Economics 203 January 2010Core Economics

Suggested Solutions to Problem Bank1

1. a) In the diagram, left branches mean heads and right branches mean tails.

N

2 1

1 1 2 2

1

2 2

(2,-2)

2

(-2, 2) (-2,2) (2,-2) (-2,2) (2,-2) (2,-2) (-2,2) (2,-2) (-2,2) (-2,2) (2,-2)

1/3 1/3 1/3

b) Each player has 4 information sets, so strategies are 4-tuples.S1 = {H,T} {H,T} {H,T} {H,T}, which contains 16 elements.S2 = {h, t} {h, t} {h, t} {h, t}, which contains 16 elements.

2. a) The extensive form is:

!"#$%&'()*

!"#$%&'()+

!"#$%&'(),

!

" # "

## # #" "

*-. *-.*-.

/

/ /

/ 0

0 0

0

% %

%

%%11 1 1

1

+23+ +23+ +23+ +23+ +23+ +23+3+2)+ 3+2)+ 3+2)+ 3+2)+ 3+2)+ 3+2)+

"

# ##

!!

.2+ .2*425 *2**26 *2*52+ +25 ,2.

7

*-+

8

/ 0

*-+

*-+ *-+

9 : 8 ; 8 ;

"

# # #

!!! "" "

< #-+ #

3. A strategy includes one action from each information set. Thus, the number of strategies isjust

QNn=1Mn.

4. a) The extensive form is:

b) S1 = {0, v/2, v}S2 = {(a, a, a), (a, a, r), (a, r, a), (a, r, r), (r, a, a), (r, a, r), (r, r, a), (r, r, r)}

c)

aaa aar ara arr raa rar rra rrr0 0, v 0, v 0, v 0, v 0, 0 0, 0 0, 0 0, 0v/2 v/2, v/2 v/2, v/2 0, 0 0, 0 v/2, v/2 v/2, v/2 0, 0 0, 0v v, 0 0, 0 v, 0 0, 0 v, 0 0, 0 v, 0 0, 0

d) No strategies are strictly dominated. One way to verify this is to note that every strategyis a best response to some opponents strategy.

e) For player 1, 0 is weakly dominated by v/2 and v. For player 2, (a, a, a) and (a, a, r)weakly dominate all other strategies.

5. a) For player 1, E is strictly dominated by A. For player 2, a is strictly dominated by anymixture of b and e, or d and e. The only weakly dominated strategies are the strictlydominated ones.

b) Iterative deletion leaves (A,C,D) for player 1 and (b, c, e) for player 2, so the game isnot dominance solvable.

c) We only need to check the strategies that survive ISD. (C, c) is the unique PSNE.d) In two player games, the set of strategies that survive iterative deletion is the same as

the set of rationalizable strategies. Thus, the answer is the same as in (b).

6. The question does not specify whether bids have to be non-negative; making either assumptionis fine. Let us assume that S1 = S2 = [0,1), but if one made the alternative assumption,the answers should not be too dierent.

a) No strategies are strictly dominated. Observe that if the opponent j makes demandxj > 100, then all demands xi yield payo 0. Thus, no strategy xi can do strictly worsethan some other strategy for every choice of opponent strategy.

b) xi = 0 is weakly dominated by any xi 2 (0, 100]. xi > 100 is also weakly dominated byany xi 2 (0, 100]. For any xi 2 (0, 100], xi is a strict best response to 100 xi, and thuscannot be weakly dominated.

c) We can solve by graphing best response functions, or by inspection:(x, 100 x) is a NE for any x 2 [0, 100].We also have equilibria of the form (x1, x2) where x1, x2 100.

2

7. Let n denote a node that is reached in equilibrium. Playing c at node n can be a best responseonly if the opponent plays c at node n+1 with positive probability. Thus, if in equilibrium cis played at node 1, then it must be played at every node with positive probability. But thisclearly is not an equilibrium, as playing c at the last node is not a best response. Thereforec cannot be played in equilibrium at node 1.Note 1: It is not hard to extend this result to mixed strategies.Note 2: It is important to distinguish between Nash equilibria and equilibria outcomes here.There are many pure strategy NE any profile where both players say stop at their respectivefirst nodes but there is a unique equilibrium outcome where stop is played at the first node.Note 3: In eect, we have shown a stronger result: viz., in any rationalizable strategy profile,player 1 says stop at the root node.

8. Let s be the unique strategy profile that survives iterated deletion of strictly dominatedstrategies.Step 1: No 0 6= s is a NE (where 0 can be a mixed strategy equilibrium).There are many ways to prove this. An indirect (but elementary) approach would recognizethat a NE is a strategy-profile that necessarily lies within the set of rationalizable strategies.Since that set is a subset of those remaining after iterated elimination of strictly dominatedstrategies, a NE could not put positive weight on a strategy deleted by ISD. Since dominance-solvability implies that ISD leads to a unique strategy profile, s is the only strategy profilethat could be a NE.One can also supply a direct proof: let s0i denote the first pure strategy in the support of 0which was deleted, and suppose this took place in round k. Then there exists a randomizationi over is remaining strategies at the start of round k that beats s0i for any si that remainsat the start of round k. In particular, because 0i remains at the start of round k, we havegi(i, 0i) > gi(s0i, 0i), so 0 cannot be a NE.Step 2: s is a NE.Suppose not. Then there exist i and si 2 argmaxsi gi(si, si) such that gi(si, si) > gi(si , si).But this implies that si 6= si could not have been deleted, a contradiction.Note: we can relax the restriction that strategy sets be finite if we include an assumptionabout existence of best responses:Assumption 1 (A1) argmaxsi gi(si, si) 6= ; for all i, si.Claim Any dominance solvable game that satisfies (A1) has a unique NE which coincideswith the iterative dominance solution s.

9. Strategy sets are S1 = S2 = S3 = [0, 1]. Let p be the price of ice cream fixed by theregulator and i designate payos for i=1,2,3. Note that in both parts of the question, thebest-response function is not always defined because of discontinuities (as in the Bertrandgame).

a) There are 3 cases to check:Case 1: s1 = s2 = 12 . Here 1 = 2 =

p2 , and any deviation yields payo p2 . Thus, this is not aNE.

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Case 3: s1 6= s2. This is clearly not a NE as either player can increase market share bymoving towards the other.

b) We will show that none of the 4 possibilities are Nash equilibria.

Case 1: s1 = s2 = s3 = 12 . Here 1 = 2 = 3 =p3 . By deviating to si + ", i secures

(1")p2 >

p3 . Thus, this is not a NE.

Case 2: s1 = s2 = s3 6= 12 . Again 1 = 2 = 3 = p3 . By moving " closer to 12 , i securesmore than half of the market and a payo > p2 . Thus, this is not a NE.

Case 3: si = sj 6= sk. Here k can clearly gain by moving closer to i and j. Thus, this isnot a NE.

Case 4: si < sj < sk. Again k can gain by moving closer to i and j. Thus, this is not aNE.

Theses cases are exhaustive, so no pure strategy NE exists!

10. We begin by converting the problem to an equivalent one where strategy spaces are compact.

Claim: There exists q such that P (q) < C 0(q) for all q > q.

This follows because P () is decreasing and concave and C 0() is increasing.Now observe that any strategy q > q is strictly dominated by q. Thus, we can throw out thestrictly dominated strategies and be sure that the set of equilibria is identical to that in theoriginal game. We now proceed in 4 steps:

a) There exists a pure strategy NE.We verify the conditions of the existence theorem given in the notes:- Strategy spaces are compact and convex by construction.- Payo function gi(q1, q2...qN ) = P (Q)qi C(qi) is continuous.- Payo function is gi(q1, q2...qN ) is concave (thus quasi-concave) in qi:

@2gi@q2i

= qiP 00(Q)| {z }

@q

@Qi= P

0(q +Qi) + qP 00(q +Qi)2P 0(q +Qi) + qP 00(q +Qi) C 00(q) > 1

4

If there were an asymmetric equilibrium, then there exist i, j such that qi 6= qj . Thenqi+Qi = qj+Qj , and hence,

qiqjQiQj = 1. But qi = q(Qi) and qj = q(Qj)

since i and j are playing best responses. Therefore q(Qi)q(Qj)

QiQj = 1, whichby the Mean Value Theorem contradicts our earlier slope calculation. Hence, allsolutions must be symmetric.

ii. q not interior ) q = (0, 0, ...0) (and thus symmetric):q not interior ) qi = 0 for some i. Then from i0s FOC, we have P (Q) C 0(0) 0.But for qj > 0,

qjP0(Q) + P (Q) C 0(qj) < P (Q) C 0(0) < 0,

so no firm j can choose qj > 0 in equilibrium.

c) The symmetric equilibrium q is unique.We know that the slope of the best-response function, q, is negative. Suppose there isanother symmetric equilibrium q0 with WLOG, Nq0i = Q0 > Q = Nqi.But

Q0 > Q ) Q0i =N 1N

Q0 >N 1N

Q = Qi

) q0i = q(Q0i) < q(Qi) = qi) Q0 < Q,

a contradiction.

d) No mixed strategy equilibrium exists.This follows from the concavity of the objective function. Suppose in equilibrium isopponents are mixing so that the sum

Pqj 6=i takes on N values (generalization to a

continuous distribution is straightforward), with probabilities 1, ...,N . Then i0s choiceof qi solves

maxqi2[0,q]

NXk=1

kqiP (qi +

Xqkj 6=i) C(qi)

.

The objective is a weighted average of strictly concave functions and is thus strictlyconcave. The choice set is compact. Thus, there is a unique optimum, so player i willnever mix!

11. The technique in this problem is to find pi(p), is best response to all other firms naming p,and then to impose symmetry (finding p such that pi(p) = p).

Let i(pi, p) be the payo to firm i when i names pi and others name p. To calculate ismarket share, we need to find the distance d to the consumer who is indierent between andi and the next firm over.

Now d is defined by

v pi td2 = v p t( 1J d)2

which yields d = J2t(p pi) + 12J .We then have

i(pi, p) = (pi c)2d = (pi c)J

t(p pi) + 1

J

.

5

is maximization problem yields FOC:

J

t(p pi) + 1

J+ (pi c)(J

t) = 0.

In the symmetric Nash equilibrium, p = pi, so the FOC reduces to p = tJ2 + c. Since all firmsname the same price, market share is 1/J for each firm.

Note: its easy to check that when all other firms name p = tJ2 + c, a firm has no profitabledeviation to a price low enough to grab all customers between him and the next firm andbeyond.

As J grows large, pi ! c and qi ! 0. With many firms, i has arbitrarily close neighbors andwe approach the Bertrand outcome with no dierentiation.

As t ! 0, pi ! c but qi ! 1/J . As transportation costs approach zero, we have eectivelyno product dierentiation and we approach the Bertrand outcome.

12. Note that the strategy set for i is Si = {Trade always, Trade i wi, Not trade i wi, Nevertrade}. With that, we can write the normal form:

1/2 Trade always Trade i w2 Not trade i w2 Never tradeTrade always 4 9p, 4 9p 2p, p 4 7p, 4 10p 0, 0Trade i w1 p,2p 0, 0 p,2p 0, 0

Not trade i w1 4 10p, 4 7p 2p, p 4 8p, 4 8p 0, 0Never trade 0, 0 0, 0 0, 0 0, 0

a) Since si = Never trade results in a 0 payo always, i is indierent among all possiblestrategies. Therefore, there are no strictly dominated strategies. The set of weaklydominated strategies for player i is given by {Not trade i wi, Never trade}.

b) Once we delete weakly dominated strategies, the game reduces to:1/2 Trade always Trade i w2

Trade always 4 9p, 4 9p 2p, pTrade i w1 p,2p 0, 0

From the above matrix, there are two NE:

i. (Trade always, Trade always), which gives each player a payo of 4 9p. Tradeoccurs in all states and the expected total surplus is 8 18p.

ii. (Trade i w1, Trade i w2), which gives each player a payo of 0. Trade never occursand the expected total surplus is 0.

c) From the above matrix, there is only one equilibrium: (Trade i w1, Trade i w2), whichgives each player a payo of 0. Trade never occurs, and total surplus is 0.

d) The socially optimal strategy profile is (Not trade i w1, Not trade i w2). Under suchstrategies, trade occurs i the state is !3, so that total surplus is 816p. These strategiesare not an equilibrium because each player would deviate to {Trade always}, given theopponents strategy. That is, the socially optimal profile involves players not trading ina state under which they would profit from trade: it requires an act of honesty whenthere is an incentive to lie.

13. a) i. Any strategy with (wi, fi) >> 0 is strictly dominated since player i could be bettero, regardless of player js strategy choice, by reducing both wi and fi by some

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amount > 0. Since wj 1600, any strategy of the form (0, fi) with fi > 1100is strictly dominated by (0, 1100). All of player is remaining pure strategies arebest responses to some pure strategy by player j: (wi, 0) is a best response to(wj , fj) = (0, wi 500) for 500 wi 1600, (wi, 0) is a best response to (wj , fj) =(500 wi, 0) for 0 wi 500, and (0, fi) for fi 1100 is a best response to(wj , fj) = (500+fi, 0). Thus, the set of strictly dominated pure strategies for playeri is precisely those with (wi, fi) >> 0 and those with (0, fi) such that fi > 1100.

ii. Since each strategy for player i that is not strictly dominated is a best response to apure strategy by player j that is not strictly dominated for player j, this is also theset of strategies that survive iterated elimination of strictly dominated strategies.

iii. The set of pure strategy NE are pairs [(w1, 0), (w2, 0)] such that w1 + w2 = 500 aswell as pairs [(wi, 0), (0, fj)] such that 500 wi 1600 and fj = wi 500.

b) i. The unique NE is (w1, f1) = (0, 1100) and (w2, f2) = (1600, 0). This strategy-profileis a NE since each strategy is a best-response to the other. To establish uniqueness,it suces to show that the game is dominance-solvable and this is the unique profilesurviving iterated elimination of strictly dominated strategies. To see this, performthe following iterative procedure (note that this actually also constitutes a proofthat this is the unique NE):Step 1: Eliminate for each player i any (wi, fi) >> 0 .Step 2-a: Eliminate for player 2 any strategy (0, f2) with f2 > 1100 ".Step 2-b: Given step 2-a, eliminate for player 1 any strategy (w1, 0)with w1 >1600 ".REPEAT STEPS 2-a AND 2-b (now with k" for k = 2, 3, ...) UNTIL ALL STRATE-GIES FOR PLAYER 2 WITH f2 > 0 HAVE BEEN ELIMINATED AND ALLSTRATEGIES FOR PLAYER 1 WITH w1 > 500.Step 3-a: Given the above eliminations, eliminate for player 2 any (w2, 0)with w2 0. Naming p > v guarantees payo 0. Thus, we rule outall p > v.

There are two cases to consider: everyone charges the same price, or not everyone chargesthe same price.

Case 1: p1 = p2 = . . . = pN = p 2 (0, v]. Here equilibrium payo is pQN for each firm. Butthen any firm can deviate to p", secure market shareK > QN , and earn payo (p")K > pQNfor small enough ".

Case 2: Let pL and pH be the lowest and highest prices named in equilibrium with pL < pH .But then a firm naming pL could deviate to pH ", maintaining a market share of K, andearn payo (pH ")K > pLK small enough ".Thus, we have no PSNE.

Finding the Symmetric MSEWe look for a symmetric MSE represented by distribution function F () where the support isgiven by [p, v] with p > 0.4

Let the (expected) equilibrium payo be given by . Suppose players j 6= i are mixing usingF (). Then for i to be mixing over [p, v], i must be indierent over every price in this range.Observe that by naming p 2 [p, v], with probability F (p)N1, is price will be the highest andhe will secure residual market share Q (N 1)K. With probability 1 F (p)N1, is pricewill not be the highest and he will secure residual market share K. Thus, expected payofrom naming p is given by

(p) = p F (p)N1(Q (N 1)K) + (1 F (p)N1)K = . (5)3The problem can also be solved with any other assumption4The proof that this is the unique symmetric MSE is a simple extension of the 2 firm case in the notes.

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We pin down by noting F (v) = 1) (v) = v(Q (N 1)K) = .Now we can solve equation (5) for F (p) to find

F (p) =(Q (N 1)K)v pK(Q (N 1)K)p pK

1N1

.

Finally, we pin down p by noting that F (p) = 0) p = (Q(N1)K)K v.That completes our characterization of the symmetric MSE.

Comparative StaticsSuppose we let N get large, adjusting K (lower) so that the assumptions hold. Observe thatp = (Q(N1)K)K v, where the coecient on v is given by the ratio of residual demand to K.Thus, even for large N , we can have p lie anywhere in the range (0, v).

However, observe that F (p) ! 1 for all p 2 [p, v], so that the equilibrium approaches thePSNE where all firms name p. Thus, the limiting case depends crucially on how we adjust Kas N gets large.

18. a)b) The normal form of the game is given by

r y daa 0, 2 10, 0 -100,-100an 3, 2 5, 5 -50,-45na 3, 10 5, 5 -50,-45nn 0, 10 0, 10 0,10

c) The unique PSNE is (nn, d).To calculate the mixed strategy Nash equilibria, observe that although no strategies arestrictly dominated, d is weakly dominated for B. We exploit this fact.Consider NE in which A places positive probability on aa, an or na (ie any strategyother than all weight on nn.) In such a NE, we observe that B cannot be placing anyprobability on d. It follows that nn is strictly dominated by an and that na is strictlydominated by a mixture of aa and an. Thus, we have a reduced 2x2 matrix. Solving forthe mixed NE in the standard way yields (35 ,

25 , 0, 0), (

58 ,

38 , 0).

Now consider NE in which A places all probability on nn. That is, A uses the purestrategy nn. Any strategy B for B is a best response to this. Therefore, we needrestrictions on Bs strategy such that nn is a best response to B.Let r,y,d represent the weights on r, y and d in B, where r = 1 y d.

A(aa) A(nn) ) 10y 100d 0A(an) A(nn) ) 3r + 5y 50d 0A(an) A(nn) ) 3r + 5y 50d 0

Observe that if the middle inequality is satisfied, then so too are the first and third. Thus,we have NE (0, 0, 0, 1), (1yd,y,d) for all y,d 2 [0, 1] satisfying 3+2y53d 0 and y + d 1.

19. a) Observe that since it is not immediately obvious what player 1 might play, LL, L andR are very risky, yielding very small gains if you guess right, and significant losses ifyou guess wrong. Thus, only M seems reasonable.

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b) The PSNE are (U,LL) and (D,R). The only non-pure MSE is given by (5152U +152D,

12LL+

12L).

To see this, observe that since no player has multiple best responses to any opponentspure strategy, there are no pure/mixed NE.Any potential 23 or 24 MSE can be ruled out because requiring indierence between> 2 strategies for player 2 yields an insoluble system with more equations than variables.The 2 2 cases must be analyzed individually.

c) M is not a part of any of the above strategies. M is, however, rationalizable. One wayto see this is that no strategies can be removed by ISD.

d) With preplay communication, players could agree on one of the NE. The self-enforcingnature of NE would make this agreement credible. The PSNE are payo equivalent andpareto dominate the 2 2 MSE, so we would expect agreement on either of these.

20. a) There is no PSNE. There are three configurations to consider:

i. Both firms quote a price: optimality implies that they must quote the same price(otherwise the payo for the one quoting the higher price would be k, and adeviation to no price quote would increase its payo). Moreover, this price pmust be at least c + 2k (otherwise each would earn (p c)/2 k < 0, in whichcase a deviation to no price quote would increase its payo). But then eitherfirm could increase its profit by slightly undercutting p (earning nearly p ck >(p c)/2 k 0).

ii. Only one firm quotes a price. In that case, the quoted price must be v (otherwisethe quoting firm could increase its profits by deviating to a higher price). Butthen the other firm, which earns zero, could obtain a strictly positive payo (nearlyv c k > 0) by quoting a price slightly below v.

iii. Neither firm quotes a price. In that case, a deviation to pi = v would yield a payoof v c k > 0, instead of 0.

b) In a mixed strategy equilibrium, a player must be indierent between all sets of strategiesselected with strictly positive probability. Here, a firm quotes no price with strictlypositive probability. When this occurs, its payo is zero; hence the equilibrium expectedpayo is zero. When a firm names a price less than c + k or greater than v, its payois strictly negative. It cannot be indierent between this and no price quote (whichyields a payo of zero). Consequently, it cant make both of these choices in an MSE.The probability that the other firm quotes price lower than p is F (p), so the firm makesthe sale with probability 1F (p). In that case, it earns profits pc. Since it incurs thecost k regardless of whether it makes the sale, its expected payo is (1 F (p)) (pc)k.In light of this, for any price in the support of F we must have (1 F (p)) (pc)k = 0.This implies F (p) = 1

pckpc

. Notice that this is non-negative and strictly increasing

on [c+ k, v], with F (c+ k) = 0. Since the support of F is the entire interval [c+ k, v],we choose so that F (v) = 1. This implies = vckvc . So F is flat up to c+k, strictlyincreasing between c+ k and v, and flat beyond v.

21. a) Strictly positive profits. A firm not earning strictly positive profits could gain simply bysetting its price equal to v (since it would sell to the loyal customers).

b) No, there isnt a PSNE for this game. We know both firms earn strictly positive profits,which means both must quote prices in (c, v]. If the prices are dierent, the firm with

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the lower price could gain by increasing it. If the prices are the same, either firm couldgain by slightly undercutting the other.

c) When quoting p, the expected payo is (p c) [K + (1 F (p))N ]. Setting the ex-pression for expected payo equal to a constant, C, and solving, we get: F (p) =1 KN

C/K(pc)

pc. Using F (v) = 1 gives us C/K = v c, so F (p) = 1 KN

vppc

.

Note that this equals zero when p = Nc+KvN+K = p00. Obviously, p00 2 (c, v). So F equals

zero up to p00 (which exceeds costs), and then increases continuously, reaching a value ofunity at p = v. As KN ! 0, p00 ! c, and F (p)! 1 for all p > c, so we have convergenceto the standard Bertrand outcome. As KN !1, p00 ! v, and F (p)! 0 for all p < c, sowe have convergence to the monopoly outcome.

22. Let v1 = v and v2 = v0 > v.

a) Any bid bi > vi for Bidder i is strictly dominated by a bid bi = 0. Removing thesestrategies yields a game where bidding b2 > v is strictly dominated by bidding b2 = v forBidder 2, given the tie-breaking assumption (namely, since Bidder 2 wins in the eventof a tie, and Bidder 1 will not bid greater than v, there is no reason for Bidder 2 to doso). All bids bi 2 [0, v] can be rationalized by some bid of the opponents, bj 2 [0, v],and therefore, no other strategies can be removed by iterative elimination of strictlydominated strategies.

b) There are no PSNE. To establish this, let us consider the following profiles (b1, b2) 2[0, v]2 to see whether there is a strictly profitable deviation:

i. bi < bj : Bidder j can improve her payos by decreasing her bid tobi+bj

2 whileensuring that she still wins the auction

ii. 0 < bi = bj : Then Bidder 2 wins the object for sure, and Bidder 1 can strictlyimprove her payos by bidding 0 instead.

iii. bi = bj = 0: Then Bidder 1 loses the object for sure and can do strictly better bybidding any b1 2 (0, v).

Observe that the preceding three cases exhaust the profiles in [0, v]2. Since all otherprofiles use strategies that are not rationalizable, this establishes that there are no PSNE.

c) By (a), all b1 > v are strictly dominated and so it could not be a best response to putpositive probability on (v,1). Similarly, for bidder 2, b2 > v is iteratively dominated,and so this too could not be part of a best response.

d) i. If b2 = x, then regardless of whether she wins, Bidder 2 pays x. She wins if x b1,and therefore, wins the object with probability F1 (x). Her expected payo frombidding x is F1 (x) v0 x.

ii. Observe that F1 (v) = 1. Therefore, from b2 = v, Bidder 2s expected payo isv0 v.

iii. Since Bidder 2 will mix according to F2 only if she is indierent between all strategiesin its support,

F1 (x) v0 x = v0 vF1 (x) =

v0 v + xv0

Observe that F1 (0) = v0vv0 .

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iv. If Bidder 1 bids x, she wins if and only if b1 > b2. Assuming that F2 is continuous(we will verify this later), her payo from b1 = x is F2 (x) v x. Since Bidder 1mixes over the interval [0, v], and F2 (v) = 1, her expected payo is 0.

v. From above

F2 (x) v x = 0F2 (x) =

x

v

Observe that F2 (0) = 0.

e) E[b1] =R v0

xv0dx =

v2

2v0 .E[b2] =

R v0xvdx =

v2 .

Hence, expected revenue is v2

2v0 +v2 and is decreasing in v

0; since increases in v0 simplyserve to depress 1s bid but do not increase 2s bid, the expected revenue is strictlydecreasing in v0.

23. a) It is strictly dominated to bid bi > v. No other strategies can be iteratively eliminated.b) There are no PSNE.

i. Case 1: 0 bi < bj v. This cannot be an equilibrium because Student j wouldlike to decrease his bribe slightly, thereby getting a payo of v bj + " > v bj .

ii. Case 2: 0 bi = bj < v. This cannot be an equilibrium because Student j wouldlike to increase his bribe slightly, thereby getting an expected payo of v bj "instead of 12v bj .

iii. Case 3: bi = bj = v. This cannot be an equilibrium because Student j would like tochange his bribe to 0, thereby getting a payo of 0 rather than 12v v < 0.

c) No, since this would be strictly dominated.d) F (bi) v bie) From (d), the payo from a bribe of v is F (v)v v=0 because F (v) = 1 by assumption.f) Student i must be indierent between all of the bribes he pays with positive probability.

Therefore, from (e), F (bi)v bi = 0 for all bi 2 [0, v] and F (bi) = biv .g) The Social Chair expects to make 2

R v0

bvdb = v.

h) The payo from a bribe bi < b0 is (1 )G (bi) v bi. The payo from a bribe b is2 + (1 )

v b.

i) When bi = 0, is payo is 0 because G(0) = 0 by assumption.j) Again, Student i must be indierent between all of the bribes he pays with positive

probability. So2 + (1 )

v b = 0) = 2(vb)v . Also, (1 )G (bi) v bi = 0)

G(bi) = bi(1)v =bi

2bv . We know G(b0) = 1) b0 = 2b v.

k) The Social Chair expects to make

2

1 2

v bv

!Z 2bv0

b

2b vdb+2v bv

b

!=

2b v2b v

v

+

4v b bv

=1v

4b2 4bv + v2 + 4bv 4b2 = v

Therefore, the spending cap left the total amount of bribing unchanged.

15

24. a) The pure strategy set is the interval [0, bmax]. A mixed strategy is a CDF on that in-terval specifically, a monotonically increasing uppersemicontinuous (right-continuous)function F with F (0) 0 and F (bmax) = 1. The mixed strategy set is the set of all suchfunctions.

b) There is no pure strategy equilibrium. If there is a single high bidder, the winner couldlower his bid. If there is more than one high bidder below the upper bound, then eitherplayer would benefit from increasing his bid slightly. If there is more than one bidder atthe upper bound, each has a negative expected payo (since the upper bound is above5) and would benefit from changing his bid to 0.

c) We need to solve for F , , and b0. Because F is atomless, the probability of winning,and hence the expected payo, when bidding 0 is 0. Because 0 is in the support of F ,the equilibrium expected payo is zero by the mixed strategy indierence condition. Abid at the cap (bmax) earns: h

2+ (1 )

i10 bmax.

Setting that expression equal to zero (the equilibrium expected payo), we see that

=10 bmax

5.

A bid of b0 earns:(1 )10 b0.

Setting that expression equal to zero (the equilibrium expected payo) and substitutingfor , we see that

b0 = 2bmax 10.A bid of b 2 [0, b0] earns:

(1 )F (b)10 b.Setting that expression equal to zero (the equilibrium expected payo) and substitutingfor , we see that

F (b) =b

2bmax 10 .To verify that the solution is a mixed strategy equilibrium, we need to check that aplayer cannot gain by deviating to a bid outside the support of F . The complement ofthe support of F is b 2 (b0, bmax). For all such b, the expected payo is:

(1 )10 b (1 )10 b0 = 0.

d) The case of no bid cap is equivalent to bmax = 10, so it is a special case of the above.As weve shown, the expected payo for each student is zero regardless of bmax. For thisgame, the payos of all players sum to 0 in every realization (and hence in expectation),so Bernheims expected payo is also 0. This means his expected revenue must be10 (just enough to compensate for the $10 bill) irrespective of bmax. Intuitively, thestudents will always compete away all positive rents, if not by bidding close to 10, thenby matching each other with the maximum bid.

16

25. a) The pure strategy set is the set [bmin,1] [ {}. A mixed strategy is a probability of playing , plus a CDF F (b) on [bmin,1] that governs choice with probability 1 . Specifically, F is a monotonically increasing uppersemicontinuous (right-continuous)function F with F (b) = 0 for b < bmin and limb!1 F (b) = 1. The mixed strategy set isthe set of all such functions.

b) There is no pure strategy equilibrium.Case 1: There is only one active bidder. Either that bidder bids less than $10, in whichcase another student would have an incentive to bid higher, or he bids $10, in whichcase he has an incentive to lower his bid.Case 2: There are at least two active bidders. If there is only one high bidder, hemust be bidding more than bmin and has the incentive and opportunity to lower his bid.If there is more than one high bidder below $10, then any one of them would benefitfrom increasing his bid slightly. If more than one bidder ties at $10, each has a negativeexpected payo, so a deviation to is better.

c) Assume F has an atom at bmin. Then a players probability of winning (and hence hisexpected payo) rises discontinuously when increasing his bid from bmin to bmin + ", acontradiction. Now assume F has an atom at 10. Then a player receives a negativeexpected payo from playing 10, and could improve his expected payo by shifting thatatom to .

d) In light of part (3), the probability of winning when bidding 10 is unity; hence theexpected payo when bidding 10 is 0. Because 10 is in the support of F , the equilibriumexpected payo is zero by the mixed strategy indierence condition.A bid at the floor earns an expected payo of:

10 bminSetting that expression equal to zero (the equilibrium expected payo), we see that

=bmin10

A bid of b earns:10 [ + (1 )F (b)] b

Setting that expression equal to zero (the equilibrium expected payo), we see that

F (b) =1

1 b

10

Substituting for , we obtain

F (b) =b bmin10 bmin

To verify that the solution is a mixed strategy equilibrium with the desired properties,we need to check that F (b) is a legitimate atomless CDF on [bmin, 10]. Substituting intothe above expression for F , we see that F (bmin) = 0 and F (10) = 1, as required; wealso see that F 0(b) = 110bmin > 0, also as required. We do not have to check deviationsoutside the support of F (because there are none), and we have already assured (byconstruction) that a player receives the same expected payo from as well as from allb in the support of F . Thus we have a mixed strategy equilibrium.

17

e) The case of no bid floor is equivalent to bmin = 0, so it is a special case of the above.As weve shown, the expected payo for each student is zero regardless of bmin. For thisgame, the payos of all players sum to 0 in every realization (and hence in expectation),so Bernheims expected payo is also 0. This means his expected revenue must be 10(just enough to compensate for the $10 bill) irrespective of bmin. Intuitively, the studentswill always compete away all positive rents with bmin = 0, and setting a higher minimumwill only cause them to compete less aggressively by opting out.

26. a) Using iterative deletion, we find that 3 must play A and 2 must play L. There are 2PSNE: (U,L,A) and (D,L,A).

b) No.

c) (U,L,A) is trembling hand perfect, but (D,L,A) is not.To show this, we note that in this problem, "-constrained equilibria must take a particularform: Note that because A is a dominant strategy for 3, in any "-constrained equilibrium,3 must be placing no more that the required " weight on B. Similarly, 2 must be placingno more than " on R. But then1(U) = (1 ")2 1 + (1 ")" 1 + (1 ")" 1 + "2 0 and1(D) = (1 ")2 1 + (1 ")" 0 + (1 ")" 0 + "2 1so that 1(U) > 1(D). Therefore, 1 must be placing no more than the required " weighton D.Thus, for "n ! 0, we have a sequence of "-constrained equilibria that converges to(U,L,A):

n = ((1 "n)U + "nD, (1 "n)L+ "nR, (1 "n)A+ "nB).Thus, (U,L,A) is trembling hand perfect.On the other hand, since all "-constrained equilibria take this form, it is impossible tofind a sequence of "-constrained equilibria converging to (D,L,A), so (D,L,A) is nottrembling hand perfect.Moral: In 3+ player games, the trembling-hand refinement can rule out more than justNE where one strategy is weakly dominated. (In 2 player games, these two refinementsare equivalent.) Ask yourself what happened here that couldnt in a 2 player setting.

27. The unique profile surviving ISD is (C, c), which is a Nash equilibrium. Therefore, it is also acorrelated equilibrium. On the other hand, it is easy to show that no pure strategy eliminatedby ISD can be a part of a correlated equilibrium. It follows that (C, c) is the unique correlatedequilibrium.

28. a) The NE are (B,L), (T,R) and (23T+13B,

23L+

13R). Thus, the set of payos is

(7, 2) , (2, 7) ,

143 ,

143

.

b) The convex hull of the payos is drawn below:

18

0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

(14/3,14/3)

(7,2)

(2,7)

c) Lets look for payos outside the convex hull in (b) within the class of equilibria describedin the notes. (Of course, these are not the only correlated equilibria. But if we find oneoutside of the convex hull, then we are done.)

L RT (1 )/2B (1 )/2 0

The symmetry of this class makes verification easy. We only need to verify that 1 wouldagree to this distribution of outcomes. If 1 is told to play B , he believes that 2 willplay L with certainty, in which case B is in fact a best response. If 1 is told to play T ,then he believes that 2 will play L with probability

+ 12= 21+ and R with probability

12

+ 12= 11+ . For T to in fact be a best response, we require 1(T ) 1(B) or

6

21 +

+ 2

1 1 +

7

2

1 +

which reduces to 12 . Thus, when 12 , the distribution over outcomes describedabove is a correlated equilibrium. Choosing = 12 yields the highest payo equilibriumin this class, with payos (514 , 5

14). This falls outside the convex hull of the NE.

29. Here a strategy is a mapping si : {1, 2, . . . ,M}! {trade, not-trade}.Mixed strategy NE: Observe that in no equilibrium will i use a strategy placing positiveprobability on trading M . Suppose the contrary, that there exists an equilibrium s in whichplay i places positive probability on trading M . Player is strategy can be a best responseonly if all the other players use strategies in which 1, . . . , (M 1) are never traded. However,this cannot be the case in a NE because player j 6= i would do better to deviate to the strategytrade only when card reads 1.

19

Iterating this proof rules out placing positive probability on any number 2 through M 1.Thus, the set of candidate strategies is reduced to si(1) = trade with probability p andsi(m) = not-trade with probability 1 for all m = 2, . . . ,M .

It is easy to check that any profile where each i uses a strategy of this form is in fact a NE.

We can not use iterative deletion here. No strategy is strictly dominated. To see this, notethat if j 6= i all use the strategy never trade, then all strategies for i yield identical payos.Correlated NE: An M is never traded in a correlated equilibrium. Suppose, towards contra-diction, that there is a correlated equilbrium in which the correlation device assigns a positiveprobability to player i using a strategy that involves trading an M . Suppose the state of theworld is such that the correlation device suggests using such strategy to player i. In orderfor him to follow the suggestion in equilibrium, it must be the case that all other playersare instructed, by the correlation device, in such state of the world, to place zero probabilityon any strategy that involves trading anything other than an M . But then, in this state ofthe world, any j 6= i will have an incentive to deviate to the strategy that trades a 1 withprobability 1, and everything else with probability zero, a contradiction.

Repeating this argument rules out trading 2 through M in a correlated equilibrium. Weare left with the set of correlation devices that that assign probability zero to any strategyinvolving trading any card greater than 1, all of which are easily verified to be correlatedequilibria.

Remark : For pedagogical purposes, we separated the 2 cases, but in fact the NE result followsfrom the correlated equilibrium result.

30. We can calculate the normal form:

aa an na nn

aa M2w2s4 ,M2w2s

42Mws

4 ,M2s

43Mws

4 ,2w4 M, 0

an M2s4 ,2Mws

4Ms4 ,

Ms4

2Ms4 ,

Mw4

2M4 , 0

na 2w4 ,3Mws

4Mw

4 ,2Ms

4Mw

4 ,Mw

42M4 , 0

nn 0,M 0, 2M4 0,2M4 0, 0

It is easily verified that the squares marked with an can never be NE for any M, s, and wthat satisfy the given conditions.

aa an na nnaa an na nn

For the remaining squares, verifying the conditions for a NE yields:

(aa, an) and (an, aa) are NE when M 2s and M > w;(an, an) is a NE when M 2 (s, w);

20

(aa, nn) and (nn, aa) are NE when M 2s.Finally, the NE of this game correspond to Bayesian Nash equilibria of the associated Bayesiangame.

31. To avoid confusion with quantity, we replace the probability q in the problem with . Inversedemand is given by p(Q) = a bQ.A BNE involves firm i choosing a quantity qic when its costs are c and a quantity qid whencosts are d. For this to be a BNE, is choices of q must be optimal, taking js strategy asgiven:

qic 2 argmaxq [a b(qjc + q) c)]q + (1 ) [a b(qjd + q) c)]q

qid 2 argmaxq [a b(qjc + q) d)]q + (1 ) [a b(qjd + q) d)]q

This yields FOCS:

qic =a b(qjc + (1 )qjd) c

2band qid =

a b(qjc + (1 )qjd) d2b

.

The symmetry of the problem implies qic = qjc = qc and qid = q

jd = qd. Solving the two FOCs

simultaneously for qc and qd yields

qc =a 12 (c d) c

3band qd =

a 2 (d c) d3b

.

32. a) A BNE where the project does not go forward definitely: 2 students play n regardlessof their type, and the rest of the students do anything. Then everyone gets a payo ofzero, and no deviation improves this.A BNE where the project goes forward with positive probability: each student con-tributes i his valuation is greater than or equal to 10. Aside from resolving indierencedierently when the students valuation is exactly 10, there is no other equilibrium ofthis type. If the probability that everyone else plays y is strictly positive, then a studentwill definitely choose y if his valuation is greater than 10, and definitely choose n if it isless than 10.

b) i. Once again, if both students choose n regardless of their valuations, this will leadto a BNE where the project never goes forward.

ii. Claim: In every BNE, students will use a cuto strategy: i(vi) = y if vi ci andi(vi) = n if vi < ci for some cuto ci.Proof : In any equilibrium, there is some probability pi that student i choosesy. Student i benefits from making a contribution i pivi 10. For pi > 0, thismeans setting a cuto ci = 10pi is optimal. For pi = 0, setting ci > 45 is optimal.Let student i use cuto ci. Then the probability that i plays y is 1 ci45 ; therefore cj =10

1 ci45= 45045ci . At a symmetric equilibrium, ci = cj = c, and hence c

245c+450 = 0.This leads to c 2 {15, 30}. Consequently, there are two symmetric equilibria: onewhere both students use 15 as a cuto, and the other where both use 30 as a cuto.

iii. At an asymmetric equilibrium, we would have c1 (45 c2) = 450 = c2 (45 c1).This yields c1 = c2. Therefore, there are no asymmetric equilibria.

c) i. As before, there is a BNE where the project never goes forward if all students choosen regardless of their valuations.

21

ii. In every BNE, students still must use a cuto strategy. This follows from theargument in part (b), treating pi as the probability that all other students choosey. At a symmetric BNE (if it exists), all students use the same cuto c. Theprobability that all other students choose y is

1 c45

N1, so the best cuto touse in response is c0 = 10

(1 c45)N1 . This means that any symmetric equilibrium has

a cuto that solves c1 c45

N1 = 10. For a project to go forward with positiveprobability, we must have c < 45. We now show that this is impossible if N 3.Let f (c,N) = c

1 c45

N1. Let c (N) = argmaxc2[0,45] f (c,N). The FOC yields1 c

45

N1 c45

(N 1)1 c

45

N2= 0.

Since for c 45, f (c,N) 0, and for c < 45, f (c,N) > 0, we can verify thatc (N) = 45N (you can verify that this is the maximum without checking the 2

nd

order condition by simply checking that for c < c (N), dfdc > 0, and for c > c (N),

dfdc < 0).

Now, let g (N) = f (c (N) , N) = 45(N1)N1

NN. Recognize that g (3) = 203 < 10.

Thus, there is no symmetric equilibrium where the project goes forward with positiveprobability with 3 agents.For N > 3, note that

dg

dN= 45

(N 1)N1NN

(log (N 1) log (N)) < 0.

Therefore, for every N 3, g (N) g (3) < 10, and hence, the unique symmetricequilibrium is the one where the project never goes forward.

33. a) Each i has a dominant strategy: Name the highest xi, namely xi = 1, regardless of type.Thus, we have a unique BNE in which each i names xi = 1 regardless of type.

b) This maximization problem is

maxx1,x22[0,1]

x1 x

22

21

+x2 x

21

22

,

which has solution x1 =12, x2 =

11.

c) Method 1: We use the Groves mechanism described in the notes. Note that whilethis implementation guarantees that truth-telling is a dominant strategy, it does notguarantee budget balance. So well have to check specifically if we can achieve budgetbalance in this case.The Groves mechanism calls for:

x1(A1 , A2 ) = x

1(

A1 ,

A2 ) =

1A2

x2(A1 , A2 ) = x

2(

A1 ,

A2 ) =

1A1

Hence TiA1 ,

A2

= Uj

x1

A1 ,

A2

, x2

A1 ,

A2

, Aj

+Ki

Aj

= 1

Ai 1

2Aj+Ki

Aj

.

22

Budget balance requires T1(A1 , A2 ) + T2(A1 , A2 ) = 0 8(A1 , A2 ), so1A1 1

2A2+K1(A2 ) +K2(

A1 ) +

1A2 1

2A1= 0.

The simplest choices for K1,K2 that satisfy this are

K1(A2 ) =1

2A2K2(A1 ) =

12A1

.

This leads to T1A1 ,

A2

= 1

A1 1

A2= T2

A1 ,

A2

.

Method 2 We derive this from first principles. Consider a dierentiable direct-revelation mechanism that implements the first best outcome with budget balance. Weassume that player 2 tells the truth. Let W (1, , 2) represent the payos of player 1when his true type is 1, his announcement is type , and player 2s true type is 2. I.e.,

W (1, , 2) =12 1

22+ T1 (, 2)

Since player 1 would want to maximize his payos, we take the FOC wrt :

dW

d=

13

+dT1 (, 2)

d

Since truth-telling must be a dominant strategy [as the mechanism satisfies incentivecompatibility conditions], it must be that for each 2, dWd |=1 = 0. Making the substi-tution implies that

dT1 (1, 2)d

= 121

Integrating up, we obtain, T1 (1, 2) = 11 + C1 (2). To establish budget balance, letT1 (1, 1) = 0. Hence, C1 (1) = 11 , and therefore letting C1 (x) = 1x assures budgetbalance. Thus, using Ti (i, j) = 1i 1j implements the optimal mechanism.

34. Let x denote the amount in Lilos box and y denote the amount in Tims box. The pairof functions sL(x) = x, and sT (y) = y constitute a Bayes-Nash eqm. Conditional on Timfollowing this strategy, then Lilos problem is to choose z tomaxz

R 1000

1100

z+y2

1{z+yx+y}dy.

If z > x, then this expression is 0, and for all z < x, this expression is increasing in z.Therefore, Lilo would not want to deviate from eqm strategies. The analysis is symmetric forTim.

35. a) A pure strategy is a mapping from [0, 1] to R+. An example: b(s) = s for any s 2 [0, 1].b) An example: b(s) = 3 for any s 2 [0, 1] is weakly dominated by the strategy b(s) = 2

for any s 2 [0, 1]. It is not strictly dominated because when the other bidder bidsaccording to b(s) = 0 for any s 2 [0, 1] there is no other strategy that yields strictlyhigher (expected) payo.

c) Suppose bidder 2 bids according to b2(s2) = s2. When observing signal s1, bidder 1s

expected payo by bidding b1 isR b1

0 (s1+ s2s2)ds2 if b1 (so that the upper limit

of the integral does not exceed 1) andR 10 (s1 + s2 s2)ds2 if b1 . Note that in the

second case the payo does not depend on b1.

23

d) Bidder 1s maximization problem is

max0b1Z b1

0(s1 + s2 s2)ds2

Ignoring the b1 constraint, FOC yields b1(s1) = 1s1.e) Unlike in the example from the notes, the value of the item to one player is dependant

on the other players signal. If the other player bids is high, that suggests that thevalue of the object is higher. Therefore, it is a best response to bid more than onesown valuation.

f) Symmetry implies 1 = , therefore = 2. Now check when = 2, b1(s1) = 2s1

= 2 for all s1, therefore the inequality constraint is satisfied.So a symmetric BNE is bi(si) = 2si for i = 1, 2.

g) From the previous part, if bidder 1s strategy is b1(s1) = s1, bidder 2s best response is

b2(s2) =

1s2 if

1s2 2 [,1) if 1s2

In particular, b2(s2) = 1s2 is a best response for bidder 2. Let =

1 , similarly wecan check that given b2(s2) = s2, bidder 1s best response is b1(s1) = s1.Therefore, there exist a continuum of asymmetric equilibria, of the form b1(s1) = s1and b2(s2) = s2 where = 1 , > 1.

h) In an equilibrium where b1(s1) = s1 and b2(s2) = 1s2, bidder 1 wins if and only ifs1 11s2.If < 2, < 1 , i.e. bidder 1 is less aggressive than bidder 2. For any s1, the expectedpayo to bidder 1 is

1(s1) =Z (1)s10

(s1 + s2 1s2)ds2 =

12( 1)s21

If > 2, > 1 , i.e. bidder 1 is more aggressive than bidder 2. For any s1, theexpected payo to bidder 1 is

1(s1) =

( R (1)s10 (s1 + s2 1s2)ds2 = 12( 1)s21 if s1 11R 1

0 (s1 + s2 1s2)ds2 = s1 12(1) if s1 > 111(s1) is increasing in . A larger means bidder 1 is more aggressive in bidding. Insuch an asymmetric equilibrium, the more aggressive is bidder 1, the less aggressivebidder 2 is. And by being more aggressive bidder 1 increases his chance of winning andreduces the payment upon winning.

36. a) Yes there is; if every other individual discloses her grade, an individual is indierentbetween disclosing her grade and not doing so since in either case, she would receive thesame marks.

b) Yes it is. If this student believes that all other students are disclosing their grades, sheis indierent between disclosing and not disclosing. As such, it is 1-rationalizable forher to not disclose her grade. Since by (a), it is a NE for all students to disclose theirgrades, it is rationalizable for all other students to disclose their grades. Therefore, itis rationalizable for a student with 10 to not disclose her grade.

24

c) -d) Proving that there is no such mixed strategy equilibrium suces to establish thatthere is no such PSE in (c). Suppose by way of contradiction that there is some mixedstrategy equilibrium where an inidividual with a strictly positive grade does not discloseher grade with positive probability. LetM > 0 be the highest grade that is not disclosedin equilibrium with positive probability. Then all students with grades of 0 shall notdisclose their grades with probability 1 [since not disclosing their grades yields a positiveexpected grade]. Therefore, any student with a grade M shall receive a grade strictlyless than M if she does not disclose her grade. Therefore, she would strictly prefer todisclose her grade, leading to a contradiction.

37. This problem is a colossal pain. The technique is to use backward induction, carefully checkingall equilibrium outcomes in each subgame.A strategy in this game is given by ((r, p,), (q, )) where r is the probability E places on In,p and q are the probabilities placed on Fight by E and I respectively, and and are theprobabilities placed on Here by E and I respectively. To simplify notation, if e.g. the entrantplays F with probability 1 if In, we write ((r, F,), (q, )) etc.We first look at the final subgame:

H TH 3, 1 0, 0T 0, 0 x, 3

Three cases:

x < 0 Here the game is dominance solvable. The unique eqm is (H,H), with outcome (3,1).

x = 0 Here we have a continuum of eqa characterized by (, T ) where 34 and corre-sponding eqm outcomes (0, 3(1 )).

x > 0 Here we have two pure strategy eqa, as well as a mixed strategy eqm: (H,H) yieldingoutcome (3,1), (T, T ) yielding outcome (x, 3), and ( = 34 , =

33+x) yielding outcome

( 3x3+x ,34).

We now replace the H/T subgame with each of the outcomes above (backward induction).In all three cases, (H,H) is an eqm, with outcome (3,1). Replacing the final subgame withthis outcome, the F/A subgame is reduced to:

F AF 3,1 1,2A 2,1 3, 1

This subgame is dominance solvable, with eqm (A,A) and outcome (3,1). Replacing thissubgame with its outcome, we find that the Entrant will choose In. Thus, 8x we have SPNE:

((In,A,H), (A,H)). (6)

Case x = 0 The F/A subgame becomes

F AF 3,1 1,2A 2,1 0, 3(1 )

25

This game has a unique eqm a mixed strategy eqm given by (p = 4353 , q =12) with outcome

(1,1). It follows that E will play Out. Thus, when x = 0 we have SPNEOut,

4 35 3 ,

,

12, T

for any 3

4. (7)

Case x > 0, outcome = (x,3) Now the F/A subgame becomes

F AF 3,1 1,2A 2,1 x, 3

Since x is in the normal form, we have subcases.

a) x > 1 Here the F/A subgame is dominance solvable. The eqm is (A,A), with outcome(x, 3). It follows that E will play In. Thus, when x > 1, we have SPNE

((In,A, T ), (A, T )). (8)

b) x = 1 Here we have a continuum of eqa given by (p,A) where p 45 , and correspondingoutcomes (1, 3 5p). For any outcome, E will play In. Thus, when x = 1, we haveSPNE

((In, p, T ), (A, T )) for any p 45. (9)

c) x < 1 Here we have a unique eqm a mixed eqm given by (p = 45 , q =1x2x) with outcome

(3x22x ,1). Whether this outcome beats the payo 0 choice of Out depends on x.When x < 2/3 we have SPNE

Out,45, T

,

1 x2 x, T

. (10)

When x = 2/3 we have SPNEr,45, T

,

1 x2 x, T

for any r 2 [0, 1]. (11)

When x > 2/3 we have SPNEIn,

45, T

,

1 x2 x, T

. (12)

Case x > 0, outcome = ( 3x3+x ,34) Now the F/A subgame becomes

F AF 3,1 1,2A 2,1 3x3+x , 34

Since x is in the normal form, we again have subcases.

26

a) x > 32 Here the F/A subgame is dominance solvable. The eqm is (A,A), with outcome( 3x3+x ,

34). In this case, E will play In. Thus, when x >

32 we have SPNE

In,A,34

,

A,

x

3 + x

. (13)

b) x = 32 Here we have a continuum of eqa, given by (p,A) for p 711 , with correspondingoutcomes (1, 311p4 ). For every outcome, E will play In. Thus, when x =

32 we have

SPNE In, p,

34

,

A,

x

3 + x

for any p 7

11. (14)

c) x < 32 Here we have a unique eqm a mixed eqm given by (p =711 , q =

32x6x ) with

outcome (7x66x ,1). Whether this outcome beats the payo 0 choice of Out depends onx.When x < 6/7 we have SPNE

Out,711,34

,

3 2x6 x ,

x

3 + x

. (15)

When x = 6/7 we have SPNEr,

711,34

,

3 2x6 x ,

x

3 + x

for any r 2 [0, 1]. (16)

When x > 6/7 we have SPNEIn,

711,34

,

3 2x6 x ,

x

3 + x

. (17)

38. a) i. Si = {for, against} for i = A,B,C.ii. See attached.iii. Here g() maps strategies to the triplet of payos where As payo is listed first, Bs

second and Cs third.

g(f, f, f) = (1, 1,1) g(a, f, f) = (1, 1,1)g(f, f, a) = (1, 1,1) g(a, f, a) = (0, 0, 0)g(f, a, f) = (1, 1,1) g(a, a, f) = (0, 0, 0)g(f, a, a) = (0, 0, 0) g(a, a, a) = (0, 0, 0)

iv. The Nash equilibria are (f, f, f), (f, f, a) and (a, a, a). The bill passes in the firsttwo equilibria.

v. Observe that for A and B, voting against the bill is weakly dominated. For C,voting for the bill is weakly dominated. Thus, only in (f, f, a) does no player use aweakly dominated strategy.

vi. There is no pure strategy SPNE. There are several ways to see this. One is toreduce the game by replacing subgames with their equilibrium outcomes. There are8 subgames: the game itself, and the subgames following completion of the show/noshow game. For the subgame following sss, there are two equilibrium outcomes:(1 ", 1 ",1 ") and (",","). For the subgame following ssn, there are also

27

two equilibrium outcomes: (1 ", 1 ",1) and (",", 0). Every other subgamehas a unique equilibrium outcome. Thus, there are 2 2 = 4 reduced games tocheck. None of these has a pure strategy NE. (Although we know for sure that thereexists a mixed strategy NE!)Another, perhaps simpler method is to rule out profiles case by case. Ask, is therean equilibrium where as the entry component of their strategies A,B and C choose:

sss No. If following sss, an eqm where the bill passes is chosen, then C would do betterto not show up. If the no pass eqm is chosen in the subgame, then A or B wouldhave done better to not show up.

ssn No. Then A or B would do better by not showing up.sns No. Then A would do better by not showing up.snn No. Then C would do better by showing up.nss No. Then B would do better by not showing up.nsn No. Then C would do better by showing up.nns No. Then C would do better by not showing up.nnn No. Then A or B would do better by showing up.

b) i. Si = {for, against} for i = A,B,C.ii. SA = {f,a}

SB = {,fa,af,aa}SC = {,fa,af,aa,fa,fafa,faaf,faaa,af,aa,afaf,afaa,aa,aafa,aaaf,aaaa}

iii. Yes. This can be shown using backward induction, replacing each subgame with itsequilibrium outcome. One eqm is given by (f,,aaaa).

iv. No. Again, using backward induction to replace each subgame with its equilibriumoutcome, we find that the equilibrium outcome is that the bill passes.

39. a) Since we are looking for SPE where voters vote sincerely, their actions are easily specifiedat every information-set (they vote for the alternative they strictly prefer, and in thecase of indierence, can settle indierence any which way). As such, what one needsto determine by backward induction is what the agenda-setter shall propose takinginto account that voters shall vote sincerely. The SPE outcomes are characterized foreach initial status quo.Table 1Status Quo Alternatives majority-preferred to status quo Outcome(0, 0, 0) (0, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1) (0, 0, 0)(0, 0, 1) (0, 0, 0), (0, 0, 1), (1, 1, 0), (1, 1, 1) (0, 0, 0)(0, 1, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 1), (1, 1, 1) (0, 0, 0)(1, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0), (0, 1, 1), (1, 1, 1) (0, 0, 0)(0, 1, 1) (0, 0, 1), (0, 1, 0), (0, 1, 1) (0, 0, 1)(1, 0, 1) (0, 0, 1), (1, 0, 0), (0, 1, 1), (1, 0, 1) (0, 0, 1)(1, 1, 0) (0, 1, 0), (1, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0) (0, 1, 0)(1, 1, 1) (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1) (0, 1, 1)

b) From part A, we know that there are only four possible outcomes on the equilibriumpath: (0, 0, 0), (0, 0, 1), (0, 1, 0), and (0, 1, 1). In round 1, each z is associated with oneof these outcomes (in the sense that, if z becomes the status quo for the second round, itwill lead to the outcome indicated in the table 1). Consequently, the two round game is

28

just like a one round game where the possible choices are (0, 0, 0), (0, 0, 1), (0, 1, 0), and(0, 1, 1). For this reduced form one round game, we can make up a table like table

.

Table 2Status quo Possibilities Outcome(0, 0, 0) (0, 0, 0), (0, 1, 1) (0, 0, 0)(0, 0, 1) (0, 0, 0), (0, 0, 1) (0, 0, 0)(0, 1, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0) (0, 0, 0)(0, 1, 1) (0, 0, 1), (0, 1, 0), (0, 1, 1) (0, 0, 1)

Combining table 2 with table 1, we can make up the required table for the two roundgame.Table 3Status Quo Outcome Proposal(s)(0, 0, 0) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(0, 0, 1) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(0, 1, 0) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(1, 0, 0) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(0, 1, 1) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(1, 0, 1) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(1, 1, 0) (0, 0, 0) (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0)(1, 1, 1) (0, 0, 1) (0, 1, 1), (1, 0, 1)

For example, consider the status quo (1, 0, 1). We know from table 1 that, if thisbecomes the status quo for round 2, the outcome will be (0, 0, 1). From table 2, we seethat, in the reduced form one round game, (0, 0, 1) leads to the outcome (0, 0, 0), whichtherefore appears in the outcome column of table 3. From table 1, we see that thisoutcome is obtained if (0, 0, 0), (0, 0, 1), (0, 1, 0), or (1, 0, 0) becomes the status quo forround 2. Consequently, the agenda setter can propose any of these alternatives in round1.

c) From part B, we know that there are only two possible outcome: (0, 0, 0) and (0, 0, 1).In round 1, each z is associated with one of these outcomes (in the sense that, if zbecomes the status quo for the second round, it will lead to the outcome indicated inthe table 3). Consequently, the three round game is just like a one round game wherethe possible choices are (0, 0, 0) and (0, 0, 1). For this reduced from one round game,we can make up a table like table 1.Table 4Status quo Possibilities Outcome(0, 0, 0) (0, 0, 0) (0, 0, 0)(0, 0, 1) (0, 0, 0), (0, 0, 1) (0, 0, 1)

Combining table 4 with table 3, we can make up the required table for the three roundgame.

29

Table 5Status Quo Outcome Proposal(s)(0, 0, 0) (0, 0, 0) Anything but (1, 1, 1)(0, 0, 1) (0, 0, 0) Anything but (1, 1, 1)(0, 1, 0) (0, 0, 0) Anything but (1, 1, 1)(1, 0, 0) (0, 0, 0) Anything but (1, 1, 1)(0, 1, 1) (0, 0, 0) Anything but (1, 1, 1)(1, 0, 1) (0, 0, 0) Anything but (1, 1, 1)(1, 1, 0) (0, 0, 0) Anything but (1, 1, 1)(1, 1, 1) (0, 0, 0) Anything but (1, 1, 1)

For example, consider the status quo (1, 0, 1). We know from table 3 that, if thisbecomes the status quo for round 2, the outcome will be (0, 0, 0). From table 4, we seethat, in the reduced form one round game, (0, 0, 0) leads to the outcome (0, 0, 0), whichtherefore appears in the outcome column of table 5. From table 3, we see that thisoutcome is obtained if anything other than (1, 1, 1) becomes the status quo for round 2.Consequently, the agenda setter can propose any of these alternatives in round 1.

d) True. From part C, the statement is true with three rounds. Moreover, with morethan three rounds, we know the outcome will be (0, 0, 0) regardless of the status quoinherited by the third to last round. Therefore, all play before the third-to-last roundis irrelevant, as all status quos lead to the same outcome. If the agenda setter proposes(0, 0, 0) in the first round, all legislators are willing to vote for it, even though they prefer(1, 1, 1).

40. Recall that in an extensive form game, a (pure) strategy consists of an action taken at eachinformation set. In this game, information sets consist of monopoly nodes and duopoly nodes.That is, in the extensive form, each player has exactly t + 1 time t information sets - oneduopoly node (no one exited in time 0 . . . t 1) and t monopoly nodes (opponent exited ateach time 0 . . . t 1).Using backward induction on the subgames that begin with monopoly nodes, actions inmonopoly nodes are easily specified:Firm A:t = 1 . . . 19 ) stay int = 20 ) stay in, 1 outt > 20 ) out

Firm B:t = 1 . . . 24 ) stay int = 25 ) stay in, 1 outt > 25 ) out

We now must specify actions taken at duopoly nodes. Again we rely on backward induction.In the duopoly subgame beginning at any t > 25, for each firm, any strategy specifying outat time t strictly dominates any strategy specifying stay in. Thus, out is chosen by eachfirm at each duopoly node with t > 25.Consider the t = 25 duopoly subgame. For firm A, any strategy specifying out strictlydominates any strategy specifying in, so A plays out at t = 25. Firm B mixes betweenout and in.Consider the t = 2124 duopoly subgames. Here As dominant strategy is to exit immediatelyregardless. It follows that B stays in and earns monopoly profits.

30

Consider the t = 20 duopoly subgame. Here we see that it is a dominant strategy for B toplay in, because in the continuation subgame, A will drop out in the next period and Bwill reap 4 periods of monopoly profit. Thus, A must play out.

Consider the t = 11 19 duopoly subgames. For each t, it is a dominant strategy for B toplay in because (by backward induction) in the continuation subgame, A will drop out inthe next period. Thus A plays out.

Consider the t = 10 duopoly subgame. in is a dominant strategy for A because duopolyprofits are positive. in is also a dominant strategy for B because as before, A will drop outin the next period.

In the t = 0 9 subgames, duopoly profits are positive for both, so both play in.Thus, the SPE take the following form:

As strategy:

Monopoly node: Stay in 0-19. Any mixture at 20. Out after 20.Duopoly node: Stay in 0-10. Exit 11 or later.

Bs strategy:

Monopoly or duopoly node: Stay in 0-24. Any mixture at 25. Out 26 or later.

41. a) The planner solvesmax

ap,a1,a2[yp + y1 + y2]

which yields ap = a1 = a2 = 1. The resulting total family income is 11.5

b) Each player i has a dominant strategy of choosing ai = 3. This yields a unique NE of(3,3,3). The resulting family income is 5.5.

c) Notice that by naming action ai, i creates a (combined) negative externality ofa2i3 on

the others. By taxing i the value of this externality (a Pigouvian tax), i internalizes hiseect on the others. Thus, let is transfer be ti(ai, aj , ak) = a

2i3 . is problem is now

maxai

" ki + ai a

2i

6 a

2j

6 a

2k

6

! a

2i

3

#

which has solution ai = 1, coinciding with the social optimum.5

d) Lets start with the subgame where the parent faces incomes yp, y1, y2. The parent solves

maxcp,c1,c2

ln cp +

12ln c1 +

12ln c2

subject to cp + c1 + c2 = yp + y1 + y2.

This is a standard Cobb-Douglas maximization problem (with prices all 1 and incomeyp + y1 + y2). The parent will redistribute the income in shares 1 : 12 :

12 . Therefore, the

goal of all parties going into the subgame is to maximize total income in stage 1.Therefore, in stage 1, i solves

maxai

[yp + y1 + y2]

which yields solution ai = 1 for all i, coinciding with the social optimum.5This tax does not balance the budget, but it is a simple matter to find one that does. One possibility is

ti(ai, aj , ak) = a2i3 +

a2j+a2k

6 .

31

The SPNE is thus

Children: name ai = 1 for i = 1, 2Parent: name ap = 1; facing Y = yp + y1 + y2 ! redistribute in shares 1 : 12 : 12 .

42. a) The normal form of the game is

A/B C DC x2 , 1 x2 x, 1 xD 1 x, x 1x2 , 1+x2

ForB, neither strategy is strictly dominant. We can identify behavior using the followingtable:

x As dominant strategy PSNE2 0, 13 D (D,D)= 13 None (D,D)13 ,

23

None None

= 23 None (C,C)> 23 C (C,C)

If x 2 0, 13, (D,D) is the unique rationalizable profile, and therefore, the unique Nashequilibrium. Similarly, if x 2 23 , 1, (C,C) is the unique rationalizable profile, andtherefore the unique Nash equilibrium. If x = 13 , observe that if A plays C with positiveprobability, any best-response by B involves playing C with positive probability. Notethat for every strategy of Bs that puts positive probability on C, A does strictly betterby playing D than by putting any weight on C. Therefore, the unique NE is (D,D).Similarly, if x = 23 , the unique NE is (C,C).Now consider x 2 13 , 23: the uniqueNE is the MSNE ((1 x)C + xD, (3x 1)C + (2 3x)D).

b) Given a deployment x, any SPE involves the play of the corresponding NE above. There-fore, a choice of x results in expected payos for A as given below:

x As expected payos 13 1x22 13 , 23 3x(1x)2 23 x2

Observe thatMaxx2( 13 , 23)3x(1x)

2 =38 S (M 1). SinceB (M) > B (M 1) > S (M 1) > S (M), it must be a unique best-response for M +1to also play In leading to a contradiction.

Since both equilibria exist regardless of how S (0) relates to B (N 1), there is notendency for the equilibrium to necessarily be ecient.

b) In Version #2, i receives payos

S (i, n) = ai bn if i chooses OutB (i, n) = c+ dn ei if i chooses In

The game is indeed dominance solvable and involves every individual choosing In. LetR (i, n) = B (i, n)S (i, n), and let the abbreviation ISD stand for Iterative eliminationof strictly dominated strategies.Lemma 1: It is strictly dominated for 1 to choose Out.Pf : Observe that the incremental utility of going In for player 1 when n others chooseIn is

R (1, n) = c e a+ (d+ b)nObserve that for all n > 0, R (1, n) > R (1, 0) = c e a > 0. Therefore, regardless ofthe choice of others, it is a dominant strategy for 1 to choose In.Lemma 2: If ISD prescribes that individuals {1, ...,M} must choose In, then ISD pre-scribes that M + 1 also chooses to go In.Pf : Observe that R (M + 1, k) is strictly increasing in k; therefore, after ISD prescribesthat individuals {1, ...,M} choose In, then there are at least M players who choose In.Therefore, for every n M ,

R (M + 1, n) R (M + 1,M)= R (M,M 1) + (d e) (a b) R (M,M 1) > 0

where the last inequality follows from the fact that ISD prescribed that individual Mchoose In. Therefore, ISD prescribes that individual M + 1 chooses to go In.Therefore, by Induction, in any rationalizable strategy profile, all individuals choose togo In. Thus, this is what happens also in every CE and NE.

c) For the case of N = 3, observe that by b., on the path of play, all 3 individuals shallchoose to go In. We have to specify o-path behavior for a SPE, and this involves thecomputation of multiple tedious cases.

44. Public Good Game

a) Normal Formc n

c 2a, 2a a, 1 + an 1 + a, a 1, 1

Contributing yields a < 1, so not contributing is a dominant strategy. Thus, the uniqueNE is (n, n).

b) Notice that punishing is costly, so no player will choose to punish his opponent in thesubgame. The SPNE involves (n, n), followed by no punishment in the second period.

33

i. Now players care about equity. Inequity in monetary payos is not desirable, whereinequity in favor of your opponent is worse than inequity in your favor.

ii. Normal form:c n

c 2a, 2a a 1, 1 + a 2n 1 + a 1, a 2 1, 1

Notice that (n, n) is still a NE, as deviating to c reduces payo and decreases equity.(c, n) and (n, c) can never be equilibria, as deviating from c to n increases payo anddecreases inequity. However, (c, c) can be sustained provided players have enoughdistaste for inequity in their favor. This condition is given by 2a > 1 + a i, or

i > 1 a for i = 1, 2.iii. Imagine that players start the sub-game with wealth levels Y1 and Y2.

Suppose Y1 > Y2. There is no eqm in which only 1 punishes, as this reduces payofor 1 and decreases equity. For 2, punishing 1 dollar reduces payo by 1, but via thereduction in inequity, increases payo by 2(K1). This will be profitable provided2(K 1) > 1 or 2 > 1K1 . In this case, 2 will punish until the players have equalmonetary payo:

Y1 Kp = Y2 p) p = Y1 Y2K 1 .

The resulting payos are 1 = 2 = KK1Y2 1K1Y1.If instead we have 2 < 1K1 , no player punishes. If 2 =

1K1 , 2 picks any p p.

The case Y1 < Y2 is symmetric to the case above.If Y1 = Y2, neither player punishes.There are other equilibria. For example, if both players have i > 1K1 and Y1 > Y2,then (p1, p2) is a NE when

p2 = p1 +Y1 Y2K 1 .

Call i an enforcer if i has i > 1K1 .If neither player is an enforcer, then there will be no punishment in the subgames,and the equilibria are as in (d).If both players are enforcers, then (n, n) can be sustained in equilibrium, but so toocan (c, c). Notice that i would never deviate to n, as that would reduce js stage1 payo, and following punishment, players will have equal payos, which will belower than js payo from the first stage. (n, c) and (c, n) can never be sustained,as deviating from c to n increases payo, and reduces punishment.If only 1 is an enforcer, (n, n) can be sustained and (c, n) and (n, c) cannot, forreasons identical to the two enforcers case. (c, c) can be sustained provided 1 hasstrong distaste for inequity in his favor: 1 > 1 a, the condition from part (d).If only 2 is an enforcer, the analysis is identical.

45. a) i. See attached figure.ii. In period 1, each player chooses to clap (C) or not clap (N). Should both players

have clapped in period 1, each player chooses C or N in 2. Thus, we have S1 =S2 = {CC,CN,NC,NN}.

34

Normal form:

CC CN NC NN

CC d2 2c, d2 2c d 2c,c d c, 0 d c, 0 CN c, d 2c c,c d c, 0 d c, 0NC 0, d c 0, d c 0, 0 0, 0NN 0, d c 0, d c 0, 0 0, 0

Marked with a * above.iii. We look at behavioral mixed strategies, keeping Kuhns theorem in mind.

There is a unique proper subgame which is reached when both players clap in period1. Notice that in the final period, cost c is sunk for each player. Clapping costsc and yields at worst d2 > c, so clapping is a dominant strategy. To find the firstperiod action, we have reduced normal form:

C N

C d2 2c, d2 2c d c, 0N 0, d c 0, 0

This reduced game has 3 equilibria. (C,N) and (N,C) are pure equilibria.(C+(1)N,C+(1)N) is the mixed strategy equilibrium, where = 2(dc)2c+d .Thus, we have three SPNE, corresponding by the actions given above, combinedwith C,C in the subgame. Corresponding payos are (d c, 0), (0, d c) and (0, 0).

b) i. Starting in the third period, observe that clapping is a dominant strategy. Movingback to the second period, clapping is again a dominant strategy. This is the casesince d2 > 4c, clapping guarantees a positive payo, even if one is forced to split theprize in the end. We now find actions in the first period by looking at the reducednormal form:

C N

C d2 3c, d2 3c d c, 0N 0, d c 0, 0

This reduced game has 3 equilibria. (C,N) and (N,C) are pure equilibria.(C + (1 )N,C + (1 )N) is the mixed strategy equilibrium, where = .Thus, we have three SPNE, corresponding by the actions given above, combinedwith C,C in periods 2 and 3. Corresponding payos are (d c, 0), (0, d c) and(0, 0).

ii. Starting in period 3, we see that clapping is a dominant strategy. To find period2 actions, we solve the same reduced normal form game as in (a) iv, which yieldedthree possible outcomes.6 To find first period actions, we examine the reducednormal form using each of these three second period outcomes:

6Note that in this case, we have a sunk cost of c, so all payos are reduced by c. But this does not aect ourequilibria.

35

Second period actions (C,N)! outcome (d c, 0)!

C N

C d 2c,c d c, 0N 0, d c 0, 0

This game is dominance solvable with unique NE (C,N). Thus, we have

SPNE 1 = (CCC,NNC) (18)

Similarly, second period actions (N,C)! outcome (0, d c)!SPNE 2 = (NNC,CCC) (19)

The third outcome involved payos (0, 0), yielding reduced normal form:

C NC c,c d c, 0N 0, d c 0, 0

This reduced game has 3 equilibria. (C,N) and (N,C) are pure equilibria.(C+(1)N,C+(1)N) is the mixed strategy equilibrium, where = 1 cd .Thus, we have three SPNE, corresponding by the actions given above, combinedwith the mixture in period 2, and (C,C) in period 3. Corresponding payos are(d c, 0), (0, d c) and (0, 0). To sum up,

SPNE 3 = ((C,C,C), (N,C,C)) (20)SPNE 4 = ((N,C,C), (C,C,C)) (21)SPNE 5 = ((C,C,C), (C,C,C)) (22)

c) i. Costs (T 1)c are sunk. Clapping yields an additionalc+ (1 p)d+ p p [c+ (1 p)d+ p p [c+ (1 p)d+ p [. . .

which is a geometric series, reducing to

11 p2 [c+ (1 p)d] .

ii. We look for a stationary equilibrium in which players clap with probability p inevery period. For A to be indierent between clapping and not clapping, we musthave

11 p2 [c+ (1 p)d]| {z }

payo from clapping

= 0|{z}payo from not clapping

which yields

p =d cd

.

Equilibrium payo is (0, 0).

36

iii. One asymmetric equilibrium is that 1 chooses to clap in every period and 2 choosesnot to clap in every period. This yields payos (d c, 0). There are other eqa, forexample, asymmetric equilibria with a tail in which players clap with probability p.

46. a) i. Candidate 1s strategy set is who he will attack: S1 = {2, 3}. Player 2 has twoinformation sets in which he can make a decision (1 attacked 2 unsuccessfully, 1attacked 3 unsuccessfully). In the other situations, he is not in the game or canonly attack 1 (which we will take as a set move rather than a choice). Therefore,his strategy set is S2 = {11, 13, 31, 33}. Finally, player 3 has four information setsin which he can make a decision (1 attacked 2 unsuccessfully then 2 attacked 1 un-successfully, 1 attacked 2 unsuccessfully then 2 attacked 3 unsuccessfully, 1 attacked3 unsuccessfully then 2 attacked 1 unsuccessfully, 1 attacked 3 unsuccessfully then2 attacked 3 unsuccessfully). So, his strategy set contains 16 elements. Note thateven though player 3 has 4 information sets, the actions and payos are the samefor each.

ii. Two things to note about this game: (1) Whenever it is a candidates turn to attack,if the number of surviving opponents is less than 2, he has no choice to make; (2)For each i = 1, 2, 3, all subgames starting with candidate i attacking while bothopponents still surviving are the same in terms of payo, disregarding the history.So this game essentially consists of three (classes of) subgames, namely, for eachi = 1, 2, 3, Gi denotes the (class of) subgame starting with candidate i decidingbetween attacking one of his two opponents.First consider subgame G3. Candidate 3s survival probability is now 1. Giventhis, he would like to attack the opponent with higher pi. Therefore, candidate 3sstrategy is to attack 1 if p1 > p2, and to attack 2 if p1 < p2.Now consider subgame G2. If p1 > p2, candidate 2s survival probability is p2(1 p3)+(1p2) if he attacks 1 and p2+(1p2) if he attacks 3; if p1 < p2, candidate 2ssurvival probability is p2(1p3)+(1p2)(1p3) if he attacks 1 and p2+(1p2)(1p3)if he attacks 3. In both cases, attacking 3 yields strictly higher survival probabilitythan attacking 1 for candidate 2. Therefore, candidate 2s strategy is to attack 3.Finally consider subgame G1 (the whole game). Note that whomever candidate 1attacks, if his attack is not successful, we are in subgame G2 which has a uniqueNE as shown above; therefore 1s surviving probability conditional on his attack notsuccessful is the same whether he attacks 2 or 3. Now if his attack is successful,attacking 2 gives him surviving probability of 1 p3 (since now that 2 is out, 3will for sure attack 1), and attacking 3 gives him surviving probability of 1 p2.Therefore, candidate 1s strategy is to attack 2 if p2 > p3 and to attack 3 if p2 < p3.To summarize, the unique SPNE is: Candidate 1 attacks 2 if p2 > p3 and attacks 3if p2 < p3; If candidate 1s attack is not successful, candidate 2 attacks 3; If neither1 nor 2s attack is successful, candidate 3 attacks 1 if p1 > p2, and attacks 2 ifp1 < p2; In all other events, no choice needs to be made.

iii. If p3 < p2, candidate 3 survives with probability p1 + (1 p1)(1 p2); if p3 > p2,candidate 3 survives with probability (1 p1)(1 p2).

b) As in the previous part, we only need to consider subgames G1, G2 and G3.First consider subgame G3. Candidate 3s survival probability is now 1. Given this, hewould like to minimize the danger posed by the survival of other opponents and thereforewill not want to spend on the charity cause. So he will attack the opponent with higherpi (just like in the previous part). Now given that p1 < p2, he will attack 2.

37

Now consider subgame G2. Since p1 < p2, candidate 2s survival probability is p2(1 p3) + (1 p2)(1 p3) if he attacks 1, p2 + (1 p2)(1 p3) if he attacks 3, and 1 p3 ifhe spends on charity. Attacking 3 yields strictly higher survival probability. Therefore,candidate 2s strategy is to attack 3.Finally consider subgame G1 (the whole game). Given the outcomes of the subgames,if candidate 1 spends on charity his survival probability is 1. If he attacks either of thetwo opponents, his survival probability is strictly less than 1. Therefore candidate 1sstrategy is not to attack but instead to spend on the charity cause.To summarize, the unique SPNE is:Candidate 1 does not attack;If candidate 1 does not attack or his attack is not successful, candidate 2 attacks 3;If there has been no successful attack, candidate 3 attacks 2;In all other events, no choice needs to be made.It diers from the SPNE in part (1) only in that candidate 1 now does not attack. Thiswill ensure that 2 and 3 fight between themselves and therefore ensure himself a survivalprobability of 1.

47. ?

48. a) The extensive form is straightforward. Pure strategies: each player makes a decision atonly one information set, so the pure strategy set for each is {perform,wait}, except forthe leader at t = 4, who has no choices (and who therefore is not really a player we willhenceforth ignore him). There are eight strategy profiles of the form {s1, s2, s3}, wheresi is either perform or wait. All strategy profiles with s1 = perform lead to payos of(4,8,8). All strategy profiles with s1 = wait and s2 = perform lead to payos of (6,2,6).All strategy profiles with s1 = s2 = wait and s3 = perform lead to payos of (4,4,0).Finally, the strategy profile (wait,wait,wait) leads to payos of (1,1,1).

i. Leader 3 will wait (0 1). Therefore, leader 2 will perform (2 1). Therefore,leader 1 will wait (4 6). The task will be performed at t = 2.

ii. Payos are (6,2,6). Only performing at t = 1 makes leader 3 better o, but thatmakes leader 1 worse o. Therefore the outcome is Pareto ecient.

iii. Yes, the outcome must be Pareto ecient. The proof is simple: There cant be anearlier performance date that makes everyone better o; if there was, the leader atthat date would break the equilibrium by performing the task. Nor can there be alater date that makes everyone better o, because vt ct is assumed to be strictlydecreasing in t.

b) i. Leader 3 will wait (0 1). Leader 2 will wait (2 4). Leader 1 will wait (4 6). Sothe task will be performed at t = 4.

ii. Payos are (1,1,1), compared with (6,2,6) in the subgame perfect Nash equilibrium,so everyone is worse o when they behave naively. Clearly, the outcome is notPareto optimal.

iii. The task is performed later with naive players than in the subgame perfect Nashequilibrium (t = 4 instead of t = 2). More generally, the task is performed weaklylater when agents are naive. Let V et be the continuation utility that leader t expectsto receive in equilibrium if period t has arrived and he decides to wait. Let V nt bethe continuation utility that a naive leader t expects to receive if period t has arrivedand he decides to wait. Obviously, V nt V et . Therefore, if wait is the equilibrium

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strategy for leader t, it must also be the naive choice for leader t. Consequently,the team cannot perform the task sooner if all leaders are naive.

49. Hide and Seek

50. a) A strategy for Player 1 is a mapping from all the dierent time and position (t, k)configurations at which he might be called to drive to the set {e, w}. In particular notethat Player 1 only drives on odd days, and if we denote the driving positions k on themap as:

M = {(x 1), ... 1, 0, 1, ..., x 1}then Player 1 can only ever be called to drive at cells with an even label (and is alwayscalled to act at such cells). So if we let K1 = {(t, k) : t 1, t is odd, |k| x 1, k is even, k < t}, then a strategy for Player 1 is a function s1 : K1 ! {e, w}. Notethat the last inequality in the definition of K1 just addresses the fact that some cellsare too far away for the Player to be called on to drive at them in the early days of thetrip. The set of strategies for Player 1 is the set of all such mappings. Similarly if we letK2 = {(t, k) : t 2, t is even, |k| x 1, k is odd, k < t}, then a strategy for Player 2is a function s2 : K2 ! {e, w}.

b) Player 2 is never called on to act in this case, so the problem is just an optimizationproblem in solitude for Player 1. Since v < v by assumption, the only Nash equilibriumand the only subgame perfect Nash equilibrium of the game is given by the strategys1(0) = w, leading to payos (v, v).

c) i. No. Under any strategy profile either x = 1 or x = 1 is reached. If the playersare to remain oscilating forever their payos are (0, 0) and it must be the case thatat whichever of these positions is reached, Player 2s strategy prescribes going backto 0. He could deviate and prescribe going to the nearest endpoint an get strictlypositive profits.

ii. Consider the strategies given by:

s1(t, 0) = e, 8ts2(t, 1) = s2(t,1) = e 8tPlayer 2 is getting his prefered outcome so it is clear that he has no incentives todeviate. By deviating Player 1 can only postpone the arrival of the party at the fareast, thereby hurting himself, so he has no incentives to deviate either.

iii. For a strategy profile s to be a Nash equilibrium in which the Far east is reached(and Nash is a necessary condition for subgame perfect Nash) it must be the casethat s2(2,1) = e, since otherwise Player 1 could set s1(1, 0) = w and they wouldreach the Far West at t = 2 and for Player 1 this is strictly the best outcome. Forplayer 2 to set s2(2,1) = e in a subgame perfect equilibrium it must be the casethat the payo that he gets by doing so is at least 2v, which is what he would get bysetting s2(2,1) = w. The most that he could hope to get by setting s2(2,1) = eis 4v (if they went all the way to the Far East after his move). So a necessarycondition for an equilibrium of this kind is v 2v. So we have that if v < 2v thereis no such equilibrium. For the case in which v 2v consider the strategies thatwe proposed in part d). It can be verified that they constitute a subgame perfectNash equilibrium.

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iv. Let t be the largest even t such that v t2v and consider the strategies givenby:

s1(t, 1) = w 8t < t 1, s1(t, 0) = e 8t t 1s2(t,1) = e 8t, s2(t, 1) = w 8t < t and s2(t, 1) = e 8t tNote that in this proposed equilibrium the Far East is reached whent t = t, andthe payos are (tv, tv). Player 1 has no profitable deviation, since by alteringhis strategy either going west (w) at t t 1 he only weakly delays the arrival ofthe party at the Far East, and by going east (e) at t < t 1 does not aect theoutcome given the self-flagelating strategy of player 2. Player 2 has no profitabledeviation either. It is clear that he cant possibly improve by switching s2(t,1)from e to w for t t 2, since this can only delay the arrival of the party tothe east. Moreover, given that position k = 1 is never reached when t < t onthe equilibrium path (given Player 1s strategy), switching s2(t, 1) from w to e fort < t, has no eect. Since on the equilibrium path, the last time that the partyvisits 1 is at t 2, switching s2(t,1) from e to w has no eects for t > t 2.Now lets check that Player 2 prefers to oscilate rather than to concede goinf to theFar West, by setting s2(t,1) = e at some point t t 2. It is clear that if hefinds it profitable to do this, then the sooner the better. The soonest at which hecould break the oscilation is at t = 2 whne they frist arrive at k = 1. If Player 2were to deviate at x = 1 when t = 2 by setting s2(2,1) = w he would get 2v,which is strictly less than his equilibrium payo tv given that v t2v.So we have established that there is a Nash equilibrium in which the party reachesthe Far East at time T = t. No such equilibrium exists for T > t and we proveit by contradiction. Suppose that there exists equilibrium s existed in which theFar East (FE) was reached at even time T > t (T has to be even simply becuaseendpoints are always reached in even periods). In any Nash equilibrium it mustbe true that if position k = 1 is reached at time t on the equilibrium path, thens2(t, 1) = e (i.e. Player 2 ceases the first actual opportunity to get to the Far East).This implies that in s, the first time that k = 1 is reached is actually T . Butthis implies that s1(t, 0) = w 8t T 1. In particular s1(1, 0) = w. This inturn implies that s2(2, 1) = e (otherwise the game would end in the Fasr East whent = 2). However setting s2(2, 1) = w would yield 2v to Player 2 and this is greaterthan T v, since T v < 2v. This last inequality is true since by assumption t is thelargest even t such that v t2v.

v. In any subgame perfect equilibrium it must be true that s2(t, 1) = e 8t. So if theFar East is reached with delay (T > 2), it must be that Player 1 is delaying thearrival. In particular it must be true that s2(1, 0) = w. Player 1 would be strictlybetter o accelerating the process, by setting s1(1, 0) = e and getting 2v instead ofT v. So the answer is T = 2: If the Far East is the outcome in a subgame perfectequilibrium, then it must be reached with no delay.

d) i. Consider the strategies given by:

s1(t, 0) = w, 8ts2(t, 1) = s2(t,1) = w 8tPlayer 1 is getting his preferred outcome so it is clear that he has no incentives todeviate. By deviating Player 2 can only postpone the arrival of the party at the FarWest, thereby hurting himself, so he has no incentives to deviate either.

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ii. Consider the strategies given by:

s1(t, 0) = w, 8ts2(t, 1) = e 8t and s2(t,1) = w 8t

In any subgame in which Player 1 is called to drive, he gets his prefered outcome,so he never has an incentive to deviate. At subgames which start at position k = 1,player 2 is acting optimally by setting s2(t, 1) = e. So we just need to check subgamesin ehich player 2 is asked to drive at position k = 1. Given t1s strategy, bydeviating in such a subgame he can only delay the arrival to the Far West, so he isstrictly better o not deviating.

iii. There are no Nash equilibria in which the party arrives to the Far West at timeT > 2. The reason is that in a Nash equilibrium that ends at the Far West,k = 1 cant be reached in the equilibrium path, since at that point Player 2 woulddeviate by driving (e). This means s1(t, 0) = w 8t. But this means t