of its limit, at which either the $i$-th or $j$-th component of
-    $\lambda_\varepsilon$ becomes integral.
-\end{proof}
-
-\begin{lemma}\label{lemma:relaxation-ratio}
-    The following inequality holds:
-    \begin{displaymath}
-        \forall\lambda\in[0,1]^n,\; 
-        \frac{\log\big(1+\mu\big)}{2\mu}
-        \,L_\mathcal{N}(\lambda)\leq
-        F_\mathcal{N}(\lambda)\leq L_{\mathcal{N}}(\lambda)
-    \end{displaymath}
-\end{lemma}
-
-\begin{proof}
-
-    We will prove that:
-    \begin{displaymath}
-        \frac{\log\big(1+\mu\big)}{2\mu}
-    \end{displaymath}
-    is a lower bound of the ratio $\partial_i F_\mathcal{N}(\lambda)/\partial_i
-    L_\mathcal{N}(\lambda)$.
-
-    This will be enough to conclude, by observing that:
-    \begin{displaymath}
-        \frac{F_\mathcal{N}(\lambda)}{L_\mathcal{N}(\lambda)}
-        \sim_{\lambda\rightarrow 0}
-        \frac{\sum_{i\in \mathcal{N}}\lambda_i\partial_i F_\mathcal{N}(0)}
-        {\sum_{i\in\mathcal{N}}\lambda_i\partial_i L_\mathcal{N}(0)}
-    \end{displaymath}
-    and that an interior critical point of the ratio
-    $F_\mathcal{N}(\lambda)/L_\mathcal{N}(\lambda)$ is defined by:
-    \begin{displaymath}
-        \frac{F_\mathcal{N}(\lambda)}{L_\mathcal{N}(\lambda)}
-        = \frac{\partial_i F_\mathcal{N}(\lambda)}{\partial_i
-        L_\mathcal{N}(\lambda)}
-    \end{displaymath}
-
-    Let us start by computing the derivatives of $F_\mathcal{N}$ and
-    $L_\mathcal{N}$ with respect to
-    the $i$-th component.
-
-    For $F$, it suffices to look at the derivative of
-    $P_\mathcal{N}^\lambda(S)$:
-    \begin{displaymath}
-        \partial_i P_\mathcal{N}^\lambda(S) = \left\{
-            \begin{aligned}
-                & P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})\;\textrm{if}\; i\in S \\
-                & - P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\;\textrm{if}\;
-                i\in \mathcal{N}\setminus S \\
-            \end{aligned}\right.
-    \end{displaymath}
-
-    Hence:
-    \begin{multline*}
-        \partial_i F_\mathcal{N} =
-        \sum_{\substack{S\subset\mathcal{N}\\ i\in S}}
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})V(S)\\
-        - \sum_{\substack{S\subset\mathcal{N}\\ i\in \mathcal{N}\setminus S}}
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S)\\
-    \end{multline*}
-
-    Now, using that every $S$ such that $i\in S$ can be uniquely written as
-    $S'\cup\{i\}$, we can write:
-    \begin{multline*}
-        \partial_i F_\mathcal{N} =
-        \sum_{\substack{S\subset\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S\cup\{i\})\\
-        - \sum_{\substack{S\subset\mathcal{N}\\ i\in \mathcal{N}\setminus S}}
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S)\\
-    \end{multline*}
-
-    Finally, by using the expression for the marginal contribution of $i$ to
-    $S$:
-    \begin{displaymath}
-        \partial_i F_\mathcal{N}(\lambda) = 
-        \sum_{\substack{S\subset\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)
-        \log\Big(1 + \mu x_i^*A(S)^{-1}x_i\Big)
-    \end{displaymath}
-
-    The computation of the derivative of $L_\mathcal{N}$ uses standard matrix
-    calculus and gives:
-    \begin{displaymath}
-        \partial_i L_\mathcal{N}(\lambda)
-        = \mu x_i^* \tilde{A}(\lambda)^{-1}x_i
-    \end{displaymath}
-    
-    Using the following inequalities:
-    \begin{gather*}
-        \forall S\subset\mathcal{N}\setminus\{i\},\quad
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq
-        P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\})\\
-        \forall S\subset\mathcal{N},\quad P_{\mathcal{N}\setminus\{i\}}^\lambda(S) 
-        \geq P_\mathcal{N}^\lambda(S)\\
-        \forall S\subset\mathcal{N},\quad A(S)^{-1} \geq A(S\cup\{i\})^{-1}\\
-    \end{gather*}
-    we get:
-    \begin{align*}
-        \partial_i F_\mathcal{N}(\lambda) 
-        & \geq \sum_{\substack{S\subset\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
-        P_\mathcal{N}^\lambda(S)
-        \log\Big(1 + \mu x_i^*A(S)^{-1}x_i\Big)\\
-        & \geq \frac{1}{2}
-        \sum_{\substack{S\subset\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
-        P_\mathcal{N}^\lambda(S)
-        \log\Big(1 + \mu x_i^*A(S)^{-1}x_i\Big)\\
-        &\hspace{-3.5em}+\frac{1}{2}
-        \sum_{\substack{S\subset\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
-        P_\mathcal{N}^\lambda(S\cup\{i\})
-        \log\Big(1 + \mu x_i^*A(S\cup\{i\})^{-1}x_i\Big)\\
-        &\geq \frac{1}{2}
-        \sum_{S\subset\mathcal{N}}
-        P_\mathcal{N}^\lambda(S)
-        \log\Big(1 + \mu x_i^*A(S)^{-1}x_i\Big)\\
-    \end{align*}
-
-    Using that $A(S)\geq I_d$ we get that:
-    \begin{displaymath}
-        \mu x_i^*A(S)^{-1}x_i \leq \mu
-    \end{displaymath}
-    
-    Moreover:
-    \begin{displaymath}
-        \forall x\leq\mu,\; \log(1+x)\geq
-    \frac{\log\big(1+\mu\big)}{\mu} x
-    \end{displaymath}
-
-    Hence:
-    \begin{displaymath}
-        \partial_i F_\mathcal{N}(\lambda) \geq
-        \frac{\log\big(1+\mu\big)}{2\mu}
-        x_i^*\bigg(\sum_{S\subset\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)^{-1}\bigg)x_i
-    \end{displaymath}
-    
-    Finally, using that the inverse is a matrix convex function over symmetric
-    positive definite matrices:
-    \begin{align*}
-        \partial_i F_\mathcal{N}(\lambda) &\geq
-        \frac{\log\big(1+\mu\big)}{2\mu}
-        x_i^*\bigg(\sum_{S\subset\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)\bigg)^{-1}x_i\\
-        & \geq \frac{\log\big(1+\mu\big)}{2\mu}
-        \partial_i L_\mathcal{N}(\lambda)
-    \end{align*}
-\end{proof}
-
-\begin{lemma}
-    Let us denote by $C_\mu$ the constant of which appears in the bound of the
-    previous lemma: $C_\mu = \frac{\log(1+\mu)}{2\mu}$, then:
-    \begin{displaymath}
-        OPT(L_\mathcal{N}, B) \leq \frac{1}{C_\mu}\big(2 OPT(V,\mathcal{N},B)
-        + \max_{i\in\mathcal{N}}V(i)\big)
-    \end{displaymath}
-\end{lemma}
-
-\begin{proof}
-    Let us consider a feasible point $\lambda^*\in[0,1]^n$ such that $L_\mathcal{N}(\lambda^*)
-    = OPT(L_\mathcal{N}, B)$. By applying lemma~\ref{lemma:relaxation-ratio}
-    and lemma~\ref{lemma:rounding} we get a feasible point $\bar{\lambda}$ with at most
-    one fractional component such that:
-    \begin{equation}\label{eq:e1}
-        L_\mathcal{N}(\lambda^*) \leq \frac{1}{C_\mu}
-        F_\mathcal{N}(\bar{\lambda})
-    \end{equation}
-
-    Let $\lambda_i$ denote the fractional component of $\bar{\lambda}$ and $S$
-    denote the set whose indicator vector is $\bar{\lambda} - \lambda_i e_i$.
-    Using the fact that $F_\mathcal{N}$ is linear with respect to the $i$-th
-    component and is a relaxation of the value function, we get:
-    \begin{displaymath}
-        F_\mathcal{N}(\bar{\lambda}) = V(S) +\lambda_i V(S\cup\{i\})
-    \end{displaymath}
-
-    Using the submodularity of $V$:
-    \begin{displaymath}
-        F_\mathcal{N}(\bar{\lambda}) \leq 2 V(S) + V(i)
-    \end{displaymath}
-
-    Note that since $\bar{\lambda}$ is feasible, $S$ is also feasible and
-    $V(S)\leq OPT(V,\mathcal{N}, B)$. Hence:
-    \begin{equation}\label{eq:e2}
-        F_\mathcal{N}(\bar{\lambda}) \leq 2 OPT(V,\mathcal{N}, B)
-        + \max_{i\in\mathcal{N}} V(i)
-    \end{equation}
-
-    Putting \eqref{eq:e1} and \eqref{eq:e2} together gives the results.
-\end{proof}
-
-\subsection{The mechanism}
-
-\begin{algorithm}\label{mechanism}
-    \caption{Budget feasible mechanism for ridge regression}
-    \begin{algorithmic}[1]
-    \State $i^* \gets \argmax_{j\in\mathcal{N}}V(j)$
-    \State $x^* \gets \argmax_{x\in[0,1]^n} \{L_{\mathcal{N}\setminus\{i^*\}}(x)
-                                    \,|\, c(x)\leq B\}$
-        \Statex
-        \If{$L(x^*) < CV(i^*)$}
-            \State \textbf{return} $\{i^*\}$
-        \Else
-            \State $i \gets \argmax_{1\leq j\leq n}\frac{V(j)}{c_j}$
-            \State $S \gets \emptyset$
-            \While{$c_i\leq \frac{B}{2}\frac{V(S\cup\{i\})-V(S)}{V(S\cup\{i\})}$}
-                \State $S \gets S\cup\{i\}$
-                \State $i \gets \argmax_{j\in\mathcal{N}\setminus S}
-                \frac{V(S\cup\{j\})-V(S)}{c_j}$
-            \EndWhile
-            \State \textbf{return} $S$
-        \EndIf
-    \end{algorithmic}
-\end{algorithm}
-
-The constant $C$ which appears in the mechanism is unspecified for now. We can
-make the analysis regardless of its value. The optimal value will be specified
-in the theorem.
-
-\begin{lemma}
-The mechanism is monotone.
-\end{lemma}
-
-\begin{proof}
-    We assume by contradiction that there exists a user $i$ that has been
-    selected by the mechanism and that would not be selected had he reported
-    a cost $c_i'\leq c_i$ (all the other costs staying the same).
-
-    If $i\neq i^*$ and $i$ has been selected, then we are in the case where
-    $L(x^*) \geq C V(i^*)$ and $i$ was included in the result set by the greedy
-    part of the mechanism. By reporting a cost $c_i'\leq c_i$, using the
-    submodularity of $V$, we see that $i$ will satisfy the greedy selection
-    rule:
-    \begin{displaymath}
-        i = \argmax_{j\in\mathcal{N}\setminus S} \frac{V(S\cup\{j\})
-        - V(S)}{c_j}
-    \end{displaymath}
-    in an earlier iteration of the greedy heuristic. Let us denote by $S_i$
-    (resp. $S_i'$) the set to which $i$ is added when reporting cost $c_i$
-    (resp. $c_i'$). We have $S_i'\subset S_i$. Moreover:
-    \begin{align*}
-        c_i' & \leq c_i \leq
-        \frac{B}{2}\frac{V(S_i\cup\{i\})-V(S_i)}{V(S_i\cup\{i\})}\\
-        & \leq \frac{B}{2}\frac{V(S_i'\cup\{i\})-V(S_i')}{V(S_i'\cup\{i\})}
-    \end{align*}
-    Hence $i$ will still be included in the result set.
-
-    If $i = i^*$, $i$ is included iff $L(x^*) \leq C V(i^*)$. Reporting $c_i'$
-    instead of $c_i$ does not change the value $V(i^*)$ nor $L(x^*)$ (which is
-    computed over $\mathcal{N}\setminus\{i^*\}$). Thus $i$ is still included by
-    reporting a different cost.
-\end{proof}
-
-\begin{lemma}
-The mechanism is budget feasible.
-\end{lemma}
-
-\begin{lemma}
-    Let us denote by $S_M$ the set returned by the mechanism. Let us also
-    write:
-    \begin{displaymath}
-        C_{\textrm{max}} = \max\left(1+C,\frac{e}{e-1}\left( 3 + \frac{12e}{C\cdot
-        C_\mu(e-1) -5e  +1}\right)\right)
-    \end{displaymath}
-
-    Then:
-    \begin{displaymath}
-        OPT(V, \mathcal{N}, B) \leq
-        C_\text{max}\cdot V(S_M) 
-    \end{displaymath}
-\end{lemma}
-
-\begin{proof}
-
-    If the condition on line 3 of the algorithm holds, then:
-    \begin{displaymath}
-        V(i^*) \geq \frac{1}{C}L(x^*) \geq
-        \frac{1}{C}OPT(V,\mathcal{N}\setminus\{i\}, B)
-    \end{displaymath}
-
-    But:
-    \begin{displaymath}
-        OPT(V,\mathcal{N},B) \leq OPT(V,\mathcal{N}\setminus\{i\}, B) + V(i^*)
-    \end{displaymath}
-
-    Hence:
-    \begin{displaymath}
-        V(i^*) \geq \frac{1}{C+1} OPT(V,\mathcal{N}, B)
-    \end{displaymath}
-
-    If the condition of the algorithm does not hold:
-    \begin{align*}
-        V(i^*) & \leq \frac{1}{C}L(x^*) \leq \frac{1}{C\cdot C_\mu}
-        \big(2 OPT(V,\mathcal{N}, B) + V(i^*)\big)\\
-        & \leq \frac{1}{C\cdot C_\mu}\left(\frac{2e}{e-1}\big(3 V(S_M)
-        + 2 V(i^*)\big)
-        + V(i^*)\right)
-    \end{align*}
-    
-    Thus:
-    \begin{align*}
-        V(i^*) \leq \frac{6e}{C\cdot C_\mu(e-1)- 5e + 1} V(S_M)
-    \end{align*}
-
-    Finally, using again that:
-    \begin{displaymath}
-        OPT(V,\mathcal{N},B) \leq \frac{e}{e-1}\big(3 V(S_M) + 2 V(i^*)\big)
-    \end{displaymath}
-    
-    We get:
-    \begin{displaymath}
-        OPT(V, \mathcal{N}, B) \leq \frac{e}{e-1}\left( 3 + \frac{12e}{C\cdot
-        C_\mu(e-1) -5e  +1}\right) V(S_M)
-    \end{displaymath}
-\end{proof}
-
-The optimal value for $C$ is:
-\begin{displaymath}
-    C^* = \arg\min C_{\textrm{max}}
-\end{displaymath}
-
-This equation has two solutions. Only one of those is such that:
-\begin{displaymath}
-    C\cdot C_\mu(e-1) -5e  +1 \geq 0
-\end{displaymath}
-which is needed in the proof of the previous lemma. Computing this solution,
-we can state the main result of this section.
-
-\begin{theorem}
-    The mechanism is individually rational, truthful and has an approximation
-    ratio of:
-    \begin{multline*}
-        1 + C^* = \frac{5e-1 + C_\mu(4e-1)}{2C_\mu(e-1)}\\
-        + \frac{\sqrt{C_\mu^2(1+2e)^2
-        + 2C_\mu(14e^2+5e+1) + (1-5e)^2}}{2C_\mu(e-1)}
-    \end{multline*}
-\end{theorem}
-\end{document}