Answers to Problem Set 7
Economics 703
Spring 2016
1. From question 3:
(a) Any separating equilibrium with 100 − 35eI < 40 can be eliminated by Cho–
Kreps if we take the unsent message to be any e slightly smaller than eI . As long as
this e is strictly smaller than eI , type II will not deviate even if believed to be type I
in response. Then type I will gain if believed to be type I, so Cho–Kreps eliminates the
equilibrium. Hence the only surviving separating equilibrium is where eI = 12/7. As for
pooling, suppose we have a pooling equilibrium at e∗ . Consider a deviation to e0 such
that 100 − 35e0 < 70 − 35e∗ and 100 − 25e0 > 70 − 25e∗ or 6/7 < e0 − e∗ < 6/5. The type
II worker prefers the equilibrium to the deviation even if the beliefs are that he is type I
in response. The type I worker will deviate if the belief in response is that he is type I.
Hence Cho–Kreps rules out all these equilibria.
(b) Cho–Kreps cannot eliminate any equilibria here. Since both types have exactly
the same preferences, there is no way to find a deviation that one type strictly prefers to
the equilibrium, while the other type strictly prefers the equilibrium to the deviation.
(c) For Cho–Kreps, consider any separating equilibrium with 40 − 35eII > 100 − 35eI .
These are analogous to the separating equilibria Cho–Kreps eliminates in the two–type
case, so it’s natural to suppose they will get eliminated here. To see that they will, fix a
deviation to e0 slightly less than eI . If we pick it closely enough, we’ll continue to have
40 − 35eII > 100 − 35e0 . Also, note that this implies
35(e0 − eII ) > 60
which implies 45(e0 − eII ) > 60. Hence
40 − 45eII > 100 − 45e0 .
Since incentive compatibility already told us that 30 ≥ 40 − 45eII , we see that neither
type II nor type III would deviate to e0 even if this leads to the belief that he is type I.
Since type I gains from the deviation if this is the belief in response, we can eliminate
any such separating equilibrium with Cho–Kreps.
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All other separating equilibria survive Cho–Kreps. To see why, fix such a separating
equilibrium and try to find a useful deviation. A deviation to e0 for which 30 ≤ 100 − 45e0
can’t help break an equilibrium. Such a deviation would attract type III for some beliefs
(namely, beliefs which put probability 1 on type I). Hence to use Cho–Kreps, we’d have
to find a type that would deviate even if the deviation led to a belief that the type
is III. But no type would prefer this. So the only deviations we need to consider are
those with 30 > 100 − 45e0 . Because 30 ≤ 100 − 45eII (because this inequality says
eII ≤ 70/45 and we know that eII < 2/7 in any separating equilibrium — see above),
all such deviations have e0 > eII . Since this is larger than eII , we cannot attract any
type to this deviation unless it leads to the belief that the type is type I. Hence it must
be a deviation that neither type II nor type III would consider for any possible belief.
Because 30 > 100 − 45e0 , we’ve already made sure of this for type III. To make sure type
II wouldn’t deviate to e0 for any belief, we’ll need 40 − 35eII > 100 − 35e0 . But the only
separating equilibria we’re still considering have 40 − 35eII = 100 − 35eI , so this requires
e0 > eI . But then type I won’t want to deviate to this. Hence we can’t use Cho–Kreps
to eliminate any of the other separating equilibria.
So consider pooling equilibria. Let e∗ be the level of education chosen by all three
types in a pooling equilibrium. Pick any e0 satisfying 100 − 25e0 > (170/3) − 25e∗ ,
(170/3) − 35e∗ > 100 − 35e0 , and (170/3) − 45e∗ > 100 − 45e0 . The third inequality is
implied by the second, so it’s actually not needed. So we need e0 for which 25(e0 − e∗ ) <
(130/3) < 35(e0 − e∗ ). Clearly, such e0 exist. Neither type II nor type III would deviate
from the equilibrium to e0 even if this led to the belief that they were type I. However,
type I would deviate if deviation leads to this belief. Hence Cho–Kreps eliminates all the
pooling equilibria.
In case you’re curious, there are stronger refinements than Cho–Kreps which eliminate
all equilibria except the analog of the Riley outcome.
For question 4, first,
equilibrium in which eA 6= 4. Recall that
√
√ consider a separating
e = 4 maximizes 4 e − e. Since 4 4 = 8, the best possible belief in response to a
deviation to e = 4 is that this is type tA . However, even with this belief, the payoff to
type tB from this level of education is 8 − 12 = −4. Since type tB ’s equilibrium payoff
is always at least 5/4 (as shown above), type tB would never deviate to e = 4 even with
this belief. Hence the belief in response to e = 4 must put probability 1 on type tA . But
then type tA would deviate to this since it gives his highest possible payoff. Hence any
equilibrium satisfying Cho–Kreps must have eA = 4.
Given this consider any equilibrium with eA = 4 and eB > 0. The best belief in
response to e = 0 is that this is type B since this leads to a wage of 2. Note that type
tA ’s equilibrium payoff is 4, so he would not deviate to e = 0 even for this belief. Hence
the belief in response to this must be that this is type tB . Given this, type tB would
deviate to e = 0. Hence the only separating equilibrum to satisfy the Cho–Kreps criterion
is eA = 4 and eB = 0.
2
2. (a) First we have to determine the cost of each effort level. Given the utility function
used and given ū = 0, the wage that the principal has to pay if he wants effort level e is
(g(e))2 . Hence his profits if he induces e1 are
2
2
5
(10) −
3
3
=
35
≈ 3.89.
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Similar calculations show that the profits to e2 are 61/25 ≈ 2.44 and the profits to e1 are
14/9 ≈ 1.56. Hence the optimal effort to induce is e1 . The wage payment is 25/9.
(b) Now suppose effort is not observable. In this case, it is impossible to induce the
agent to choose effort level e2 . To see this, let’s suppose we can induce this effort level.
Suppose w(·) is a wage contract which leads the agent to choose effort e2 . Following the
hint, let v1 = v(w(πH )) and let v2 = v(w(πL )). Then if this induces e2 , it must be true
that
1
1
1
2
v1 + v2 − g(e1 ) ≤ v1 + v2 − g(e2 )
3
3
2
2
and
1
2
1
1
v1 + v2 − g(e3 ) ≤ v1 + v2 − g(e2 ).
3
3
2
2
Rearranging the first inequality yields
1
(v1 − v2 ) ≤ g(e1 ) − g(e2 ),
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while the second one gives
1
(v1 − v2 ) ≥ g(e2 ) − g(e3 ).
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(Intuitively, this just says that the marginal product of extra effort is below the marginal
cost when moving from e2 up to e1 but is worth it in moving from e3 up to e2 . The
symmetry of the problem makes these marginal products the same in the two cases.)
Clearly, then, we must have g(e1 ) − g(e2 ) ≥ g(e2 ) − g(e3 ) or g(e2 ) ≤ (g(e1 ) + g(e3 ))/2.
But (g(e1 ) + g(e3 ))/2 = 1.5, while g(e2 ) = 1.6. Hence effort e2 cannot be induced. If
g(e2 ) were less than 1.5, then it could be implemented.
(c) Now that we know that e2 cannot be achieved, finding the optimal contract is a
matter of characterizing the choice between e1 and e3 . We know that a constant wage will
lead the agent to choose e3 , so the principal’s profits from inducing this effort level are
the same as those calculated in (a). Hence we only need to consider the cost of inducing
effort level e1 . Using the same notation as in part (b), we see that we need to choose the
wage so that
1
1
1
2
v1 + v2 − g(e1 ) ≥ v1 + v2 − g(e2 )
3
3
2
2
and
2
1
1
2
v1 + v2 − g(e1 ) ≥ v1 + v2 − g(e3 ).
3
3
3
3
3
Rearranging the first inequality yields
1
1
(v1 − v2 ) ≥
6
15
or v1 − v2 ≥ 2/5. Rearranging the second one gives
1
1
(v1 − v2 ) ≥
3
3
or v1 − v2 ≥ 1. Because this constraint is more severe, we can ignore the first one and
focus only on this one. This means that the incentive constraint that e1 be better than
e2 cannot be binding. Hence we can treat this just like a two effort problem with just
effort levels e1 and e3 . Hence we can use the first order condition from class:
"
#
1
f (π | e3 )
=
γ
+
µ
1
−
.
v 0 (w(π))
f (π | e1 )
Since v(w) = w1/2 , 1/v 0 (w) = 2w1/2 . Hence, letting w1 = w(πH ) and w2 = w(πL ),
1/2
2w1
or
"
1/3
=γ+µ 1−
2/3
γ µ
w1 =
+
2 4
#
2
.
Similar calculations yield
γ−µ 2
.
2
Recall that both constraints are binding. Hence v1 − v2 = 1. That is,
w2 =
γ µ γ µ
+ − + =1
2 4
2 2
so µ = 4/3. The individual rationality constraint is
2
1
5
v1 + v2 ≥ .
3
3
3
Since it must bind, 2v1 + v2 = 5. Hence
γ µ
γ µ
2
+
+ − =5
2 4
2
2
So γ = 10/3. Hence w(πH ) = 4 and w(πL ) = 1.
Which effort should the principal induce? His profits to inducing e3 , as calculated
above, are about 1.56. His profits to inducing e1 are approximately 3.67, which is still
better than the profits to inducing e3 .
4
3. (a) With a risk neutral agent, we know we can achieve the first best. There are many
ways to do this. One is to “sell the firm to the agent” as we discussed in class. More
specifically, the principal can set a wage contract of w(π) = π − k for each π ∈ {πS , πF }
where
k = max{πS − c, pπS + (1 − p)πF }.
Note that k is the maximum surplus to be divided since the agent can either choose eH
and generate πS with probability 1 at a cost of c or choose eL and generate πS with
probability p at zero cost. The agent then will choose eH if πS − c > pπS + (1 − p)πF , eL
if the reverse strict inequality holds, and be indifferent otherwise.
(b) Now the specific contract we used in part (a) is no longer feasible. To see this,
note that given the function above, we have
w(πF ) = πF − max{πS − c, pπS + (1 − p)πF } ≤ πF − (pπS + (1 − p)πF ) = p(πF − πS ) < 0.
Let calculate the optimal contract by finding the best way to induce each level of effort
and then find the best effort.
To induce the low effort, the principal can simply pay a flat wage of 0. This will
induce the agent to choose low effort and leaves him indifferent between accepting and
rejecting the contract. Clearly, the principal cannot induce eL at a lower cost.
Let’s turn to the high effort then. So the principal chooses wS (short for w(πS )) and
wF (or w(πF )) to minimize wS subject to wS ≥ 0, wF ≥ 0,
wS − c ≥ 0
and
wS − c ≥ pwS + (1 − p)wF
The non–negativity constraints together with the last constraint imply wS − c ≥ 0 so
we can ignore this constraint. The last constraint says wS ≥ wF + c/(1 − p). Since
c/(1 − p) > 0 and we have to have wF ≥ 0, we see that the constraint that wS ≥ 0 also
must hold and so can be ignored. In short, the principal minimizes wS subject to
wS ≥ wF +
c
1−p
wF ≥ 0.
Clearly, the solution is wF = 0 and wS = c/(1 − p).
Just as in the model discussed in class, we see that the cost of inducing the low effort
is the same with or without the limited liability constraint, while the cost of inducing
the high effort is higher with limited liability since it goes from c to c/(1 − p) > c. Thus
if the induced effort changes when we add the constraint, it changes from eH to eL .
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