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\begin{center}

 \begin{Large}

 {\bf   Biased Experts in a Sender-Receiver Model  }  \\

\end{Large}


\bigskip



 \bigskip

November 11,  2008   \\

\bigskip

Eric Rasmusen \\

\bigskip

{\it Abstract}

\end{center}

  \vspace*{-12pt}


  If  someone biased is using an intermediary to collect and convey
information, he does best by  not being able to  secretly convey his
bias to the intermediary. He may do just as badly if there is a chance
he might be caught imposing the bias, and he does best if he is caught
with probability one.


\bigskip

\begin{small}

\noindent
  \hspace*{20pt} Dan R. and Catherine M. Dalton Professor, Department
of
Business
Economics and Public Policy, Kelley School of
Business, Indiana University.
Erasmuse@indiana.edu.   \url{http://www.rasmusen.org}. This paper:
 \url{http://www.rasmusen.org/papers/bias-rasmusen.pdf}.

\noindent
Keywords: sender-receiver game, bias, experts.

\noindent
 I thank Marco Ottaviani for helpful discussion.


\end{small}

\newpage




\noindent
  {\bf  1.   Introduction}







 In a sender-receiver game, the sender knows some item of information,
a number
$x$,  and sends a message, possibly false, about it to the receiver,
who
takes an action. The sender would like the receiver to believe that
the value of the information is $x+k$ and the receiver knows that the
sender has that motivation. The receiver therefore does not trust the
sender's message, and ignores it. This is worse for both receiver  and
sende than if the sender were forced to send only true messages.

   One solution to the problem is for the pair of players to hire a
third person, whom we might call an expert, to discover the
intormation and send the message. (Equivalently, the sender could at
some cost show the agent the information directly.) It is important
that the expert be unbiased and truthful, however. If the sender
chooses the expert and chooses an expert with bias or bribes a neutral
expert, the receiver will ignore what the expert says , just as he
ignored the sender himself.  Then, the sender incurs the cost of the
expert and the cost of the bribe uselessly. The sender would prefer
not to be able to bribe an expert.

  In this note, I explore  a version of this game. The difference is
that the sender may  be able to secretly bribe the expert, but there
is some probability that the receiver discovers whether the agent has
been bribed or not.

 More generally, this would apply whenever the sender tries to make
his message more credible at some cost, so long as his method might be
duplicitous.


\bigskip

\noindent
  {\bf  2. The Model}

 The two players are the sender and the receiver. In addition, Nature
and a hired expert  make  deterministic  choices. The order of play is
as follows, where $x$, $b$, $m$, $k$,   and $a$ are scalars.


(0) Nature chooses the value of   $x$.   The sender and  receiver
do  not know $f(x)$,
and place
a diffuse prior with mean 0 on the distribution of $x$.


(1)The sender pays a bribe of   $b$ to the expert for him to bias his
message upwards by amount $k$,  or does not
pay it.

(2) If the sender pays a bribe, the agent observes the bribery  with
probability  $\theta$.  If the sender does not offer a bribe, the
agent observes the ``restraint''  with probability $\gamma$ (for
discussion of this
odd statement, see below) .

(3)   If the sender does not bribe him, the expert  sends the
message $m=x$.  If the sender does bribe him, he sends the message
$m =x+u$.

(4) The receiver picks an action $a$ after observing the message.





The sender's payoff   is
\begin{equation}\label{e1}
\pi_p =  - (x+k- a)^2 -  I(b) b,
\end{equation}
where $I(b) $ an indicator variable that equals 1 if a bribe is paid
and 0 otherwise.

The receiver's payoff   is
\begin{equation}\label{e1}
 \pi_a =  - (x - a)^2
\end{equation}



\bigskip

Discussion

 The payoff functions have quadratic loss from the action not being at
the player's ideal point, but quadratic loss is unimportant to the
qualitative results.

We do not need to specify a payoff function for the expert, since he
reacts very  simply to the sender's action.

 In this model  the sender does not know the
information before he decides whether to bribe the expert. If he does,
we are in the world of Lanzi \&  Mathis (2004).

    The assumption that  ``If the sender does not offer a bribe,
the agent observes this with probability $\gamma$''  allows for the
possibility that the agent might or might not become  sure that the
sender did not pay a bribe. We might imagine that the agent posts a
guard on the expert to see if a bribe is paid, but the guard falls
asleep (and is honest enough to admit it) with probability $1-\gamma$.
If this scenario seems uninteresting, set $\gamma=0$  for the rest of
the paper.

 I do not have eq. 3 well in mind. If the sender soetimes pays a
bribe, if he can make t he message as big as he likes, he will make it
bigger than x+k, knowing the receiver will scale it back. Ah-- that's
not hard to work in. It means that the agent WILL choose $a=x+k$ when
the expert is bribed and the bribery is not detected. Or will it?




\bigskip


\noindent
{\bf Equilibrium}

 In the table below the action is shown conditional on whether the
receiver observes a bribe being paid, is sure no bribe was paid, or is
uncertain. There are three  kinds of equilibria. In a Restraint
Equilibrium, the sender does not attempt to bribe the expert. In a
Bribery Equilibrium he bribes him with probability one. In a Mixed
Equilibrium, he bribes him with some probability between zero and one.

\begin{tabular}{l |llll}
 Equilibrium &     $a$|bribe & $a$|restraint  & $a$|nothing & Payoff
\\
 & & & & \\
\hline
 & & & & \\
(1) Restraint &    $m-k$    &   m &  m  &$-k^2$  \\
 & & & & \\
 (2)Bribery &    0  &   $x$  & 0 &$-k^2-b$  \\
 & & & & \\
(3) Mixed  &   $m-k$   &   $x$ &  $m-z $ &$-(1-\theta) z^2 -  \theta
k^2- b$  \\
 & & & &\\
\hline
\end{tabular}

The mixed-strategy equilibrium payoff is between the other two. That
makes sense.

What is interesting is that there can be multiple equilibria,  maybe
(?) Pareto ranked. Suppose receivers think the sender will bribe. Then
he will-- the Bribery equilibrium. Suppose  receivers think the sender
will maybe bribe. Then we can find ourselves in the mixed eq. This
needs to be shown.







  There  usually exists  a babbling equilibrium in sender-receiver
games. Here  there does not, because it is not the sender who sends
the message directly, but the expert, who tells the truth even if he
knows the receiver will ignore the message.

We will analyze each equilibrium in turn and see when multiple
equilibria exist.






\bigskip

\noindent
{\bf The Restraint Equilibrium}


The sender's expected payoff in equilibrium is
\begin{equation}\label{e1}
\pi_p =  - ( x+k  - x)^2 = -k^2
\end{equation}

 If the sender deviated  and paid the bribe  his expected payoff
would become
\begin{equation}\label{e1}
\pi_p =      - (1-\theta)( x+k - (x+k))^2  - \theta ( x+k  - x)^2 - b
= -\theta k^2 -b.
\end{equation}

Deviation is not profitable if
\begin{equation}\label{e1}
 - k^2 \geq  -\theta k^2 -b.
\end{equation}
i.e.
\begin{equation}\label{e1}
 b \geq (1- \theta) k^2
  \end{equation}



\bigskip

\noindent
{\bf The Bribery Equilibrium}


The sender's expected payoff in equilibrium is
\begin{equation}\label{e1}
\pi_p =  - (x+u_1- (x+k-k))^2 -    b = -k^2-b
\end{equation}

 If the sender deviated  and did not pay the bribe  his expected
payoff would become
\begin{equation}\label{e1}
\pi_p =  - (1-\gamma) (x+k- (x-k))^2  - \gamma  (x+k -x)^2   =
-4(1-\gamma)k^2 - \gamma k^2.
\end{equation}

Deviation is not profitable if
\begin{equation}\label{e1}
  -k^2-b  \geq -4(1-\gamma)k^2 - \gamma k^2,
\end{equation}
i.e. if
\begin{equation}\label{e1}
  [4   -4\gamma) - 1+ \gamma] k^2 \geq  b
\end{equation}
i.e. if
\begin{equation}\label{e1}
  b \leq  3(1  -  \gamma)   k^2
\end{equation}


\bigskip

\noindent
{\bf  The Mixed Equilibrium}

  The sender pays the bribe with probability $p$.
If the agent observes neither bribery nor restraint, he chooses action
$m-z$.






The sender's expected payoff in equilibrium  must equate his two pure-
strategy payoffs, so
\begin{equation}\label{e1}
\begin{array}{lll}
\pi_p (bribery) & =  - (1-\theta)(x+k- (x+k-z))^2 -  \theta (x+k-
(x+k-k)^2-  b \\
& = \pi_p (restraint) &= - (1-\gamma)(x+k- (x -z))^2 -  \gamma (x +k -
 x )^2 .  \\
\end{array}
\end{equation}



 This means that
\begin{equation}\label{e1}
  (1-\theta) z^2 +  \theta k^2- b =
    (1-\gamma)( k+z )^2 +  \gamma  k^2
\end{equation}
so
\begin{equation}\label{e1}
     0=
    (1-\gamma)k^2  - 2(1-\gamma)kz   +(1-\gamma) z  ^2   - (1-\theta)
z^2  +  \gamma  k^2  -\theta  k^2 -b
\end{equation}

By the quadratic formula,
\begin{equation}\label{e1}
z =
\end{equation}


The receiver's utility from choosing $a= m-z$ must be maximal for  the
$z$ we just found, which means the expected value of $x$ must be $m-z$
when the receiver observes nothing.

\begin{equation}\label{e1}
E(x) = p (x+k) + (1-p)x =z.
\end{equation}

Solving this yields
\begin{equation}\label{e1}
  p  x+pk  + x  -px   =z
\end{equation}
so
\begin{equation}\label{e1}
 pk      =z  +x
\end{equation}
so
so
\begin{equation}\label{e1}
 p      = \frac{z  +x}{k}
\end{equation}

 It is interesting that $p$ does not depend on $\theta$ or $\gamma $
directly, only through $z$.




\bigskip

\noindent
  {\bf  3. Extension: Making the Message Distortion Endogenous}


Now let us think what happens when the message distortion is not fixed
at $k$, but can take any value $u$. We will first assume that if the
bribery is observed, so is $u$, and then ask what happens when $u$ is
never observed.

xxx

\begin{tabular}{lllll}
 Equilibrium &  $u$(bias)  &  $a$|bribe & $a$|restraint  & $a$|nothing
\\
 & & & &\\
\hline
 & & & &\\
(1) Restraint &  $u_1$  &  $m-u_1$    &   m &  m  \\
 (2)Bribery &   ---  &  0  &   $x$  & 0  \\
(3) Mixed  &  $u_3$ &  $m-u_3$   &   $x$ &  $m $  \\
 & & & &\\
\hline
\end{tabular}



\bigskip



\noindent
{\bf The Restraint Equilibrium}


The sender's expected payoff in equilibrium is
\begin{equation}\label{e1}
\pi_p =  - ( x+k  - x)^2 = -k^2
\end{equation}

 If the sender deviated  and paid the bribe  his expected payoff
would become
\begin{equation}\label{e1}
\pi_p =      - (1-\theta)( x+k - (x+k))^2  - \theta ( x+k  - x)^2 - b
= -\theta k^2 -b.
\end{equation}

Deviation is not profitable if xxx


Out of equilibrium, what happens if the sender does pay a bribe? He
will follow this deviation by choice of $u_1$ as a bias.




\bigskip

\noindent
{\bf The Bribery Equilibrium}

 How much is the bias? If it were always k, then the sender would
deviate to 2k.  But it always has to be the same. Or does it? Does he
have ot mix ovef the bias?

 The receiver has to ignore the message in making his choice.

 I'm very confused now.

The sender's expected payoff in equilibrium is
\begin{equation}\label{e1}
\pi_p =  - (x+u_1- (x+k-k))^2 -    b = -k^2-b
\end{equation}

 If the sender deviated  and did not pay the bribe  his expected
payoff would become
\begin{equation}\label{e1}
\pi_p =  - (1-\gamma) (x+k- (x-k))^2  - \gamma  (x+k -x)^2   =
-4(1-\gamma)k^2 - \gamma k^2.
\end{equation}

Deviation is not profitable if xxx

\bigskip

\noindent
{\bf  The Mixed Equilibrium}

  The sender pays the bribe with probability $p$.
If the agent observes neither, he chooses action $m-z$, where

The sender's expected payoff in equilibrium is
\begin{equation}\label{e1}
\begin{array}{lll}
\pi_p (bribe) & =  - (x+k- a)^2 -  I(b) b \\
& = \pi_p (no\; bribe) &= - (x+k- a)^2 -  I(b) b \\
\end{array}
\end{equation}

 If the sender deviated  and paid the bribe  his expected payoff
would become
\begin{equation}\label{e1}
xs
\end{equation}

\bigskip



\begin{equation}\label{e1}
sdf
\end{equation}


\begin{equation}\label{e1}
sdf
\end{equation}


\begin{equation}\label{e1}
sdf
\end{equation}



\begin{equation}\label{e1}
sdf
\end{equation}



\noindent
{\bf 9. Concluding Remarks}



\newpage




\noindent
{\bf References}

Crawford, P. V. and J. Sobel (1982), “Strategic Information
Transmission,”
Econometrica, 50, 1431-51.

Gilligan, T. W. and K. Krehbiel (1989), “Asymmetric Information and
Legislative
Rules with a Heterogeneous Committee,” American Journal of Political
Science, 33, 459-90.

Krishna, V. and J. Morgan (2001), “Asymmetric Information and
Legislative
Rules: Some Amendments,” American Political Science Review, 95,
435-52.

Lanzi , Thomas  and Jérôme Mathis
 Consulting an Expert with Potentially Conflicting Preferences
Journal	Theory and Decision
Publisher	Springer Netherlands
ISSN	0040-5833 (Print) 1573-7187 (Online)
Issue	Volume 65, Number 3 / November, 2008
DOI	10.1007/s11238-007-9070-2
Pages	185-204

Ottaviani, M. and P. N. Sørensen (2005), “Professional Advice,”
Journal

Argumentation in Sender-Receiver Games (2004)
Lanzi , Thomas  and Jérôme Mathis
September 16, 2004.
Abstract


B.L. LIPMAN and D.J. SEPPI, Robust Inference in Communication
Games with Partial Provability, J. Econ. Theory 66 (1995), 370-405.
[10] P. MILGROM, Good news and bad news: representation theorems and



D.J. SEIDMAN and E. WINTER, Strategic Information Transmission
with Veri…able Messages, Econometrica 65 (1997), 163-170.
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