Statistics 155 Homework Assignment 5 (due Tuesday, October 18, 2016)
1. (Series and parallel games) Karlin and Peres, Exercise 3.1, p72.
(Note: As well as giving the value of the game, specify optimal strategies for the Troll and the Traveler.)
Solution:
(a)
Note that the value of the game is the effective resistance.
Let’s break the graph into three sub-graphs G1 , G2 , G3 as described above. The effective resistance
of G1 is 1+ 1+1+1 1 +1 = 1+2/7 = 9/7, the effective resistance of G2 is 1+ 1 1 + 1+1 1 = 3/5+2/3 =
2
1+1/2
2
19/15, and the effective resistance of G3 is 1. The graph of game consists of three graph G1 , G2 , G3
1
= 171/439.
in parallel, and thus the effective resistance of the graph is 7/9+15/19+1
(b)
Let’s label vertices and edges as described above. From vertex 1, both troll and traveler choose one
of the graphs G1 , G2 , G3 . Note that the optimal strategy of troll and traveler are the same. We
have the effective conductance of the graph G1 ( G2 , G3 respectively) is 7/9(15/19, 1, respectively).
It follows from [Claim 3.1.6, Kalin & Peres] that
7/9
= 133/439,
7/9 + 15/19 + 1
15/19
P(G2 |vertex 1) = P(move along either edge b or edge c|vertex 1) =
= 135/439,
7/9 + 15/19 + 1
1
P(G3 |vertex 1) = P(move along edge d|vertex 1) =
= 171/439.
7/9 + 15/19 + 1
P(G1 |vertex 1) = P(move along edge a|vertex 1) =
1
Once a traveler choose G2 , he/she has 2 choices of edges to move along.
1
= 3/5,
1
1 + ( 1+1/2
)
P(move along b| vertex 1 and G2 is chosen) =
P(move along c| vertex 1 and G2 is chosen) =
1
1
1+1/2
1
)
+ ( 1+1/2
= 2/5.
Combining these conditional probabilities and P(G2 |vertex 1), it is clear that
P(move along edge b|vertex 1) = 3/5 ∗ 135/439 = 81/439,
P(move along edge c|vertex 1) = 2/5 ∗ 135/439 = 54/439.
We can apply the same logic to other vertices and derive the following conditional probabilities.
From vertex 2,
P(move along edge e|vertex 2) = 1/2,
P(move along edge f |vertex 2) = 1/2.
From vertex 3,
P(move along edge l|vertex 3) =
P(move along edge m|vertex 3) =
1
1+
1
2
1
2
1+
1
2
= 2/3,
= 1/3.
From vertex 4,
P(move along edge n|vertex 4) = 1.
From vertex 5,
P(move along edge g|vertex 5) =
P(move along edge h|vertex 5) =
P(move along edge j|vertex 5) =
P(move along edge k|vertex 5) =
1+
1
2
1
= 2/7,
+1+1
1
2
1+
1
2
1+
1
2
1+
1
2
= 1/7,
+1+1
1
= 2/7,
+1+1
1
= 2/7.
+1+1
From vertex 6,
P(move along edge i|vertex 6) = 1.
2
2. (Two-player general-sum games) Karlin and Peres, Exercise 4.2, p96.
Solution: The payoff matrix is as follows.
W ork
P arty
W ork
(8, 8)
(10, 3)
P arty (3, 10)
(0, 0)
The pure NEs are (W, P ) and (P, W ). To find a mixed NE which is not pure, we can use equalizing
strategy. If x = (p, 1 − p) is a strategy for the player 1, then the equalizing strategy gives that
8p + 3(1 − p) = 10p
⇔ 5p + 3 = 10p
⇔ 3 = 5p
⇔ 3/5 = p
Thus, x = (3/5, 2/5). Since the game is symmetric, the mixed Nash Equilibria that is not pure is
((3/5, 2/5), (3/5, 2/5)).
3. (Multiplayer general-sum games) Karlin and Peres, Exercise 4.8, p98.
Solution: If the firm 3 advertises on evening TV, then the payoff matrix is described as follows.
F irm1\F irm2 M orning
M orning
(0, 0, 300)
Evening
(0, 200, 0)
Evening (200, 0, 0)
(0, 0, 0)
If the firm 3 advertises on morning TV, then the payoff matrix is described as follows.
F irm1\F irm2 M orning
M orning
(0, 0, 0)
Evening
(300, 0, 0)
Evening (0, 300, 0)
(0, 0, 200)
It is evident that there are 6 pure symmetric Nash equilibria: (M,M,E),(E,E,M),(M,E,E),(E,M,M),
(E,M,E),(M,E,M). Denote the symmetric Nash equilibria by ((p, 1 − p), (p, 1 − p), (p, 1 − p)). Since we
know that 0 < p < 1, we can use the equalizing strategy.
200(1 − p)2 = 300p2
⇔200(1 − 2p + p2 ) = 300p2
⇔100p2 + 400p − 200 = 0
⇔p2 + 4p − 2 = 0
√
−4 + 24
⇔p =
2√
⇔p = −2 + 6
Therefore, the symmetric mixed Nash Equilibrium is ((−2 +
3
√
6, 3 −
√
6), (−2 +
√
6, 3 −
√
6))