Northwestern
University
ECON 410-3: Spring 2015
TA Session 4
Prof. Jeff Ely
TAs: Arjada Bardhi
Egor Starkov
Problem 1
See solutions to TA session 3.
Problem 2
In Congress, before voting on a certain topic it is often the case the a Congressional committee is formed to
investigate the subject. Then, after the committee investigates the subject and suggests a policy to the entire
Congress a vote is held. However, it has been noted that if the committee has different preferences than the entire
congress it might not have incentives to report truthfully. This problem is known in the economic literature as
a Cheap Talk problem.
To analyze this setup we can use the following setup (implicitly thinking of the median committee member
and the median congress member and a median voter theorem):
Assume the policy is chosen from the real line, the state of nature ω ∼ U (0, 1). Denote the policy suggested
by the committee (the Sender) as m and the policy implemented by congress (the Receiver) as a.
The timing of the game is as follows: First the committee learns the state of nature, then the committee
suggest a policy, m ∈ R. After learning m the congress selects a policy a ∈ R to implement, this is known as an
open rule as congress may select any policy regardless of the message m.
The payoffs are given by us (a, ω) = −(a − (ω + b))2 ur (a, ω) = −(a − ω)2 , where b ∈ (0, 1).
1. Draw the extensive form of the game, and write down the strategy sets.
The extensive form of the game is such that: Nature (N) moves first by choosing ω; Sender (S), after
observing the chosen ω plays m; the Receiver (R) observes m but not ω (so all nodes m after different ω
are in the same information set for the Receiver) and chooses a. For a drawing of the game tree, see your
section notes. A strategy for S is: m : (0, 1) → R, while a strategy for R is a : R → R.
2. Show that there always exists a ’babbling’ equilibrium in which m is uninformative about the realized state
of nature.
Let m(ω) = 1 for all ω ∈ (0, 1). By Bayes’ rule, the belief µ(ω|m) follows the uniform distribution U (0, 1).
The strategy of R after observing m = 1 is to choose a that maximizes:
Z
max
a
1
−(a − ω)2 dω
0
Then, a∗ (1) = 1/2. We also need to specify the behavior of players off the equilibrium path. Let a(m) = 21
for off-path messages (m 6= 1) as well. For beliefs, suppose that R believes that ω ∼ U (0, 1) for all m. Such
a PBE is completely uninformative, as we wanted.
3. Look for informative Perfect Bayesian Equilibria of the form that if ω ∈ [0, ω ∗ ] the committee reports ’low’,
and otherwise it reports ’high’. Are such equilibria unique? Do they have a unique outcome distribution?
Such an informative PBE is of the form: m(ω) = mL for ω ∈ (0, ω ∗ ) and m(ω) = mH 6= mL for ω ∈ [ω ∗ , 1).
The on-path beliefs and on-path response of R are such that:
µ(mL ) ∼ U (0, ω ∗ ) ⇒ a(mL ) =
µ(mH ) ∼ U [ω ∗ , 1) ⇒ a(mH ) =
∗
ω∗
2
1 + ω∗
.
2
∗
1
∗
∗
∗
To solve for ω ∗ , we need to set us ( ω2 , ω ∗ ) = us ( 1+ω
2 , ω ). This gives ω = 2 − 2b. For ω to∗ be strictly
positive, it must be that b < 41 . For off-path belief, let µ(m) = µ(mH ) (hence, a(m) = 1+ω
2 ) for any
Problem 2 continued on next page. . .
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Northwestern
University
Problem 2 (continued)
ECON 410-3: Spring 2015
TA Session 4
Prof. Jeff Ely
TAs: Arjada Bardhi
Egor Starkov
m 6= mL . S has no incentives to deviate. The informative equilibrium of this form (with two on-path
messages) exists for b < 41 and is unique where it exists.
4. Find a condition on b, such the equilibrium found in (b) is not the most informative equilibrium of the
game.
Let us attempt to construct a more informative equilibrium, i.e. an equilibrium where S pools over more
than two intervals. Consider a more informative equilibrium of the form: m(ω) = mL for ω ∈ (0, ω ∗ ),
m(ω) = mM for ω ∈ [ω ∗ , ω ∗∗ ), m(ω) = mH for ω ∈ [ω ∗∗ , 1). Following a similar analysis as in part 3,
we solve for the threshold points: (ω ∗ , ω ∗∗ ) = (1/3 − 4b, 2/3 − 4b). But, we need: 0 < ω ∗ < ω ∗∗ < 1, so
b < 1/12.
Suppose now that instead of using an open rule, congress may chose between a status quo policy αs = 21
or the policy suggested by the committee αc , this is known as a closed rule. Assume b < 61 .
5. Find a non-babbling equilibrium of this new game. Hint: look for equilibrium with four different regions.
A low region where S’s optimal action is played, a medium region where the status quo is played, a high
region where a constant action is taken, and a very high region in which S’s optimal action is played.
A non-babbling equilibrium of this new game is of the form:


ω + b if ω ∈ (0, ω1 )


 1 if ω ∈ [ω , ω )
1
2
αs (ω) = 2

c if ω ∈ [ω2 , ω3 )




ω + b if ω ∈ [ω3 , 1)
(
αr (αs ) =
αs on path
αc off path


P r(ω = αs − b) = 1 if αs ∈ (b, ω1 + b)





U [ω1 , ω2 ) if αs = 21

µ(αs ) = U [ω2 , ω3 ) if αs = c




P r(w = αs − b) = 1 if αs ∈ [ω3 + b, 1 + b)



P r(ω = 1 ) = 1 if otherwise
2
We use the indifference conditions for S to solve for (ω1 , ω2 , ω3 ) = ( 12 − b, 2c − b + 14 , c − b). To solve for c,
we use an indifference condition for R between accepting αc = c and αs = 1/2:
Z
ω3
−(c − ω)2
ω2
1
dω =
ω3 − ω2
Z
ω3
ω2
1
1
−( − ω)2
dω
2
ω3 − ω2
To get rid of the integrals, let us observe that for z ∼ U [x, y]:
Z
y
x
1
(a − z)
dz =
y−x
2
Z
y
x
(a2 − 2az + z 2 )
1
dz = a2 − 2aE[z] + E[z 2 ] =
y−x
= a2 − 2aE[z] + (E[z])2 + V ar(z) = (a − E[z])2 + V ar(z)
From here,
c−
ω2 + ω3
1 ω2 + ω3
1
=− +
⇔ c = 4b + .
2
2
2
2
Hence, (ω1 , ω2 , ω3 ) = ( 12 − b, 12 + b, 3b + 12 ). Having constructed this PBE, we can check that neither S nor
Problem 2 continued on next page. . .
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Northwestern
University
Problem 2 (continued)
ECON 410-3: Spring 2015
TA Session 4
Prof. Jeff Ely
TAs: Arjada Bardhi
Egor Starkov
R have a profitable deviation.
Problem 3
A worker is currently employed by a firm, called firm 2. There is also another firm, called firm 3 which is trying
to hire the worker away from firm 2. The worker’s productivity is θ, a number in the unit interval [0, 1]. This
means that he generates revenue equal to iθ if firm i = 2, 3 employs him and that firm will earn profits iθ − wi
where wi is the wage paid to the worker. Note that the worker would be more productive working for firm 3
than for firm 2. It is common-knowledge that, based on their employment experience, firm 2 and the worker
both know the true value of θ. The worker’s current wage is 0.
Firm 3 is trying to hire the worker away from firm 2 by making a wage offer w3 ≥ 0 to the worker. Firm 3
does not know the true value of θ and it is common knowledge that firm 3 believes it is distributed uniformly
on the unit interval. The offer made by firm 3 is observable by both the worker and firm 2. If firm 3 makes an
offer, then firm 2 will make a competing offer to the worker in attempt to retain him. Let w2 ≥ 0 denote any
competing offer made by firm 2.
After the offer is made by firm 3 and the competing offer, the worker observes both offers and decides which
one to accept. The payoffs are as follows. If the worker accepts the offer of firm i = 2, 3, then the worker’s payoff
is wi , and firm i’s payoff is iθ − wi , with the other firm getting a payoff of zero.
1. Draw the extensive form of this game, indicating a move by nature selecting the worker’s type and all of
the appropriate information sets.
First nature chooses θ, then firm 3 makes an offer w3 , then firm 2 makes a counteroffer w2 (θ, w3 ), and
finally worker makes a choice between offers given w2 , w3 , θ.
2. What is the set of strategies for firm 2?
Firm 2 sets a wage offer given worker’s type and firm 3’s offer, so a strategy for firm 2 is a mapping
[0;1]×R+
σ2 : [0; 1] × R+ → R+ . Set of strategies of firm 2 is the set of such mappings, i.e. R+
.
3. Find a weak Perfect Bayesian Equilibrium of this game. Remember to describe fully the strategies of all
players. Show that it is a weak PBE.
Firm 3 offers w3 = 0. Firm 2’s strategy is
(
2
3
w (w ) =
w3
if 2θ ≥ w3
0
otherwise
And the worker accepts the highest offer, breaking ties by accepting the offer of firm 2. This is a weak
PBE because the worker’s acceptance decision is sequentially rational, and given the worker’s acceptance
decision, the strategy of firm 2 is sequentially rational. Then the strategy of firm 3 is sequentially rational
because any offer w3 greater than zero will be accepted if and only if the worker’s productivity is θ < w3 /2,
giving a conditionally expected productivity of w3 /4 and a conditional expected profit of 3w3 /4 − w3 which
is worse than the zero utility obtained in equilibrium.
4. Describe as completely as you can the set of all outcomes consistent with weak PBE in this game.
In any weak PBE the worker stays with firm 2 and earns a wage of zero.
Problem 4
Roy is coming to plant flowers in Zoe’s garden. Zoe loves flowers, her utility for a garden with x flowers is
z(x) = x
Problem 4 continued on next page. . .
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Northwestern
University
Problem 4 (continued)
ECON 410-3: Spring 2015
TA Session 4
Prof. Jeff Ely
TAs: Arjada Bardhi
Egor Starkov
Roy plants a unit mass of seeds and the fraction of these that will bloom into flowers depends on how attentive
Roy is as a gardener. Roy’s attentiveness is his type θ. In particular when Roy’s type is θ, absent any sabotage
by Zoe, there will be θ flowers in Zoe’s garden in Spring. Roy’s attentiveness is unknown to everyone and it is
believed by all to be uniformly distributed on the unit interval.
Jane, Zoe’s neighbor, is looking for a gardener for the following Spring. Jane has high standards, she will
hire Roy if and only if he is sufficiently attentive. In particular, Jane’s utility for hiring Roy when his true type
is θ is given by
2
j(θ) = θ − .
3
(Her utility is zero if she does not hire Roy.) Roy tends to one and only one garden per year. Therefore Roy will
continue to plant flowers in Zoe’s garden for a second year if and only if Jane does not hire him away from her.
Consequently, Zoe is contemplating sabotaging Roy’s flowers this year. If Zoe destroys a fraction 1 − α of
Roy’s seeds then the total number of flowers in Zoe’s garden when Spring arrives will be x = αθ. Of course
sabotage is costly for Zoe because she loves flowers. There will be no sabotage in the second year because after
two years of gardening Roy goes into retirement. Therefore, if Zoe destroys 1 − α in the first year and Roy
continues to work for Zoe in the second year, Zoe’s total payooff will be
z(αθ) + z(θ)
whereas if Roy is hired away by Jane, then Zoe’s total payoff is just z(αθ).
This is a two-player (Zoe and Jane) extensive-form game with incomplete information. The timing is as
follows. First, Roy’s type is realized. Nobody observes Roy’s type. Zoe moves first and chooses α ∈ [0, 1]. Jane
does not observe α but does observe the number of flowers in Zoe’s garden. Then Jane chooses whether or not
to hire Roy away from Zoe. Then the game ends.
This question is about the set of Perfect Bayesian Equilibria.
1. Draw the extensive form of this game.
2. What is the (pure) strategy space for Jane?
3. Show that Jane always has a best-reply which is of the threshold variety: there is a threshold T ∈ [0, 1]
such that she hires Roy if and only if there are more than T flowers in Zoe’s garden.
4. Find the set of all Perfect Bayesian equilibria in which Jane uses a threshold strategy.
(a) Describe each equilibrium fully, and
(b) Prove that there are no other threshold equilibria.
5. Are there any other Perfect Bayesian equilibria?
Solution:
The set of pure strategies for Zoe is AZ = [0, 1] and for Jane is AJ = {h : [0, 1] → {0, 1}} which specifies for
each level of αθ if Jane hires Roy or not.
Let α ∈ [0, 1] and lets look for the best reply of Jane. Given an observed t1 = αθ1 , if it is optimal for Jane to
hire then it must be optimal to hire for any observed t2 > t1 (this is because it is optimal to hire iff θ1 = tα1 ≥ 32 ).
Therefore is Zoe’s strategy is α ∈ [0, 1] the BRJ (α) is a set of threshold strategies, equal up to measure zero
transformations.
The claim is not actually true for mixed strategies (as pointed out by Bernard in the TA session), for the
counterexample let
(
1 with probability 21
α= 1
with probability 12
2
Problem 4 continued on next page. . .
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Northwestern
University
Problem 4 (continued)
ECON 410-3: Spring 2015
TA Session 4
Prof. Jeff Ely
TAs: Arjada Bardhi
Egor Starkov
then
11 1
; 2 ]) = 12 z + 12 2z =
E(θ|αθ = z ∈ [ 24
3z
2
3
∈ [ 33
48 ; 4 ] > 2/3
While
7
7
E(θ|αθ = z ∈ ( 12 , 12
]) = z ∈ [ 21 ; 12
] < 2/3.
11 1
7
In particular the best response includes hiring Roy for some αθ ∈ [ 24
, 2 ] but not for αθ ∈ ( 12 , 12
] and so it is
not of the threshold kind. However, the claim is that even though Jane’s BR to a mixed strategy of Zoe is not
necessarily a threshold strategy, it is still true that Zoe is not willing to mix in equilibrium1 so we can restrict
our attention to pure strategies of Zoe.
Now lets see what is Zoe’s best response to a given threshold strategy by Jane h(x) = 1[x > T ]. The problem
is to maximize the expected value
Z
max Eθ [αθ + θ(1 − h(αθ))] =
α
1
[αθ + θ(1 − h(αθ))]dθ
0
Z min{ Tα ,1}
α
= +
θdθ
2
0
α
T2 1
= + min{ 2 , }
2
2α 2
1+T
=
2
where the max is attained at α∗ = T . (The last step is easily seen from the fact that the objective function is
increasing in [0, T ] and it is strictly convex in [T, 1], so it is just a matter of comparing α = T, 1).
Observe also that the optimal threshold for Jane for a given α > 0 is just 32 α, and for α = 0 Jane’s BR is
anything where she does not hire when observing t = 0 (so no hiring over the eq. path).
Combining this facts the unique PBE strategies are for Zoe to set α = 0 and for Jane to hire iff getting a
positive signal, so h(x) = 1[x > 0]. There is multiplicity in the PBE beliefs for Jane which just require her to
believe θ ≥ 23 whenever she observes αθ > 0 (although this never happens because its off the equilibrium path).
1 Given Zoe’s mixed strategy α, define its upper bound ᾱ = sup (support (α)) and observe that any h ∈ BR (α) should prescribe
J
h(αθ) = 1 if αθ ≥ 23 ᾱ. Define α0 = α · 1[α ≤ 23 ᾱ] and see Z(α0 , h) > Z(α, h) for all h ∈ BRJ (α). This means if ᾱ > 0 then α is not
part of any equilibrium. [This argument might be incorrect.]
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