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authorPaul <Paul@Pauls-MacBook-Air.local>2014-10-19 18:50:58 -0400
committerPaul <Paul@Pauls-MacBook-Air.local>2014-10-19 18:50:58 -0400
commitac476dfbfae989f9ed11edcc473ff0bd4d2c33de (patch)
tree5d7e948311d046b34bf872d194fb9aadad3dd404
parent6a316d49e546569de3730552e4d50d8b82f80459 (diff)
downloadecon2099-ac476dfbfae989f9ed11edcc473ff0bd4d2c33de.tar.gz
[ps2] Wrote up Exercise 4.19
-rw-r--r--ps2/main.tex45
1 files changed, 38 insertions, 7 deletions
diff --git a/ps2/main.tex b/ps2/main.tex
index 17f7693..9f547d6 100644
--- a/ps2/main.tex
+++ b/ps2/main.tex
@@ -34,6 +34,7 @@
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
+\newtheorem*{claim}{Claim}
\author{Thibaut Horel \and Paul Tylkin}
\title{Economics 2099 Problem Set 2 -- Solutions}
@@ -46,6 +47,36 @@
\section*{Exercise 4.14*}
\section*{Exercise 4.19}
+\begin{claim} In regular, matroid environments, the revenue of the surplus maximization mechanism with monopoly reserves is a 2-approximation to the optimal mechanism revenue.
+
+\begin{proof} Let REF denote the optimal mechanism and its expected revenue and let APX denote the surplus maximization mechanism with monopoly reserves and its expected revenue. Similarly to the single-item case, we know that REF $\geq$ APX and want to show that REF $\leq$ 2APX. Let $I$ denote the set of agents allocated by the optimal mechanism and let $J$ denote the set of agents allocated by the surplus maximization mechanism with monopoly reserves. Then, $$REF = \E\left[\sum_{i \in I}\phi_i(v_i)\right]$$ and $$APX = \E\left[\sum_{j \in J}\phi_j(v_j)\right].$$ By linearity of expectation, we can write this as $$REF = \sum_{i \in I}\E\left[\phi_i(v_i)\right]$$ and $$APX = \sum_{j \in J}\E\left[\phi_j(v_j)\right].$$ Now, we can write the expected revenue of the optimal mechanism by conditioning on whether $i = j$ for $i \in I$, $j \in J$. (i.e. whether agent $i$ is allocated the same item by the optimal mechanism as agent $j$ is by the surplus maximization mechanism; an alternate way of phrasing this is that the map $I \rightarrow J$ is an injection). So we can write $$REF = \underbrace{\sum_{\substack{i \in I \\ j \in J}}\E[\phi_i(v_i) | i = j] \Pr[i=j]}_{REF_=} + \underbrace{\sum_{\substack{i \in I \\ j \in J}}\E[\phi_i(v_i) | i \neq j] \Pr[i\neq j]}_{REF_{\neq}}.$$ We will prove the desired result by showing both $REF_=$ and $REF_{\neq}$ are bounded from above by APX.
+
+\begin{align}
+REF_= &= \sum_{\substack{i \in I \\ j \in J}}\E[\phi_i(v_i) | i = j] \Pr[i=j] \\
+&= \sum_{\substack{i \in I \\ j \in J}}\E[\phi_j(v_j) | i = j] \Pr[i=j] \\
+&\leq \sum_{\substack{i \in I \\ j \in J}}\E[\phi_j(v_j) | i = j] \Pr[i=j] + \sum_{\substack{i \in I \\ j \in J}}\E[\phi_j(v_j) | i \neq j] \Pr[i\neq j] \label{42}\\
+&= \sum_{j \in J}\E\left[\phi_j(v_j)\right] \\
+&= APX.
+\end{align}
+
+Note that Equation $\ref{42}$ follows from Lemma 4.2 (i.e. from the monotonicity of virtual value functions).
+
+Next, we consider $REF_{\neq}$.
+
+\begin{align}
+REF_{\neq} &= \sum_{\substack{i \in I \\ j \in J}}\E[\phi_i(v_i) | i \neq j] \Pr[i\neq j] \\
+&\leq \sum_{\substack{i \in I \\ j \in J}}\E[v_i | i \neq j] \Pr[i\neq j] \\
+&\leq \sum_{\substack{i \in I \\ j \in J}}\E[\hat{v}_i | i \neq j] \Pr[i\neq j] \\
+&= \sum_{\substack{i \in I \\ j \in J}}\E[v_j | i \neq j] \Pr[i\neq j] \label{428} \\
+&\leq \sum_{\substack{i \in I \\ j \in J}}\E[v_j | i = j] \Pr[i=j] + \sum_{\substack{i \in I \\ j \in J}}\E[v_j | i \neq j] \Pr[i\neq j] \label{429} \\
+&= APX.
+\end{align}
+
+Note that Equation $\ref{428}$ follows from Lemma 4.28 and Equation $\ref{429}$ follows from Theorem 4.29.
+
+\end{proof}
+
+\end{claim}
\section*{Additional Problem}
@@ -83,7 +114,7 @@ distribution over subsets of $A$:
\end{cases}
\end{displaymath}
and similarly for $\Pr^A_{\mathbf{F^I}}$. That is, the only subsets of $A$ which
-have a non-zero probability are the ones which ``correspond'' to a $n$-tuple in
+have a non-zero probability are the ones which ``correspond'' to an $n$-tuple in
$A_1\times\ldots\times A_n$ (they are in the image of the canonical injection
given above).
@@ -115,13 +146,13 @@ simply $a$. Hence we can apply Lemma 4.13 to $g$ for subsets of $A$ and obtain:
\paragraph{General distributions} We first look at a continuous distribution
$\mathbf{F}$ with bounded support. On the support of $\mathbf{F}$, it is
-possible to approximate its CDF arbitrarily well in $L^\infty$ norm (hence in
-$L^2$ norm since we are on a bounded set) by simple functions (convex
+possible to approximate its CDF arbitrarily well in the $L^\infty$ norm (hence in
+the $L^2$ norm since we are on a bounded set) by simple functions (convex
combination of step functions). But a step function is simply the CDF of
-a dirac distribution. Hence, we can approximate $\mathbf{F}$ arbitrarily well
-in $L^2$ norm with finite support distributions. But convergence in $L^2$ norm
-implies weak convergence by Cauchy-Schwarz inequality. That is, we have
-a sequence $F_n$ of finite support distribution such that:
+a Dirac distribution. Hence, we can approximate $\mathbf{F}$ arbitrarily well
+in the $L^2$ norm with finite support distributions. But convergence in the $L^2$ norm
+implies weak convergence by the Cauchy-Schwarz inequality. That is, we have
+a sequence $F_n$ of finite support distributions such that:
\begin{displaymath}
\E_{F_n}[\max_i v_i] \xrightarrow[n\rightarrow \infty]{} \E_F[\max_i v_i]
\end{displaymath}