From 4bae70d0319bc980375f8c1974785cbf12b370de Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Tue, 23 Sep 2014 20:10:57 -0400
Subject: Some typo fixes

---
 ps1/main.tex | 51 +++++++++++++++++++++++++++++----------------------
 1 file changed, 29 insertions(+), 22 deletions(-)

(limited to 'ps1/main.tex')

diff --git a/ps1/main.tex b/ps1/main.tex
index 55c4971..a503471 100644
--- a/ps1/main.tex
+++ b/ps1/main.tex
@@ -26,6 +26,7 @@
 \DeclareMathOperator*{\Pr}{\mathbb{P}}
 
 \newcommand{\inprod}[1]{\left\langle #1 \right\rangle}
+\newcommand{\R}{\mathbb{R}}
 \newcommand{\eqdef}{\mathbin{\stackrel{\rm def}{=}}}
 \newcommand{\llbracket}{[\![}
 
@@ -39,7 +40,7 @@
 
 \maketitle
 
-\section{Exercise 2.5}
+\section*{Exercise 2.5}
 \begin{enumerate}[(a)]
 \item We are analyzing the scenario when agent 1 has values in $U[0,1]$ and agent $U[0,\frac{1}{2}]$. Informally, we are considering two different scenarios: one in which agent 1's value exceeds $\frac{1}{2}$, and one in which his value is less than $\frac{1}{2}$. In the first scenario, agent 1 will always win.
 
@@ -86,7 +87,7 @@ We will show a counterexample to the requirement that the item it always allocat
 
 \end{enumerate}
 
-\section{Exercise 2.10}
+\section*{Exercise 2.10}
 
 We first try to find a symmetric Bayes-Nash equilibrium. Let us assume that
 such a symmetric equilibrium exists and let us denote by $s$ the common
@@ -124,11 +125,18 @@ Rearranging the right hand side, we get:
     - w_2)F(v_i)}
 \end{equation}
 Finally, using integration by parts, we see that:
-\begin{displaymath}
+\begin{equation}\label{eq:parts}
     \int_0^{v_i} F(z)dz = v_iF(v_i) - \int_0^{v_i}zf(z)d(z)
-\end{displaymath}
+\end{equation}
 where $f(z) = F'(z)$ denotes the density function of $F$. This, in addition to
-\cref{eq:strag2} leads to the following expression for our candidate strategy:
+\cref{eq:strag2} leads to the following expression for our candidate
+strategy\footnote{The quantity $\int_{0}^{v_i}zf(z)dz$ can be naturally
+    interpreted as the expected value of the other agent conditioned on the
+    fact that is has value less than $v_i$, which is the payment in a second
+    price auction. In fact, an alternative way to derive our candidate strategy
+    would be to use revenue equivalence and equate the payments of our
+    first-price mechanism for the position auction to the payments of the VCG
+mechanism.}:
 \begin{displaymath}
     \boxed{
     s(v_i) = \frac{(w_1 - w_2)\big(\int_0^{v_i} zf(z)dz\big)}{w_2 + (w_1
@@ -141,36 +149,36 @@ Let us now verify that $s$ is indeed increasing. We compute:
     s'(v_i) = \frac{(w_1-w_2)v_if(v_i)\big(w_2 + (w_1-w_2)F(v_i)\big)
     - (w_1-w_2)\big(\int_0^{v_i}zf(z)dz\big)(w_1-w_2)f(v_i)}{\big(w_2+(w_1-w_2)F(v_i)\big)^2}
 \end{displaymath}
-We only care about the sign of the numerator, because the denominator is always
-positive. Rearranging the terms, it is
-equal to:
+We only care about the sign of the numerator because the denominator is always
+positive. Rearranging the terms, the numerator is equal to:
 \begin{displaymath}
     (w_1-w_2)w_2v_if(v_i) + (w_1-w_2)^2f(v_i)\left(v_iF(v_i)
     - \int_0^{v_i}zf(z)dz\right)
 \end{displaymath}
 It is easy to see that this quantity is non-negative since the term inside the
-rightmost parentheses is equal to $\int_0^{v_i}F(z)dz$ using the same integration by
-parts as above.
+rightmost parentheses is equal to $\int_0^{v_i}F(z)dz$ using the integration by
+parts of \cref{eq:parts}.
 
 As noted above, $s$ being increasing implies that the allocation rule of
 the position auction with strategy $s$ is also increasing. By design, $s$
-satisfies the payment identity of Theorem 2.10. But we cannot conclude that $s$
+satisfies the payment identity of Theorem 2.10. We cannot conclude that $s$
 is a Bayes-Nash equilibrium yet because it is not onto. However, we can show
 that bids which are not attained by $s$ are dominated. Since $s$ is
 non-decreasing, its maximum value is:
 \begin{displaymath}
-    s(1) = \frac{(w_1-w_2)\int_0^1 zf(z)dz}{w_1}
+s^* = \frac{w_1-w_2}{w_1}\int_{\R^+} zf(z)dz}
 \end{displaymath}
-Let us show that bids above $s(1)$ are dominated by $s(1)$. The utility of an
-agent with value $v$ when bidding $s(1)$ is $u = w_1\big(v-s(1)\big)$ since he will be
-allocated to the first position in this case. Then consider a bid $b>s(1)$; the
-utility in this case will be $u' = w_1(v-b)$ which is less than $u$.
+Let us show that bids above $s^*$ are dominated by $s^*$. The utility of an
+agent with value $v$ when bidding $s^*$ is $u = w_1\big(v-s^*\big)$ since he
+will be allocated to the first position in this case. Then consider a bid
+$b>s^*$; the utility in this case will be $u' = w_1(v-b)$ which is less than
+$u$.
 
 We have now established the uniqueness of a symmetric Bayes-Nash equilibrium.
-Furthermore we can rule out the possibility of assymetric strategy profiles by
+Furthermore we can rule out the possibility of asymmetric strategy profiles by
 applying verbatim the proof of Theorem 2.10.
 
-\section{Exercise 3.1}
+\section*{Exercise 3.1}
 
 We recall that the virtual value function for agent $i$ is defined as $$\phi_i(v_i) = v_i - \frac{1 - F_i(v_i)}{f_i(v_i)}$$ for a cumulative density function $F_i(v_i)$ and density function $f_i(v_i)$, with $F_i'(v_i) \stackrel{\text{def}}{=} f_i(v_i)$. The virtual function satisfies the following relationship:
 \begin{equation}\label{eq:virt}
@@ -193,11 +201,10 @@ where we defined $h_i(v_i)$ to be the inverse of the hazard rate function:
 \end{displaymath}
 
 Note that by assumption, $h_i$ is non-increasing (since the harzard rate function
-is non-decreasing). Hence, we need to consider $\bar{h_i}$, the function obtained
+is non-decreasing). Hence, we need to consider $\bar{h}_i$, the function obtained
 by ironing $h_i$. Since $h_i$ is non-increasing, we have to apply the ironing
 procedure to the whole interval of values, leading to a constant ironed
-function $\bar{h_i} = c_i$. We will call $c_i$ the \emph{ironed constant} of
-agent $i$.
+function $\bar{h}_i = c_i$. 
 
 By construction, the mechanism maximizing residual surplus is the VSM mechanism
 where we use the $(c_1,\ldots,c_n)$ as virtual functions. Note that it is easy
@@ -207,7 +214,7 @@ ironed constants
 
 
 
-\section{Exercise 3.4}
+\section*{Exercise 3.4}
 
 \begin{enumerate}[(a)]
 
-- 
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