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\begin{LARGE}

Eric Rasmusen, erasmuse@indiana.edu, Nov 9, 2006

 
\section*{   13 Auctions}

This chapter is a big one. It really would take 3 75-minute sessions-- and don't try to use all these slides even for 3 sessions. Pick and choose which ones you want to cover. 

 I plan  to cover, in one 75-minute session,  parts of 13.1, 13.2, 13.5. I will pick just a few derivations to do, e.g., the first-price private value auction optimal strategy. 

 I don't think I'll use overheads--I will write on the board. I will use selected overheads as notes for myself. 

 Sell a soft drink using the 5 auction  rules. Say I will collect the money from the lowest winning bid. 

  \newpage

\begin{center}
{\bf  13.1  Values Private and Common, Continuous and Discrete  }
\end{center}
 
 

 

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

Call the dollar value of the utility that bidder $i$
receives from an object its  {\bf  value }  to him, $v_i$, and we will denote
his estimate of the value by  $\hat{v_i}$.


  {\bf Private-value auction}: a bidder can learn nothing about his value
from
knowing the values of the other bidders. (antique chairs-- not for resale)

  {\bf Independent private-value
auction}:   knowing his own value tells him nothing about OTHER  bidders'
values.

 {\bf Affiliated private-value auction}:
  he can
  use  knowledge of his  own value to deduce something about other
players' values.

  {\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
  

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

\noindent
 {\bf Order of Play:}\\
0. Nature chooses   Bidder $i$'s value for the object  to be either $v_i=10$ or
$v_i=16$, with equal probability. (The seller's value is zero.)  \\
 
 
 

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

\noindent
 {\bf Order of Play:}\\
0. Nature chooses   Bidder $i$'s value for the object,  $v_i$, using the
strictly positive, atomless  density $f(v)$ on the interval $[\underline{v},
\overline{v}]$.  \\

\newpage


\begin{center}
{\bf  A Mechanism Interpretation   }
\end{center}

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.). \\
  

\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}

 

  The  mechanism could allocate the good with 70\%
probability to the high bidder and with 30\%  probability to the lowest bidder.


Each bidder could be made to pay the amount he bids, even if he
loses.

 The payment $t$ could include  an entry fee.

There could be a ``reserve
price,''   a minimum bid for which the seller will surrender the good. 

 

\newpage

\begin{center}
{\bf   13.2   Optimal Strategies under Different   Rules  in Private-Value
Auctions    }
\end{center}
 
 
 

 \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

 \begin{center}
{\bf Types of Ascending Auctions    }
\end{center}


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

  (2)   The {\bf open-exit} auction.

 (3)   The {\bf silent-exit} auction.


 (4) The {\bf Ebay auction}.

  (5) The {\bf Amazon auction}.
   
The
ascending auction   can be seen as a mechanism in which each bidder announces his value (which becomes his bid), the object is awarded to whoever announces the highest value (that is, bids highest), and he pays the second-highest announced value (the second-highest bid).

\noindent
{\bf Discussion}\\
    A bidder's dominant strategy in a private-value  ascending auction is to
stay
in the bidding   until  bidding higher would require him to exceed his   value
and then to stop. 

 
\newpage
 
 \begin{center}
{\bf First-Price  (first-price sealed-bid) }
\end{center}
 
 

\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.

 

\newpage

 \begin{center}
{\bf  Strategies in the First-Price Auction   }
\end{center}

    In the first-price auction what the winning bidder wants to do is to have
submitted a sealed bid  just enough higher than the second-highest bid to win.

If all the bidders' values are common knowledge and he can predict the second-
highest bid perfectly, this is a simple problem. 

 If the values are private
information, then he has to guess at the second-highest bid, however, and take a
gamble.

 His tradeoff is between bidding high--thus winning more often--and
bidding low--thus benefiting more if the bid wins.  

 His optimal strategy
depends on his degree of risk aversion and beliefs about the other bidders, so
the equilibrium is less robust  to mistakes in the assumptions of the model than
the equilibria of ascending and second-price auctions. 

 \newpage

\begin{center}
{\bf    The First-Price Auction with a Continuous Distribution of Values }
\end{center}

 

   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}

Thus,  
\begin{equation} \label{e10c}
  p(v) = v - \frac{ \pi(v, p(v))}{   G(p(v))}.
  \end{equation}

 \noindent
   Lemma 1: If a  player's  equilibrium bid function is differentiable, it    is strictly increasing in  his value: $p'(v) >0$. 

  Lemma 1 implies that
  the bidder with the greatest $v$ will bid highest  and win.

  \newpage


\begin{center}
{\bf  Using the Envelope Theorem   }
\end{center}


  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}

 
 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}

 
Then  
\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}
    Integrate   over  all possible values  from
zero to $v$   and include the base value of $\pi(0)$ (=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}
 

\newpage

\begin{center}
{\bf   The Bid Function  }
\end{center}


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}
  
 
 

  

\newpage


\begin{center}
{\bf  The First-Price Auction:   A Mixed-Strategy Equilibrium  in the 
 Ten-Sixteen   Auction       }
\end{center}

 
   When the value distribution does not have a continuous support,   the
equilibrium in a first-price auction may not even  be in pure strategies.

   Now
let  each  of two  bidders' private  value  $v$  be  either 10 or 16  with equal
probability  and   known only to himself.

 In a first-price auction,  a bidder's  optimal strategy is to   bid  $p(v=10) =
10$,  and if $v=16$ to use a mixed strategy, mixing   over the support
$[\underline{p},\bar{p}]$, where it will turn out that $\underline{p}=10$ and
$\bar{p}= 13$, and the   expected payoffs will be:
\begin{equation} \label{e18}
 \begin{array}{ll}
   \pi (v=10)  &=0 \\
 &\\
   \pi (v=16)  &=  3  \\
&\\
   \pi_s  &=  11.5.    \\
  \end{array}
\end{equation}
 These are the same payoffs  as in the ascending auction, an equivalence  we
will come back to in a later section.

\newpage

 
\begin{center}
{\bf  The Equilibrium   }
\end{center}



     $ p(v =10) = 10$. If either bidder used
the bid $p<10$,  the other player would  deviate to  $(p+\epsilon)$,
and a   bid above 10 exceeds the object's value.

 
 
The bid  $p(v=16)$  will be between  10 (so the bidder can
win if his rival's value is 10)  and   16 (which would always win,
but unprofitably). 

The  pure strategy of $(p=10)|(v=16)$ will win with
probability of at least 0.50,     yielding   payoff     $0.50 (16-10) =3$.  This rules
out bids in $(13, 16]$, because their payoff is less than
3.
 

 

   The upper bound $\bar{p}$  must be exactly 13. If it were any less, then the
other player would respond by using the pure strategy of $(\bar{p}+\epsilon)$,
which would   win with probability one and yield  a payoff of greater than the
payoff of  3 ($=0.5 (16-10)$) from $p=10$.  


  When a player mixes over a continuum, the modeller must be careful to check
for \\
 (a)
  atoms (some particular  point  which has positive   probability, not just
positive density), and \\
(b)  gaps  (intervals  within the mixing range with  zero
probability of  bids). Are there any atoms or gaps  within  the interval
[10,13]?  

\newpage

  
\begin{center}
{\bf  The Mixing Density   }
\end{center}


 
The mixing density $m(p)$ is positive over the entire
interval $[10, 13]$, with no atoms.  
Since if our player has value $v=16$  there is probability 0.5 of winning
because the other player has $v=10$ and probability $0.5 M(p)$ of winning
because the other player has $v=16$ too but bid less than $p$, the payoff is
  \begin{equation} \label{e21}
0.5  (16-p) + 0.5   M(p)   (16-p)  =3.
 \end{equation}

This implies that $  (16-p) +  M(p)  (16-p)  = 6$, so
\begin{equation} \label{e22}
   M(p)    =  \frac{6}{16-p}-1,
\end{equation}
   which has the density
 \begin{equation} \label{e22a}
   m(p)    =     \frac{6}{(16-p)^2}
\end{equation}
 on the support $[10,13]$, rising from $m(10) = \frac{1 }{6}$ to $m(13) =
\frac{4 }{6}$.

 
  

\newpage


\begin{center}
   {\bf Second-Price  Auctions  (Second-price sealed-bid, Vickrey) }   
\end{center}

 

\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


\begin{center}
{\bf  Asymmetric Equilibria   }
\end{center}



Consider a variant of the Ten-Sixteen Auction,  in which each of two bidders'
values  can be 10 or 16, but  where the realized  values    are  common
knowledge.

 Bidding one's value is  a {\bf symmetric equilibrium},  meaning that
the bid function $p(v)$ is the same for both bidders: $ \{p(v=10) = 10, p(v=16)=
16\}$.  

\noindent
But 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}

 
 

\newpage

\begin{center}
 {\bf Descending Auctions (Dutch)}

\end{center}

 
 
\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.



 The   descending auction is {\bf strategically equivalent} to the
first-price   auction.

 
 

\newpage

\begin{center}
 {\bf All-Pay  Auctions}
\end{center}

 

\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.

\noindent
{\bf Discussion}:
  The winning bid will be lower in the all-pay auction than  under the other
rules,    because bidders  need a bigger payoff when they do win to make up for
their negative payoffs when they lose.    At the same time, since  even the
losing bidders pay something to the seller  it is not obvious that the seller
does badly (and in fact, it turns out to be just as good an auction rule as the
others, in this simple risk-neutral context).

  


\newpage

\begin{center}
{\bf The Equal-Value All-Pay Auction }
\end{center}

 

  Suppose  each of the $n$ bidders has the same value, $v$.

  Under the all-pay auction rule,   this game is quite
interesting.

 The equilibrium is  in mixed strategies.

 Either the maximum bid is less than $v$, in which case
someone could deviate to $p=v$ and increase his payoff;

 or one bidder bids $v$
and the rest bid  at most $p'<v$, in which case the high bidder will deviate to
bid  just above $p'$.

\newpage

\begin{center}
{\bf   A Symmetric Equilibrium   }
\end{center}


 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.   The payoff function for each  bidder is
the
probability of winning times the value of the prize minus the bid, which is paid
with probability one,  and if we equate that to zero we get
 \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}
 
   

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


\newpage

 
\begin{center}
  {\bf The Continuous-Value All-Pay Auction  }    
\end{center}



\noindent


 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.


\newpage

\begin{center}
{\bf  Finding the Equilibrium    }
\end{center}


 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.

 \newpage

\begin{center}
{\bf  The Outcome   }
\end{center}


   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 $\pi = v(v^9) -  (9/10)v^{10}  =  v^{10}/(10)$.  




\newpage

\begin{center}
 {\bf  The Dollar Auction }
\end{center}

 

Consider an  ascending   auction  to sell a  dollar bill in which
the players offer higher and higher bids, and the highest bidder wins-- but both
the   first- and   second-highest bidders pay their bids. 

What happens? 
  
 
\newpage


\begin{center}
 THE  REVENUE EQUIVALENCE THEOREM.  
\end{center}

 
   

\noindent
  
 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{center}
{\bf   Proving the Theorem   }
\end{center}


\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. 

\noindent
{\it  Proof.}
     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. 

 \newpage


\begin{center}
{\bf   A Second Step in the Proof   }
\end{center}


 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}

\newpage
 
\begin{center}
{\bf   Finishing the Proof  }
\end{center}


 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. 


   \newpage


\begin{center}
{\bf   A REVENUE EQUIVALENCE COROLLARY    }
\end{center}


   
   
 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

\begin{center}
{\bf  Risk Aversion in Private-Value Auctions  } 
\end{center}

 
  

When bidders are risk averse, the Revenue Equivalence Theorem fails. 

 Consider
Bidder 1 in The   Ten-Sixteen Auction   when he knows his own  value is  $v_1=
16$
but does not know $v_2$.  In the  second-price auction,  he has an equal chance
of a payoff of either 0 (if $v_2=16$) or 6 (if $v_2=10$), regardless of whether
the bidders are risk averse or not, because bidding    one's value  is a weakly
dominant strategy.

Compare that with his payoff in the first-price  auction, in which the
equilibrium is in mixed strategies.   If the bidders are risk neutral, then  as
we found earlier,  if the   bidder has value 16    he wins using  a bid in  the
mixing support  [10,13]  and achieves a payoff in [3,6]    with probability
0.75,  and    he loses and earns payoff of zero with probability 0.25.  The
(0,6) gamble of the second-price auction is riskier than the  (0, 3 to 6) gamble
of the first-price auction.  The  (0,6) gamble is simpler, but it has more
dispersion.

\newpage

\begin{center}
{\bf  Risk Aversion in the First Price Auction   }
\end{center}


If the bidders are risk averse, then the optimal strategies in the first-price
auction change. It remains true that the bidders mix on an  interval
$[10,\bar{p}]$. We derived $\bar{p}$ and the optimal mixing distribution  by
equating expected payoffs, however, and a certain win at a price of 10 will now
be worth more to a bidder than a 50\% chance of winning at a price of 13. Let us
denote the concave utility function of each bidder by $U(v-p)$ and normalize by
defining $U(0) \equiv 0$. The expected payoff from $p=10$, which wins with
probability 0.5,   must  equal the  expected payoff from the upper bound
$\bar{p}$ of the mixing support, so
 \begin{equation} \label{e44a}
 0.5 U(6) =  U( 16- \bar{p}).
 \end{equation}
   Since $0.5   U(6) < U( 16- 13)$   by concavity of $U$, it must be that
$\bar{p} >13.$  

\newpage


\begin{center}
{\bf  The Mixing Density   }
\end{center}


We found the mixing distribution function $M(p)$ by equating
$\pi(p)$ to the payoff from bidding 10, which is $0.5 U(6)$,  so
 \begin{equation} \label{e44b}
  \pi(p) =  0.5 U(16-p) +  0.5 M(p) U(16-p) = 0.5 U(6),
 \end{equation}
 which can be solved to yield
\begin{equation} \label{e44c}
   M(p)  = \frac{U(6)} {U(16-p)} -1,
 \end{equation}
 which has the density
\begin{equation} \label{e44d}
     m(p)  = \frac{U(6)} U'(16-p) {U(16-p)^2},
 \end{equation}
 compared with the risk-neutral density $ m(p)     = \frac{6}{(16-p)^2}  $ from
equation (\ref{e22a}). Thus,  risk aversion of the bidders actually spreads out
their equilibrium bids (the support is broader than  [10,13]), but it remains
true that the first-price auction is less risky than the second-price auction.

\newpage

\begin{center}
{\bf  Risk Aversion in the Continuous-Value Auction   }
\end{center}


What happens in the Continuous-Value Auction? In  the  second-price auction ,the
optimal strategies are unchanged, so seller revenue does not change if bidders
are risk averse.

 To solve for the equilibrium of the first-price auction,  
look at a given bidder's incentive to report his true type $v$ as $z$ in an
auction in which the payment is $p(z)$ and the probability of winning the object
is $G(z)$.

 The bidder maximizes by choice of $z$
 \begin{equation} \label{e45}
\begin{array}{ll}
 \pi(v,z)& =
   G(z ) U[v  - p(z)]  \\
 & \\
 &= F(z)^{n-1} U[v  - p(z)], \\
 \end{array}
 \end{equation}
where   $\pi(v,0) = 0$ because $F(0) =0$. 

  At the optimum,
\begin{equation} \label{e46}
\begin{array}{ll}
\frac{ \partial \pi(v,z ) }{\partial z} &  =
 (n-1)  F(z)^{n-2}  f(z)   U[v  - p(z)]\\ & 
  +  F(z)^{n-1} U'[v  - p(z)] [ -p'(z)]=0,\\
\end{array}
 \end{equation}

\newpage


\begin{center}
{\bf  The Ultimate Effect of Risk Aversion   }
\end{center}

 

 In equilibrium,   $z=v$. Using that fact, for all $v>\underline{v}$ (since
$F(\underline{v}) =0$) we can solve  equation (\ref{e46}) for
 $p(z)$ to get
\begin{equation} \label{e47}
 p(v) =  \left( \frac{(n-1)     f(v)}{ F(v) } \right) \left(  \frac{ U[v  -
p(v)] }{  U'[v  - p(v)]}\right)
 \end{equation}
  Now let's  look at the effect of risk aversion on  $p(v)$. If $U$ is linear,
then
\begin{equation} \label{e47a}
\frac{ U[v  - p(v)] }{  U'[v  - p(v)]}= v  - p(v),
\end{equation}
but if  the bidder is risk averse, so $U$ is strictly concave,
\begin{equation} \label{e47b}
\frac{ U[v  - p(v)] }{  U'[v  - p(v)]}> v  - p(v).
\end{equation}
 Thus, for a given $v$,  the bid function in (\ref{e47})  makes the    bid
higher if
the bidder is risk averse than if he is not.  The bid for every value of $v$
except $v=\underline{v}$  increases  ($p(\underline{v}) =\underline{v}$,
regardless of risk aversion). 

 By
  increasing his bid from the level optimal for a risk-neutral bidder, the risk-
averse bidder insures himself. If he wins, his surplus is slightly less because
of the higher price, but he is more likely to win and avoid a surplus of zero.


\newpage

\begin{center}
{\bf   Seller Revenue and Risk Aversion  }
\end{center}


As a result the seller's revenue is greater in the first-price   than in the
second-price auction if bidders are risk averse.

 But since under risk neutrality
the   first-price  and second-price auctions yield  the same revenue,   under
risk aversion the first-price auction must yield greater revenue, both in
expectation and conditional on the highest $v$ present in the auction.

 The
seller, whether risk neutral or risk averse,  will prefer the first-price
auction when
bidders are risk averse.


\newpage


\begin{center}
{\bf     Uncertainty over   One's Own Value }
\end{center}

 

 
 {\it  If the seller can reduce bidder uncertainty over the value of the object
being auctioned, should he do so? }

 

  Suppose there are $n$ bidders, each with a private value, in an ascending
auction.  Each measures his private value  $v$ with an independent error
$\epsilon>0$. This error is with equal probability $-x, +x$ or 0. 

  The bidders
have diffuse priors, so they take all values of $v$ to be equally likely, ex
ante.  

Let us denote a bidder's  measured value by $\hat{v}= v + \epsilon$,
which
is an unbiased estimate of $v$.

 In the ascending auctions we have been studying
so far, where $\epsilon=0$, the optimal bid ceiling was $v$.

 Now, when $\epsilon
>0$,  what bid ceiling should be used by  a  bidder with utility function $U(v-
p)
$?

\newpage


\begin{center}
{\bf  Optimal Strategies for Uncertain Bidders   }
\end{center}


 If the  bidder  wins the auction and pays $ p$ for the object, his  expected
utility  at that point is
\begin{equation} \label{e43}
 \pi ( p )  =  \frac{U([\hat{v}-x] - p)  }{3}+ \frac{ U (\hat{v}  - p) }{3}
+\frac{U([\hat{v}+x]  - p)    }{3}
\end{equation}

If he is risk neutral, this yields him a payoff of zero if $p = \hat{v}$, and
winning at any lower price would yield a positive payoff. Under risk neutrality uncertainty over one's own value does not affect
the optimal strategy.  



 If the bidder  is risk averse, however,   then the utility function $U$  is
concave and
\begin{equation} \label{e44}
\frac{ U([\hat{v}-x] - p)  }{3}  +\frac{U( [\hat{v}+x] - p)  } {3}< \left(
\frac{2}{3} \right) U(\hat{v}-p),
\end{equation}
so his expected payoff  in equation (\ref{e43}) is less than $U( \hat{v}-p)$,
and if $p=\hat{v}$ his payoff is less than $U(0)$.

 
\newpage

\begin{center}
{\bf 13.4  Reserve Prices and  the  Marginal Revenue    Approach  }
\end{center}


  
  

 A {\bf reserve price} $p^*$ is a bid put in by the seller, secretly or openly,
before the auction begins, which commits him not to sell the object if nobody
bids more than  $p^*$. 

 The seller will often find that a reserve price can
increase his payoff.  If he does,  it turns out that he will choose a reserve
price strictly greater than his own value: $p^*>v_s$.

 To see this, we will use
the {\bf marginal revenue approach}  to auctions [
Bulow \& Roberts (1989)]  

 This approach compares the seller
in an auction to an ordinary monopolist who sells using a posted price. 

 \newpage

\begin{center}
{\bf   An Auction with One Bidder and a Reserve Price  }
\end{center}


\noindent
  The seller will do badly
in
any of the auction rules we have discussed so far.   
 
 What should the  seller's offer $p^*$  be? 

Let the bidder have value distribution $F(v)$
on $[\underline{v}, \bar{v}]$ which is differentiable and strictly increasing,
so the density  $f(v)$ is always positive. Let the seller value the object at
$v_s \geq  \underline{v}$.  The seller's payoff is
\begin{equation} \label{e36}
\begin{array}{ll}
 \pi(p^*) & =     Pr (  p^*< v) (p^* - v_s) +Pr (  p^*> v) (0) \\
 & \\
 & =    [1-F(p^*)] (p^* - v_s).\\
\end{array}
 \end{equation}
 This has first-order-condition
 \begin{equation} \label{e37}
 \frac{d\pi(p^*)}{d p^*}  =  [1-F(p^*)]  - f(p^*) [p^* - v_s]=0.
 \end{equation}
 On  solving (\ref{e37}) for for $p^*$ we get
 \begin{equation} \label{e38}
{\displaystyle p^*=  v_s + \left( \frac{1-F(p^*)}{f(p^*)} \right).  }
 \end{equation}
   The reserve price is strictly greater than the seller's
value  for the object ($p^*> v_s$) unless the solution is such that $F(p^*)=1$
because the optimal reserve price is the greatest possible bidder value, in
which
case the object has probability zero of being sold. 

\newpage

\begin{center}
{\bf 2.  Multiple Bidders.}
\end{center}


\noindent
   Now let there be $n$ bidders, all  with  values
distributed independently by   $F(v)$. Denote the bidders with the highest and
second-highest values as Bidders 1 and 2.

The seller's payoff in a second-price auction  is
\begin{equation} \label{e39}
\begin{array}{ll}
 \pi(p^*) & =  Pr (  p^*> v_1) (0)+   Pr ( v_2< p^*< v_1 ) (p^*  - v_s)\\
 &  +  Pr (
p^*< Ev_2 < v_1) (v_2 - v_s)     \\
 & \\
 &  {\displaystyle =  \int_{v_1=\underline{v}}^{p^*}  f(v_1) (0) dv_1 +
\int_{v_1=p^*}^{\bar{v}}   \bigg( \int_{v_2=\underline{v}}^{p^*} (p^* - v_s)
f(v_2)dv_2  }  \\
 &  {\displaystyle  +   \int_{v_2=p^*}^{v_1} (v_2 - v_s) f(v_2)dv_2  \bigg) f(v_1)
dv_1   }  \\
 & \\
\end{array}
 \end{equation}
  This expression integrates over two random variables. First, it matters
whether $v_1$ is greater than or less than $p^*$,  the outer integrals. Second,
it matters whether  $v_2$  is less than  $p^*$ or not, the inner integrals.
 
 




 It turns out that
 \begin{equation} \label{e40a}
 p^*=  v_s + \frac{1-F(p^*)}{f(p^*)},
 \end{equation}
 just what we found in    equation (\ref{e38}) for the one-bidder case.
Remarkably,  the optimal reserve price is unchanged!

  


\newpage

\begin{center}
{\bf 3. A Continuum of Bidders: The Marginal Revenue Interpretation }
\end{center}

 

  Now think of  a firm with a constant marginal cost of $c$  facing  a
continuum of bidders      along the same distribution $F(v)$ that we have been
using.  The quantity of bidders with values above $p$ will be $(1-F(p))$, so the
demand equation  is
\begin{equation} \label{e40b}
 q(p)   = 1-F(p),\;\; Revenue \equiv  pq  = p(1-F(p))
\end{equation}
  

 The marginal revenue is then (keeping in mind that $\frac{dq}{dp} = -f(p)$)
\begin{equation} \label{e41}
 \begin{array}{ll}
Marginal \; Revenue \equiv  \frac{dR}{dq} & = p + \left( \frac{dp}{dq} \right)
q\\
  &=   p + \left( \frac{1}{  \frac{dq}{dp}} \right)q\\
& =  p-    \frac{ 1-F(p)}{   f(p)}     \\
\end{array}
\end{equation}
 Profit-maximizing  monopoly
price is the one at which the marginal revenue    equals $c$.

\newpage

 
\begin{center}
 {\bf Figure 3a:  Auctions and Marginal Revenue: Reserve Price Needed or Not Needed}
 \end{center}



\includegraphics[width=150mm]{fig13-03a.jpg}




\includegraphics[width=150mm]{fig13-03b.jpg}


 

\newpage

 
\begin{center}
{\bf  Monopoly and Auctions   }
\end{center}



    If output is reduced below the competitive level, the outcome is
inefficient, as in conventional monopoly.   
This happens if   no sale takes place of the one unit even though
$v> v_s$ for some bidder.   

 Unlike a conventional monopoly, there is a possibility of inefficient
``overproduction''  in an auction. 

That happens if the sale takes place even
though no bidder values the good as much as the seller: $v<v_s$ for the winning
bidder. 

 A positive reserve price, therefore, can help efficiency rather than
hurt it.

All the five auction forms --- first-price, second-price, descending, ascending,
and all-pay---  can be efficient in a private-value setting, but  only  if the
reserve price is set  not at the profit-maximizing level  but at $p^*= v_s$.

  
 




\newpage


\begin{center}
{\bf Hindering Bidder Collusion}
\end{center}
 
 

  Robinson (1985)  has  pointed out that whether the auction is private-value or
common-value, the first-price  auction is superior to the second-price   or
ascending auctions for deterring collusion among bidders.


 
  Consider a bidder's cartel in which bidder Smith has a private value of 20,
the
other bidders' values are each 18, and they agree that everybody will bid 5
except Smith, who will bid 6.  

 In an
ascending auction this is self-enforcing, because if somebody cheats and bids
7, Smith is willing to go all the way up to 20 and the cheater will end up with
no gain from his deviation.

 

In
 a first-price   auction  the   bidders have a strong temptation to cheat. The
bid $p'$ that  the colluders would choose for Smith would be lower than $p'=20$,
since he would have to pay his bid, but if  $p'$
  is  anything less than the other bidders' value of 18  any one of them could
gain by  deviating to bid more than $p'$ and win.

 



 


 \newpage


\begin{center}
{\bf  Common Value Auctions and the Winner's Curse   }
\end{center}

   In a pure common-value auction, all players have the same value, but they estimate it with different errors. 

What happens if everybody bids his best estimate of the value? 


Thus: shade your bid till you expect zero profit even if you overestimated the most. 

    If Smith is more risk averse than
Brown, then Smith should be more cautious for two reasons. 

 The gamble is worth less to Smith-- the reason analyzed above in the
private-value setting. 

Also,  when Smith wins against a
rival like Brown who regularly bids more, Smith probably overestimated the
value.  

 If there is a private value component, and it is bigger for Brown than for Smith, Smith should also be extra-cautious. 
  
 \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 

\begin{center}
  {\bf  Strategies in Common-Value Auctions}
\end{center}

 


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. 

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

\begin{center}
{\bf  Signals from a Uniform Distribution    }
\end{center}


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 is 
\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}

    

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.


\newpage

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


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.  

 

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

 

\newpage

\begin{center}
{\bf  The Uniform-Signal  Common-Value Auction}
\end{center}


   

\noindent
 {\bf Order of Play:} 
0. Nature chooses  the common value for the object $v$ using the uniform density
on $[-\infty, \infty]$ (the limit of $[-x,x]$ as $x$ goes to infinity), and sends signal $s_i$  to Bidder $i$ using the uniform
distribution on $ [v- m, v+m]$.\\
1. The seller chooses a mechanism  that allocated the object and payments based
on each player's choice of $p$.   He also chooses the procedure in which bidders
select $p$ (sequentially, simultaneously, etc.). \\
  2. Each bidder  simultaneously chooses   to participate in the auction or  to
stay  out. \\
 3. The bidders  and the  seller choose value of $p$ according to the mechanism
procedure. \\
  4. The  object is allocated and transfers are paid according to the
mechanism.\\

 \newpage  

\begin{center}
{\bf  The Ascending Auction (open-exit)  }
\end{center}

 
 
\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.

 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

\begin{center}
{\bf   Seller Revenue  }
\end{center}


  The price paid by the winner will be the price at which the second-highest
bidder drops out, which is $\frac{s_{(n)} +s_{(2)}}{2}$.  


   
 \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). 

That this is an open-exit auction is crucial. Other bidders need to learn   the lowest signal. 

 \newpage

\begin{center}
{\bf    Common Values:  The   Second-Price Auction     }
\end{center}


 


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

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

On average, the  second-highest bidder actually has the signal $E s_{(2)} = v
+\left(  \frac{ n-3}{n+1} \right) m$, so 
 \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. (We  found revenue of  $(v-10)$ with $n=4$ in the
 ascending auction.) 

 If $n=2$, however,  the expected price is the same.  $v_{(n)}= v_{(2)}$, so the winning price is based on the same information in both auctions. 


\newpage

\begin{center}
{\bf Common Values:  The  First-Price Auction  }
\end{center}

 
\noindent
{\bf  Equilibrium:}   Bid $(s_i - m)   $.

\noindent
{\bf  Explanation:}    Bidder $i$ bids $(s_i-z)$ for some amount $z$ that does not
depend on his signal, because given the assumption of diffuse priors, he does
not know whether his signal is a high one or a low one. 

  Define $T_i$ to be how
far the signal $s_i$ is above its minimum possible value, $(v-m)$, so
\begin{equation} \label{e57c}
 T_i  \equiv s_i - (v - m)
 \end{equation}
and   $s_i \equiv v - m +T_i$. Bidder $i$ has the highest signal and wins the
auction if  $T_i$ is big enough, which has probability $\left(\frac{T_i}{2m}
\right)^{n-1} $, which we will define as $G(T_i)$,  because it is the
probability that the $(n-1)$ other signals are all less than $s_i = v -m +T_i$.
He earns $v$  minus his bid of $ (s_i- z)$ if he wins, which equals $(z+m-
T_i)$.

\newpage

\begin{center}
{\bf  The Epsilon Argument    }
\end{center}


 If  Bidder $i$ deviated and bid   a small amount $\epsilon$ higher,
he
would win    with a higher probability, $G(T_i+\epsilon)$, but he would lose
$\epsilon$ whenever he would have won with the lower bid.     Using a Taylor
expansion, $G(T_i+\epsilon) \approx  G(T_i) + G'(T_i) \epsilon$, so
 \begin{equation} \label{e58}
 G(T_i+\epsilon)-  G(T_i)  \approx  (n-1) T_i^{n-2}  \left(   \frac{1  }{2m}
\right)^{n-1}  \epsilon.
 \end{equation}
  The  benefit from bidding higher is the higher probability,
$[G(T_i+\epsilon)-  G(T_i)]$ times the  winning surplus $(z+m-T_i)$. The
loss from    bidding higher is that the bidder  would pay an additional
$\epsilon$ in the $\left(   \frac{T_i }{2m} \right)^{n-1}$ cases in which he
would have won anyway.

\newpage

\begin{center}
{\bf A Bidder's Optimal Strategy    }
\end{center}


 In equilibrium, he is indifferent about this
infinitesimal deviation, taking the expectation across all possible values of
his ``signal height'' $T_i$,    so
 \begin{equation} \label{e59}
\int_{T_i=0}^{2m}  \left[ \left( (n-1) T_i^{n-2}  \left(   \frac{1  } {2m}
\right)^{n-1}  \epsilon   \right)   \left( z+m-T_i  \right)   - \epsilon \left(
\frac{T_i }{2m} \right)^{n-1}  \right] dT_i =0.
 \end{equation}
 This implies that
 \begin{equation} \label{e60}
\begin{array}{l}
\epsilon  \left(   \frac{1 }{2m} \right)^{n-1}     \int_{T_i=0}^{2m} \left[
\left( (n-1) T_i^{n-2} \right)   \left( z+m  \right) \right. \\
 \left.   -  (n-1) T_i^{n-1} -
T_i^{n-1}  \right] dT_i =0.
\end{array}
\end{equation}
which in turn implies that
 \begin{equation} \label{e61}
\begin{array}{l}
{\displaystyle  \epsilon \bigg(   \frac{1}{2m} \bigg)^{n-1}   
  \bigg|_{T_i=0}^{2m}  \bigg( T_i^{n-1}       \left( z+m \right)}  \\
\left.  -    T_i^n  \right) =0 ,\\
\end{array} 
 \end{equation}
so $ (2m)^{n-1}(z+m) -(2m)^n -0+0 =0$ and  $z=m$. Bidder $i$'s optimal strategy
in the symmetric equilibrium is to bid $p_i = s_i - m$.


\newpage

\begin{center}
{\bf  Winning Bid and Expected Revenue   }
\end{center}


 The winning bid is set by the bidder with the highest signal, and that highest
signal's expected value is
\begin{equation} \label{e62}
 \begin{array}{lll} E s_{(1)}& = &   \underline{s}  + \left(\frac{ n+1-1}{n+1}
\right) \left(    \overline{s}- \underline{s}  \right)\\
 & & \\
&= &  v-m +
\left(\frac{ n }{n+1}\right) \left(2m  \right)\\
\end{array}
 \end{equation}
 The expected revenue is therefore
 \begin{equation} \label{e63}
 E p_{(1)} =       v -  \left(1 \right)  \left(  \frac{  1}{n+1} \right)2 m.
  \end{equation}
 If $m=50$ and $n=4$, then
\begin{equation} \label{e64}
E p_{(1)} =v -    \left( \frac{1}{5}  \right) (100) = v-20.
\end{equation}
 Here, the revenue is even lower than in  the second-price auction, where it was
$(v-15)$ (and the revenue is lower even if $n=2$).    


 

 Revenue ranking: Ascending highest, then second-price, then first-price. 

 



 \newpage

 



\begin{center}
 {\bf The Wallet Game }
\end{center}
 

\noindent
 {\bf Order of Play}\\ 
 (0) Nature  chooses the amounts $s_1$ and $s_2$ of the money in each  player's
wallet  using density functions $f_1(s_1)$ and $f_2(s_2)$. Each player observes
only his own wallet's contents.  \\
 (1)  Each player chooses a bid ceiling   $p_1$ or $p_2$.   An auctioneer
auctions off the two wallets by gradually raising the price until  either $p_1$
or $p_2$ is reached.

\noindent
  {\bf Payoffs:} \\
The player who bids less has a payoff of zero. The winning player pays the bid
ceiling of the loser and hence  has a payoff
of
\begin{equation} \label{e64a}
  s_1+s_2 -  Min (p_1, p_2)
\end{equation}

  


A symmmetric equilibrium is for Bidder $i$ to choose bid ceiling $p_i = 2s_i$.


This is an equilibrium because if he wins at exactly that price,  Bidder $j$'s
signal must be $s_j   = s_i $ and the value of the wallets is    $2s_i$.


If  Bidder $i$ bids any lower,  he might pass up a chance to buy the wallet for
less than its value. 

If he bids any higher,  he would only  win  if $p>2s_j$
too, which implies that $p>s_i+s_j$.

\newpage

 
\begin{center}
{\bf   Affiliation  }
\end{center}

 
 
\noindent
 {\it Definition.} The signals $x_1$ and $x_2$ are  affiliated  if for all
possible realizations  $Small < Big$ of $x_1$  and $Low < High$ of $x_2$, the
joint probability  $f(x_1, x_2)$ is such that
$z_1$ and $z_2$,
 \begin{equation} \label{e65}
\begin{array}{l}
 f(x_1=Small, x_2=Low) f(x_1=Big, x_2=High)\\
 \\ 
  \geq  f(x_1=Small, x_2=High)f(x_1=
Big, x_2=Low).
\end{array}
\end{equation}

   Thus, affiliation says that  the probability  the values of $x_1$ and $x_2$
move in the same direction is
greater than the probability   they move oppositely. 

 

 The   implication of two signals being affiliated is that the expected value
of the winning bid conditional on the signals is  increasing in all the signals.


When one signal rises, that has the positive direct effect of increasing the
bid of the player who sees it, and non-negative indirect effects  once the other
players see his bid increase and deduce that he had a high signal.

\newpage

\begin{center}
{\bf   The Monotone Likelihood-Ratio Property  }
\end{center}


 The Monotone Likelihood-Ratio Property  is the
same thing  expressed in terms of the conditional densities, the posteriors.


 \noindent
 {\it Definition.} The conditional probability $g(x_1|x_2)$ satisfies the
Monotone
Likelihood Ratio Property  if  the likelihood ratio is weakly decreasing in
$x_1$, that is,  for all  possible realizations  $Small < Big$ of $x_1$  and
$Low < High$ of $x_2$,
 \begin{equation} \label{e70}
   \frac{ g(Big| Low)} {g(Big| High)}  \leq  \frac{ g(Small| Low)} {g(Small|
High)}.
\end{equation}

 The Monotone Likelihood Ratio Property says that as $x_2$  goes from Low to
High,  the   Big value of $x_1$ becomes relatively more likely.

 It can be shown
that this implies that for any value $z$,    the conditional cumulative
distribution of $x_1$ up to $x_1=z$ given $x_2$ weakly increases with $x_2$,
which is to say that the distribution $G(x_1|x_2)$  conditional on a larger
value of $x_2$ stochastically dominates the  distribution conditional  on a
smaller value of $x_2$.

 
    
\newpage


\begin{center}
{\bf   The Linkage Principle:   }
\end{center}


 
  {\bf  The linkage principle}:  when the amount of   affiliated
information available to bidders increases, the equilibrium sales price  becomes
greater.  

The seller should have a policy of disclosing    any affiliated
information  he possesses.

Auction rules which reveal affiliated
information in the course of the auction (e.g., open-exit auctions)  or use it
in determining the winner's payment  (e.g., the second-price auction) will
result in
higher prices. 

 
   
 
 \end{LARGE}

\end{document}
