Answers to Problem Set 6
Economics 703
Spring 2016
1. The expected quality conditional on the wage is
E[θ | r(θ) ≤ w] = E[θ | θ ≤ w/a].
Let H(w) denote the function on the right–hand side. One has to be careful to distinguish
cases where w/a is in the interval [θ, θ̄] from those cases where it is outside this range.
First, suppose w/a ≤ θ. When this inequality is strict, the expectation is not well defined
since we’re conditioning on a zero probability event. When w/a = θ, the expectation is
θ. That is, H(aθ) = θ. Since a < 1, aθ < θ. So w < H(w) at w = aθ. Hence at this
w, demand is infinite and supply is zero, so we don’t have an equilibrium. Note that the
same conclusion would follow if we defined H(w) = θ for w < aθ.
Next consider the possibility that w/a ≥ θ̄. For any such w, H(w) is the unconditional
expectation of θ which is 21 (θ + θ̄). So is H(w) above or below w at w = aθ̄? It is above
the diagonal if and only if
1
aθ̄ ≤ (θ + θ̄).
(1)
2
When this holds, H(w) > w for all w ∈ [aθ, aθ̄]. H then stays at the unconditional
expectation for all higher w, so the unique equilibrium in this case is where w = 12 (θ + θ̄),
the unconditional expectation. Here all workers are hired.
So suppose that (1) does not hold. In this case, the H function crosses the diagonal
somewhere between w = aθ and w = aθ̄. In this range, we have
H(w) =
1
2
w
+θ .
a
So H(w) = w only at the point where
2aw = w + aθ
or w = aθ/(2a − 1). How do we know this is well–defined — that is, that a 6= 1/2? By
hypothesis, (1) does not hold, so we must have
1
aθ̄ > (θ + θ̄)
2
1
or (2a − 1)θ̄ > θ. Because both θ̄ and θ are strictly positive, this requires 2a > 1.
In short, we have two possibilities. If (1) holds, the unique equilibrium wage is the
unconditional expectation. If (1) does not hold, it is aθ/(2a − 1). It is not hard to show
that the equilibrium wage is weakly decreasing in a. It’s strictly decreasing in a in the
region where (1) does not hold. But once a is small enough that (1) holds, the wage no
longer depends on a, so it’s only weakly decreasing in a in general. Similarly, the wage
is weakly increasing in θ̄. It is strictly increasing in θ.
If we set θ = 0, our first step changes. Above, we said that aθ < θ, but this is not true
when θ = 0. Hence we see that w = 0 is an equilibrium since H(0) = θ = 0. Also, in this
case, (1) holds iff a ≤ 1/2. If a < 1/2, H is strictly above the diagonal for all w ∈ (0, aθ̄).
As a result, we have two equilibria: w = 0 and w equal to the unconditional expectation.
If a = 1/2, H is the diagonal, so every wage between 0 and the unconditional expectation
of θ is an equilibrium. If a > 1/2, H is strictly below the diagonal for all w > 0, so the
unique equilibrium is at w = 0.
Finally, suppose θ = 0, θ̄ = 1, and r(θ) = aθ − b. Now the conditional expectation
we are interested in is
#
"
w+b
.
H(w) = E θ | θ ≤
a
For all w ≥ 0, (w + b)/a > 0. For all w ≤ 1/2, (w + b)/a < 1 by the assumption that
a > b + .5. Since 1/2 is the unconditional expectation, this tells us that (w + b)/a ≤ 1 for
the “relevant range” of values of w. Since H(0) > 0 and H(1/2) < 1/2, there is a unique
w ∈ (0, 1) satisfying the equilibrium condition. This w is defined by w = (1/2)(w + b)/a,
so w = b/(2a − 1). Now the wage is increasing in b and decreasing in a.
2. (a) In the complete information case, the interest rate will depend on the type of
the entrepreneur. Let ri be the interest rate for type i, where i is either h or `. By
assumption, this interest rate is determined by pi (1 + ri )L − L = 0 or 1 + ri = 1/pi . The
entrepreneur takes the loan if
pi [S − (1 + ri )L] + (1 − pi )D ≥ 0.
Substituting for 1 + ri and rearranging gives
pi S + (1 − pi )D − L ≥ 0.
By assumption, p` S + (1 − p` )D − L ≥ D > 0, so bad entrepreneurs take the loan. By
assumption, ph > p` and S > D, so good entrepreneurs take the loan also.
(b) There are three possibilities for equilibrium. First, it could be that only the h types
take the loan. In this case, we know from the above that 1 + r = 1/ph . Second, it
could be true that both types take the loan. In this case, we have 1 + r = 1/p̄ where
p̄ = λph + (1 − λ)p` . Finally, it could be true that only the ` types take the loan. In this
case, 1 + r = 1/p` .
2
Is the first an equilibrium? From the above, we know that the h types would take
this loan. What about the ` types? TheyWhen the costs are the same, there is still a
trivial separating equilibrium obtained by using indifference. That is, we can have type
II picking 0 education and type I picking eI such that 100 − 30eI = 40 or eI = 2. Both
types would be indifferent between deviating and not, so it’s an equilibrium. Of course,
it’s not particularly plausible.
As to pooling, we can have both types picking any e∗ such that 70 − 30e∗ ≥ 40 or
e∗ ≤ 1. This case is essentially the same as in (a).
would take the loan if
p` S + (1 − p` )D −
p`
L ≥ 0.
ph
Because p` < ph , the left–hand side is strictly larger than p` S + (1 − p` )D − L which we
know is positive. Hence the ` types take the loan also, contradicting the hypothesis that
only the h types do. Hence this is not an equilibrium.
When is the second an equilibrium? Low types take the loan if
p` S + (1 − p` )D ≥
p`
L.
p̄
Because p` < ph implies p` < p̄, again, low types take the loan. High types take the loan
if
ph
ph S + (1 − ph )D ≥ L.
p̄
Rearranging, this is
λph + (1 − λ)p` ≥
ph L
.
ph S + (1 − ph )D
By assumption, this does not hold. Hence only low types take the loan, contradicting
the hypothesis that both types do. Hence this is not an equilibrium either.
Finally, is the last an equilibrium? From (a), we know that the low types would take
the loan. The high types would not take the loan in part (b) and the interest rate here
is higher. Hence they will not take the loan. So this is the unique equilibrium.
3. (a) Let ei be the amount of education a type i worker gets in equilibrium. For a
separating equilibrium, we must have eII = 0. Given this, incentive compatibility holds
iff
100 − 25eI ≥ 40
40 ≥ 100 − 35eI
or 12/5 > eI > 12/7. For every eI in this range, we get a separating equilibrium by
having the beliefs assign probability 1 to type II in response to any level of education
other than eI .
3
For pooling equilibria, we know eI = eII . Let this amount of education be e∗ . Then
we know that 70 − 35e∗ ≥ 40 or e∗ ≤ 6/7. For every such e∗ , we can get a pooling
equilibrium by having the beliefs assign probability 1 to type II in response to any other
level of education.
(b) When the costs are the same, there is still a trivial separating equilibrium obtained
by using indifference. That is, we can have type II picking 0 education and type I picking
eI such that 100−30eI = 40 or eI = 2. Both types would be indifferent between deviating
and not, so it’s an equilibrium. Of course, it’s not particularly plausible.
As to pooling, we can have both types picking any e∗ such that 70 − 30e∗ ≥ 40 or
e ≤ 1. This case is essentially the same as in (a).
∗
(c) Let ei be the education of type i as before. Just as before, the lowest productivity
type, which is now type III, must choose 0 education. So the incentive compatibility
constraints are
100 − 25eI ≥ 40 − 25eII , 30
40 − 35eII ≥ 100 − 35eI , 30
30 ≥ 100 − 45eI , 40 − 45eII .
You can rearrange these to
14/5 ≥ eI ≥ 14/9
2/7 ≥ eII ≥ 2/9
12/5 ≥ eI − eII ≥ 12/7
For any eI and eII satisfying these inequalities, we can get a fully separating equilibrium
by setting the beliefs so that type III is inferred in response to any level of education
other than eI or eII .
For pooling, we’d have eI = eII = eIII . Let e∗ be the common education level. For
this to be an equilibrium, we require (170/3) − 45e∗ ≥ 30 or e∗ ≤ 16/27. Again, for any
such e∗ , we can construct a pooling equilibrium.
4. Just as in the standard Spence model, the employers must both get zero√expected
profits. Because the beliefs are degenerate, this means that they must pay 4 e to the
worker if he is believed to be type tA and 2 if he is believed to be type tB . Note that this
is true both on and off the equilibrium path.
Let eA and eB be the education choices by the worker of type tA and tB respectively
√
in a separating equilibrium. By definition, then, eA 6= eB , so eA must lead to wage 4 eA
and eB must lead to a wage of 2. Hence the equilibrium conditions we require are
√
4 eA − eA ≥ 2 − eB
√
2 − 3eB ≥ 4 eA − 3eA .
4
As long as eA and eB satisfy these conditions, we know that neither type will deviate to
the other type’s education level.
So the only other conditions we require are those that make sure that neither type
will deviate to any other education level. Clearly, the way to block a deviation is to
make the worst √
possible inference in response to that deviation. Since the wage must
be either 2 √
or 4 e, this means that the wage in response to a deviation to e should
be min{2, 4 e}. Note that this means that we cannot conclude that eB must be 0 in
equilibrium: a deviation to zero education can lead to a zero wage if the employers infer
that the worker is type tA .
Given this, we see that the type tA worker cannot get a payoff lower than
√
e} − e.
max
min{2,
4
e
√
√
Note that min{2, 4 e} = 2 for e ≥ 1/4 and is 4 e otherwise. So the√maximum cannot be
at e > 1/4 — e = 1/4 would necessarily be better.
√ Note also that 4 e − e is increasing in
e up to its maximum at e = 4. Hence min{2, 4 e}−e is increasing in e for all e ∈ [0, 1/4].
Hence the value of the maximum is at e = 1/4 and so is 2 − (1/4) = 7/4.
Similarly, type tB cannot get a payoff lower than
√
max min{2, 4 e} − 3e.
e
Just as above, this is increasing in e for all e ∈ [0, 1/4] so the maximum is again at
e = 1/4. So the value of the maximum is 2 − (3)(1/4) = 5/4.
Summarizing, the set of separating equilibria is all the (eA , eB ) pairs satisfying the
incentive compatibility conditions above and
√
7
4 eA − eA ≥
4
5
2 − 3eB ≥ .
4
You can reduce this further, but this is the answer.
5