Test 2. ECON3161, Game Theory. Tuesday, November 6 th

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1 Test 2 ECON36, Game Theory Tuesday, November 6 th Drectons: Answer each queston completely. If you cannot determne the answer, explanng how you would arrve at the answer may earn you some ponts.. (20 ponts) Consder the followng sequental game between three players, represented by the followng game tree, wth P s payo lsted rst, P2 s second, and P3 s thrd. Pay careful attenton to whch player makes a decson at each nformaton set. a (5 ponts) How many strateges does each player have? All players have 2 nformaton sets. Player has 3 actons at both nformaton sets so has 3x3 = 9 strateges. Player 2 has 2 actons at one nformaton set and 3 at the other, so 3x2 = 6 strateges. Player 3 has 2 actons at both nformaton sets so 2x2 = 4 strateges. b (0 ponts) Fnd the subgame perfect Nash equlbrum (SPNE) to ths game. Player 3 chooses J over I because 7 > 6 and Player 3 also chooses L over K because 6 > 4.

2 Player chooses M over N and O because 5 > 2 and 5 > 3. Now, Player 2 chooses E over D because 8 > 7. Player 2 also chooses F over G and H because Player 2 receves 2 from choosng F and wll only receve from choosng ether G (because Player 3 chooses L) or H (because Player chooses M). Now, Player chooses B because he receves 6 from B (because Player 2 s choosng F ) and ths s (concdentally) hs hghest payo n the game. So the SPNE s: P: B, M P2: E, F P3: J, L c (5 ponts) What s the outcome of the game f the players use the SPNE? The outcome s that Player wll choose B and Player 2 wll choose F and the game wll end wth Player recevng 6, Player 2 recevng 2, and Player 3 recevng (30 ponts) Consder a smultaneous quantty choce game between 2 rms. Each rm chooses a quantty, q and q 2 respectvely. The nverse market demand functon s gven by P (Q) = 980 6Q, where Q = q + q 2. Frm has total cost functon T C (q ) = 24q and Frm 2 has total cost functon T C (q 2 ) = 8 (q 2 ) 2 + 2q The pro t functons for each rm are: = (980 6q 6q 2 ) q (24q + 624) 2 = (980 6q 6q 2 ) q 2 8 (q 2 ) 2 + 2q a (0 ponts) Fnd the best response functons for Frms and 2. For Frm we have: = 980q 6q 2 6q q 2 24q 624 = 956q 6q 2 6q q = 956 2q 6q 2 0 = 956 2q 6q 2 For Frm 2 we have: 2q = 956 6q 2 q = 956 6q = 980q 2 6q q 2 6q 2 2 8q 2 2 2q = 968q 2 24q2 2 6q q 2 2 = 968 q 2 6q 0 = 968 q 2 6q q 2 = 968 6q q 2 = 968 6q So the best response functons for Frm and 2 are, respectvely: q = 956 6q q 2 and q 2 =. Techncally, the best responses are: q = Max 0; 956 6q2 2 and q2 = Max 0; q 2 = 968 6q because nether rm can produce a negatve quantty. 2

3 b (0 ponts) Fnd the Nash equlbrum to ths game. Usng the best response functons we have: Now usng that q = 52 we have: q = 956 6q q q = q 2q = q 2q = q = q 90q = 3680 q = 52 q 2 = 968 6q q 2 = q 2 = 22 So the PSNE to ths game s: q = 52 and q 2 = 22. c (0 ponts) Fnd () the total market quantty, (2) the market prce, and (3) each rm s pro t. The total market quantty s Q = q + q 2 = 74. The prce s P (Q) = 980 6Q = = 936. Frm s pro t s: = P q (24q + 624) = ( ) = = Frm 2 s pro t s: 2 = P q 2 2 = q q = ( ) 2 = = (20 ponts) There are three ratonal prates, A, B, and C, who nd 00 gold cons and must decde how to dstrbute them. The prates have a strct order of senorty: A s superor to B and C, and B s superor to C. The prate world s rules of dstrbuton are as follows:. The most senor prate proposes a dstrbuton of cons among the prates. 3

4 2. The prates, ncludng the proposer, then vote on whether to accept ths dstrbuton. 3. If the proposed allocaton s approved by a majorty or a te vote, then the proposal s mplemented. 4. If the proposed allocaton s NOT approved, the proposer s thrown overboard from the prate shp and left at sea, and the next most senor prate makes a new proposal to begn the system agan. Prates base ther acceptance of the proposal on three factors. Frst, each prate wants to reman on the shp (not be thrown overboard). Second, each prate wants to maxmze the number of gold cons he receves. Thrd, each prate would prefer to throw another overboard, f all other results would otherwse be equal (so that f Prate C s o ered 0 cons he prefers to vote aganst the proposal as he gets to throw the proposer overboard f the proposal s not approved, and C s guaranteed to get at least 0 cons from any other proposal made). a (5 ponts) Suppose that Prate C s the only remanng prate. the cons? Hnt: Ths s easy, do not overthnk t. How does he propose to dstrbute Prate C proposes that he keeps all 00 cons because there are no other prates wth whom to share. b (5 ponts) Now suppose that Prate A has been thrown overboard and that Prates B and C reman. How does Prate B propose to dstrbute the cons? Prate B proposes that he keeps all 00 cons because he (B) wll vote to accept the dstrbuton whle C wll vote to reject. But Prate B has the tebreaker so the dstrbuton s accepted. c (0 ponts) Now suppose all three prates are stll on the shp. equlbrum to ths game. Fnd a subgame perfect Nash Ths s a lttle more nvolved. We can use parts a and b here as we know what Prates B and C wll do f they get to make a proposal. Now, look at the outcome f A s tossed overboard: Prate B wll receve all 00 cons and Prate C wll receve 0. If A o ers B any amount of cons (even 00) wll Prate B accept the dstrbuton no, because f B votes aganst the dstrbuton he knows he wll get all the 00 cons PLUS he gets the satsfacton of throwng A overboard. So Prate A wll never o er Prate B any cons. What about Prate C? If A s thrown overboard C wll get zero cons Prate A knows that f he o ers C 0 cons C wll vote aganst hm and he (A) wll be thrown overboard. But what f A o ers C a sngle con? Prate C wll accept because he knows that he wll get 0 cons f A s thrown overboard actually, Prate C wll accept any amount of cons above 0, but Prate A wll only o er con to C because A wants to maxmze hs own share of gold cons. So the SPNE s: Prate A o ers Prate B 0 cons and Prate C con (keepng 99 for hmself). Prate B rejects ANY o er by Prate A and Prate C votes to accept any o er by Prate A n whch C receves at least con. Prate A votes for hs own dstrbuton because he wants to stay on the shp. If A s thrown overboard then Prate B wll propose 00 cons for hmself and 0 for C. Prate C wll vote aganst B s proposal and Prate B wll vote for hs own proposals. If A and B are thrown overboard then Prate C wll propose that he keep all 00 cons and he wll, obvously, vote for hs own proposal. And then he wll nd another crew. 4. (30 ponts) Consder the followng two smultaneous games, denoted G and G2: G P2 C D P A 8 ; 4 4 ; 7 B 6; 0 2; 2 G2 P8 S T P7 Q 8 ; 2 5; 0 R 6; 0 7 ; 4

5 a (0 ponts) Fnd all PSNE to each game. In G there s only one PSNE, P choose A and P2 choose D. In G2 there are two PSNE: () P7 choose Q and P8 choose S and (2) P7 choose R and P8 choose T. b (5 ponts) Do any players have a strctly domnant strategy n ether game? and whch strategy s strctly domnant? If so, whch players In game G both players have a strctly domnant strategy. and P2 has a strctly domnant strategy of D. P has a strctly domnant strategy of A c (5 ponts) Suppose that each game s repeated 7 tmes and that all players know ths. Is there a unque subgame perfect Nash equlbrum to each of these repeated games? If there s what s the SPNE to the ntely repeated game? If there s more than one SPNE to the ntely repeated game, explan why there are multple SPNE. Accordng to the theorem game G has a unque SPNE because both players have a strctly domnant strategy n the stage game AND the game s ntely repeated. Game G2 does not have a unque SPNE because there are multple PSNE to the stage game there are many, many SPNE to ths game. It could be that the players chooose (Q; S) every round, or that they choose (R; T ) every round, or that they alternate between those two outcomes, or... well, there are many other possbltes. d (0 ponts) Now, consder only G (the game on the left). Suppose that the game s n ntely repeated. Suppose that Player uses the followng strategy: Choose B the rst tme the game s played. Choose B unless a defecton occurs by Player 2, where a defecton s Player 2 choosng D. If a defecton s observed, choose A forever regardless of what Player 2 chooses after the defecton. Suppose that Player 2 uses the followng strategy: Choose C the rst tme the game s played. Choose C unless a defecton occurs by Player, where a defecton s Player choosng A. If a defecton s observed, choose D forever regardless of what Player chooses after the defecton. Fnd the mnmum dscount rate,, needed for each player to support these strateges as a SPNE to the n ntely repeated game. We know that we need the cooperaton, devaton, and defecton payo s for both players. have Coop = 6, Devate = 8, and Defect = 4. So for P we wll need: For P we For P2 we have 2 Coop = 0, 2 Devate = 2, and 2 Defect = 7. So for P2 we wll need: Thus, the mnmum dscount rate for P s = 2 and for P2 t s =

6 Bonus: (0 ponts) Consder game G n queston 4. Suppose that Player has a = 0:500 and that Player 2 knows ths. Rather than use a grm trgger strategy and punsh Player forever f he devates, Player 2 wshes to punsh Player only the mnmum number of perods n order to ensure that both players cooperate on the (B; C) outcome. So, the strateges the players use are the same as n part d of queston 4 only now Player 2 uses a punshment strategy of If a defecton s observed, choose D for k perods regardless of what Player chooses after the defecton, then return to choosng C regardless of what Player has chosen durng those k perods. What s the mnmum number of perods, k, that Player 2 must punsh Player n order to guarantee that Player wll not devate? Note: Player knows the strategy that Player 2 s usng and wll act optmally durng the punshment phase. There s a brute force method of answerng ths queston and a more elegant method. The brute force method s farly smple create a table and see when devatng stops beng more bene cal than cooperatng. For example, f Player 2 were to punsh for two perods then Player s payo s from devatng would be 8, then 4, then 4, then back to 6 forever. But f Player cooperates then Player receves 6 forever. Ths means we really only care about the rst 3 payo s because both devatng once and cooperatng have the same payo s (6 forever) once we get past 3 perods, so even dscounted they are the same. Here s how to look at the table below. The rst column, k, s the perod. Say that Player devates n Perod 0 (f he devates). Then look at the next k perods. The column for k s just 0:500 k. The column for coop s just Player s payo f he cooperates n ths round (so t s always 6). The column P unshed s just how much Player receves when punshed (so an 8, because he gets one perod of devaton payo, and then a strng of 4s). The column k Coop s just the product of k and Coop for that row and the same for k P unshed. The columns Sum k Coop and Sum k P unshed are the cumulatve sum of Player s payo s n each round. So when k = Player has receved both hs payo for k = 0 and k = so hs "Sum" payo s f he has cooperated and f he devated, so clearly he would lke to devate f Player 2 s only punshng for perod. When k = 2 Player s cumulatve payo s f he had cooperated the entre tme and f he cheated and Player 2 punshed 2 perods. And so on we see that when k = (so that Player 2 s punshng for perods) Player s stll better o cheatng ( f cooperate s slghtly less than f he cheats). Fnally, f Player 2 punshes for 2 perods then Player would prefer to cooperate the entre tme rather than cheat (because 2: > 2: ). One thng we know from answerng part d n queston 4 s that there has to be some number k for whch ths wll be true because we know that f Player 2 punshes forever (uses the grm trgger strategy) then Player needs a dscount rate greater than or equal to 2 n order to want to cooperate f, for example, = 0:4, then there s no amount of perods that Player 2 can punsh Player to get hm to agree to the "cooperatve" or Pareto mprovng outcome of A; D. k k Coop P unshed k Coop k P unshed Sum k Coop Sum k P unshed Now, the more elegant method. Ths nvolves usng the results on the n nte sum of. Let s thnk about what Player receves f he cooperates: Coop = X =0 6 = 6 6

7 Now, what does he receve f he defects and Player 2 punshes for k perods? He receves hs devaton payo n the rst perod, then he receves k perods of hs defecton (or equlbrum) payo, then they go back to cooperatng so he receves hs cooperaton payo from k + to. Wrtng that out n an equaton we have: The last sum, P =k+ 6, s just: P unshed = 8 + X =k+ 6 = k+ 4 + = X =0 X =k+ 6 6 = k+ 6 The d cult pece to deal wth s P k = 4. I never remember what ths s, but we can always solve for t. I m gong to drop o the 4 term for a moment because t s a constant and focus on P k =. We know that: X = =0 X =k+ = k+ And we also know: just by ts de nton. Substtutng we have: X X = + + =0 = =k+ k+ k+ + k+ k+ = + = = = = = = = = = + k+ So now we know what P k = s. Don t forget what we are tryng to do we want to nd k so that: Coop P unshed X =0 6 X =0 = + 6 = X =k+ X 6 =k+ 7

8 We now know what all these summaton terms are, so substtute n and we have: k+ 6 8 ( ) k+ k+ + 6 k k+ + 6 k k k+ 2 k+ At ths pont we have two varables, and k, but we know that = 0:500. If we take the natural log (ln) of both sdes we get: 2 :500 0:500 k+ 0:0002 0:500 k+ ln (0:0002) (k + ) ln (0:500) ln (0:0002) ln (0:500) k + Note that ln (0:500) s NEGATIVE so when dvdng we need to swtch the nequalty sgn from to. ln (0:0002) k ln (0:500) :292 k So we know that Player 2 would need to punsh for at least :292 perods, but snce they must punsh n nteger amounts they have to punsh for at least 2 (whch s what our prevous "brute force" method wth the table also tells us). 8

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