Answers to Problem Set 2
Economics 703
Spring 2016
1. Recall that the Cournot duopoly game had Si = [0, ∞) for i = 1, 2 and payoff functions
of
u1 (s1 , s2 ) = [a − s1 − s2 ]s1 − cs1
and analogously for player 2. This game is strictly dominance solvable.
First, we have to identify strictly dominated strategies. Note that
∂u1
= a − s2 − c − 2s1 .
∂s1
Since s2 ≥ 0,
∂u1
≤ a − c − 2s1 .
∂s1
Hence at any s1 > (a − c)/2, the derivative is negative, no matter what s2 is. In other
words, for any s2 , the payoff to s1 = (a − c)/2 must be larger than the payoff to any
larger s1 . Hence any s1 > (a − c)/2 is strictly dominated. The analogous statement is
true for player 2 as well. Hence we can eliminate all such strategies. (Remember that
with strict dominance solvability, we will be okay if we miss some eliminations, as long
as we get them eventually. For this reason, we don’t need to make sure that no other s1
is strictly dominated.) Let s̄1 = (a − c)/2.
So now we’ve reduced the game to one with strategy sets [0, s̄1 ] for each player. Now
we see that for any s2 ≤ s̄1 ,
1−c
∂u1
≥ a − s̄1 − c − 2s1 =
− 2s1 .
∂s1
2
Hence for any s1 < (a − c)/4, the derivative is definitely positive for any s2 in the relevant
range. That is, given the strategies we’ve eliminated for player 2, setting s1 = (a − c)/4
strictly dominates any smaller s1 . Hence we can eliminate strategies below (a − c)/4. Let
s2 = (a − c)/4.
Now we’ve reduced the game to having strategy sets [s2 , s̄1 ]. The pattern should be
becoming clear: given an upper bound on s2 of s̄, player 1 will always set s1 > (a−c−s̄)/2.
1
Given a lower bound on s1 of s, player 1 will always set s1 < (a − c − s)/2. We started
with s = 0 and derived from that an upper bound of (a − c − 0)/2 = (a − c)/2. Taking
this as an upper bound, we get a new lower bound of (a − c − (1/2)(a − c))/2 = (a − c)/4,
etc. So letting s̄k+1 denote our (k + 1)st upper bound and sk our k th lower bound, we
have
a − c − sk
s̄k+1 =
2
and
a − c − s̄k−1
sk =
.
2
We can substitute from one equation into the other to get
s̄k+1 =
a − c s̄k−1
+
4
4
sk+1 =
a − c sk−1
+
.
4
4
and similarly
Let s̄∗ be the limit of the sequence of upper bounds and let s∗ be the limit of the
sequence of lower bounds. (Technically, we ought to check that the sequences converge.
However, it is not hard to show that they are bounded, monotone sequences and therefore
they must converge.) Clearly, the limits will have to satisfy
s̄∗ =
a − c s̄∗
+
4
4
and exactly the same equation for s∗ . Hence the two limits are the same. Therefore, the
strategy sets are converging to single points, where the single point is s̄∗ . Solving, we see
that
a−c
,
s̄∗ =
3
which is the Nash equilibrium we calculated in class.
2. Fix any s01 < v1 . I show that s1 = v1 dominates s1 = s01 . To see this, note that for
any s2 < s01 , u1 (s01 , s2 ) = u1 (v1 , s2 ) = v1 − s2 . For s2 = s01 , the fact that s01 < v1 implies
1
u1 (s01 , s2 ) = (v1 − s01 ) < v1 − s01 = u1 (v1 , s2 ).
2
For s2 ∈ (s01 , v1 ), we have
u1 (s01 , s2 ) = 0 < v1 − s2 = u1 (v1 , s2 ).
Finally, for s2 ≥ v1 , we have u1 (s01 , s2 ) = u1 (v1 , s2 ) = 0. Hence s1 = v1 dominates s1 = s01 .
2
Fix any s01 > v1 . I show again that s1 = v1 dominates s1 = s01 . For any s2 ≤ v1 ,
u1 (s01 , s2 ) = u1 (v1 , s2 ) = v1 − s2 . (Note that if s2 = v1 , 1 only wins with probability 1/2
when he bids v1 , but his payoff is 0 whether he wins or not, so this statement is still
correct for that case.) For s2 ∈ (v1 , s01 ), we have
u1 (s01 , s2 ) = v1 − s2 < 0 = u1 (v1 , s2 ).
For s2 = s01 , we have
u1 (s01 , s2 ) = (1/2)[v1 − s2 ] < 0 = u1 (v1 , s2 ).
Finally, for s2 > s01 , we have u1 (s01 , s2 ) = u1 (v1 , s2 ) = 0. Hence s1 = v1 dominates s1 = s01 .
Obviously, the same is true for player 2. To generalize to I players, replace the s2
above with the largest of the sj ’s for j 6= 1.
3. (a) Suppose to the contrary that there is a pure strategy Nash equilibrium, say (b∗1 , b∗2 ).
Suppose that b∗2 > b∗1 . This means that player 2 wins and so his payoff is v2 − b∗2 . Since
player 2 could always bid zero and get a payoff of zero, we know that v2 ≥ b∗2 or else 2
would deviate. But then player 1 could bid b∗2 + ε < v1 and he would be strictly better
off. Hence b∗1 ≥ b∗2 .
Suppose b∗1 = b∗2 . Then each gets the object with probability 1/2. (Actually, the
1/2 isn’t important for this part as long as each has a nonzero probability of getting the
object.) As argued above, this means that we must have v2 ≥ b∗2 . Again, this means that
1 could increase his bid by ε and get the object for sure. For a sufficiently small ε, this
must be strictly better for him.
Hence we conclude that b∗1 > b∗2 . But then any deviation by 1 to a bid strictly between
these two must make 1 better off. It does not affect his winning and lowers his payment.
Since this covers all the cases, we see that there is no pure strategy equilibrium.
(b) Suppose (b∗1 , b∗2 ) is a pure Nash equilibrium. First, suppose b∗1 = b∗2 . Just as above,
player 1 could increase his bid by δ and win outright. For him to not want to do this, it
must be true that v1 − b∗1 − δ ≤ (1/2)[v1 − b∗1 ] or v1 − 2δ ≤ b∗1 . It also must be true that 2
does not want to cut his bid. Hence we must have b∗1 ≤ v2 . So v1 − 2δ ≤ b∗1 ≤ v2 < v1 − δ
where the last inequality comes from our assumption. Hence the only possibility for such
an equilibrium is b∗1 = b∗2 = v1 − 2δ. This is an equilibrium iff v2 = v1 − 2δ or n2 = n1 − 2.
It is easy to see that there are no equilibria where b∗1 < b∗2 , so let’s look for equilibria
where b∗1 > b∗2 . I claim that for any b∗2 between v2 − δ and v1 − 2δ, we have an equilibrium
with bids (b∗2 +δ, b∗2 ). To see that this is an equilibrium, note that 1’s payoff is v1 −b∗2 −δ ≥
δ. If he deviates to a higher bid, his payoff falls. If he deviates to a bid below b∗2 , his
payoff falls to zero. If he deviates to bidding b∗2 , he ties the second bidder and gets the
object with probability 1/2. His payoff from this is lower than his payoff to bidding b∗2 + δ
3
iff
(1/2)(v1 − b∗2 ) ≤ v1 − b∗2 − δ
or δ ≤ (1/2)(v1 − b∗2 ). Since b∗2 ≤ v1 − 2δ, we have
(1/2)(v1 − b∗1 ) ≥ (1/2)(2δ) = δ.
So the payoff to deviating to b∗2 is either the same as or below the payoff to bidding b∗2 + δ.
Clearly, 2’s strategy is optimal since going up means that he ties or wins the object at a
bid greater than or equal to his valuation and going down doesn’t change his payoff.
To see that there is no equilibrium with a higher winning bid, note that 1’s bid here
is v1 − δ at most. If we try to get an equilibrium where he bids v1 and player 2 bids
v1 − δ, we find that 1 would deviate to match 2’s bid. After all, he gets zero from bidding
v1 and gets (1/2)(δ) > 0 if he matches.
To see that there is no equilibrium with a lower winning bid, notice that 1’s bid is
always at least v2 here. If we try to construct an equilibrium where 1 wins with a bid of
v2 − δ, note that player 2 would prefer to match this in order to get (1/2)(δ) instead of
0. (Having them both bid v2 − δ would not be an equilibrium either since 1 would rather
go up to v2 .)
Now let’s turn to equilibria where no player uses a dominated strategy. It is easy to
see that it is dominated to bid above your valuation since this gets you 0 at best and
could get you a negative payoff. Bidding your valuation would dominate this since it
ensures a payoff of zero. Bidding your valuation, though, is also dominated since it gives
zero for sure while bidding your valuation minus δ always gives either zero or something
strictly positive. Hence if we remove dominated strategies, we rule out all the equilibria
above where b∗2 ≥ v2 . This leaves us with a unique equilibrium of (v2 , v2 − δ).
4. a) There are N pure strategy equilibria. Pick any player and suppose he is the only
contributor. He clearly doesn’t want to deviate since his payoff, 1 − c, is strictly bigger
than the payoff to deviating, 0. Similarly, given that there is a contributor, none of
the noncontributors want to deviate to contributing. There are no equilibria with no
contributors (each person wants to deviate to contributing) and no equilibria with more
than one contributor (each contributor wants to deviate to not contributing).
b) Let p be the probability of contributing. The payoff to contributing is 1 − c. The
expected payoff to not contributing is the probability that no one else contributes ((1 −
p)N −1 ) times 0 plus the probability that at least one other person contributes (1 −
(1 − p)N −1 ) times 1. For this to be a mixed strategy equilibrium, each person must be
indifferent between contributing and not, so
1 − (1 − p)N −1 = 1 − c
or
1 − p = c1/(N −1) .
4
Hence p = 1−c1/(N −1) . The probability that at least one person contributes is 1−(1−p)N
or 1 − cN/(N −1) . As N → ∞, this converges to 1 − c.
5