2016-1 Final Solutions
1a) Clearly, (T,L) and (B,R) are the two pure-strategy NEs.
There are no other NEs here: suppose player 1 mixes. Then player 2’s only BR would be L.
But player 1’s BR to L is T, so player 1 can’t mix in NE. A symmetric argument shows that
player 2 can’t mix either.
1b) Clearly (k,k) for k=0,1,…,n are pure-strategy NEs. They are also the only ones: if the
players play a pure strategy profile where they pick different numbers, then the player
picking the lower number could gain by deviating to a higher number.
Once again, there are no non-pure NEs here. Suppose player 1 mixes in a NE, and let 𝑘1 be
the highest number on which he puts positive probability. Then any number below 𝑘1 cannot
be a best response for player 2: deviating to 𝑘1 leads to a higher payoff with positive
probability, and cannot lead to a lower payoff. (In fact, any number greater than or equal to
𝑘1 is a best response.) Therefore, player 2 will not pick any number below 𝑘1 with positive
probability in this NE. Thus, player 1 would profit from deviating to never picking a number
below 𝑘1 , which violates our assumption. A symmetric argument shows that player 2 can’t
mix either.
2a) The inverse demand function is P = 20-Q/3.
Firm 1’s profit function is (P-8)q1 = (12-q2/3)q1 – q12/3.
Firm 1’s f.o.c. is therefore 12-q2/3 = 2q1/3.
Similarly, firm 2’s f.o.c. is 14-q1/3 = 2q2/3.
Solving this system of equations gives q1=10 and q2=16.
Because there is only one subgame here, by definition, SPE and NE are the same.
2b) We proceed by backward induction.
Firm 1’s f.o.c. in the last step is 16-q2/3 = 2q1/3, or q1=24-q2/2. Given the capacity constraint,
we actually have q1=min{k,24-q2/2}.
Firm 2’s profit function is still (14-q1/3)q2 – q22/3.
- If q2 ≥ 48-2k, then q1 = 24-q2/2. The profit function then becomes (6+q2/6)q2 – q22/3, or
6q2 - q22/6. The f.o.c. is 6 = q2/3, so q2 = 18. This is valid if k ≥ 15, and yields profit 54.
- If q2 ≤ 48-2k, then q1 = k is a constant. The f.o.c. is the same as in part a: q2 = 21-k/2.
This is valid if k ≤ 18, and yields profit (14-k/3)q2 – q22/3 = (21-k/2)2/3.
The two cases above identified local maxima. When 15 ≤ k ≤ 18, there are two of them, and
we need to compare the profit in each case to find the global maximum. The profit in the first
case is greater if and only if 54 > (21-k/2)2/3, or 9√2 > 21-k/2, or k > 42-18√2 ≈ 16.544.
Therefore, if k > 42-18√𝟐, then q2 = 18; if k < 42-18√𝟐, then q2 = 21-k/2.
(For k = 42-18√2, you can pick either or even let firm 2 mix.)
Note: An unfortunate mark on the exam sheet made the demand look like Q = 60-.3P instead of
Q = 60-3P. In that case:
2a) The inverse demand function is P = 200-10Q/3.
Firm 1’s profit function is (P-8)q1 = (192-10q2/3)q1 – 10q12/3.
Firm 1’s f.o.c. is therefore 192-10q2/3 = 20q1/3.
Similarly, firm 2’s f.o.c. is 194-10q1/3 = 20q2/3.
Solving this system of equations gives q1=19 and q2=19.6.
2b) Firm 1’s f.o.c. in the last step is 196-10q2/3 = 20q1/3, or q1=29.4-q2/2. Given the capacity
constraint, we actually have q1=min{k,29.4-q2/2}.
Firm 2’s profit function is still (194-10q1/3)q2 – 10q22/3.
- If q2 ≥ 58.8-2k, then q1 = 29.4-q2/2. The profit function then becomes (96+5q2/3)q2 –
10q22/3, or 96q2 - 5q22/3. The f.o.c. is 96 = 10q2/3, so q2 = 28.8. This is valid if k ≥ 15,
and yields profit 1382.4.
- If q2 ≤ 58.8-2k, then q1 = k is a constant. The f.o.c. is the same as in part a: q2 = 29.1-k/2.
This is valid if k ≤ 19.8, and yields profit (194-10k/3)q2 – 10q22/3 = 10(29.1-k/2)2/3.
The two cases above identified local maxima. When 15 ≤ k ≤ 19.8, there are two of them, and
we need to compare the profit in each case to find the global maximum. The profit in the first
case is greater if and only if 1382.4 > 10(29.1-k/2)2/3, or 14.4√2 > 29.1-k/2, or k > 58.228.8√2 ≈ 17.471. Therefore, if k > 58.2-28.8√𝟐, then q2 = 28.8; if k < 58.2-28.8√𝟐, then q2
= 29.1-k/2.
(For k = 58.2-28.8√2, you can pick either or even let firm 2 mix.)
2c) If firm 1 builds a large capacity, then we essentially have a Stackelberg game where firm 2
moves first, and firm 2 has an incentive to produce more in order to cut firm 1’s production
by reducing the demand left for firm 1. Because the capacity is non-binding (and its cost is
sunk), q2 does not depend on k when k is large enough.
If firm 1’s capacity is relatively low, then it becomes binding, so that firm 1’s production is
fixed (at least for reasonable quantities by firm 2). In that case, firm 2 simply best responds
to firm 1’s capacity as if firm 1 has already picked quantity, and therefore no longer has an
incentive to increase production in order to reduce firm 1’s quantity.
3a) No. Suppose we’re at the start of stage 2, and (C,D) was played in stage 1. According to this
strategy profile, player 1 should now play D forever, while player 2 should play C in stage 2.
This is not a best response for player 2: knowing that, going forward, player 1 defects no
matter what happens, player 2 should also defect every period.
(Note: This is why I really insisted in class about watching out for the relevant “categories”
of history. In grim trigger, (C,D) and (D,D) were in the same category (i.e. they lead to the
same play in subsequent periods), while in this strategy profile, they aren’t.)
3b) It suffices to find all values of 𝛿 for which “grim trigger” is a SPE: since grim trigger
specifies the harshest possible punishment for deviating, if it cannot sustain cooperation, then
no other strategy profile can.
10
3𝛿
𝟏𝟎
We need: 1−𝛿 ≥ 20 + 1−𝛿, i.e. 𝜹 ≥ 𝟏𝟕.
4) Nothing. Two possible justifications:
- All we’re assuming are the four axioms of expected utility theory (completeness,
transitivity, continuity and independence).
-
Utility already captures different attitudes towards risk. If we assumed players maximize
expected wealth, we’d be assuming risk-neutrality, but that is not the case.
5a) Insurance companies tend to be large because they work by pooling risk. Therefore, each
shareholder usually only owns a small part of the firm, and thus is only subject to a small
share of the risk associated with each policy. If shareholders have a smooth utility function,
then over small stakes, the function is almost linear, and thus shareholders are almost riskneutral.
5b) First, we know that the firm’s first best must involve full insurance: otherwise, since Sue is
risk-averse, the firm could take on any remaining risk and charge Sue more than the expected
loss associated with that risk.
i)
Here, the firm simply makes Sue indifferent between being insured and uninsured.
o If c=0, Sue’s expected utility without insurance is 0.25(-5) + 0.75 (-1) = -2, so her
certainty equivalent is $5,000. Therefore, she is willing to pay $10,000-$5,000 =
$5,000. The firm’s expected profit is $5,000 – 0.25($8,000) = $3,000.
o If c=1, Sue’s expected utility without insurance is 0.15(-5.3) + 0.85 (-1.3) = -1.9,
so her certainty equivalent is $6,250. Therefore, she is willing to pay $10,000$6,250 = $3,750. The firm’s expected profit is $3,750 – 0.15($8,000) = $2,550.
ii)
Here, the firm gets to decide whether Sue has to choose c=1, and again makes Sue
indifferent between being insured and uninsured.
First, we find Sue’s outside option: her expected utility is lower with c=0 (-2) than
with c=1 (-1.9). Thus, the firm needs to let Sue have expected utility -1.9.
Now, Sue is allowed to choose c=0, she would be willing to pay up to $10,000(1 1/1.9) = $90,000/19 ≈ $4,737. The firm’s expected profit is $4,737 – 0.25($8,000) =
$2,737.
If the firm requires c=1, then like in part i, it will make $2,550.
Thus, the firm chooses c=0 and makes profit approximately $2,737.
5c) No. As explained in part b, the first best involves offering full insurance at price $5,000 for
type c=0, and price $3,750 for type c=1. If the firm cannot observe c, then type c=0 will
pretend being type c=1.
5d) This is second-degree (or quantity; or indirect) price discrimination. The firm does not
necessarily achieve its second best by selecting the optimal value of a because, in the
second best, the premium does not have to be proportional to x.
5e) On the one hand, the firm wishes to insure Sue as much as possible, for the same reason as in
the first best. On the other hand, if Sue is insured too much, she will not have enough
incentive to pick c=1. The contract must ensure that Sue is at least as well off as being
uninsured (IR constraint) and that she has enough incentive to pick c=1 (IC constraint). Both
these constraints are binding: if the IR doesn’t bind, the firm could charge a higher
premium1; if the IC doesn’t bind, the firm could increase both the premium and the payout in
1
Note that a higher premium, by making Sue poorer, increases her marginal utility of wealth, and therefore makes
her more likely to pick c=1 to avoid a loss. Alternatively, the firm can decrease the policy’s payout while keeping
case of loss in such a way as to maintain Sue’s expected utility (so as to not violate IR). The
latter increases the firm’s expected profit because it reduces the risk borne by Sue, which she
is willing to pay for since she is risk-averse.
6a) Transfers/welfare do depend on who is assigned the property rights. “Quantity” would be a
better word.
6b) By the Coase theorem, the efficient quantity is achieved. If there were income effects, this
quantity would depend on who is assigned the property rights.
7) This statement is likely false: by creating a uniform price for carbon emissions, tradable
permits will, in theory, lead to exactly the actions reducing emissions whose costs are below
that price. That is, costly initiatives to reduce emissions will be taken if and only if their cost
is below a certain threshold. By contrast, with industry-specific regulations, there is no
guarantee that this will occur: unless the regulations are exactly right, high-cost initiatives
may be taken while low-cost initiatives are overlooked.
the same premium (this corresponds to “lowering the wage for bad outcome” in the principal-agent problem from
class).