SF2972 Game Theory Exam with Solutions March 19, 2015

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1 SF2972 Game Theory Exam with Soltions March 9, 205 Part A Classical Game Theory Jörgen Weibll an Mark Voornevel. Consier the following finite two-player game G, where player chooses row an player 2 chooses colmn: a b c e a, 2, 0 2, 2, 0 0, b 0, 0, 0 0, 0 0, 0 0, 0 c 0, 0 0, 0 3, 3 6, 0 6, 0 0, 0 0, 0 0, 6 4, 4 7, 0 (a) Fin all pre strategies that are weakly ominate (by a pre or mixe strategy). [.5 pts] (b) Fin all pre strategies that are strictly ominate (by a pre or mixe strategy). [.5 pts] (c) Fin all pre-strategy Nash eqilibria. [.5 pts] () Fin all pre-strategy perfect eqilibria. [ pt] (e) Fin all pre-strategy proper eqilibria [ pt] Soltion (a) a, a, b, b an e (b) a an a (c) (b, b ) an (c, c ) () (c, c ) (e) (c, c )

2 2 SF2972 Game Theory Exam with Soltions March 9, There are n inivials who together maintain a forest. Each inivial i makes effort x i 0, reslting in tility i (x,..., x n ) = 00y + y x i for the inivial, where y = x + + x n is the total amont of efforts to maintain the forest. Hence, everyboy gets more tility from the forest the better it is maintaine. All inivials choose their efforts simltaneosly. Each inivial strives to maximize his or her tility. (a) Show that this game has exactly one symmetric Nash eqilibrim (in pre strategies), that is, an eqilibrim in which all inivials make the same effort, x. Fin x an the associate total amont of effort, y = nx. (b) Sppose that the inivials can pre-commit to a common effort level, the same for all. Let ˆx be the common effort level that maximizes the sm of all inivials tilities. Fin ˆx an the associate total amont of effort, ŷ = nˆx. Compare these (socially optimal) levels with those in Nash eqilibrim. Is there a ifference when n =? When n >? Explain! [.5 pts] Soltion (a) Strictly concave payoff fnctions. FOC: 00 = ( + y) 2 y = 9 x = 9/n. (b) Strictly concave welfare fnction. FOC: 00n = ( + y) 2 ŷ = 0 n ˆx = (0 n ) /n. Ths ˆx/x = (0 n ) /9, increasing in n from at n =. 3. Consier the following extensive form game: a, 0 e c f b 2 e f, 0 0, 3 0,, (a) Fin the corresponing strategic (i.e., normal form) game. (b) Fin all sbgame perfect eqilibria in behavioral strategies. (c) Fin all seqential eqilibria. [ pt] [3 pts] Soltion (a) c (a, e), 0, 0 (a, f), 0, 0 (b, e), 0 0, (b, f) 0, 3, (b) There are two sbgames: the game as a whole an a proper sbgame starting at the ecision noe of player 2.

3 SF2972 Game Theory Exam with Soltions March 9, (c) The latter game c e, 0 0, f 0, 3, has a niqe Nash eqilibrim where e is chosen with probability 2/3 an c is chosen with probability /2. Player s expecte payoff in this eqilibrim is /2. So in the game as a whole, it is optimal for player to choose a with probability. Conclsion: There is a niqe sbgame perfect eqilibrim in behavioral strategies where player chooses a with probability an e with probability 2/3 an player 2 chooses c with probability /2. Strategies: The behavioral strategies in a seqential eqilibrim mst be sbgame perfect, so (b) gives only one caniate. Belief system: Any completely mixe profile of behavioral strategies in which the probability of c approaches /2 mst assign eqal probability to both noes in player s information set. Conclsion: there is a niqe caniate for a seqential eqilibrim: the profile of behavioral strategies in (b) with a belief system assigning probability /2 to both noes in s information set. Given the existence theorem of seqential eqilibria, this is the game s niqe seqential eqilibrim. 4. Consier the signaling game below:, L R 2, 3, 0 t /2 4, 0 2 chance 2, 2 t /2 2, 2, 0 L R 4, 0 (a) Fin, if any, the separating eqilibria where chooses R if chance selects t, bt L if chance selects t. (b) Fin, if any, the pooling eqilibria where chooses R both if chance selects t an if chance selects t. Soltion (a) Denoting s strategy as (R, L), player 2 s best reply is to choose in the left information set an in the right, enote (, ). Bt s strategy (R, L) is not a best reply to (, ): it is better to choose R if chance selects t. Conclsion: no sch eqilibria. (b) If player plays (R, R), 2 assigns eqal probability to both noes in the right information set, making the niqe best reply there. So there are two cases:

4 4 SF2972 Game Theory Exam with Soltions March 9, 205 Is ((R, R), (, )) a pooling eqilibrim? The strategy of pl. is seqentially rational an so is 2 s strategy in the right information set. Bayesian consistency imposes no restrictions on beliefs in the left information set, bt seqential rationality of action reqires that the probability assigne to the top noe there lies in [0, 2/3]. Is ((R, R), (, )) a pooling eqilibrim? No! Pl. s best response to (, ) is to choose L after chance selects t. Conclsion: pooling eqilibria ((R, R), (, )) where 2 assigns probability α [0, 2/3] to the top noe in the left information set an probability /2 to the noes in the right information set. Part B Combinatorial Game Theory Jonas Sjöstran 5. For any game G, we efine the game + G := 0 0 G. (a) Show that + ++G = 0 for any game G. (b) Fin all games G sch that + G = G. [ pt] Soltion (a) The game trees for + ++G an 0 look like this: a b j h l c f k e i G g A position in the sm H = + ++G 0 will be enote by a two-letter combination xx where x is a noe in the game tree of + ++G an X is a noe in the game tree of 0. The game starts at aa. The following iagram shows that the secon player has a winning strategy. For the first player every option is shown, an for the secon player only one option from each position is shown a winning move. In all terminal positions, the first player is at trn. aa ab cd fd ba B ed bc cb E id ac ce fb D B ca eb ib fe he le ke A E C

5 SF2972 Game Theory Exam with Soltions March 9, (b) If + G = G then + ++G = + +G = + G = G so, by (a), G mst be eqal to 0. To see that + 0 = 0 we raw the game trees of + 0 an 0 an o the same reasoning as in (a). In fact, we only have to omit the noes j, k, l an G from the tree in (a) an the noes ke an le from the strategy iagram! 6. The game Toppling ominoes is playe with a row of black an white ominoes. In each move, the player chooses a omino an topples it either left or right. Every omino in that irection also topples an is remove from the game. Left may only choose a black omino an Right may only choose a white one. Here is an example of a game of Toppling ominoes: L R L Prove that the following eqalities hol for any nonnegative integers m an n. Soltion Let s introce the following notation. = m n m+ n+ = n 2n = 2 n 2n+ = 0 0 n n+2 A m,n := m+ n+ B n := 2n C n := 2n+ D n := n+2 We will prove all eqalities by inction. They clearly hol for m = n = 0. [4 pts] A m,n = A m,n,..., A 0,n, (n + ), m, m,..., 0 A m,n,..., A m,0, m +, n, (n ),..., 0 which by inction is eqal to m n,..., 0 n, (n+), m, m,..., 0 m (n ),..., m 0, m+, n, (n ),..., 0. Deleting all ominate options yiels A m,n = m n. Similarly, B n = C n,..., C 0, B n,..., B 0 C n,..., C 0, B n,..., B 0 in = 2,..., n 2, (n ),..., 0 0 2,...,, (n ),..., 0 n 20 = (n ),..., 0 (n ),..., 0 = n

6 6 SF2972 Game Theory Exam with Soltions March 9, 205 an C n = B n,..., B 0 C n,..., C 0 in = n,..., 0 2,..., n 2 0 = n,..., 0 2 n = 2, n where the last eqality follows from the simplicity theorem. Finally, D n = A 0,n+, 0 A 0,n,..., A 0,0, in = 0 (n + ), 0 0 n,..., 0 0 = 0 0 n, where the last eqality follows from eleting ominate options. 7. Let G = an H = 0 4. What are the temperatres of G, H an G + H? [4 pts] Soltion We have G t = 5 3 t t t nless this expression is a nmber. Clearly, 5 t 3 + t if 0 t, 5 3 t = 4 if t >, so, by the translation theorem, an 5 3 t t t = 5 2t 3 t if 0 t, 4 t t if t >, 5 2t 3 t if 0 t, G t = 4 t t if < t 2, 2 if t > 2. We concle that the temperatre of G is 2. The game H is even easier to cool: t t 4 if 0 t 2, H t = 2 if t > 2, an its temperatre is 2. Cooling is linear, so (G + H) t = G t + H t = 5 2t 3 t + t t 4 if 0 t, 0 if t >. It is easy to check that (G + H) = = 0, so the temperatre of G + H is. 8. The impartial game Kayles is playe with a row of pins, possible containing some gaps. A move consists in throwing a ball that removes either one or two ajacent pins. Example: Compte the Grny vale of K 5, a row of 5 pins:.

7 SF2972 Game Theory Exam with Soltions March 9, Soltion We let K n enote a row of n ajacent pins an compte the Grny vales of K 0, K,..., K 5 by sing the mex rle an nim aition. K 0 = 0, K = K 0 = 0 =, K 2 = K 0, K = 0, = 2, K 3 = K, K 2, K + K =, 2, + =, 2, 0 = 3, K 4 = K 2, K + K, K 3, K 2 + K = 2, 0, 3, 2 + = 2, 0, 3, 3 =, K 5 = K 3, K 2 + K, K 4, K 3 + K, K 2 + K 2 = 3, 3,, 2, 0 = Compte the vale of the following Ble-Re Hackenbsh position. (Soli eges are ble an ashe eges are re.) Soltion The sign-expansion of a Hackenbsh path is obtaine by walking along the path, starting at the gron, an writing a pls for each ble ege an a mins for each re ege. Ths, Left has the following options from the epicte position: ( ) = , (+) + ( + + +) = + ( ), (++) + ( + + ) = 2 + ( ), (+ + + ) + ( + ) = (3 2 ) + ( ), ( ) + ( ) = ( ) 2. The largest of these is (3 2 ) + ( ) = 3 4. Right s options are (+ + +) + ( + +) = 3 + ( ), ( ) + ( +) = ( ) + ( ), ( ) + ( ) = ( ), ( ) = , of which (3 + ) + ( 2 + ) = 5 is the smallest one We concle that the vale of the position is 3 5 =. 4 4