From 7357385b5fcd9a564bcb545cfd0def6893e62052 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Wed, 24 Sep 2014 01:18:01 -0400
Subject: Remaining typos in P1

---
 ps1/main.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

(limited to 'ps1/main.tex')

diff --git a/ps1/main.tex b/ps1/main.tex
index 0605359..84a6e7a 100644
--- a/ps1/main.tex
+++ b/ps1/main.tex
@@ -73,12 +73,12 @@ 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 $$\E[v_2|v_1\geq v_2] =  \begin{cases} \frac{1}{2}\big(v_1+\frac{1}{2}\big) &\text{ if } v_1 > \frac{1}{2} \\ 0 &\text{ if } v_1  \leq \frac{1}{2} \end{cases}$$
-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, 
+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 $U\big[\frac{1}{2},v_1\big]$. 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 $$\E[v_1|v_2\geq v_1] =  \int_0^{v_2} x\,dx = \frac{v_2}{2}$$
+Computing $$\E[v_1|v_2\geq v_1]  = \frac{v_2}{2}$$
 since conditioned on $v_2\geq v_1$, $v_1$ is $U\big[0,v_2\big]$.
 
 So $$s_2(v_2) = \frac{v_2}{2}.$$
-- 
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