From 95f22fe1d61dcf93ab2c88b4e54e99569579ef21 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Tue, 23 Sep 2014 10:39:25 -0400
Subject: End of Problem 2

---
 ps1/main.tex | 30 ++++++++++++++++++------------
 1 file changed, 18 insertions(+), 12 deletions(-)

(limited to 'ps1/main.tex')

diff --git a/ps1/main.tex b/ps1/main.tex
index e45246d..f888998 100644
--- a/ps1/main.tex
+++ b/ps1/main.tex
@@ -106,11 +106,10 @@ conditioning on whether agent $1$ wins, we get:
 \begin{displaymath}
     x_1(v_1) = w_1 \Pr[s(v_1)\geq s(v_2)] + w_2\Pr[s(v_1)\leq s(v_2)]
 \end{displaymath}
-But note that if $s$ exists, it has to be non-decreasing (otherwise, this would
-contradict the fact that the allocation rule is non-decreasing by Theorem
-2.10). Hence, $\Pr[s(v_1)\geq s(v_2)] = \Pr[v_1 \geq v_2]$ and similarly,
+By assumption, $s$ is strictly increasing. Hence, $\Pr[s(v_1)\geq s(v_2)]
+= \Pr[v_1 \geq v_2]$ and similarly,
 $\Pr[s(v_1) \leq s(v_2)] = \Pr[v_1\leq v_2]$. Now, denoting by $F$ the
-distribution of agent's values, we can rewrite:
+distribution of the agent's values, we can rewrite:
 \begin{displaymath}
     x_1(v_1) = w_1F(v_1) + w_2\big(1-F(v_1)\big) = w_2 + (w_1-w_2)F(v_1)
 \end{displaymath}
@@ -137,30 +136,37 @@ where $f(z) = F'(z)$ denotes the density function of $F$. This, in addition to
 }
 \end{displaymath}
 
-Let us now verify that $s$ is indeed non-decreasing. We compute:
+Let us now verify that $s$ is indeed increasing. We compute:
 \begin{displaymath}
     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. Rearranging the terms, it is
+We only care about the sign of the numerator, because the denominator is always
+positive. Rearranging the terms, it 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
-paranthesis is equal to $\int_0^{v_i}F(z)dz$ using the same integration by
+rightmost parentheses is equal to $\int_0^{v_i}F(z)dz$ using the same integration by
 parts as above.
 
-As noted above, $s$ being non-decreasing implies that the allocation rule of
-the position auction with strategy $s$ is non-decreasing. By design, $s$
+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 $s$
 is a Bayes-Nash equilibrium yet because it is not onto. However, we can show
-that bids which are not in the image of $s$ are dominated. Since $s$ is
+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}
 \end{displaymath}
-
-By Theorem 2.10, there are no asymmetric strategy profiles (there is a unique symmetric Bayes-Nash equilibrium).
+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 $u = w_1(v-s(1))$ 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 $u$.
+
+We have now established the uniqueness of a symmetric Bayes-Nash equilibrium.
+We can now conclude by applying Theorem 2.10 which states that there are no
+asymmetric strategy profiles.
 \end{document}
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