Bayesian Games – Nature & Common Values
A Bayesian Game (with nature and common values) consists of
� a set of players N = {1, . . . , n},
� a set of states of nature Ω,
� a set of actions Ai available to player i,
� a set of possible types Ti of player i,
� a type function τi : Ω → Ti assigning a type of player i to every
state of nature,
� a payoff function ui for every player i
ui : A1 × · · · × An × Ω → R
� a probability distribution P over Ω called common prior.
As before, a pure strategy for player i is a function si : Ti → Ai.
314
Bayesian Games – Nature & Common Values
Given a pure strategy si of player i and a state of nature ω ∈ Ω, we
denote by si(ω) the action si(τi(ω)) chosen by player i when the state
is ω.
We denote by s(ω) the action profile (s1(τ1(ω)), . . . , sn(τn(ω))).
Given a set A ⊆ Ω of states of nature and a type ti ∈ Ti of player i, we
denote by P(A | ti) the conditional probability of A conditioned on
the event that player i has type ti.
We define the expected payoff for player i by
ui(s1, . . . , sn; ti) = Eω∼P [ui(s(ω); ω) | τi(ω) = ti]
Here the right hand side is the expected payoff of player i with respect
to the probability distribution P conditioned on his type ti.
Definition 93
A pure strategy profile s = (s1, . . . , sn) ∈ S in the Bayesian game is
a pure strategy Bayesian Nash equilibrium if for each player i and
each type ti ∈ Ti and every pure strategy s�
i
of player i we have that
ui(si, s−i; ti) ≥ ui(s�
i , s−i; ti)
315
Adverse Selection
� A firm C is taking over a firm D.
� The true value d of D is not known to C, assume that it is
uniformly distributed on [0, 1].
This is of course a bit artificial, more precise analysis can be done with
a different distribution.
� It is known that D’s value will flourish under C’s ownership:
it will rise to λd where λ > 1.
� All of the above is a common knowledge.
Let us model the situation as a Bayesian game (with common
values).
316
Adverse Selection (Cont.)
� N = {C, D},
� Ω = [0, 1] where d ∈ Ω expresses the true value of D,
� AC = [0, 1] where c ∈ AC expresses how much is the firm C
willing to pay for the firm D,
AD = {yes, no} (sell or not to sell),
� TC = {t1} (a trivial type) and TD = Ω = [0, 1],
� τC (d) = t1 and τD(d) = d for all d ∈ Ω,
� uC (c, yes; d) = λd − c and uC (c, no; d) = 0
uD(c, yes; d) = c and uD(c, no; d) = d,
� P is the uniform distribution on [0, 1].
Is there a BNE?
317
Adverse Selection (Cont.)
What is the best response of firm D to an action c ∈ [0, 1] of firm C?
Such a best response must satisfy:
� say yes if d < c
� say no if d > c
So the expected value of the firm D (in the eyes of C) assuming that
D says yes is c/2.
Indeed, assuming that the firm D says yes, the value d is uniformly
distributed between 0 and c, so the average is c/2.
Therefore, the expected payoff of C is
λ(c/2) − c = c
�
λ
2
− 1
�
which is negative for λ ≤ 2. So it is not profitable (on average) for
the firm C to buy unless the target D more than doubles in value after
the takeover!
318
Committe Voting
Consider a very simple model of a jury made up of two players
(jurors) who must collectively decide whether to acquit (A), or to
convict (C) a defendant who can be either guilty (G) or innocent (I).
Each player casts a sealed vote (A or C), and the defendant is
convicted if and only if both vote C.
A prior probability that the defendant is guilty is q > 1
2 (i.e., P(G) = q)
and is common knowledge.
Assume that each player gets payoff 1 for a right decision and 0 for
incorrect decision. We consider risk neutral players who maximize
their expected payoff.
We may model this situation using a strategic-form game:
A C
A 1 − q, 1 − q 1 − q, 1 − q
C 1 − q, 1 − q q, q
Is there a dominant strategy?
319
Committee Voting (Cont.)
Let’s make things a bit more complicated.
Assume that each juror has a different expertise and, when observing
the evidence, gets a private signal ti ∈ {θG, θI} that contains
a valuable piece of information. That is if the defendant is guilty, θG is
more probable, if innocent, θI is more probable. For i ∈ {1, 2} :
P(ti = θG | G) = P(ti = θI | I) = p >
1
2
P(ti = θG | I) = P(ti = θI | G) = 1 − p <
1
2
We also assume that the players get their signals independently
conditional on the defendants condition:
P(t1 = θX ∧ t2 = θY | Z) = P(t1 = θX | Z) · P(t2 = θY | Z)
for all X, Y, Z ∈ {G, I}.
320
Committe Voting (Cont.)
We obtain a Bayesian game:
� N = {1, 2}
� A1 = A2 = {A, C}
� Ω = {(Z, θX , θY ) | Z, X, Y ∈ {G, I}}
� T1 = T2 = {θG, θI}
� τ1(Z, θX , θY ) = θX and τ2(Z, θX , θY ) = θY
� For arbitrary U, V ∈ {A, C} and X, Y ∈ {G, I} we have that
ui(U, V; (G, θX , θY )) =



1 if U = V = C,
0 otherwise.
ui(U, V; (I, θX , θY )) =



0 if U = V = C,
1 otherwise.
� P(Z, θX , θY ) = P(Z)P(t1 = θX | Z)P(t2 = θY | Z)
I.e., P(Z, θX , θY ) is the probability of choosing (Z, θX , θY ) as follows:
First, Z ∈ {G, I} is randomly chosen (Z = G has probability q). Then,
conditioned on Z, θX and θY are independently chosen.
321
Committee Voting (Cont.)
Now consider just one player i. If the player i would be able to decide
by himself, how does his decision depend on his type ti ∈ {θG, θI}?
If ti = θG, then how probable is that the defendant is guilty?
P(G | ti = θG) =
P(ti = θG | G)P(G)
P(ti = θG)
=
pq
qp + (1 − q)(1 − p)
> q
so that the posterior probability of G is even higher.
If θI is received, then how probable is that the defendant is guilty?
P(G | ti = θI) =
P(ti = θI | G)P(G)
P(ti = θI)
=
(1 − p)q
q(1 − p) + (1 − q)p
< q
which means, clearly, that the player is less sure about G.
In particular, player i chooses I instead of G if
P(G | ti = θI) =
q(1 − p)
q(1 − p) + (1 − q)p
<
1
2
which holds iff p > q.
322
Committee Voting (Cont.)
So if p > q each player would choose to vote according to his signal.
Denote by XY the strategy of player i in which he chooses X if ti = θG
and Y if ti = θI.
Question: Is (CA, CA) BNE assuming that p > q ?
u1(CA, CA; θI) = P(I | t1 = θI)
= P(I | t1 = θI ∧ t2 = θG)P(t2 = θG | t1 = θI)
+ P(I | t1 = θI ∧ t2 = θI)P(t2 = θI | t1 = θI)
u1(CC, CA; θI) = P(G ∧ t2 = θG | t1 = θI) + P(I ∧ t2 = θI | t1 = θI)
= P(G | t1 = θI ∧ t2 = θG)P(t2 = θG | t1 = θI)
+ P(I | t1 = θI ∧ t2 = θI)P(t2 = θI | t1 = θI)
Note that the blue expressions are equal, so the payoff depends only on
the red ones, where player 2 is assumed to consider the defendant guilty.
Intuitively, if player 2 chooses A, then the decision of player 1 does
not have any impact. On the other hand, if player 2 chooses C, then
the decision is, in fact, up to player 1 (we say that he is pivotal).
323
Committee Voting (Cont.)
So what is the probability that the defendant is guilty assuming that
the vote of player 1 counts? That is, assuming t2 = θG and t1 = θI ?
P(G | t1 = θI ∧ t2 = θG) =
P(t1 = θI ∧ t2 = θG | G)P(G)
P(t1 = θI ∧ t2 = θG)
=
(1 − p)pq
p(1 − p)
= q >
1
2
> (1 − q)
= P(I | t1 = θI ∧ t2 = θG)
which means that player 1 is more convinced that the defendant is
guilty contrary to the signal! This means that even though individual
decision would be "innocent", taking into account that the vote should
have some value gives "guilty".
Hence u1(CA, CA; θI) < u1(CC, CA; θI) and thus playing CC is
a better response to CA.
By the way, is (CC, CA) a BNE?
324
Winner’s Curse
An auction for a new oil field (of unknown size), assume only two
firms competing (two players).
The field is either small (worth $10 million), medium (worth $20
million), large (worth $30 million).
That is, the real value v of the field satisfies v ∈ {10, 20, 30}.
Assume some prior information about the size of the filed:
P(v = 10) = P(v = 30) =
1
4
P(v = 20) =
1
2
The government is selling the field in the second-price sealed-bid
auction, so that in the case of a tie, the winner is chosen randomly
(and pays his bid). That is, in effect, in case of a tie, the payoff of
each player is (v − b)/2 where v is the value, b the (common) bid.
Using the same argument as for the "ordinary" second-price auction with
private values one may show that playing the true private value weakly
dominates all other bids.
325
Winner’s Curse (Cont.)
Each of the firms performs a (free) exploration that will provide
the type ti ∈ {L, H} (low or high), correlated with the size as follows:
� If v = 10, then t1 = t2 = L
� If v = 30, then t1 = t2 = H
� If v = 20, then for i ∈ {1, 2}, conditioned on v = 20,
the exploration results are uniformly distributed:
There are four possible results, (L, L), (L, H), (H, L), (H, H),
each with probability 1
4 .
Given the signal ti, player i may estimate the true value of the field:
P(v = 10 | ti = L) =
1
2
P(v = 10 | ti = H) = 0
P(v = 20 | ti = L) =
1
2
P(v = 20 | ti = H) =
1
2
P(v = 30 | ti = L) = 0 P(v = 30 | ti = H) =
1
2
Thus E(v | ti = L) = 1
2 10 + 1
2 20 = 15.
and E(v | ti = H) = 1
2 20 + 1
2 30 = 25
326
Winner’s Curse (Cont.)
Is it a good idea to bid the expected value?
Define a strategy si for player i by
� si(L) = E(v | ti = L)
� si(H) = E(v | ti = H)
Is (s1, s2) a Nash equilibrium?
Consider t1 = L. Then player 1 bids 15. What is his expected payoff?
Note that if t2 = H, then player 2 bids 25 and wins, which means that
player 1 gets payoff 0. So player 1 can get a non-zero value only if
t2 = L. This implies that
u1(s1, s2; L) = P(v = 20 ∧ t2 = L | t1 = L) · (20 − 15)/2
+ P(v = 10 ∧ t2 = L | t1 = L) · (10 − 15)/2
= P(v = 20 ∧ t2 = L | t1 = L) · 5/2
+ P(v = 10 ∧ t2 = L | t1 = L) · (−5)/2
327
Winner’s Curse (Cont.)
In what follows we show that
P(v = 20 ∧ t2 = L | t1 = L) =
1
4
(31)
P(v = 10 ∧ t2 = L | t1 = L) =
1
2
(32)
which means that
u1(s1, s2; L) = P(v = 20 ∧ t2 = L | t1 = L) · 5/2
+ P(v = 10 ∧ t2 = L | t1 = L) · (−5)/2
=
1
4
5
2
+
1
2
(−5)
2
=
−5
8
< 0
and player 1 would be better off by bidding 0 and always losing!!
Intuition: Player 1 wins only if the signal of player 2 is L, which in
effect means, that assuming win, the effective expected value of
the field is lower than the predicted expected value.
In the rest of the proof we heavily use the Bayes’ theorem and the law of total
probability.
328
Winner’s Curse (Cont.) : Proof of Equation (31)
P(v = 20 ∧ t2 = L | t1 = L) =
= P(v = 20 ∧ t2 = L | t1 = L ∧ t2 = L) · P(t2 = L | t1 = L)
+ P(v = 20 ∧ t2 = L | t1 = L ∧ t2 = H) · P(t2 = H | t1 = L)
= P(v = 20 | t1 = L ∧ t2 = L) · P(t2 = L | t1 = L)
Here
P(t2 = L | t1 = L) =
= P(t2 = L | t1 = L ∧ v = 10) · P(v = 10 | t1 = L)
+ P(t2 = L | t1 = L ∧ v = 20) · P(v = 20 | t1 = L)
= 1 ·
1
2
+
1
2
·
1
2
=
3
4
(Here we used the fact that t1 and t2 are independent assuming a fixed v)
We show (see next slide) that
P(v = 20 | t1 = L ∧ t2 = L) =
1
3
and thus
P(v = 20 ∧ t2 = L | t1 = L) =
1
3
·
3
4
=
1
4 329
Winner’s Curse (Cont.) : Proof of Equation (31)
First, note that
P(t1 = L ∧ t2 = L | v = 10) = 1
P(t1 = L ∧ t2 = L | v = 20) =
1
4
Now by Bayes’ theorem
P(v = 20 | t1 = L ∧ t2 = L) =
= [ P(t1 = L ∧ t2 = L | v = 20) · P(v = 20) ] / P(t1 = L ∧ t2 = L) =
=
1
4 · 1
2
P(t1 = L ∧ t2 = L)
=
1
8 · P(t1 = L ∧ t2 = L)
But by the law of total probability
P(t1 = L ∧ t2 = L) =
= P(t1 = L ∧ t2 = L | v = 10)P(v = 10)+
+ P(t1 = L ∧ t2 = L | v = 20)P(v = 20)
= 1 ·
1
4
+
1
4
·
1
2
=
3
8
which gives P(v = 20 | t1 = L ∧ t2 = L) = 1
3 . 330
Winner’s Curse (Cont.) : Proof of Equation (32)
Finally, similarly as for (31),
P(v = 10 ∧ t2 = L | t1 = L) =
= P(v = 10 ∧ t2 = L | t1 = L ∧ t2 = L) · P(t2 = L | t1 = L)
+ P(v = 10 ∧ t1 = L | t1 = L ∧ t2 = H) · P(t2 = H | t1 = L)
= P(v = 10 | t1 = L ∧ t2 = L) · P(t2 = L | t1 = L)
=
2
3
·
3
4
=
1
2
Here P(v = 10 | t1 = L ∧ t2 = L) = 2
3 follows from
P(v = 20 | t1 = L ∧ t2 = L) = 1
3 and P(v = 30 | t1 = L ∧ t2 = L) = 0.
331