From 52d821568a337e4bfa226fe127e2e592c42b3457 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Wed, 24 Sep 2014 01:14:23 -0400
Subject: Just realized there was a bug in Problem 1

---
 ps1/main.tex | 14 +++++++-------
 1 file changed, 7 insertions(+), 7 deletions(-)

(limited to 'ps1/main.tex')

diff --git a/ps1/main.tex b/ps1/main.tex
index 5fd6d6a..0605359 100644
--- a/ps1/main.tex
+++ b/ps1/main.tex
@@ -72,18 +72,18 @@ For agent 1,$$\Pr[v_1 \geq v_2] = \begin{cases} \int_{\frac{1}{2}}^{v_1} 2\,dx =
 
 For agent 1, $P_1^{SP}(v_1)$ equals $\E[v_2|v_1\geq v_2]\cdot \Pr[v_1 \geq v_2]$. By Revenue Equivalence (Corollary 2.5), for $i \in \{1,2\}$, $P_i^{SP}(v_i) = P_i^{FP}(v_i)$. 
 
-We directly compute $$\E[v_2|v_1\geq v_2] =  \begin{cases} \int_{\frac{1}{2}}^{v_1} 2x\,dx = v_1^2 - \frac{1}{4} &\text{ if } v_1 > \frac{1}{2} \\ 0 &\text{ if } v_1  \leq \frac{1}{2} \end{cases}$$
-
-Thus, 
-$$s_1(v_1) = v_1^2-\frac{1}{4} \text{ if } v_1 > \frac{1}{2}.$$
+We directly compute $$\E[v_2|v_1\geq v_2] =  \begin{cases} \frac{1}{2}\big(v_1+\frac{1}{2}\big) &\text{ if } v_1 > \frac{1}{2} \\ 0 &\text{ if } v_1  \leq \frac{1}{2} \end{cases}$$
+Indeed, in expectation a uniform random variable evenly divides the interval it is over, and conditioned on $v_1\geq v_2$, $v_2$ is $U\big[\frac{1}{2},1\big]$. Thus, 
+$$s_1(v_1) =  \begin{cases} \frac{1}{2}\big(v_1+\frac{1}{2}\big) &\text{ if } v_1 > \frac{1}{2} \\ 0 &\text{ if } v_1  \leq \frac{1}{2} \end{cases}.$$
 
 For agent 2,  $$\Pr[v_2 \geq v_1] =  \int_{0}^{v_2}\,dx = v_2$$
 
-Computing $$\E[v_1|v_2\geq v_1] =  \int_0^{v_2} x\,dx = \frac{v_2^2}{2}$$
+Computing $$\E[v_1|v_2\geq v_1] =  \int_0^{v_2} x\,dx = \frac{v_2}{2}$$
+since conditioned on $v_2\geq v_1$, $v_1$ is $U\big[0,v_2\big]$.
 
-So $$s_2(v_2) = \frac{v_2^2}{2}.$$
+So $$s_2(v_2) = \frac{v_2}{2}.$$
 
-We will show a counterexample to the requirement that the item it always allocated to the agent with the highest value. Specifically, we will exhibit a scenario where $v_1 < v_2$ but $s_1(v_1) > s_2(v_2)$ and so agent 1 will be allocated the item despite having a lower valuation. Let $v_1 = \frac{7}{8}$ and let $v_2 = 1$. Then, $$s_1(v_1) = \frac{49}{64} - \frac{16}{64} = \frac{33}{64}$$ but $$s_2(v_2) = \frac{1}{2.}$$ Therefore, no Bayes-Nash Equilibrium with the desired properties can exist.
+We will show a counterexample to the requirement that the item it always allocated to the agent with the highest value. Specifically, we will exhibit a scenario where $v_1 < v_2$ but $s_1(v_1) > s_2(v_2)$ and so agent 1 will be allocated the item despite having a lower valuation. Let $v_1 = \frac{1}{2}$ and let $v_2 = \frac{3}{4}$. Then, $$s_1(v_1) = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}\right) = \frac{1}{2}$$ but $$s_2(v_2) = \frac{3}{8}<\frac{1}{2}.$$ Therefore, no Bayes-Nash Equilibrium with the desired properties can exist.
 
 \end{enumerate}
 
-- 
cgit v1.2.3-70-g09d2

