From 5f60239117c0b86488f87beba2057782fd1d223b Mon Sep 17 00:00:00 2001
From: Stratis Ioannidis <stratis@stratis-Latitude-E6320.(none)>
Date: Mon, 5 Nov 2012 13:34:29 -0800
Subject: appendix

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+\begin{lemma}\label{lemma:monotone}
+The mechanism is monotone.
+\end{lemma}
+\begin{proof}
+    Consider an agent $i$ with cost $c_i$ that is
+    selected by the mechanism, and suppose that she reports
+    a cost $c_i'\leq c_i$ while all  other costs stay the same.
+    Suppose that when $i$ reports $c_i$, $L(\xi) \geq C V(i^*)$; then, as $s_i(c_i,c_{-i})=1$, $i\in S_G$.
+     By reporting a cost $c_i'\leq c_i$, $i$ may be selected at an earlier iteration of the greedy algorithm.
+    %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.
+ Denote by $S_i$
+    (resp. $S_i'$) the set to which $i$ is added when reporting cost $c_i$
+    (resp. $c_i'$). We have $S_i'\subseteq S_i$; in addition, $S_i'\subseteq S_G'$, the set selected by the greedy algorithm under $(c_i',c_{-i})$; if not, then greedy selection would terminate prior to selecting $i$ also when she reports $c_i$, a contradiction. Moreover, we have
+    \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*}
+    by the monotonicity and submodularity  of $V$. Hence  $i\in S_G'$. As $L(\xi)$,  is the optimal value of \eqref{relax} under relaxation $L$ when $i^*$ is excluded from $\mathcal{N}$, reducing the costs can only increase this value, so under $c'_i\leq c_i$ the greedy set is still allocated and $s_i(c_i',c_{-i}) =1$.
+    Suppose now that when $i$ reports $c_i$, $L(\xi) < C V(i^*)$. Then $s_i(c_i,c_{-i})=1$ iff $i = i^*$. 
+    Reporting $c_{i^*}'\leq c_{i^*}$ does not change $V(i^*)$ nor
+    $L(\xi) \leq C V(i^*)$; thus $s_{i^*}(c_{i^*}',c_{-i^*})=1$. 
+\end{proof}
+\begin{lemma}\label{lemma:budget-feasibility}
+The mechanism is budget feasible.
+\end{lemma}
+\begin{proof}
+Suppose that $L(\xi) < C V(i^*)$. Then the mechanism selects $i^*$. Since the bid of $i^*$ does not affect the above condition, the threshold payment of $i^*$ is $B$ and the mechanism is budget feasible.
+Suppose  that $L(\xi) \geq C V(i^*)$. 
+Denote by $S_G$ the set selected by the greedy algorithm, and for $i\in S_G$,  denote by
+$S_i$ the subset of the solution set that was selected by the greedy algorithm just prior to the addition of $i$---both sets determined for the present cost vector $c$. 
+%Chen \emph{et al.}~\cite{chen} show that, 
+Then for any submodular function $V$, and for all $i\in S_G$:
+%the reported cost of an agent selected by the greedy heuristic, and holds for
+%any submodular function $V$:
+\begin{equation}\label{eq:budget}
+   \text{if}~c_i'\geq \frac{V(S_i\cup\{i\}) - V(S)}{V(S_G)} B~\text{then}~s_i(c_i',c_{-i})=0
+\end{equation}
+In other words, if $i$ increases her cost to a value higher than $\frac{V(S_i\cup\{i\}) - V(S)}{V(S_G)}$, she will cease to be in the selected set $S_G$. As a result,
+\eqref{eq:budget}
+implies that the threshold payment of user $i$ is bounded by the above quantity.
+%\begin{displaymath}
+%\frac{V(S_i\cup\{i\}) - V(S_i)}{V(S_G)} =  B
+%\end{displaymath}
+Hence, the total payment is bounded by the telescopic sum:
+\begin{displaymath}
+    \sum_{i\in S_G} \frac{V(S_i\cup\{i\}) - V(S_i)}{V(S_G)} B = \frac{V(S_G)-V(\emptyset)}{V(S_G)} B=B\qed
+\end{displaymath}
+\end{proof}
+
+
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