Game Theory with Applications to Finance and Marketing

Game Theory with Applications to Finance and
Marketing
Lecture 0: A Quick Introduction and Some Examples
Instructor: Chyi-Mei Chen
Room 1102, Management Building 2
(02) 3366-1086
(email) cchen@ccms.ntu.edu.tw
1. Definition 1. A game can be described by:
• Who are the players
• What strategies are available to each player
• What does each player get given players’ choices of strategies
A game described by (i) the set of players, (ii) the strategies available
to each player, and (iii) the payoff of each player as a function of
the vector of all players’ strategic choices is called a game depicted in
normal form. A game described in normal form is called a strategic
game.
A game can also be described in extensive form, which needs to specify
the timing of players’ moves and what they know when they move, in
addition to the things specified in normal form. A game described in
extensive form is called an extensive game.
2. Example 1. The following is a two-player normal-form game.
player 1/player 2 L R
U 0,1 -1,2
D 2,-1 -2,-2
• Who are the players Players 1 and 2.
• What strategies are available to player 1 Player 1 has two pure
strategies, U and D, and player 2 has two pure strategies, L and
R. A mixed strategy for player 1 is a probability distribution over
1

U and D, and a mixed strategy for player 2 is a probability distribution
over L and R. For example, playing U and D with probabilities
1 and 1 is one mixed strategy for player 1, and playing
2 2
U and D with probabilities 1 and 2 is another mixed strategy for
3 3
player 1. Apparently, each player has an infinite number of different
mixed strategies. In its broad sense, mixed strategies include
pure strategies.
• What does each player get given players’ choices of pure strategies
Players 1 and 2 get respectively 0 and 1, if the vector of the two
players’ pure strategies (called a strategy profile) is (U, L); that is,
player 1 plays U and player 2 plays L. Let us write
Similarly, we have
u 1 (U, L) = 0, u 2 (U, L) = 1. (1)
u 1 (U, R) = −1, u 2 (U, R) = 2, u 1 (D, L) = 2, (2)
u 2 (D, L) = −1, u 1 (D, R) = −2, u 2 (D, R) = −2. (3)
The functions u 1 (·, ·) and u 2 (·, ·) are referred to as the two players’
payoff functions. Now, what are the two players’ payoffs when,
respectively, player 1 plays U and D with probabilities 1 2 and 1 2
and player 2 plays L and R with probabilities 1 3 and 2 3 Here
we assume that players are “expected-utility” maximizers, so that
player 1’s payoff is
1
2 · 1
3 · u 1(U, L) + 1 2 · 2
3 · u 1(U, R)
+ 1 2 · 1
3 · u 1(D, L) + 1 2 · 2
3 · u 1(D, R),
and player 2’s payoff is
1
2 · 1
3 · u 2(U, L) + 1 2 · 2
3 · u 2(U, R)
+ 1 2 · 1
3 · u 2(D, L) + 1 2 · 2
3 · u 2(D, R).
2

3. Example 2. Consider the following Cournot game: firms 1 and 2
producing the same product must compete in supply quantities. The
inverse demand curve is
where
P (Q) = 1 − Q,
Q = q 1 + q 2
is the total supply of the product. For simplicity, assume that firms
have no costs.
• Who are the players Firms 1 and 2.
• What strategies are available to player 1 Any non-negative real
number q 1 , standing for the supply quantity chosen by firm 1, is
a pure strategy for firm 1. Any cumulative distribution function
F 1 : [0, +∞) → [0, 1] is a mixed strategy for firm 1. By symmetry,
any non-negative real number q 2 , standing for the supply quantity
chosen by firm 2, is a pure strategy for firm 2, and any cumulative
distribution function F 2 : [0, +∞) → [0, 1] is a mixed strategy for
firm 2.
• What does each player get given players’ choices of pure strategies
(q 1 , q 2 ) Firm 1 gets profit u 1 (q 1 , q 2 ) = q 1 (1 − q 1 − q 2 ), and firm
2 gets profit u 2 (q 1 , q 2 ) = q 2 (1 − q 1 − q 2 ). (We generally will write
π 1 and π 2 instead of u 1 and u 2 , if the latter actually represent
firms’ profits.) What do the two players get if player 1 adopts
the mixed strategy F 1 and player 2 adopts the mixed strategy F 2
The answers are respectively
and
∫ ∞ ∫ ∞
0
0
∫ ∞ ∫ ∞
0
0
u 1 (x, y)dF 1 (x)dF 2 (y)
u 2 (x, y)dF 1 (x)dF 2 (y).
4. Definition 2. A (pure strategy) Nash equilibrium (NE) for a twoplayer
normal-form game {X, Y, u 1 (x, y), u 2 (x, y)}, where X and Y are
respectively the sets of pure strategies for the two players (hereafter,
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the two players’ strategy spaces), is a pair (x ∗ , y ∗ ) such that x ∗ ∈ X,
y ∗ ∈ Y , and
u 1 (x ∗ , y ∗ ) ≥ u 1 (x, y ∗ ), ∀x ∈ X, (4)
u 2 (x ∗ , y ∗ ) ≥ u 2 (x ∗ , y), ∀y ∈ Y. (5)
(The last two inequalities are called incentive compatibility conditions.)
Given y ∈ Y , we say that x ∗ ∈ X is player 1’s best response against y
if
u 1 (x ∗ , y) ≥ u 1 (x, y), ∀x ∈ X. (6)
If for all y ∈ Y , there exists a unique player 1’s best response against
y, then we can write x ∗ = r 1 (y), and refer to r 1 (·) as player 1’s reaction
function. Similarly, given x ∈ X, we say that y ∗ ∈ Y is player 2’s best
response against x if
u 2 (x, y ∗ ) ≥ u 2 (x, y), ∀y ∈ Y. (7)
If for all x ∈ X, there exists a unique player 2’s best response against
x, then we can write y ∗ = r 2 (x), and refer to r 2 (·) as player 2’s reaction
function. Apparently, if the two-player game has a unique Nash
equilibrium (x ∗ , y ∗ ), then it must be that
r 1 (y ∗ ) = x ∗ , y ∗ = r 2 (x ∗ ). (8)
That is, an NE must appear at the intersection of the two reaction
functions.
More generally, let r 1 (y) contain the elements of X that maximize the
function u 1 (·, y) and r 2 (x) contain the elements of Y that maximize
the function u 2 (x, ·). Then (x ∗ , y ∗ ) is a pure-strategy NE of the game
{X, Y, u 1 (x, y), u 2 (x, y)} if and only if
x ∗ ∈ r 1 (y ∗ ), y ∗ ∈ r 2 (x ∗ ). (9)
Again, this requires that x ∗ be one of player 1’s best responses against
y ∗ , and y ∗ be one of player 2’s best responses against x ∗ .
A mixed-strategy NE is defined similarly, but with players playing
mixed strategies rather than pure strategies.
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5. Now we solve for the pure-strategy NEs for the game described in
Example 1. It is easy to see that (U,R) and (D,L) are the two purestrategy
NEs of the game.
6. Now we solve for the mixed-strategy NEs for the game described in
Example 1. A mixed strategy Nash equilibrium (p, q) of this game is
a pair of the two players’ mixed strategies, such that player 1 plays
U with probability p ∈ (0, 1) and player 2 plays L with probability
q ∈ (0, 1), and such that player 2’s mixed strategy makes player 1
feel indifferent about U and D, and player 1’s mixed strategy makes
player 2 feels indifferent about L and R; that is, the following incentive
compatibility conditions hold:
qu 1 (U, L) + (1 − q)u 1 (U, R) = qu 1 (D, L) + (1 − q)u 1 (D, R), (10)
pu 2 (U, L) + (1 − p)u 2 (D, L) = pu 2 (U, R) + (1 − p)u 2 (D, R). (11)
Solving, we have p = 1 and q = 1 . There is an obvious reason for
2
3
the above two equations: if given his rival’s mixed strategy, a player
strictly prefers one pure strategy to the other, then he will assign zero
probability to the latter; that is, a mixed strategy can never be his best
response. Thus in a mixed strategy Nash equilibrium, a player has to
feel indifferent about his two pure strategies.
7. Now we solve for the pure-strategy NEs for the game described in
Example 2. By definition, it is a pair (q1, ∗ q2), ∗ such that given q2, ∗ q1 ∗ is
profit maximizing for firm 1, and given q1 ∗, q∗ 2 is profit maximizing for
firm 2. The procedure is first to find the best response r i for firm i given
any possible q j , for i, j = 1, 2, i ≠ j. Then, the NE can be obtained by
finding the intersection of the two reaction functions. So, consider step
1. To solve for r i (·), given any q j , consider firm i’s problem of finding
its profit-maximizing supply quantity:
max
q i
q i (1 − q i − q j ),
and the (necessary and sufficient) first-order condition gives
r i (q j ) = 1 − q j
.
2
5

(The reaction function is well-defined because given each q j , there exists
exactly 1 optimal q i for firm i.) Now, consider step 2. By definition,
an NE (q1, ∗ q2) ∗ is located at the intersection of the two firms’ reaction
functions. Thus (q1 ∗, q∗ 2 ) must satisfy
r 1 (q ∗ 2) = q ∗ 1, r 2 (q ∗ 1) = q ∗ 2. (12)
Solving, we obtain the Nash equilibrium for the above Cournot game:
(q ∗ 1 , q∗ 2 ) = (1 3 , 1 3 ).
Note that this game has no mixed-strategy NEs.
8. Example 3. Consider a two-player game, where the two must pick an
integer from the set {1, 2, · · · , 100} at the same time. If they pick the
same number, then they each get 1; or else, they each get zero. Find
the pure strategy NE’s. Find the mixed strategy NE’s.
9. Definition 3. The games appearing in the above two examples are
called simultaneous games, because players move at the same time in
those games. In a sequential game, on the other hand, players take turn
to move. Sequential games are usually described in extensive form and
represented by a game tree. A game tree is composed of a collection of
decision nodes and a set of branches connecting those nodes. The first
mover’s decision node is the root of the tree, and each pure strategy
available to the first mover is represented by a branch emanating from
that decision node. These branches lead to the second mover’s decision
nodes. For example, consider the following sequential game:
⎡
Up—
(I)—
⎢
⎣ Down—
(II)—
.
(II)—
⎡
⎢
⎣
⎡
⎢
⎣
Right— (0, 1)
Left— (−1, 2)
Right— (2, −1)
Left— (−2, −2)
Note that player II’s two decision nodes are connected by a dotted line,
which says that player II, when determining her own moves, does not
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know whether player I has moved up or down. Thus these two decision
nodes define player II’s information set. Formally, an information set
is a set of decision nodes for a player, who, while knowing that he is
sitting on one of those nodes contained in the information set, cannot
tell which node in the information set he is exactly sitting on.
The remaining game tree starting from a singleton information set is
called a subgame of the original game. A subgame perfect Nash equilibrium
(SPNE) is an NE for a game in extensive form, which specifies
NE strategies in each and every subgame.
10. Example 4. M is the owner-manager of a firm. The debt due one year
from now has a face value equal to $10. The total assets in place are
worth only $8. Just now, a new investment opportunity with NPV=x >
1 became available, which requires no addition investment. M comes to
his creditor and says that he will give up the investment project unless
the creditor agrees to reduce the face value of debt by $1.
(i) Suppose x > 2. Show that there is an NE in which the creditor
agrees to reduce the face value of debt and M makes the investment.
(ii) Show that the NE in (i) is not an SPNE. Find an SPNE.
(iii) How may your conclusion about (ii) change if x ∈ (1, 2]
(iv) Define bankruptcy as a state where the firm’s equity value drops
to zero. Explain why bankruptcy does not take place in (iii).
11. Example 5. A monopolistic firm F selling product X at its store is
faced with 2 consumers A and B, where A is an expert in computer
science, while B has no knowledge in computer-related knowledge. F
cannot distinguish A from B, but F knows that A is willing to pay 2
dollars for 1 unit of X, while B’s reservation value for X is 5 dollars. A
and B both have unit demand for X.
(i) First suppose that e-commerce is unavailable. Suppose that production
is costless. What is the price for X What is F’s profit
(ii) Next suppose that F can also open up a retail outlet on the net,
which will cost t > 0. Should F sells exclusively through the on-line outlet
Should F ignore the chance to open up an on-line outlet Should
F sell its product through its original store as well as the on-line outlet
(iii) What if B’s reservation value is 3.9 dollars
(iv) What is missing in the above analysis
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Solution. For part (i), F should price X at 5 dollars, which is also F’s
profit. For part (ii), F can price X at 2 dollars on the net and at 5
dollars at the original store, which raises F’s profit by 2 dollars. This is
beneficial as long as t < 2. 1 For part (iii), giving up the on-line trading
opportunity implies a profit of 4 dollars for F; and selling through both
channels yields a profit of 3.9 + 2 − t, so that F should sell through
both channels if t < 1.9. 2
As for part (iv), many important issues are left out. For example, conventional
outlets usually provide in-store services that on-line retailers
cannot afford to, and transactions at the conventionl outlets usually
incur a transportation cost for consumers. Furthermore, on-line transactions
may imply a higher probability of commerical frauds.
Note that the assumption that B has a higher reservation value than
A does is important in the above analysis. If the reservation values are
reversed, F cannot price discriminate between A and B.
Finally, let me ask the following question: Other things being equal,
does an increase in demand encourage F to adopt the dual channels
strategy
Verify that, for t ∈ (1.9, 2), other things being equal, an increase in
A’s demand (here, the reservation value) from 2 dollars to 3.1 dollars
actually decrease F’s incentive to use the on-line outlet. We know that
when A’s reservation value is 2 dollars, F will use the on-line outlet
because t < 2; but when A’s reservation value becomes 3.1 dollars, F
will use the on-line outlet if and only if (5 − 3.1) > t, that is 1.9 > t.
1 Note that A is not served without on-line markets, and t is the cost that F incurs in
order to extract the 2 dollars from A in the presence of on-line markets.
2 Note that selling to A on the net actually creates a loss: F still charges A 2 dollars,
but now F has to spend t > 0. What is beneficial is that F can now charge B 3.9 instead of
2 dollars. Why With A being directed to the on-line markets, B is identified as the only
segment left to be served at the original store, and hence F can fully extract the consumer
surplus from B. The extra 1.9=3.9-2 dollars that F can make from serving B must be
greater than the loss incurred when F moves A from the store to the on-line markets in
order for F to adopt this dual channels strategy, and hence the condition is t < 1.9. From
here it becomes clear that F’s incentive to spend t may be reduced when A’s reservation
value increases: if A is already served without the internet, then the benefit from spending
t is equal to the difference in A’s and B’s reservation values, and this difference is decreasing
in A’s reservation value.
8

Thus F will give up the on-line outlet when A’s valuation rises from 2
dollars to 3.1 dollars. The intuition has been spelt out in the preceding
footnote, where we emphasized that F’s benefit from spending t is
decreasing in A’s reservation value if F would choose to serve A in
the absence of the internet.
On the other hand, an increase in B’s demand does weakly encourage
F to use the on-line outlet: it does not affect F’s preference about
spending t if F would choose not to serve A in the absence of the
internet, but it increases F’s benefit from spending t if F would choose
to serve A in the absence of the internet.
So, we conclude that an increase in demand may not encourage F to
use the on-line outlet; it depends on which segment’s demand increases
more and which segment’s demand increases less. In particular, an increase
in the valuation of the low-valuation segment (which also has expertise
in computer science) need not encourage the firm to sell through
the internet.
12. Definition 4. In the above we have assumed that players know the
payoff functions of each other. Such a game is a game with complete (or
symmetric) information. What if at least one player in the game does
not know for sure another player’s payoff function We call it a game
with information asymmetry, or a game with incomplete information,
or simply a Bayesian game. In a Bayesian game, at least one player Z
has more than one possible payoff function. We say that this player Z
has more than one type. At least one other player W cannot be sure
which type player Z has. In this case, we shall look for an equilibrium
called Bayesian equilibrium (BE). This is nothing but a Nash equilibrium
of an enlarged version of the original game, where each different
type of Z is treated as a distinct player.
To give a concrete example, suppose that we have a two-player game
where player 1’s strategy space is X and player 2’s strategy space is Y ,
and player 2 has two possible types (or two possible payoff functions), θ 1
and θ 2 , which, from player 1’s perspective, may occur with probabilities
π 1 and π 2 respectively. Certainly, player 2 knows his own type. A
Bayesian equilibrium for this two-player game is nothing but the Nash
equilibrium of the three-player game where the two types of player 2
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are treated as two different players. Thus a Bayesian equilibrium is a
triple (x ∗ , y1 ∗, y∗ 2 ) such that x∗ ∈ X, y1 ∗ ∈ Y , y∗ 2 ∈ Y , and the following
three incentive compatibility conditions hold:
π 1 u 1 (x ∗ , y ∗ 1) + π 2 u 1 (x ∗ , y ∗ 2) ≥ π 1 u 1 (x, y ∗ 1) + π 2 u 1 (x, y ∗ 2), ∀x ∈ X; (13)
u 2 (x ∗ , y ∗ 1; θ 1 ) ≥ u 2 (x ∗ , y; θ 1 ), ∀y ∈ Y ; (14)
u 2 (x ∗ , y ∗ 2 ; θ 2) ≥ u 2 (x ∗ , y; θ 2 ), ∀y ∈ Y. (15)
In words, x ∗ is player 1’s best response, which is on average the optimal
strategic choice of player 1. It is not really player 1’s best response
against player 2 if player 1 is sure that player 2 will use y1 ∗ . Neither is
it player 1’s best response against player 2 if player 1 is sure that player
2 will use y2 ∗ . Since player 1 can only choose one x in X to play against
two possible types of player 2, given his conjecture of (y1 ∗, y∗ 2 ), the choice
x ∗ must be on average optimal. On the other hand, player 2 knows
his own type, and his best response against player 1’s average optimal
choice x ∗ depends on his type. Note that θ denotes player 2’s type, and
it determines u 2 (x, y)! This is why we say that incomplete information
in this game is equivalent to player 1 not knowing player 2’s payoff
function. Again, player 1’s average optimal choice x ∗ , player 2’s best
response y1 ∗ when his type is θ 1, and player 2’s best response y2 ∗ when his
type is θ 2 , must altogether form a Nash equilibrium. This three-player
Nash equilibrium is what we defined as the Bayesain equilibrium.
13. Definition 5. Given a game, an event is mutual knowledge if every
player knows it. Given a game, an event is called the players’ common
knowledge if every player knows it, everyone knows that everyone knows
it, everyone knows that everyone knows that everyone knows it, and so
on. Anything which is not common knowledge is some player’s private
information. A player’s private information is also called his type. A
game where no players have private information (everything relevant is
common knowledge) is a game with complete information. Otherwise,
the game is one with incomplete information, or one with information
asymmetry.
14. Example 6. Three boys are locked in a room without a mirror and
are unallowed to talk to each other. Each of them is wearing a hat
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which is either red or black. Each boy can see the hats the other two
boys are wearing, but not his own hat. At times 0, 1, 2, and so on,
their teacher will come ask each of them in private whether the boy
can figure out the color of his hat, and if he can, then he is free to go
home. Suppose that the three boys are all wearing red hats.
(i) At what time will the room become empty
(ii) Now suppose that at time zero the teacher tells all the boys that
at least one of them is wearing a red hat. At what time will the room
become empty
15. Definition 6. An incomplete-information game where the uninformed
players can move after observing the informed players’ moves is called
a dynamic game, and otherwise a static game.
16. Example 7. Let us modify Example 2 by assuming a random demand
curve. This gives rise to a static game with incomplete information.
More precisely, let the inverse demand curve be
P (q 1 + q 2 ) = ã − q 1 − q 2 , (16)
where only firm 1 knows the outcome of the random variable ã. Firm
1
2 only knows that ã may be 2 with prob. or 4 with prob. 2
. Since
3 3
the same firm 1 will act differently when it receives different damand
information, we should regard firm 1 as different players given that it
has received different information. Interpreted this way, this game has
3 players—the firm 1 seeing ã = 2; the firm 1 seeing ã = 4; and firm 2.
An NE for this 3-player game is what we shall be looking for, which is
also called a Bayesian equilibrium (BE) of the game. Find a BE.
Solution. First observe that firm 1 has two possible types. Firm 2
has only one type (no private information). So, we should consider
a three-firm game, where the two types of firm 1 will be regarded as
two different players, and then we look for the NE of the new 3-player
game. By definition, we must find three strategies q1(2), ∗ q1(4), ∗ and q2.
∗
These 3 strategies are such that, given any two of them, the third one
is the corresponding player’s best response!
In other words, for firm 2,
q ∗ 2
1
= arg max
q 2 3 [q 2(2 − q1 ∗ (2) − q 2)] + 2 3 [q 2(4 − q1 ∗ (4) − q 2)].
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Similarly, for the firm 1 that has seen ã = 2,
q1 ∗ (2) = arg max q 1 (2 − q 1 − q
q 2 ∗ );
1
and for the firm 1 that has seen ã = 4,
q1 ∗ (4) = arg max q 1 (4 − q 1 − q
q 2 ∗ ).
1
Each of the above three maximization problems is concave, and so
the 3 first-order conditions are necessary and sufficient. Each firstorder
condition gives a reaction function for one of the player. Again,
the NE must appear at the intersection of the 3 reaction functions.
Solving, the NE of the 3-player game, or the BE of the original game
with information asymmetry, is the following:
(q ∗ 1 (2), q∗ 1 (4), q∗ 2 ) = (4 9 , 13 9 , 10
9 ).
17. Definition 7. A dynamic Bayesian game is called a signalling game if
there are only two players, informed and uninformed, and the informed
moves first as a Stackelberg leader, and the uninformed moves as a follower.
The game ends after the uninformed’s move. A perfect Bayesian
equilibrium (PBE) is defined by a set of actions (one for each type of
each player) and a set of posterior beliefs called the supporting beliefs
for the PBE that specify what the uninformed player thinks about the
informed players’ types after seeing each possible actions taken by the
informed, such that, given that all types of all players play the specified
equilibrium strategies, each player finds playing her equilibrium
strategy optimal given the specified equilibrium beliefs. Moreover, all
posterior beliefs must be updated using Bayes law whenever possible.
18. Example 8. Spence’s Signalling Model This is a game with two
players, an employer and an employee (a worker). The worker may
have high (H) or low (L) productivity, which is his private information
unknown to the employer. The employer only knows that the worker’s
productivity is H with prob. α and L with prob. (1 − α). The worker
can go to school for a couple of years. Schooling costs respectively c H
per year for the type-H worker and c L per year for the type-L worker.
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Suppose that education does not change the worker’s productivity and
that c H < c L . Further assume that the labor market is competitive in
the sense that the wage w received by a worker is equal to the expected
productivity of that worker. To summarize: The two types of worker,
H and L, first choose e H and e L in R + , which represent how many years
of education the worker is willing to receive before appearing in the job
market, and after seeing e H and e L , the employer chooses w.
This game is ‘dynamic’ in nature, because the uninformed player (the
employer) can see the informed player’s (worker’s) choice of e before
choosing w. The choice e is regarded by the uninformed employer as a
signal about the worker’s type. The employer first make a conjecture
regarding how many years of education the two types of worker would
respectively choose, and based on his conjecture, when he actually sees
the choice e made by the worker, he uses Bayes Law to form a posterior
belief regarding the worker’s type, and then chooses w based on the
posterior belief. Since seeing different choices e made by the worker, the
employer may form different beliefs and then choose different w, we can
represent equilibrium actions by (e H , e L , w(e)), such that (Nash equilibrium
concept!) (i) expecting the employer’s pricing rule w(·), e H and
e L are respectively the type-H and type-L worker’s best response; (ii)
w(e H ) and w(e L ) should make the employer just break even (because
of competitive labor market) conditional on the employer’s posterior
belief f(e) =prob.(H|e); and (iii) the posterior belief is consistent with
Bayes Law. 3
19. Now we are ready to solve the PBE’s for the Spencian signalling game.
By definition, a pure-strategy PBE of this game is defined as a tuple
(e H , e L , f(e) = prob.(H|e), w(e)), such that (i) if e H = e L = e ∗ (called
a pooling equilibrium), then
f(e ∗ ) = α, (prior beliefs are also posterior)
3 The key difference between a dynamic Bayesian game and a static Bayesian game is
that, in the latter the uninformed looks for his best responses based on his prior beliefs
about the informed’s types, whereas in the former the uninformed looks for his best responses
based on his posterior beliefs about the informed’s types. This happens because
the uninformed can see a signal before making his move, and it should use this new piece
of information to infer the informed’s type.
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f(e) ∈ [0, 1], ∀e ≠ e ∗ , (Bayes law has no bite for zero-probability event)
w(e) = f(e)H + [1 − f(e)]L, ∀e,
w(e ∗ ) − c H · e ∗ ≥ w(e) − c H · e, ∀e,
w(e ∗ ) − c L · e ∗ ≥ w(e) − c L · e, ∀e;
and such that (ii) if e H ≠ e L (called a separating equilibrium), then
f(e H ) = 1, f(e L ) = 0 (uncertainty is completely resolved)
f(e) ∈ [0, 1], ∀e ≠ e H , e L , (Bayes law has no bite)
w(e) = f(e)H + [1 − f(e)]L, ∀e,
w(e H ) − c H · e H ≥ w(e) − c H · e, ∀e,
w(e L ) − c L · e L ≥ w(e) − c L · e, ∀e.
Check if these are consistent with the definition of PBE.
20. A mixed-strategy equilibrium where there exist at least two signals e
and e ′ such that f(e)[1 − f(e)] = 0 and 0 < f(e ′ ) < 1 is called a
semi-separating equilibrium. In such an equilibrium, sometimes the
uninformed can identify the informed player’s type by seeing the signal,
but sometimes she cannot.
21. The above states that, considering both pure-strategy and mixed-strategy
PBE’s, there are generally three types of equilibria for a dynamic game
with incomplete information: pooling equilibrium (in which e H = e L ),
separating equilibrium (in which e H ≠ e L ), and semi-separating equilibrium
(where either H or L randomizes). Now let us solve for these
PBE’s for Spencian signalling model.
22. Pooling Equilibrium: Suppose both H and L choose e ∗ and f(e) =
0, ∀e ≠ e ∗ . (Note that we have arbitrarily selected a set of supporting
beliefs f(e) for off-equilibrium signals e.) Then, w H = w L = w(e ∗ ) =
αH + (1 − α)L ≡ w. We need the following incentive compatibility
conditions (hereafter IC conditions) to hold for respectively type-H and
type-L workers:
w − c H e ∗ ≥ L − c H · 0,
14

w − c L e ∗ ≥ L − c L · 0.
Thus, we have a continuum (an uncountably infinite number) of pooling
equilibria: Each e ∗ ≤ w−L
c H
corresponds to one pooling PBE.
23. Separating Equilibrium: Suppose H and L chooses respectively
E and e, E ≠ e, and f(e ′ ) = 0, ∀e ′ ≠ E. In this case, of course,
w(E) = H, w(e) = L (why). We need the following IC conditions to
hold:
H − c H E ≥ L − c H · 0,
L − c L e ≥ H − c L · E.
Immediately, we have e = 0 (why). Again, we have a continuum of
separating equilibria: ≥ E ≥ H−L
c L
.
H−L
c H
24. Semi-Separating Equilibrium: Consider the equilibrium where
one type randomizes over two signal levels and the other concentrates
on one signal level. At first, assume that in equilibrium H randomizes
over E (prob. p) and e (prob. 1 − p) and L plays e with probability
one, with E > e. For simplicity, assume that f(e ′ ) = 0, e ′ ≠ E, e.
Then, such an equilibrium exists if and only if E > e ≥ 0 and
(1 − α)(H − L)
c H
≤ E − e ≤
(H − L)
c H
.
On the other hand, assume that in equilibrium H plays some E with
prob. 1 and L randomizes over E (prob. q) and e (prob. 1 − q) with
E > e ≥ 0. Again, assume f(e ′ ) = 0, ∀e ′ ≠ E, e. It follows that e = 0!
Such an equilibrium exists if and only if
α(H − L)
c L
≤ E ≤ H − L
c L
.
25. Example 9. (The Game of Beer and Quiche) Two cowboys A
and B meet in a bar, and A may be weak (w) or strong (s), which is A’s
private information. The game proceeds as follows. A first decides to
order either a beer (b) or a quiche (q), and upon observing A’s order,
B decides to or not to fight A. We assume that in the absence of B, A
prefers beer (b) to quiche (q) if he is (s), otherwise he prefers (q) to
15

(b). The prior beliefs of B are such that A is (s) with probability 0.9.
Now the payoffs: if A orders and eats something he dislikes, he gets 0,
or else he gets 1, if B does not fight A, A gets an additional payoff of
2. On the other hand, B gets 1 if he has no chance to fight, gets 2 if
he fights A and A is of the weak type, and gets zero if he fights A and
A is of the strong type.
This game has two pooling PBE’s:
(1) Equilibrium (B): Both types of A order a beer and B’s strategy
is to fight A if and only if he sees A order a quiche. What are the
supporting beliefs Let f(s) =pro.(A is strong| A orders s), for all
s ∈ {b, q}. Then of course f(b) = 0.9. Note that s = q is a zero
probability event. Recall that Bayes Law says
P (E|F )P (F ) = P (E ⋂ F ),
where E and F are two random events. From the probability theory,
we know that for any two events C and D,
Thus we have
C ⊂ D ⇒ P (C) ≤ P (D).
P (F ) = 0 ⇒ P (E ⋂ F ) = 0,
since E ⋂ F ⊂ F . Thus given P (F ) = 0, Bayes Law requires
P (E|F ) · 0 = 0,
and hence P (E|F ) can be anything contained in [0, 1]. Let E be the
event that A is of the strong type, and F the event that B has observed
that A ordered a quiche. We conclude that any f(q) ∈ [0, 1] will be
consistent with Bayes Law in this case. We must find at least one
f(q) ∈ [0, 1] so that the above strategy profile does constitute the two
players’ best responses against each other. Note that for B to fight A
after seeing A order a quiche, it is necessary that
1 ≤ f(q) · 0 + [1 − f(q)] · 2 ⇒ f(q) ≤ 1 2 .
Now we show that given the beliefs f(b) = 0.9 and f(q) being anything
in [0, 1 ], the aforementioned A’s and B’s strategies are respectively the
2
16

two players best responses. For A, if his type is (s), he gets 1 + 2 = 3 if
he orders a beer, and if he deviates and orders a quiche, then not only
he eats something he hates but he also must fight B, yielding a payoff
of 0 + 0 = 0. Thus A will not deviate if he is of type (s). What if A is
of type (w) If he orders a beer, then he must eat something he hates,
but the good news is that he can avoid fighting B, so that his payoff
is 0 + 2 = 2; and if he deviates and orders a quiche, then he will have
to fight B, so that his payoff is 1 + 0 = 1. We conclude that the weak
type of A does not want to deviate either. What about B We have
shown that given f(q) ≤ 1 , fighting A if A dares to order the quiche is
2
really optimal for B. On the other hand, if A orders a beer, then since
B expects both types of A to do so, ordering the beer really does not
tell B anything new, and B’s posterior beliefs are identical to his prior
beliefs (A is of the strong type with prob. 0.9), and so not to fight A
is optimal for B.
To sum up, we have shown that the following is a PBE (check if it
corresponds to our definition of a PBE!):
(i) The strong type of A orders a beer;
(ii) The weak type of A also orders a beer;
(iii) B’s strategy must describe what he will do in every possible contingency:
B will fight A if A ordered a quiche, and B will not fight A
if A ordered a beer;
(iv) The supporting beliefs fully describe what B thinks of A in every
possible contingency: B thinks that A is of the strong type with prob.
0.9 if he sees A order a beer; and B thinks that A is of the strong type
with prob. f(q) if he sees A order a quiche, where f(q) is any real
number contained in [0, 1].
2
(2) Equilibrium (Q): Both types of A order a quiche and B’s strategy
is to fight A if and only if he sees A order a beer. What are the supporting
beliefs Let f(s) =pro.(A is strong| A orders s), for all s ∈ {b, q}.
Then of course f(q) = 0.9. Now for f(b) to induce B to fight A after
seeing A order a beer, it is necessary that
1 ≤ f(b) · 0 + [1 − f(b)] · 2 ⇒ f(b) ≤ 1 2 .
Now we show that given the beliefs f(q) = 0.9 and f(b) being anything
in [0, 1 ], the aforementioned A’s and B’s strategies are respectively the
2
17

two players best responses. For A, if his type is (w), he gets 1 + 2 = 3
if he orders a quiche, and if he deviates and orders a beer, then not
only he eats something he hates but he also must fight B, yielding a
payoff of 0 + 0 = 0. Thus A will not deviate if he is of type (w). What
if A is of type (s) If he orders a beer, then he must eat something he
hates, but the good news is that he can avoid fighting with B, so that
his payoff is 0+2 = 2; and if he deviates and orders a beer, then he will
have to fight B, so that his payoff is 1 + 0 = 1. We conclude that the
strong type of A does not want to deviate either. What about B We
have shown that given f(b) ≤ 1 , fighting A if A dares to order the beer
2
is really optimal for B. On the other hand, if A orders a quiche, then
since B expects both types of A to do so in equilibrium, ordering the
quiche really does not tell B anything new about A, and B’s posterior
beliefs are identical to his prior beliefs (A is of the strong type with
prob. 0.9), and so not to fight A is optimal for B.
To sum up, we have shown that the following is a PBE (check if it
corresponds to our definition of a PBE!):
(i) The strong type of A orders a quiche;
(ii) The weak type of A also orders a quiche;
(iii) B’s strategy must describe what he will do in every possible contingency:
B will fight A if A ordered a beer, and B will not fight A if
A ordered a quiche;
(iv) The supporting beliefs fully describe what B thinks of A in every
possible contingency: B thinks that A is of the strong type with prob.
0.9 if he sees A order a quiche; and B thinks that A is of the strong
type with prob. f(b) if he sees A order a beer, where f(b) is any real
number contained in [0, 1].
2
26. Example 10. Recall the game of three doors (labeled no. 1, no. 2,
and no. 3) with one hidden car. This is a TV show, where a host
is faced with a guest, who must guess behind which among the three
closed doors there is a hidden new car.
(a) The guest first chooses one closed door.
(b) Then the host offers to open another closed door for the guest,
saying that the guest would get the car if the car is hidden behind
the door selected by the host.
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(c) Then the door chosen by the host is opened, and if there is indeed
a car behind that door, it is goven as a gift to the guest; and
if there is no car hidden behind that door, then the guest will
be given an opportunity to re-make his choice between the two
remaining closed doors.
(d) After the guest re-makes his choice, the remaining two doors are
both opened, and the car will be given to the guest if it is hidden
behind the door chosen by the guest in the stage (c); or else, the
car is returned to the sponsor of the TV show. This ends the
game.
Assume that the guest’s payoff is 1 if he gets the new car and zero
otherwise. Assume that the host knows from the very beginning behind
which closed door the new car is hidden.
(1) First assume that the host’s payoff is 1 regardless of whether the
guest gets the new car. Which door should the guest select at stage
(c), if the guest does not get the new car at stage (c)
(2) Now assume instead that the host’s payoff is 1 if the guest fails to
get the new car at the end of the show, and the host’s payoff is zero if
otherwise. Which door should the guest select at stage (c), if the guest
does not get the new car at stage (c)
(3) Now assume instead that the host’s payoff is 0 if the guest fails to
get the new car at the end of the show, and the host’s payoff is 1 if
otherwise. Which door should the guest select at stage (c), if the guest
does not get the new car at stage (c)
Solution. This is clearly a signaling game with the host and the guest
being respectively the informed signal sender and the uninformed signal
receiver. The guest always feels indifferent about the three closed door
at stage (a), in all of parts (1), (2) and (3). However, the guest’s
optimal choice at stage (c) differs among parts (1), (2) and (3) as we
now demonstrate.
In part (1), the host will ignore his private information, and simply
open another door for the guest. If the guest does not get the car,
then the guest should consider the two remaining closed doors equally
likely to have the new car hidden behind them. Hence the guest can
randomize over those two doors in any fashion.
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In part (2), the host will avoid giving away the new car. Knowing
this, the guest should infer as follows: if the car is hidden behind the
door chosen by the guest at stage (a), then from the host’s perspective
opening the door selected by the host at stage (b) is as good as opening
the door that neither the guest nor the host has chosen, so that the
door chosen by the host at stage (b) is opened at stage (c) only with
probability 1 ; but it the car is not hidden behind the door chosen by the
2
guest at stage (a), then the host must select the door that he opened
at stage (c) with probability one at stage (b)! This implies that the
guest should always alter his choice at stage (c)!
In part (3), the host wants to help the guest get the new car. If the
guest does not get the car at stage (c), it proves that the car must be
hidden behind the door originally chosen by the guest at stage (a), and
hence at stage (c) the guest should always stick to his original choice
made at stage (a)! 4
4 We shall come back to reconsider this example in lecture 5, where we shall write down
the above PBE’s in a formal manner.
20