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\noindent
20 October 2009  \\
 Eric Rasmusen, Erasmuse@indiana.edu   

 
 
 \newpage

 \noindent
 {\bf Private-Value and  Common-Value Auctions }

 


In a {\bf private-value auction}, a bidder can learn nothing about his value
from
knowing the values of the other bidders.

 Knowing all the other values  in
advance would not change his estimate. 
It might change      his bidding
strategy.

SPECIAL CASE 1:  {\bf Independent private-value
auction}, in which knowing his own value tells him nothing about other bidders'
values.


SPECIAL CASE 2:   {\bf Affiliated private-value auction}
  he can use  knowledge of his  own value to deduce something about other
players' values.  (they are correlated in a certain sense)

  {\bf Pure  common-value auction}, the bidders have identical values, but
each
bidder forms his own  estimate   on the basis of his own  private information.


   
\newpage

 

\noindent
  {\bf    Auction  Rules    }


 

\noindent
 1 Ascending (English, open-cry, open-exit); \\

2 First-Price (first-price sealed-bid); \\

 3 Second-Price  (second-price sealed-bid, Vickrey); \\

 4 Descending (Dutch) \\

5 All-Pay\\


\newpage

 \noindent
{\it Ascending (English, open-cry, open-exit)}

\noindent
 {\bf Rules}\\
  Each bidder is free to revise his bid upwards.  When no bidder wishes to
revise
his bid further, the highest bidder wins the object and pays his bid.

 
\noindent
 {\bf Strategies}\\
  A bidder's strategy is his series of bids as a function of (1) his value, (2)
his prior estimate of other bidders' values, and (3) the past bids of all the
bidders. His bid can therefore be updated as his information set changes.

\noindent
  {\bf Payoffs}\\
 The winner's payoff is his value minus his highest bid ($t=p$ for him and $t=0$
for everyone else).  The losers' payoffs are zero.


\newpage
   
\noindent
Some variations: 



 

 

\noindent
 (1)  The bidders  offer new  prices  using  pre-specified {\bf increments} such as
dollars or thousands of dollars.  


  (2)   The {\bf open-exit} auction, in which the price rises continuously and
bidders show their willingness to pay the price by not dropping out, where a
bidder's dropping out is publicly announced to the other bidders. 


 (3)   The {\bf silent-exit} auction (my neologism), in which the price rises
continuously and bidders show their willingness to pay the price by not dropping
out, but a bidder's dropping out is not known to the other bidders. 


 (4) The {\bf Ebay auction}, in which a bidder  submits his  ``bid ceiling.''
There is a prespecified
ending time.  


  (5) The {\bf Amazon auction}, in which a bidder  submits his   bid ceiling.   Prespecified ending time OR ten minutes after the last
increase in the current winning bid, whichever is later.  \\


\newpage


 A mechanism $[G(\tilde{v}_i, \tilde{v}_{-i}) v_i - t
(\tilde{v}_i, \tilde{v}_{-i} )]$  takes payments $t$ and gives an  object
with probability $G$ to    player $i$     if he announces
that his value is $\tilde{v}_i$ and the other players  announce $\tilde{v}_{-i}$. 
   
An 
ascending auction   can be seen as a mechanism in which each bidder announces his value, the object is awarded to whoever announces the highest value, and he pays the second-highest announced value (the second-highest bid).


  \newpage

\begin{center}
{\bf The Continuous-Value Auction }
 \end{center}
{\bf Players:}  One  seller and two  bidders.

 
0. Nature chooses   Bidder $i$'s value for the object,  $v_i$, using the
strictly positive, atomless  density $f(v)$ on the interval $[\underline{v},
\overline{v}]$.  \\
1. The seller chooses a mechanism $[G(\tilde{v}_i, \tilde{v}_{-i}) v_i - t
(\tilde{v}_i, \tilde{v}_{-i} )]$ that takes payments $t$ and gives the object
with probability $G$ to    player $i$   (including the seller)  if he announces
that his value is $\tilde{v}_i$ and the other players  announce $\tilde{v}_{-i}
$.   He also chooses the procedure in which bidders  select $\tilde{v}_i$
(sequentially, simultaneously, etc.). \\
  2. Each bidder  simultaneously chooses   to participate in the auction or  to
stay  out. \\
 3. The bidders  and the  seller choose $\tilde{v}$ according to the mechanism
procedure. \\
  4. The  object is allocated and transfers are paid according to the mechanism,
if it was accepted by all bidders.\\

\newpage

\noindent
 {\bf Payoffs:}\\
 The seller's payoff is
 \begin{equation} \label{e3}
 \pi_s=     \sum_{i=1}^{n} t(\tilde{v}_i, \tilde{v}_{-i}  )
\end{equation}


    Bidder $i$'s payoff is zero if he does not participate, and otherwise is
 \begin{equation} \label{e4}
   \pi_i(v_i)  =   G(\tilde{v}_i, \tilde{v}_{-i}) v_i - t(\tilde{v}_i, \tilde{v}
_{-i} )
 \end{equation}


\newpage



   In the Continuous-Value Auction, denote the highest announced value by
$\tilde{v}_{(1)}$,  the second-highest by $\tilde{v}_{(2)}$, and so forth. 

 The highest bidder gets
the object with probability

$G(\tilde{v}_{(1)}, \tilde{v}_{-1}) = 1$ 

at price

 $ t
(\tilde{v}_{(1)}, \tilde{v}_{-1} ) = \tilde{v}_{(2)}$,

 and    for 
$i \neq
1$,  $G(\tilde{v}_{(i)}, \tilde{v}_{-i}) = 0$ 


 $ t
(\tilde{v}_{(1)}, \tilde{v}_{-1} )   =0.$ 

 \newpage


 The PROFIT-MAXIMIZING (optimal?) mechanism has a 

 {\bf reserve price} $p^*$ below which the object would remain
unsold.
Thus

 $G(\tilde{v}_{(1)},
\tilde{v}_{-1}) = 1$ 

 $ t
(\tilde{v}_{(1)}, \tilde{v}_{-1} ) = Max \{ \tilde{v}_{(2)}, p^*\}$ if
$\tilde{v}_{(1)} \geq  p^*$  but $G(\tilde{v}_{(1)}, \tilde{v}_{-1}) = 0$ if
$\tilde{v}_{(1)} < p^*$.


 Optimal mechanisms are
not always efficient.  The object will go unsold if
$\tilde{v}_{(1)}< p^*$.


 \newpage





 
 \noindent
{\bf First-Price  (first-price sealed-bid) }


\noindent
{\bf Rules}\\
Each bidder submits one bid, in ignorance of the other bids.  The highest bidder
pays his bid and wins the object.

\noindent
 {\bf Strategies}\\
 A bidder's strategy is his bid as a function of his value.

\noindent
  {\bf Payoffs}\\
 The winner's payoff is his value minus his bid.  The losers' payoffs are zero.

\bigskip


   \newpage


\noindent
{\bf    The First-Price Auction with a Continuous Distribution of Values }
 

   Suppose Nature independently  assigns  values to  $n$ risk-neutral bidders    using the  continuous  density $f(v)>0 $ (with cumulative probability $F(v)$)  on the  support $[0,    \bar{v}]$.

   A bidder's payoff as a function of his value $v$ and his bid function $p(v)$
is,    letting  $G(p(v))$   denote the probability of
winning with a particular $p(v)$:
\begin{equation} \label{e10}
  \pi(v, p(v)) = G(p(v)) [v - p(v)].
  \end{equation}

  
\newpage


Now go to the board. 
 
\begin{normalsize}
   
 Now let us try to find an  equilibrium bid function. From equation (\ref{e10}), it is 
\begin{equation} \label{e10c}
  p(v) = v - \frac{ \pi(v, p(v))}{   G(p(v))}.
  \end{equation}
 That is not very useful in itself,  since it has $p(v)$ on both sides. We need to find ways to rewrite 
  $\pi$  and  $G$ in terms of just $v$. 

First,  tackle 
  $G(p(v))$.   Monotonicity of the bid function  (from Lemma 1) implies that
  the bidder with the greatest $v$ will bid highest  and win. Thus,   the probability $G(p(v))$  that a  bidder with price $p_i$ will
win is the probability that $v_i$ is the highest value of all $n$ bidders.  The
probability  that a bidder's value $v$ is the highest  is  $F(v)^{n-1}$, the probability
that each of the other $(n-1)$ bidders has a value  less than $v$. Thus,
\begin{equation} \label{e13b}
 G(p(v)) = F(v)^{n-1}.
\end{equation}

   Next   think about  $\pi(v, p(v))$. 
 The Envelope Theorem says that if $\pi(v, p(v))$ is the value of a function
maximized by choice of $p(v)$  then its total derivative with respect to $v$
equals its partial derivative,  because  $\frac{\partial \pi  }
{\partial p}=0$: 
 \begin{equation} \label{e12}
\begin{array}{ll}
  \frac{d \pi(v, p(v))}{d v} & =  \frac{\partial \pi(v, p(v))}{\partial p}
    \frac{\partial p  }{\partial v}  + \frac{\partial \pi(v,
p(v))}{\partial v} 
  =  \frac{\partial \pi(v, p(v))}{\partial v}.\\
\end{array}
  \end{equation}
  We can apply  the Envelope Theorem to equation (\ref{e10}) to see how $\pi$ changes with $v$  assuming $p(v)$ is chosen optimally, which is appropriate because we are  characterizing not just any bid function, but  the optimal bid function. Thus,  
\begin{equation} \label{e13}
 \frac{d \pi(v, p(v))}{dv}  =  G(p(v)).
  \end{equation}
   Substituting from equation (\ref{e13b}) gives us $\pi$'s derivative, if not $\pi$, as a function of $v$:  
\begin{equation} \label{e13a}
 \frac{d \pi(v, p(v))}{dv}  =  F(v)^{n-1}.
  \end{equation}
  To get $\pi(v, p(v))$ from its derivative, (\ref{e13a}),    integrate   over  all possible values  from
zero to $v$   and include the a base value of $\pi(0)$ as the  constant of integration:
\begin{equation} \label{e14}
\begin{array}{ll}
 \pi(v, p(v))&{\displaystyle  = \pi(0) + \int_0^v   F(x)^{n-1} dx}=  \int_0^v   F(x)^{n-1} dx. \\
\end{array}
  \end{equation}
The last step is true because  a bidder with $v=0$ will never bid a
positive amount and so will have a payoff of $\pi(0, p(0))=0$.


We can now return to   the bid function in 
  equation (\ref{e10c})  and substitute for $G(p(v))$ and   $\pi(v, p(v))$ from  equations (\ref{e13b})  
(\ref{e14}):
\begin{equation} \label{e15}
{\displaystyle  
  p(v) = v - \frac{\int_0^v   F(x)^{n-1} dx}{    F(v)^{n-1} }.
}
  \end{equation}
  

Suppose $F(v) = v/\bar{v}, $ the uniform distribution. Then (\ref{e15}) becomes
\begin{equation} \label{e17}
\begin{array}{ll}
p(v) &= {\displaystyle v- \frac{   \int_0^v
\left(\frac{x}{\bar{v}}\right)^{n-1} dx }{ \left( \frac{v}{\bar{v}}\right)^{n-1}   }} \\
  & \\
 & ={\displaystyle  v-       \frac{  \bigg|_{x=0}
^{v}    \left(\frac{1}{\bar{v}  }\right)^{n-1}  \left(\frac{1}{n }\right)     
 x^n   }{ \left( \frac{v}{\bar{v}}\right)^{n-1} }  }  \\
& \\
 &{\displaystyle  = v-  \frac{  \left(\frac{1}{\bar{v}  }\right)^{n-1}  \left(\frac{1}{n }\right)     
 v^n   -0    } { \left( \frac{v}{\bar{v}}\right)^{n-1}   } }  \\
& \\
 &  =  v - \frac{v}{n} =     \left( \frac{n-1 }{n} \right)v.     \\
\end{array}
  \end{equation}


  What a happy  ending to a complicated derivation! If there are 
  two bidders and values are uniform on
$[0,1]$, a bidder should bid $ p   = v /2$, which since he has probability  $v $ of
winning yields an expected payoff of $  v^2/2$. If $n=10$ he should bid
$\frac{9} {10} v$, which since he has probability $v^9$ of winning yields
him an expected payoff of $  v^{10}/10,$ quite close to zero if $v  < 1$.
 
  
\end{normalsize}
  

\newpage
\noindent
 {\bf Second-Price  Auctions  (Second-price sealed-bid, Vickrey) }

\noindent
 {\bf Rules}\\
  Each bidder submits one bid, in ignorance of the other bids.  The bids are
opened, and the highest bidder pays the amount of the second-highest bid and
wins
the object.

\noindent
 {\bf Strategies}\\
 A bidder's strategy is his bid as a function of his value.


\noindent
 {\bf Payoffs}\\
The winning bidder's payoff is his value minus the second-highest bid.  The
losing bidders' payoffs are zero. The seller's payoff is the second-highest-bid.

 

\newpage

\noindent
 Consider the following equilibrium.
\begin{equation} \label{e22b}
\begin{array}{ll}
    p_1(v=10) = 10 & p_1(v= 16)=16\\
 & \\
 p_2(v=10) = 1 & p_2(v=16)=10  \\
\end{array}
\end{equation}

 Since Bidder 1    never bids less than 10,  Bidder 2 knows that if $v_2=10$  he
can never get a positive payoff, so he is willing to choose  $p_2(v=10) = 1$.
Doing so results in a sale price of 1, for any $p_1 >1$,  which is better for
Bidder 1 and worse for the seller than a price of 10, but Bidder 2 doesn't care
about their payoffs.  In the same way, if $v_2=16$, Bidder 2 knows that if he
bids    10 he will win if $v_1=10$, but if $v_2=16$ he would have to pay 16 to
win and would earn a payoff of zero.  He might as well bid   10 and earn his
zero by losing.  

  
 \newpage
  \noindent
  {\bf Descending Auctions (Dutch)}

\noindent
 {\bf Rules}\\
The seller announces a bid, which he continuously lowers until some bidder stops
him and takes the object at that price.

\noindent
 {\bf Strategies}\\
 A bidder's strategy is when to stop the bidding as a function of his value.

\noindent
{\bf Payoffs}\\
The winner's payoff is his value minus his bid. The losers' payoffs are zero.


Descending auctions   are {\bf strategically equivalent} to the
first-price   auction, which means   there is a one-to-one mapping between the
strategy sets and the equilibria of the two games. 

 
 
 


\newpage

  \noindent
  {\bf All-Pay  Auctions}

\noindent
 {\bf Rules}\\
 Each bidder places a bid simultaneously. The bidder with the highest bid wins,
and each bidder pays the amount   he bid.

\noindent
  {\bf Strategies}\\
 A bidder's strategy is his bid as a function of his value.


\noindent
{\bf Payoffs}\\
 The winner's payoff is his value minus his bid. The losers' payoffs are  the
negative of their bids.
 
\newpage


 

\noindent
{\bf The Equal-Value All-Pay Auction }

  Suppose  each of the $n$ bidders has the same value, $v$.  Why is the equilibrium in mixed strategies? 



 Suppose we have a symmetric equilibrium, so  all bidders use the same mixing
cumulative distribution $M(p)$. Let us conjecture that $\pi(p) =0$, which we
will
later verify. 
  
 \begin{equation} \label{e24}
  M(p)^{n-1}  v =    p,
\end{equation}
 so
  \begin{equation} \label{e25}
   M(p)     = \sqrt[n-1]{\frac{p}{v}} 
\end{equation} 
   At the extreme bids that a bidder with value $v$ might offer,    $M(0)  =
\sqrt[n-1]{\frac{0}{v}} = 0$ and  $M(v)  = \sqrt[n-1]{\frac{v}{v}} = 1$,  so we
have found a  valid distribution function $M(p)$.  Moreover, since the payoff
from one of the strategies between which it mixes, $p=0$, equals zero, we have
verified  our   conjecture that $\pi(p) =0$ in the equilibrium.


\newpage

\begin{equation} \label{e25}
   M(p)     = \sqrt[n-1]{\frac{p}{v}},
\end{equation}

\includegraphics[width=150mm]{fig13-02.jpg}

\begin{center}
 {\bf Figure 2: The Bid Function in an All-Pay Auction with Identical Bidders}
\end{center}

\newpage

 \begin{normalsize}

 
\noindent
{\bf The Continuous-Value All-Pay Auction  }

 Suppose  each of the $n$ bidders picks his value $v$ from the same density
$f(v)
$. Conjecture that  the equilibrium is symmetric,  in pure strategies,    and
that the bid function, $p(v)$, is strictly increasing.   The  equilibrium payoff
function for a bidder with value $v$ who pretends he has value  $z$ is
  \begin{equation} \label{e261}
  \pi(v,z) =  F(z)^{n-1} v - p(z),
\end{equation}
 since   if our bidder bids $p(z)$, that is the highest bid only if all $(n-1)$
other bidders have $v<z$, a probability of $F(z)$ for each of them.


 The function   $\pi(v,z) $ is not necessarily concave in $z$, so  satisfaction
of the  first-order condition will not be a sufficient condition for payoff
maximization, but it is  a necessary condition   since the optimal $z$ is not 0
(unless $v=0$) or infinity and from (\ref{e261})  $\pi(v,z) $  is differentiable
in $z$ in our conjectured equilibrium.  Thus, we need to find $z$ such that
 \begin{equation} \label{e27}
   \frac{\partial \pi(v,z)}{\partial z}   =  (n-1) F(z)^{n-2} f(z) v - p'(z)=0
\end{equation}
 In the equilibrium, our bidder does follow the strategy $p(v)$, so $z=v$ and we
can write
 \begin{equation} \label{e28}
 p'(v)=   (n-1) F(v)^{n-2} f(v) v
\end{equation}
Integrating up, we get
 \begin{equation} \label{e29}
 p(v)=  p(0) +  \int_0^v (n-1) F(x)^{n-2} f(x) x dx
\end{equation}
 This is deterministic, symmetric,  and  strictly increasing in $v$, so we have
verified    our conjectures.


We can verify that truthelling is a symmetric equilibrium strategy by
substituting for $p(z)$ from (\ref{e29}) into  payoff equation (\ref{e261}).
\begin{equation} \label{e30}
\begin{array}{ll}
  \pi(v,z) & =  F(z)^{n-1} v - p(z)\\
 &\\
 & {\displaystyle=  F(z)^{n-1} v -p(0) -  \int_0^z (n-1) F(x)^{n-2} f(x) x dx}\\
 &\\
 &  {\displaystyle =  F(z)^{n-1} v -p(0) - F(z)^{n-1} z +   \int_0^z   F(x)
^{n-1}   dx,} \\
\end{array}
   \end{equation}
 where the last step uses integration by parts ($\int g h' = gh - \int h g'$,
where $g = x$ and $h'=(n-1) F(x)^{n-2} f(x)$).   Maximizing  (\ref{e30}) with
respect to $z$ yields
\begin{equation} \label{e31}
   \frac{\partial \pi(v,z)}{\partial z}   =  (n-1) F(z)^{n-2} f(z) (v-z),
\end{equation}
 which is maximized by setting $z=v$. Thus, if  $(n-1)$ of the  bidders are
using
this $p(v)$ function, so will  the remaining bidder, and we have a Nash
equilibrium.

    Let's see what happens with a particular value distribution. Suppose  values
are uniformly distributed over [0,1], so $F(v) =v$. Then equation (\ref{e29})
becomes
 \begin{equation} \label{e31a}
\begin{array}{ll}
 p(v) & =  {\displaystyle  p(0) +  \int_0^v (n-1) x^{n-2}  (1) x dx} \\
 & \\
 & {\displaystyle  =  p(0) +  \bigg|_{x=0}^v    (n-1)  \frac{x^n }{n}   } \\
 & \\
 &{\displaystyle  =  0+  \left( \frac{n-1}{n}  \right) v^n, }
 \end{array}
\end{equation}
 where we can tell that $p(0)=0$ because  if $p(0)>0$ a bidder with $v=0$  would
have a negative expected payoff.  If there were $n=2$ bidders,  a bidder with
value $v$ would bid $ v^2/2$, win with probability $v$, and  have expected
payoff $\pi = v(v) -  v^2/2 = v^2/2$.  If there were $n=10$ bidders,   a bidder
with value $v$ would bid $ (9/10) v^{10} $, win with probability $v^9$, and
have expected payoff $\pi = v(v^9) -  (9/10)v^{10}  =  v^{10}/(10)$.   As we
will see when we discuss the Revenue Equivalence Theorem, it is no accident that
this is the same payoff as for the   first-price auction when values were
uniformly distributed on [0,1], in equation (\ref{e9}).

\end{normalsize}

 \newpage
 

\noindent
   THE  REVENUE EQUIVALENCE THEOREM. { 
 Let  all players be risk-neutral with private values drawn independently from the same atomless, strictly increasing 
distribution $F(v)$   on $[\underline{v},
\bar{v}]$.   If  under either     Auction  Rule $A_1$ or
  Auction Rule $A_2$  it is true that:    \\

 (a)  The winner of the object is the player with the highest value; and     \\

 (b) The lowest bidder type, $v=\underline{v}$, has an expected payment of zero;
    \\

   then the symmetric equilibria of the two auction rules have  the same expected  payoffs for each type of bidder and for the seller.}


\newpage



 \begin{normalsize}

\noindent
{\it  Proof of RET.}
     Let us  represent the auction as the truthful equilibrium of a direct mechanism in which each bidder  sends a message $z$ of his type $v$ and then pays an expected amount $p(z)$. (The Revelation Principle says that we can do this.)  By assumption (a), the  probability that a player   wins the object given that he chooses message $z$ equals  $F(z)^{n-1}$, the probability that all $(n-1)$ other players have values $v<z$.  Let us denote this winning probability by $G(z)$, with density $g(z)$. Note that $g(z)$ is well defined because we   assumed that $F(v)$ is atomless and everywhere increasing. 

 
 The  expected payoff of any  player of type $v$ is the same, since we are restricting ourselves to symmetric equilibria. It equals     
 \begin{equation} \label{e33}
 \pi(z, v) =  
   G(z) v  - p(z). 
 \end{equation}
  The  first-order condition     with respect to the player's choice of type message $z$  (which we can use because neither $z=0$  nor $z=\bar{v}$ is the optimum  if  condition (a)    is to be true) is 
 \begin{equation} \label{e34}
 \frac{ d \pi(z; v)}{d z }  = g(z) v - \frac{ d p(z)}{d z } =0,    
 \end{equation}
 so 
 \begin{equation} \label{e34a}
\frac{ d p(z)}{d z } = g(z) v.      
 \end{equation}
  We are looking at a truthful equilibrium, so we can replace $z$ with $v$: 
\begin{equation} \label{e34b}
\frac{ d p(v)}{d v} = g(v) v.      
 \end{equation}
 
 Next, we integrate (\ref{e34b}) over all values from zero to $v$, adding $p( \underline{v})$ as the  constant of integration: 
\begin{equation} \label{e33a}
  p(v)  = p( \underline{v}) + \int_{\underline{v}}^v  g(x) x dx.      
 \end{equation}
 We can use (\ref{e33a} to substitute for $p(v)$ in the payoff equation (\ref{e33}), which becomes, after replacing $z$ with $v$  and setting $p( \underline{v})=0$ because of assumption (b),  
 \begin{equation} \label{e35}
 \pi(v, v) =  
   G(v) v  - \int_{\underline{v}}^v  g(x) x dx. 
 \end{equation}
    Equation (\ref{e35}) says  the expected payoff of a bidder of type $v$ depends only on the $G(v)$ distribution, which in turn depends only  on the $F(v)$ distribution, and not on the $p(z)$ function or other details of the particular auction rule. But if the bidders' payoffs do not depend on the auction rule, neither does the seller's.   Q.E.D. 


\end{normalsize}
 

\newpage
   
 The symmetric equilibria of the   ascending, first-price,  second-price,   descending, and all-pay auctions with continuous values all satisfy the two conditions of the Theorem: (a) the winner is the bidder with the highest value, and (b) the  lowest type makes an expected payment of zero. Thus, the following corollary is true. 

 


  \noindent
 A REVENUE EQUIVALENCE COROLLARY. { 
 Let  all players be risk-neutral with private values drawn from the same
strictly increasing, atomless distribution $F(v)$.  The symmetric equilibria of
the  ascending, first-price,  second-price,   descending, and all-pay auctions
all   have the same  expected  payoffs for each type of bidder and for the
seller.}

  
 \newpage

 \noindent
 TWO CAVEATS


(1) Although the different auctions have the same expected payoff  for the seller,
they do not have the same realized payoff.

 
(2) We need INDEPENDENT private values for the Revenue Equivalence Theorem. 

 Consider what happens if there are two bidders, both with values drawn uniformly from [0,10], but interdependently, with $v_2=10-v_1$.

 If we put aside   equilibria with  weakly dominated strategies (e.g., for a player to bid 0 if his value is less than 5), the second-price auction yields revenue equal to $p=v_{(2)}$, the second-highest value. 

The seller can extract more revenue  by using the auction rule that the winner is the highest bidder, and he pays 10 minus the second-highest bid: both players bid their values then, and the winning  payment equals the highest value. 

 

 

\newpage


 \noindent
 {\bf    Common-Value Auctions and the Winner's Curse}

\noindent
{\bf  The Winner's Curse:} If bidders in a pure common-value auction  all bid up to their valuations, the winner will be the one who overestimated the value the most, and he will have a negative payoff. 

 \newpage

  One way to think about a bidder's conditional estimate is to think about it as
  a conditional bid.

 The bidder
     would
like to submit a bid of  

  [{\it $X$ if I lose, but $(X-Y)$ if I win}],

 where
$X$ is his   value estimate conditional  on losing and $(X-Y)$ is his
estimate conditional  on winning.

If he still won with a bid of
$(X-Y)$ he would be happy. If he lost, he would be relieved.

 But Smith can
achieve the same effect by simply submitting the bid $(X-Y)$ in the first place,
since  when he loses, the size of his bid is irrelevant.

\newpage

Another  way to look at  the Winner's Curse  is based on   the Milgrom (1981)
definition of ``bad news''.

  Suppose   the
government is auctioning off the mineral rights to a plot of land with common
value $v$ and that bidder $i$ has value estimate $\hat{v}_i$. Consider a symmetric equilibrium. 

  If Bidder 1 wins with a bid $p(\hat{v}_1)$ that is based
on
his prior value estimate $\hat{v}_1$, his posterior value estimate $\tilde{v}_1$
is
 \begin{equation} \label{e47c}
 \tilde{v}_1 = E(V | \hat{v}_1, p(\hat{v}_2) < p(\hat{v}_1), \ldots, p(\hat{v}
_n) < p(\hat{v}_1)).
 \end{equation}

    The news that $p(\hat{v}_2) < \infty$ would be neither good nor bad, since
it conveys no information.

  The information that $p(\hat{v}_2) < p(\hat{v}_1)$ 
  is bad news, since it rules out values of $p$ more likely to be
produced by large values of $\hat{v}_2$.    Hence,
  \begin{equation} \label{e48}
\tilde{v}_1 < E(V | \hat{v}_1) = \hat{v}_1,
 \end{equation}
   

\newpage

 
 
 

 \begin{center}
 {\bf Table 1  Bids by Serious Competitors in Oil Auctions}

 \begin{tabular}{ llll }
 \hline &   &     &    \\
 {\bf Offshore} & {\bf  Santa Barbara} &  {\bf Offshore}  &  {\bf Alaska} \\
   {\bf Louisiana} & {\bf   Channel} &  {\bf Texas}  &  {\bf  North Slope} \\
 {\bf 1967} & {\bf 1968} &  {\bf  1968}  &  {\bf 1969} \\
Tract SS 207 &  Tract 375  &  Tract 506    & Tract 253   \\
 &   &     &    \\
 \hline &   &     &    \\
32.5 & 43.5  &  43.5   &    10.5\\
17.7 & 32.1  & 15.5    & 5.2   \\
11.1 & 18.1 & 11.6    &2.1   \\ 7.1  &10.2   &8.5     & 1.4   \\
 5.6 & 6.3  & 8.1    &0.5   \\
4.1 &   &   5.6  &   0.4 \\ 3.3 &   &   4.7  &    \\
 &   &    2.8  &    \\
 &   &   2.6  &    \\
 &   &   0.7  &    \\ &   &   0.7  &    \\
 &   &   0.4  &    \\ &   &     &    \\
\hline \end{tabular}
 \end{center}

 

   


 \newpage

\noindent
  {\bf  Strategies in Common-Value Auctions}

Milgrom \& Weber (1982) found that when there is a common-value element in an
auction  and signals are ``affiliated''    then  revenue
equivalence fails.

 The first-price and descending auctions are still identical,
but they raise less revenue than the ascending or second-price auctions. If
there are more than two bidders, the ascending auction raises more revenue than
the second-price auction. (In fact, if signals are affiliated  then even in a
private value auction, in which each bidder knows his own value with certainty,
the first-price and descending auctions will do worse.)

  \newpage

Suppose   $n$ signals are independently drawn from the  uniform distribution  on
$ [\underline{s},  \overline{s} ]$. Denote the $j^{th}$  highest signal by
$s_{(j)}$.
The expectation of the $k$th highest value happens to be
\begin{equation} \label{e49}
 E s_{(k)} =  \underline{s} +\left(  \frac{ n+1-k}{n+1} \right) \left(
\overline{s} - \underline{s} \right)
 \end{equation}
      This means the expectation of the very  highest value is
\begin{equation} \label{e49a}
\begin{array}{ll}
 E s_{(1)}& =
  \underline{s} +\left(  \frac{ n }{n+1} \right) \left(  \overline{s} -
\underline{s} \right) \\
\end{array}
   \end{equation}
The expectation of the
second-highest value is
 \begin{equation} \label{e50}
\begin{array}{ll}
E s_{(2)}& =   \underline{s} +\left( \frac{ n-1}{n+1} \right) \left(
\overline{s} - \underline{s} \right)\\
\end{array}
     \end{equation}
The  expectation of the lowest value, the $n$'th highest,  is
 \begin{equation} \label{e51}
\begin{array}{ll}
E s_{(n)} &=   \underline{s} +\left( \frac{  1 }{n+1} \right) \left(
\overline{s} - \underline{s} \right) . \\
\end{array}
   \end{equation}

\newpage

Let $n$ risk-neutral bidders, $i= 1,2,...n$ each receive a signal $s_i$
independently drawn from the uniform distribution on $ [v- m, v+m]$, where $v$
is the true value of the object to each of them. Assume that they have ``diffuse
priors'' on $v$, which means they think any value   from $v=-\infty$ to $v=
\infty$ is equally likely and we do not need to make use of bayes's rule.
The best estimate of the value given the set of $n$ signals  is
 \begin{equation} \label{e52}
  E v | (s_1, s_2, ..., s_n) =     \frac{ s_{(n)}+ s_{(1)}}{2}.
  \end{equation}


The estimate depends only on two out of the $n$   signals---   a remarkable
property of the uniform distribution.   If there were five signals $\{6,7,7, 16,
24\}$, the expected value of the object  would be 15 (=[6+24]/2), well above the
mean  of 12 and the median of 7, because only the extremes of 6 and 24 are
useful information. 

  

\newpage

     Someone who saw just signals $s_{(n)}
$ and $s_{(1)}$ could deduce that $v$ could not be less than $(s_{(1)}- m)$ or
greater than  $(s_{(n)}+m)$. Learning the signals in between  would be
unhelpful, because  the only information that, for example,  $s_{(2)}$
conveys is that $v  \leq   (s_{(2)}+m)$ and $v  \geq   (s_{(2)}-m)$, facts which
our observer had already figured  out  from $s_{(n)}$ and $s_{(1)}$.

\includegraphics[width=150mm]{fig13-04.jpg}


\begin{center}
 {\bf Figure  4: Extracting Information From Uniformly Distributed Signals }
 \end{center}


\newpage

 
 

\newpage

 
\noindent
 {\bf  The Ascending Auction (open-exit)  }

\noindent
{\bf  Equilibrium:}  If no bidder has quit yet,  Bidder $i$ should drop out
when
the price rises to  $s_i$. Otherwise, he should  drop out when the price rises
to $p_i = \frac{p_{(n)} +s_i}{2}$, where $p_{(n)}$ is the price at which the
first dropout occurred.

\noindent
{\bf  Explanation:} If no other bidder   has quit yet,     Bidder $i$  is safe
in
agreeing to pay his signal, $s_i$. Either  (a) he has the lowest signal, or  (b)
everybody else has  the same signal value  $s_i$ too, and they will all drop out
at the same time. In case (a), having the lowest signal,  he will lose anyway.
In case
(b),  the best estimate of the value is $s_i$, and that is where he  should drop
out.

\newpage

 Once one bidder has dropped out at $p_{(n)}$, the other bidders can deduce that
he had the lowest signal, so they know  that  signal $s_{(n)}$ must equal
$p_{(n)}$.   Suppose Bidder $i$ has signal $s_i > s_{(n)}$.  Either (a) someone
else has a higher signal and  Bidder $i$ will lose the auction anyway and
dropping out too early does not matter, or (b)  everybody else  who has not yet
dropped out has signal $s_i$ too, and they will all drop out at the same time,
or (c) he  would be the last to drop out, so  he will win. In cases (b) and (c),
his estimate of the value is $p_{(i)}= \frac{p_{(n)} +s_i}{2}$, since $p_{(n)}$
and $s_i$ are the extreme signal values and the signals are uniformly
distributed, and that is where he  should drop out.

\newpage

  The price paid by the winner will be the price at which the second-highest
bidder drops out, which is $\frac{s_{(n)} +s_{(2)}}{2}$. The  expected values
are, from equations  (\ref{e50}) and (\ref{e51}), 
\begin{equation} \label{e53}
\begin{array}{ll}
E s_{(n)}& =  (v-m) +\left(  \frac{ n+1-n}{n+1} \right) \left( (v+m)-  (v-m)
\right) \\
 & \\
&  =v +\left(  \frac{ 1-n}{n+1} \right)  m \\
\end{array}
 \end{equation}
  and
\begin{equation} \label{e54}
\begin{array}{ll}
 E s_{(2)} &=  (v-m) +\left(  \frac{ n+1-2}{n+1} \right) \left( (v+m) -  (v-m)
\right)\\
 &\\
 & =   v +\left(  \frac{ n-3}{n+1} \right)  m.\\
\end{array}
 \end{equation}

  Averaging them yields the expected winning price,
 \begin{equation} \label{e55}
\begin{array}{ll}
E p_{(2)} &=   \frac{\left[  v +\left(  \frac{ 1-n}{n+1} \right) m \right]  +
\left[v +\left(  \frac{ n-3}{n+1} \right)  m \right]   } {2} \\
 &\\
 & =   v -    \left( \frac{1}{2} \right)\left( \frac{  1}{n+1} \right)   2m. \\
\end{array}
\end{equation}
 If $m=50$ and $n=4$, then
\begin{equation} \label{e57a}
 E p_{(2)} =v -   \left(  \frac{1}{10}  \right) (100) = v-10 .
\end{equation}

Expected seller revenue increases in $n$, the number of bidders (and thus of
independent signals) and falls in the
 uncertainty $m$ (the inaccuracy of the signals).  This will be true for all
three auction rules we examine here.

 It is not always true that the bidders can deduce the lowest signal in an
ascending auction and use that to form their bid.    Their
ability to  discover $s_{(n)}$ depended crucially on the open-exit feature of
the auction--- that the player with the lowest signal had to openly drop out,
rather than lurk quietly in the background.   A secret-exit ascending auction
would  behave   like a second-price auction instead.

\newpage

\noindent
 {\bf  The   Second-Price Auction   }

\noindent
{\bf  Equilibrium:}  Bid $p_i = s_i  - \left( \frac{n-2}{n}\right) m $.

\noindent
{\bf  Explanation:} In forming his strategy, Bidder $i$  should think  of
himself
as being tied for winner with one other bidder, and so having to pay  exactly
his
bid.  Thus, he imagines himself as  the highest of $(n-1)$ bidders drawn from
$[v-m, v+m]$ and tied with one other. On   average,  if this happens,
  \begin{equation} \label{e55a}
\begin{array}{ll}
 s_i &=   (v-m) +\left( \frac{([n-1]+1- 1)} {[n-1] +1}  \right)([v+m]- [v-m])\\
 & \\
 & =   v + \left( \frac{n-2}{n}  \right)(m).
\end{array}
 \end{equation}
   He will bid the value $v$ which solves equation (\ref{e55a}), yielding the
optimal strategy,     $p_i = s_i - \left( \frac{n-2}{n} \right) (m) $.

\newpage

On average, the  second-highest bidder actually has the signal $E s_{(2)} = v
+\left(  \frac{ n-3}{n+1} \right) m$, from equation (\ref{e54}). So the expected
price, and hence the expected revenue from the auction,  is
 \begin{equation} \label{e56}
\begin{array}{ll}
 E p_{(2)}& =    [v +\left(  \frac{ n-3}{n+1} \right)  m ]- \left(\frac{n-2}{n}
\right)  (m)\\
  & \\
& =  v- \left(\frac{ n-1}{n}\right)\left( \frac{ 1}{n+1}\right) 2m.\\
\end{array}
 \end{equation}
 If $m=50$ and $n=4$, then
\begin{equation} \label{e57b}
 E p_{(2)} =v - \left( \frac{3}{4} \right)  \left(  \frac{1}{5}\right) (100) =
v-15.
\end{equation}

  If there are at least three bidders, expected revenue is  lower in the second-
price auction.  



 
 
 

\end{Huge}

\end{document}
