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\begin{center}
{\bf G751, Spring 2011: Test 1}
\end{center}


 There are 100 points in total. Please start each of the 3  questions on a new sheet of paper.
This is a  closed-book exam.
 \bigskip


\noindent
Question 1 (30 points).
\begin{center}
\noindent
{\bf   Finding Equilibria}\\
 \begin{tabular}{lllccccc}
  &       &             &\multicolumn{5}{c}{\bf Operating System Company}\\
  &       &   &            {\it  Cheap}   &   &  $Moderate$   & &$Careful$   \\
&   && & &  & \\
   &   &  $ Lots$                &  0,0 &    &  $-1,0$&  & $3,4$\\
 &  && & &  & \\
& {\bf Appwriter:} &    {\it Medium }    &  $ 3,1$ &  &0,-2  & & $2,-1$ \\
  &   && & &  & \\
&  &    {\it Few }    &  $ 0,1$ &  &1,-1  & & $1,-2$ \\
\end{tabular}
\end{center}



(a) How far can you get using iterated deletion of weakly dominated strategies?

 
\underline{  {\it Answer.}}Moderate is dominated by Cheap for the  OS Company, so it can be dropped. Few is then dominated by
Lots for the Appwriter and can be dropped.


(b)  What are the pure-strategy  Nash equilibria?

 
\underline{  {\it Answer.}}(Medium, Cheap) and (Lots, Careful).

(c) What is the mixed-strategy Nash equilibrium?

 
\underline{  {\it Answer.}}  Appwriter choose Medium with probability  $\gamma$  and Lots  with probability $(1-\gamma)$. OS
Company chooses Cheap  with probability $\theta$ and Careful with probability $1-\theta$.
\begin{equation}
\pi_{App} (Medium) =   \theta  ( 3)   + (1-\theta) ( 2)     = \pi_{App} (Lots)   =\theta (0 )   + (1-\theta) (3 )  
\end{equation}
 so $ 3 \theta + 2 - 2\theta =3- 3\theta$ and $4\theta=1$ so $\theta = .25$. 
\begin{equation}
\pi_{OS} (Cheap) =   \gamma (1 )   + (1-  \gamma) ( 0)     = \pi_{OS} (Careful)   =  \gamma (-1 )   + (1-  \gamma) (4 )  
\end{equation}
 so $\gamma =  -\gamma + 4 - 4\gamma$ and  $ 6 \gamma =4$ so $\gamma = 2/3$. 

\noindent
Question 2 (30 points).
The ability of the VP of a company takes an integer value from 1 to 10.
Initially, the CEO and the VP   both think all values are equally probable.
Then, they each see how the VP performed in designing a new business plan, and their information
partitions become different.  We assume, following Harsanyi, that each player knows the other
player's partition, though not the particular state of the world that has been reached.

The CEO's information partition is (1,2)  (3) (4,6) (5,7)   (8) (9) (10). The VP's is
  (1,2,3) (4,5,6) (7) (8) (9,10).

 
(a) What is the finest common coarsening of the two information partitions?

 
\underline{  {\it Answer.}}  (1,2,3) (4,5,6,7) (8)  (9,10)

  
 
 (b) Suppose we define the VP's ability to be  high if it is  5,6,7, 8, 9 or 10.
For what abilities is his high ability common knowledge?


 
\underline{  {\it Answer.}} For 8, 9 or 10.

 
(c)For what abilities do both players know that the VP's ability is high?

 
\underline{  {\it Answer.}}  For abilities 7,8, 9, and  10. 





\noindent
Question 3 ( 40 points). There are two types of workers. Type 1 workers are lazy, with a cost of  1
for high
effort, and
type
2 workers are even lazier, with a cost of 2 for high effort.  Each type is half of the population.
The boss must decide each day with what probability to inspect  a worker's production, which costs
the boss 1. The boss and worker simultaneously choose effort and whether to inspect.  The boss earns
value 4 from high effort production and 0 from low effort production.
If he find the     worker has not exerted high effort, the boss fires the worker with no pay,.   If the
boss does not inspect or finds that effort was high, the boss pays the worker 3. The boss's payoff
is the value of production minus his effort cost and the amount the worker is paid. The worker's
payoff is his pay minus his effort cost.  The equilibrium
will
involve mixed strategies. Denote the type 1 worker's probability  of  high effort as $\theta_1$ and
the
type 2's as $\theta_2$, and the boss's probability of inspecting by $\gamma$.



(a) Draw the extensive form, putting the boss's decision before the worker's.

\includegraphics[width=6in]{test1-extform.jpg}

(b) Why are there no   equilibria in which mixed strategies are not used by  at least one player?

 
\underline{  {\it Answer.}} If the boss always inspects, then  both types of worker would always choose high effort, yielding
payoffs of 1 for the boss. But then the boss could deviate to Don't Inspect and raise his payoff to
2 by avoiding the cost of inspection. If the boss always chooses not to inspect, then the workers will always choose low effort, in which case the boss could increase his payoff by deviating to Inspect.  Thus it is clear that the boss, at least, will  use
mixed strategies in any Nash equilibrium. 

 The way to answer a question like this is to show that some player would deviate from the hypothesized equilibrium.  An equilibrium in which at least one player   does not use mixed strategies is one in which the boss does not use  pure strategies  and   the worker  does not use pure strategies.  Thus, start with the possible strategy profiles in which all the players use pure strategies, and show that in each of them at least one player has incentive to deviate.  If even one player deviates, then the profile is not an equilibrium, so it is enough to show that the boss would deviate from any profile in which he plays a pure strategy and the worker plays the worker's best response. 


(c)  What would be each player's pure-strategy payoff as a function of the other player's mixing probabilities?  

 
\underline{  {\it Answer.}}For the worker  the payoff functions would be
$$
 \pi_1(high) = -1 +  3 = \pi_1(low)=  \gamma (0) + (1-\gamma) (3)
$$


$$
 \pi_2(high) = -2 +  3 = \pi_2(low)=  \gamma (0) + (1-\gamma) (3)
$$


Note that no  one value of $\gamma$ can make both of these payoff equations true. Thus, one type of
worker has to play a pure strategy.   Either the Type 1 workers will always choose high effort, or the Type 2 workers will always choose low effort.  

$$
 \pi_{boss}(inspect)  = -1 + .5 (\theta_1 (4-3)+(1-\theta_1) (0)) + .5( \theta_2 (4-3) + .5 +
.5(1-\theta_2) (0)) = -1 + .5 \theta_1  + .5 \theta_2
$$

 $$
 \pi_{boss}(not\; inspect)  =   .5 (\theta_1 (4-3) +.5(1-\theta_1) (-3)) + .5 (\theta_2 (4-3) + 
.5(1-\theta_2) (-3)) =  2 \theta_1   + 2 \theta_2  - 3  
$$

I gave full credit if you instead answered with the players'  overall payoff  as a function of his  own mixing probability too. 


(d) What are the equilibrium strategies?


 
\underline{  {\it Answer.}} The key is to realize that since no one  boss's mixing probability $\gamma$ can equate both the
Type 1 and the Type 2 worker's payoffs, it must be that one type of worker plays a pure strategy. It
could be that the Lazy type 1 always chooses high effort, or that the Very Lazy Type 2 always
chooses low effort. The boss has   the very low payoff  of -3  when the worker is not inspected and chooses low effort. Let's guess that that means he will inspect enough that both types of workers have positive probability of high effort--- which means the  Type 1 workers will choose high effort with probability 1. 


If we set $\theta_1=1$, then we can equate the boss's payoffs thus:
 $$
 \pi_{boss}(  inspect)  =   -1 +  .5+  .5 \theta_2 = \pi_{boss}(not\;  inspect)  =
   2 + 2 \theta_2  -3
$$
 
  This results in  $ -.5 + .5 \theta_2=  -1+ 2 \theta_2 $ and $ \theta_2=1/3$.  

Using Type 2's equating of payoffs,  
$$
 \pi_2(high) = -2 +  3 = \pi_2(low)=  \gamma (0) + (1-\gamma) (3)
$$
  This yields $1= 3-3\gamma$ so $\gamma=2/3$. 

 The equilibrium is thus for all of the Type 1 workers to choose high effort,  for .35 of the Type 2
workers to, and for the boss to inspect with probability 2/3.

 What about the possibility of an equilibrium with $\theta_2=1$ and with type 1 workers mixing? If we set $\theta_2=1$ and try to equate the boss's pure-strategy payoffs we get: 
$$
 \pi_{boss}(inspect)  =   -1 + .5 \theta_1  +  0
=
 \pi_{boss}(not\; inspect)  =    =  2 \theta_1   + 2(0)  - 3  
$$
 This yields $ 2 = 1.5\theta_1$ and $\theta_1=4/3$, which is impossible.  The problem is that if the type 2 workers always choose low effort,  the boss will respond with $\gamma=1$, always inspecting, even if all of the type 1 workers high choose effort. 

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