Download Economics for Business

Economics for Business
Problem Set 4
Due Date: 8th December, 2005
1. Consider the following game matrix.
Player B
Player A
Left
Right
Top
a,b
c,d
Bottom
e,f
g,h
(a) If (top, left) is a dominant strategy equilibrium, then we know that a>_e____, b>__d____,
___c__>g, and ___f___>h..
(b) If (Top, left) is a Nash equilibrium, then which of the above inequalities must be satisfied?
________a>e and b>d______________________
(c) If (top, left) is dominant strategy equilibrium must it be a Nash equilibrium? Why?
Yes. This is because in dominant strategy equilibrium, no one will ever deviate no matter what the
other players does. Hence it must be an optimal response given opponent choose the dominant
strategy, hence a Nash equilibrium.
2. Consider the following game: I flip a coin, and while it is in the air, you call either heads or
tails. If you call the coin correctly, you get to keep the coin.
(1)
Suppose you know that the coin always comes up heads. What is the best strategy for
you to pursue?
Always call Heads.
(2)
Suppose that the coin is unbalanced and comes up heads 80% of the time and tails
20% of the time. Now what is your best strategy?
Call Heads brings you 80%x1+20%x(-1)=0.6. Call Tails gives you 0.2x1+0.8x(-1)=-0.6. So
best strategy is to call Heads.
(3)
What if the coin comes up heads 50% of the time and tails 50% of the time? What is
your best strategy?
In this case call Heads or Tails will yield equal expected payoff, so call Heads or Tails are all
best strategy.
(4)
Now suppose that I am able to choose the type of coin that I will toss (where a coin’s
type is the probability that it comes up heads), and that you will know my choice.
What type of coin should I choose to minimize my losses?
You should choose the coin with equal probability comes up heads and tails, i.e., the coin in
(3).
(5)
What is the Nash mixed strategy equilibrium for this game?
The Nash mixed strategy equilibrium in this game is you choose a balance coin and I call
Heads or Tails with one half chance.
3. Consider the following payoff matrix.
Kicker
Goalie
Left
Right
Jump Left
1,0
0,1
Jump Right
1-p,p
1,0
Now the probability that the kicker will score if he kicks to the left and the goalie jumps to the
right is p. We will want to see how the equilibrium probabilities change as p changes.
(1)
If the goalie jumps left with probability  G , then if the kicker kicks right, his
probability of scoring is: _  G ___.
(2)
If the goalie jumps left with probability  G , then if the kicker kicks left, his
probability of scoring is:__(1-  G )p__.
(3)
Find the probability  G that makes kicking left and kicking right lead to the same
probability of scoring for the kicker.
 G =(1-  G )p =>  G =p/(1+p)
(4)
If the kicker kicks left with probability
 K , then if the goalie jumps left, the
probability that the kicker will not score is: __  K __
(5)
If the kicker kicks left with probability  K , then if the goalie jumps right, the
probability that the kicker will not score is__  K (1-p)+(1-  K )__.
(6)
Find the probability
 K that makes the payoff to the goalie equal from jumping left
or jumping right.
 K =  K (1-p)+(1-  K ) ==>  K =1/(1+p)
(7)
The variable p tells us how good the kicker is at kicking the ball into the left side of
the goal when it is undefended. As p increases, does the equilibrium probability that
the kicker kicks to the left increase or decrease? Explain why this happen in a way
that even a TV sports announcer might understand.
As p increase, the probability that the kicker kicks to the left decreases. This is because higher p
implies the goal keeper will be more likely to jump to the left. This in turn implies the kicker
should kick to the right more often.
4. Find the mixed-strategy Nash equilibrium of the following game.
Left
Right
Top
2,1
0,2
Bottom
1,2
3,0
Let p denote the probability to choose Left, then (1-p) is the probability to choose Right. Similarly,
let q denote the probability to choose Top, and (1-q) is the probability to choose Bottom.
Mixed-strategy requires the expected payoff from choosing Left and Right is equal, i.e.
q  1  (1  q)  2  q  2  (1  q)  0  q 
2
3
Similarly we have
p  2  (1  p)  0  p  1  (1  p)  3  p 
3
4
5. Game theory and the Nash equilibrium concept can be used to study the interaction between
criminals and law-enforcement agencies. Gary Becker, a Nobel Prize-winning economist, led
the way on this kind of research and showed that economic analysis is extremely useful in this
policy arena. Now consider the game between police and thief and police. The police selects a
level of law enforcement, which is a number x  0 . The thief selects a level of crime, y  0 .
These choices are made simultaneously and independently. The police’s payoff is given by
u G   xc 4  y 2 / x with the interpretation that  y 2 / x is the negative effect of crime on
4
society and c is the cost of law enforcement, per unit of enforcement. The number c is a
positive constant. The thief’s payoff is given by uT  y
that y
1/ 2
1/ 2
/(1  xy) , with the interpretation
is the probability that the thief evades captures.
(1)
Write the first-order conditions that define the player’s best-response functions and
solve them to find the best-response functions. Graph the best-response functions.
For the police, its first order condition is dUg/dx=0, i.e.
 c 4  y 2 / x 2  0 , or  x 2 c 4  y 2  0 , or x  y / c 2 .
For the thief, its first order condition is dUt/dy=0, i.e.
1 1 / 2
y (1  xy) 1  xy1 / 2 (1  xy)  2  0 , or y=1/x.
2
y
x  y / c2
y  1/ x
x
(2)
Compute the Nash equilibrium of this game.
From (1), we can find out that x=1/c, and y=c.
The results shows that the higher of the law enforcement cost, the higher the crime level and the
lower the effort put by police.