From 3fdd3c6674acfecffa3ec734f0183f1df6a239c1 Mon Sep 17 00:00:00 2001
From: Stratis Ioannidis <stratis@stratis-Latitude-E6320.(none)>
Date: Sun, 28 Oct 2012 15:16:26 -0700
Subject: file splitting

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+
+All the budget feasible mechanisms studied recently (TODO ref Singer Chen)
+rely on using a greedy heuristic extending the idea of the greedy heuristic for
+the knapsack problem. In knapsack, objects are selected based on their
+\emph{value-per-cost} ratio. The quantity which plays a similar role for
+general submodular functions is the \emph{marginal-contribution-per-cost}
+ratio: let us assume that you have already selected a set of points $S$, then
+the \emph{marginal-contribution-per-cost} ratio per cost of a new point $i$ is
+defined by:
+\begin{displaymath}
+    \frac{V(S\cup\{i\}) - V(S)}{c_i}
+\end{displaymath}
+
+The greedy heuristic then simply repeatedly selects the point whose
+marginal-contribution-per-cost ratio is the highest until it reaches the budget
+limit. Mechanism considerations aside, this is known to have an unbounded
+approximation ratio. However, lemma TODO ref in Chen or Singer shows that the
+maximum between the greedy heuristic and the point with maximum value (as
+a singleton set) provides a $\frac{5e}{e-1}$ approximation ratio.
+
+Unfortunately, TODO Singer 2011 points out that taking the maximum between the
+greedy heuristic and the most valuable point is not incentive compatible.
+Singer and Chen tackle this issue similarly: instead of comparing the most
+valuable point to the greedy solution, they compare it to a solution which is
+close enough to keep a constant approximation ratio:
+\begin{itemize}
+    \item Chen suggests using $OPT(V,\mathcal{N}\setminus\{i\}, B)$.
+        Unfortunately, in the general case, this cannot be computed exactly in
+        polynomial time.
+    \item Singer uses using the optimal value of a relaxed objective function
+        which can be proven to be close to the optimal of the original
+        objective function. The function used is tailored to the specific
+        problem of coverage.
+\end{itemize}
+
+Here, we use a relaxation of the objective function which is tailored to the
+problem of ridge regression. We define:
+\begin{displaymath}
+    \forall\lambda\in[0,1]^{|\mathcal{N}|}\,\quad L_{\mathcal{N}}(\lambda) \defeq
+    \log\det\left(I_d + \mu\sum_{i\in\mathcal{N}} \lambda_i x_i x_i^*\right)
+\end{displaymath}
+
+We can now present the mechanism we use, which has a same flavor to Chen's and
+Singer's
+
+\begin{algorithm}\label{mechanism}
+    \caption{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}
+
+Notice, that the stopping condition in the while loop is more sophisticated
+than just ensuring that the sum of the costs does not exceed the budget. This
+is because the selected users will be payed more than their costs, and this
+stopping condition ensures budget feasibility when the users are paid their
+threshold payment.
+
+We can now state the main result of this section:
+\begin{theorem}
+    The mechanism in \ref{mechanism} is truthful, individually rational, budget
+    feasible. Furthermore, choosing:
+    \begin{multline*}
+        C = C^* =  \frac{5e-1 + C_\mu(2e+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*}
+    we get 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*}
+    where:
+    \begin{displaymath}
+        C_\mu = \frac{\log(1+\mu)}{2\mu}
+    \end{displaymath}
+\end{theorem}
+
+The proof will consist of the claims of the theorem broken down into lemmas.
+
+Because the user strategy is parametrized by a single parameter, truthfulness
+is equivalent to monotonicity (TODO ref). The proof here is the same as in TODO
+and is given for the sake of completeness.
+
+\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}
+
+The proof is the same as in Chen and is given here for the sake of
+completeness.
+\begin{proof}
+
+\end{proof}
+
+Using the characterization of the threshold payments from Singer TODO, we can
+prove individual rationality similarly to Chen TODO.
+\begin{lemma}
+The mechanism is individually rational
+\end{lemma}
+
+\begin{proof}
+
+\end{proof}
+
+The following lemma requires to have a careful look at the relaxation function
+we chose in the mechanism. The next section will be dedicated to studying this
+relaxation and will contain the proof of the following lemma:
+\begin{lemma}
+    We have:
+    \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{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.
+
+\subsection{Relaxations of the value function}
+
+To prove lemma TODO, we will use a general method called pipage rounding,
+introduced in TODO. Two consecutive relaxations are used: the one that we are
+interested in, whose optimization can be computed efficiently, and the
+multilinear extension which presents a \emph{cross-convexity} like behavior
+which allows for rounding of fractional solution without decreasing the value
+of the objective function and thus ensures a constant approximation of the
+value function. The difficulty resides in showing that the ratio of the two
+relaxations is bounded.
+
+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}
+
+We can now prove lemma TODO from previous section.
+
+\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}
+
+
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