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diff --git a/proofs.tex b/proofs.tex new file mode 100644 index 0000000..ee57762 --- /dev/null +++ b/proofs.tex @@ -0,0 +1,428 @@ +\subsection{Proof of Theorem~\ref{thm:main}}\label{sec:proofofmainthm} + +We now present the proof of Theorem~\ref{thm:main}. Truthfulness and individual +rationality follow from monotonicity and threshold payments. Monotonicity and +budget feasibility follow the same steps as the analysis of \citeN{chen}; + for the sake of completeness, we restate their proof in the Appendix. + +The complexity of the mechanism is given by the following lemma. +\begin{lemma}[Complexity]\label{lemma:complexity} + For any $\varepsilon > 0$, the complexity of the mechanism is + $O(\text{poly}(n, d, \log\log \varepsilon^{-1}))$. +\end{lemma} +\begin{proof} + The value function $V$ in \eqref{modified} can be computed in time + $O(\text{poly}(n, d))$ and the mechanism only involves a linear + number of queries to the function $V$. + The function $\log\det$ is concave and self-concordant (see + \cite{boyd2004convex}), so for any $\varepsilon$, its maximum can be found + to a precision $\varepsilon$ in $O(\log\log\varepsilon^{-1})$ of iterations of Newton's method. Each iteration can be + done in time $O(\text{poly}(n, d))$. Thus, line 3 of + Algorithm~\ref{mechanism} can be computed in time + $O(\text{poly}(n, d, \log\log \varepsilon^{-1}))$. Hence the allocation + function's complexity is as stated. + %Payments can be easily computed in time $O(\text{poly}(n, d))$ as in prior work. +\junk{ + Using Singer's characterization of the threshold payments + \cite{singer-mechanisms}, one can verify that they can be computed in time + $O(\text{poly}(n, d))$. + } +\end{proof} + +Finally, we prove the approximation ratio of the mechanism. We use the +following lemma which establishes that $OPT'$, the optimal value \eqref{relax} of the fractional relaxation $L$ under the budget constraints + is not too far from $OPT$. +\begin{lemma}[Approximation]\label{lemma:relaxation} + $ OPT' \leq 2 OPT + + 2\max_{i\in\mathcal{N}}V(i)$ +\end{lemma} +The proof of Lemma~\ref{lemma:relaxation} is our main technical contribution, and can be found in Section \ref{sec:relaxation}. + +Using Lemma~\ref{lemma:relaxation} we can complete the proof of +Theorem~\ref{thm:main} by showing that, for any $\varepsilon > 0$, if +$OPT_{-i}'$, the optimal value of $L$ when $i^*$ is excluded from +$\mathcal{N}$, has been computed to a precision $\varepsilon$, then the set +$S^*$ allocated by the mechanism is such that: +\begin{equation} \label{approxbound} +OPT +\leq \frac{10e\!-\!3 + \sqrt{64e^2\!-\!24e\!+\!9}}{2(e\!-\!1)} V(S^*)\! ++ \! \varepsilon +\end{equation} +To see this, let $OPT_{-i^*}'$ be the true maximum value of $L$ subject to +$\lambda_{i^*}=0$, $\sum_{i\in \mathcal{N}\setminus{i^*}}c_i\leq B$. Assume +that on line 3 of algorithm~\ref{mechanism}, a quantity $\tilde{L}$ such that +$\tilde{L}-\varepsilon\leq OPT_{-i^*}' \leq \tilde{L}+\varepsilon$ has been +computed (Lemma~\ref{lemma:complexity} states that this is computed in time +within our complexity guarantee). If the condition on line 3 of the algorithm +holds, then: +\begin{displaymath} + V(i^*) \geq \frac{1}{C}OPT_{-i^*}'-\frac{\varepsilon}{C} \geq + \frac{1}{C}OPT_{-i^*} -\frac{\varepsilon}{C} +\end{displaymath} +as $L$ is a fractional relaxation of $V$. Also, $OPT \leq OPT_{-i^*} + V(i^*)$, +hence: +\begin{equation}\label{eq:bound1} + OPT\leq (1+C)V(i^*) + \varepsilon +\end{equation} +Note that $OPT_{-i^*}'\leq OPT'$. If the condition does not hold, from Lemmas +\ref{lemma:relaxation} and \ref{lemma:greedy-bound}: +\begin{align*} + V(i^*) & \stackrel{}\leq \frac{1}{C}OPT_{-i^*}' + \frac{\varepsilon}{C} + \leq \frac{1}{C} \big(2 OPT + 2 V(i^*)\big) + \frac{\varepsilon}{C}\\ + & \leq \frac{1}{C}\left(\frac{2e}{e-1}\big(3 V(S_G) + + 2 V(i^*)\big) + 2 V(i^*)\right) + \frac{\varepsilon}{C} +\end{align*} +Thus, if $C$ is such that $C(e-1) -6e +2 > 0$, +\begin{align*} + V(i^*) \leq \frac{6e}{C(e-1)- 6e + 2} V(S_G) + + \frac{(e-1)\varepsilon}{C(e-1)- 6e + 2} +\end{align*} +Finally, using again Lemma~\ref{lemma:greedy-bound}, we get: +\begin{equation}\label{eq:bound2} + OPT(V, \mathcal{N}, B) \leq + \frac{3e}{e-1}\left( 1 + \frac{4e}{C (e-1) -6e +2}\right) V(S_G) + + \frac{2e\varepsilon}{C(e-1)- 6e + 2} +\end{equation} +To minimize the coefficients of $V_{i^*}$ and $V(S_G)$ in \eqref{eq:bound1} +and \eqref{eq:bound2} respectively, we wish to chose for $C=C^*$ such that: +\begin{displaymath} + C^* = \argmin_C + \max\left(1+C,\frac{3e}{e-1}\left( 1 + \frac{4e}{C (e-1) -6e +2} + \right)\right) +\end{displaymath} +This equation has two solutions. Only one of those is such that $C(e-1) -6e ++2 \geq 0$. This solution is: +\begin{equation}\label{eq:constant} + C^* = \frac{8e-1 + \sqrt{64e^2-24e + 9}}{2(e-1)} \label{eq:c} +\end{equation} +For this solution, $\frac{2e\varepsilon}{C^*(e-1)- 6e + 2}\leq \varepsilon.$ +Placing the expression of $C^*$ in \eqref{eq:bound1} and \eqref{eq:bound2} +gives the approximation ratio in \eqref{approxbound}, and concludes the proof +of Theorem~\ref{thm:main}.\hspace*{\stretch{1}}\qed + +\subsection{Proof of Lemma~\ref{lemma:relaxation}}\label{sec:relaxation} + +We need to prove that for our relaxation $L$ given by +\eqref{eq:our-relaxation}, $OPT'$ is close to $OPT$ as stated in +Lemma~\ref{lemma:relaxation}. Our analysis follows the \emph{pipage rounding} +framework of \citeN{pipage}. + +This framework uses the \emph{multi-linear} extension $F$ of the submodular +function $V$. Let $P_\mathcal{N}^\lambda(S)$ be the probability of choosing the set $S$ if we select each element $i$ in $\mathcal{N}$ independently with probability $\lambda_i$: +\begin{displaymath} + P_\mathcal{N}^\lambda(S) \defeq \prod_{i\in S} \lambda_i + \prod_{i\in\mathcal{N}\setminus S}( 1 - \lambda_i). +\end{displaymath} +Then, the \emph{multi-linear} extension $F$ is defined by: +\begin{displaymath} + F(\lambda) + \defeq \mathbb{E}_{S\sim P_\mathcal{N}^\lambda}\big[V(S)\big] + = \sum_{S\subseteq\mathcal{N}} P_\mathcal{N}^\lambda(S) V(S) +\end{displaymath} + +For \EDP{} the multi-linear extension can be written: +\begin{equation}\label{eq:multi-linear-logdet} + F(\lambda) = \mathbb{E}_{S\sim + P_\mathcal{N}^\lambda}\bigg[\log\det \big(I_d + \sum_{i\in S} x_i\T{x_i}\big) \Big]. +\end{equation} +Note that the relaxation $L$ that we introduced in \eqref{eq:our-relaxation}, +follows naturally from the \emph{multi-linear} relaxation by swapping the +expectation and the $\log\det$ in \eqref{eq:multi-linear-logdet}: +\begin{displaymath} + L(\lambda) = \log\det\left(\mathbb{E}_{S\sim + P_\mathcal{N}^\lambda}\bigg[I_d + \sum_{i\in S} x_i\T{x_i} \bigg]\right). +\end{displaymath} + +The proof proceeds as follows: +\begin{itemize} +\item First, we prove that $F$ admits the following rounding property: let +$\lambda$ be a feasible element of $[0,1]^n$, it is possible to trade one +fractional component of $\lambda$ for another until one of them becomes +integral, obtaining a new element $\tilde{\lambda}$ which is both feasible and +for which $F(\tilde{\lambda})\geq F(\lambda)$. Here, by feasibility of a point +$\lambda$, we mean that it satisfies the budget constraint $\sum_{i=1}^n +\lambda_i c_i \leq B$. This rounding property is referred to in the literature +as \emph{cross-convexity} (see, \emph{e.g.}, \cite{dughmi}), or +$\varepsilon$-convexity by \citeN{pipage}. This is stated and proven in +Lemma~\ref{lemma:rounding} and allows us to bound $F$ in terms of $OPT$. +\item Next, we prove the central result of bounding $L$ appropriately in terms +of the multi-linear relaxation $F$ (Lemma \ref{lemma:relaxation-ratio}). +\item Finally, we conclude the proof of Lemma~\ref{lemma:relaxation} by +combining Lemma~\ref{lemma:rounding} and Lemma~\ref{lemma:relaxation-ratio}. +\end{itemize} + +\begin{comment} +Formally, if we define: +\begin{displaymath} + \tilde{F}_\lambda(\varepsilon) \defeq F\big(\lambda + \varepsilon(e_i + - e_j)\big) +\end{displaymath} +where $e_i$ and $e_j$ are two vectors of the standard basis of +$\reals^{n}$, then $\tilde{F}_\lambda$ is convex. Hence its maximum over the interval: +\begin{displaymath} + I_\lambda = \Big[\max(-\lambda_i,\lambda_j-1), \min(1-\lambda_i, \lambda_j)\Big] +\end{displaymath} +is attained at one of the boundaries of $I_\lambda$ for which one of the $i$-th +or the $j$-th component of $\lambda$ becomes integral. +\end{comment} + +\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 components is + fractional %, that is, lies in $(0,1)$ and: + and $F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})$. +\end{lemma} +\begin{proof} + We give a rounding procedure which, given a feasible $\lambda$ with at least + two fractional components, returns some feasible $\lambda'$ with one less fractional + component such that $F(\lambda) \leq F(\lambda')$. + + 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{equation}\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{equation} + 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{displaymath} + \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{displaymath} + 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: +For all $\lambda\in[0,1]^{n},$ + %\begin{displaymath} + $ \frac{1}{2} + \,L(\lambda)\leq + F(\lambda)\leq L(\lambda)$. + %\end{displaymath} +\end{lemma} +\begin{proof} + The bound $F_{\mathcal{N}}(\lambda)\leq L_{\mathcal{N}(\lambda)}$ follows by the concavity of the $\log\det$ function. + To show the lower bound, + we first prove that $\frac{1}{2}$ is a lower bound of the ratio $\partial_i + F(\lambda)/\partial_i L(\lambda)$, where + $\partial_i\, \cdot$ denotes the partial derivative with respect to the + $i$-th variable. + + Let us start by computing the derivatives of $F$ and + $L$ with respect to the $i$-th component. + Observe that: + \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{displaymath} + \partial_i F(\lambda) = + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})V(S) + - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in \mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S) + \end{displaymath} + Now, using that every $S$ such that $i\in S$ can be uniquely written as + $S'\cup\{i\}$, we can write: + \begin{displaymath} + \partial_i F(\lambda) = + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\big(V(S\cup\{i\}) + - V(S)\big) + \end{displaymath} + The marginal contribution of $i$ to + $S$ can be written as +\begin{align*} +V(S\cup \{i\}) - V(S)& = \frac{1}{2}\log\det(I_d + + \T{X_S}X_S + x_i\T{x_i}) + - \frac{1}{2}\log\det(I_d + \T{X_S}X_S)\\ + & = \frac{1}{2}\log\det(I_d + x_i\T{x_i}(I_d + +\T{X_S}X_S)^{-1}) + = \frac{1}{2}\log(1 + \T{x_i}A(S)^{-1}x_i) +\end{align*} +where $A(S) =I_d+ \T{X_S}X_S$. +% $ V(S\cup\{i\}) - V(S) = \frac{1}{2}\log\left(1 + \T{x_i} A(S)^{-1}x_i\right)$. +Using this, + \begin{displaymath} + \partial_i F(\lambda) = \frac{1}{2} + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S) + \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big) + \end{displaymath} + The computation of the derivative of $L$ uses standard matrix + calculus. Writing $\tilde{A}(\lambda) = I_d+\sum_{i\in + \mathcal{N}}\lambda_ix_i\T{x_i}$: + \begin{displaymath} + \det \tilde{A}(\lambda + h\cdot e_i) = \det\big(\tilde{A}(\lambda) + + hx_i\T{x_i}\big) + =\det \tilde{A}(\lambda)\big(1+ + h\T{x_i}\tilde{A}(\lambda)^{-1}x_i\big) + \end{displaymath} + Hence: + \begin{displaymath} + \log\det\tilde{A}(\lambda + h\cdot e_i)= \log\det\tilde{A}(\lambda) + + h\T{x_i}\tilde{A}(\lambda)^{-1}x_i + o(h) + \end{displaymath} + Finally: + \begin{displaymath} + \partial_i L(\lambda) + =\frac{1}{2} \T{x_i}\tilde{A}(\lambda)^{-1}x_i + \end{displaymath} + +For two symmetric matrices $A$ and $B$, we write $A\succ B$ ($A\succeq B$) if +$A-B$ is positive definite (positive semi-definite). This order allows us to +define the notion of a \emph{decreasing} as well as \emph{convex} matrix +function, similarly to their real counterparts. With this definition, matrix +inversion is decreasing and convex over symmetric positive definite matrices. +In particular, +\begin{gather*} + \forall S\subseteq\mathcal{N},\quad A(S)^{-1} \succeq A(S\cup\{i\})^{-1} +\end{gather*} + Observe that: + \begin{gather*} + \forall S\subseteq\mathcal{N}\setminus\{i\},\quad + P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq + P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\})\\ + \forall S\subseteq\mathcal{N},\quad P_{\mathcal{N}\setminus\{i\}}^\lambda(S) + \geq P_\mathcal{N}^\lambda(S) + \end{gather*} + Hence: + \begin{align*} + \partial_i F(\lambda) + % & = \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} P_\mathcal{N}^\lambda(S) \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\ + & \geq \frac{1}{4} + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S) + \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\ + &\hspace{-3.5em}+\frac{1}{4} + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\}) + \log\Big(1 + \T{x_i}A(S\cup\{i\})^{-1}x_i\Big)\\ + &\geq \frac{1}{4} + \sum_{S\subseteq\mathcal{N}} + P_\mathcal{N}^\lambda(S) + \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big) + \end{align*} + Using that $A(S)\succeq I_d$ we get that $\T{x_i}A(S)^{-1}x_i \leq + \norm{x_i}_2^2 \leq 1$. Moreover, $\log(1+x)\geq x$ for all $x\leq 1$. + Hence: + \begin{displaymath} + \partial_i F(\lambda) \geq + \frac{1}{4} + \T{x_i}\bigg(\sum_{S\subseteq\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{displaymath} + \partial_i F(\lambda) \geq + \frac{1}{4} + \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)\bigg)^{-1}x_i + = \frac{1}{4}\T{x_i}\tilde{A}(\lambda)^{-1}x_i + = \frac{1}{2} + \partial_i L(\lambda) + \end{displaymath} + +Having bound the ratio between the partial derivatives, we now bound the ratio $F(\lambda)/L(\lambda)$ from below. Consider the following cases. + First, if the minimum of the ratio + $F(\lambda)/L(\lambda)$ is attained at a point interior to the hypercube, then it is + a critical point, \emph{i.e.}, $\partial_i \big(F(\lambda)/L(\lambda)\big)=0$ for all $i\in \mathcal{N}$; hence, at such a critical point: + \begin{equation}\label{eq:lhopital} + \frac{F(\lambda)}{L(\lambda)} + = \frac{\partial_i F(\lambda)}{\partial_i + L(\lambda)} \geq \frac{1}{2} + \end{equation} + Second, if the minimum is attained as + $\lambda$ converges to zero in, \emph{e.g.}, the $l_2$ norm, by the Taylor approximation, one can write: + \begin{displaymath} + \frac{F(\lambda)}{L(\lambda)} + \sim_{\lambda\rightarrow 0} + \frac{\sum_{i\in \mathcal{N}}\lambda_i\partial_i F(0)} + {\sum_{i\in\mathcal{N}}\lambda_i\partial_i L(0)} + \geq \frac{1}{2}, + \end{displaymath} + \emph{i.e.}, the ratio $\frac{F(\lambda)}{L(\lambda)}$ is necessarily bounded from below by 1/2 for small enough $\lambda$. + Finally, if the minimum is attained on a face of the hypercube $[0,1]^n$ (a face is + defined as a subset of the hypercube where one of the variable is fixed to + 0 or 1), without loss of generality, we can assume that the minimum is + attained on the face where the $n$-th variable has been fixed + to 0 or 1. Then, either the minimum is attained at a point interior to the + face or on a boundary of the face. In the first sub-case, relation + \eqref{eq:lhopital} still characterizes the minimum for $i< n$. + In the second sub-case, by repeating the argument again by induction, we see + that all is left to do is to show that the bound holds for the vertices of + the cube (the faces of dimension 1). The vertices are exactly the binary + points, for which we know that both relaxations are equal to the value + function $V$. Hence, the ratio is equal to 1 on the vertices. +\end{proof} + +\begin{proof}[of Lemma~\ref{lemma:relaxation}] +Let us consider a feasible point $\lambda^*\in[0,1]^{n}$ such that $L(\lambda^*) + = OPT'$. 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(\lambda^*) \leq 2 + F(\bar{\lambda}) + \end{equation} + Let $\lambda_i$ denote the fractional component of $\bar{\lambda}$ and $S$ + denote the set whose indicator vector is $\bar{\lambda} - \lambda_i e_i$. + By definition of the multi-linear extension $F$: + \begin{displaymath} + F(\bar{\lambda}) = (1-\lambda_i)V(S) +\lambda_i V(S\cup\{i\}) + \end{displaymath} + By submodularity of $V$, $V(S\cup\{i\})\leq V(S) + V(\{i\})$, hence: + \begin{displaymath} + F(\bar{\lambda}) \leq V(S) + V(i) + \end{displaymath} + Note that since $\bar{\lambda}$ is feasible, $S$ is also feasible and + $V(S)\leq OPT$. Hence: + \begin{equation}\label{eq:e2} + F(\bar{\lambda}) \leq OPT + + \max_{i\in\mathcal{N}} V(i) + \end{equation} + Together, \eqref{eq:e1} and \eqref{eq:e2} imply the lemma. \hspace*{\stretch{1}}\qed +\end{proof} + +\subsection{Proof of Theorem \ref{thm:lowerbound}} + +\begin{proof} +Suppose, for contradiction, that such a mechanism exists. Consider two +experiments with dimension $d=2$, such that $x_1 = e_1=[1 ,0]$, $x_2=e_2=[0,1]$ +and $c_1=c_2=B/2+\epsilon$. Then, one of the two experiments, say, $x_1$, must +be in the set selected by the mechanism, otherwise the ratio is unbounded, +a contradiction. If $x_1$ lowers its value to $B/2-\epsilon$, by monotonicity +it remains in the solution; by threshold payment, it is paid at least +$B/2+\epsilon$. So $x_2$ is not included in the solution by budget feasibility +and individual rationality: hence, the selected set attains a value $\log2$, +while the optimal value is $2\log 2$. +\end{proof} + |