SC 631: Games and Information
Autumn 2014
Lecture 8: August 26
Instructor: Ankur A. Kulkarni
Scribes: Shubham Jain, Karan Ganju, Palash Kala, Royal Jain
Note: LaTeX template courtesy of UC Berkeley EECS dept.
Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications.
They may be distributed outside this class only with the permission of the Instructor.
8.1
8.1.1
Mixed Strategies
Introduction
For a Z-S game matrix A ∈ Rm×n , mixed strategy is defined as the probability distribution on pure strategies
while the pure strategies are represented by the rows and columns.
Row player is represented by a vector y ∈ Rm , s.t. y ≥ 0 & 1T y=1
Column player is represented as vector z ∈ Rm , s.t. z ≥ 0 & 1T z=1
Here the vector 1 = (111....1)T
Now, let Y be the set of mixed strategies for row player and Z be the set of mixed strategies for column
player.
8.1.2
Saddle Point
Saddle point for a Z-S game is defined as the tuple of strategies (y ∗ , z ∗ ) ∈ Y ∗ Z s.t.
y ∗T Az ≤ y ∗T Az ∗ ≤ y T AZ ∗ ∀y ∈ Y, ∀ z ∈ Z
8.1.3
Security Strategy
Security Strategy for a row player is given by y ∗ ∈ Y s.t.
max y ∗T Az ≤ max y T Az
z∈Z
z∈Z
∀y ∈ Y
Similarly for a column player, the security strategy is z ∗ ∈ Z s.t.
min y T Az ∗ ≥ min y T Az
y∈Y
y∈Y
∀z ∈ Z
According to utility theory, a rational player should look at the expected value and hence must maximize
the same to maximize his utility.
P
Expected value of utility for the given mixed strategy = i,j aij yi zj = y T Az
8-1
8-2
Lecture 8: August 26
8.1.4
Solving Security Strategy Equations
max y ∗T Az = min(max y T Az)
z∈Z
y∈Y
z∈Z
Let maxz∈Z y T Az be f (y).
So we need to find miny∈Y f (y).
Three reasons when minimum cannot be defined:
• Function decreases to −∞
• Infimum is not attainable in the domain (Hole in the domain)
• Function is discontinuous (Hole in the range)
Lecture 8: August 26
8.1.5
8-3
Weierstrass’ Theorem
Theorem 8.1 Let f : K → R be a continuous function and K ⊆ Rn be a compact set. Then f attains it’s
infimum on K. ⇒ ∃x ∈ K st. f (x) = inf f (z)
z∈K
Lemma 8.2 Set S ⊆ Rn is compact ⇔ S is closed and bounded.
Recall: S ⊆ Rn is bounded ⇔ ∃∞ > M > 0 s.t. kxk ≤ M ∀x ∈ S
S ⊆ Rn is closed if for any convergent sequence x1 , x2 ... s.t xi ∈ S and x = lim xi , we have x ∈ S.
i→∞
8.1.6
Existence of Mixed Strategy
We know that the mixed strategy will exist if we can obtain min f (y). Hence we only need to show that Y is
y∈Y
compact and f is continuous. By lemma 8.2 it can be shown that Y is compact by proving that Y is closed
and bounded.
Y is boundedpas y ≥ 0 & 1T y = 1. Hence for any y ∈ Y, yi ≤ 1 for all i = 1, . . . , m. Hence for any
2 ≤ m. Thus Y is bounded (we can take M above to be m)
y ∈ Y, kyk = y12 + y22 + . . . ym
To show closedness, consider a sequence y (1) , y (2) , y (3) ...y = lim y (k) and y (i) ∈ Y ∀i
k→∞
To prove Y is closed we have to prove that y ∈ Y : Since each y (k) ∈ Y , we must have y (k) ≥ 0 ⇒ lim
k→∞
y (k) ≥ 0 ⇒ y ≥ 0
Again since each y (k) ∈ Y, 1T y (k) = 1
1 = lim 1T y (k) = 1T lim y (k) = 1T y
k→∞
k→∞
Therefore, y ∈ Y . Hence, Y is closed and Y is also bounded. ⇒ Y is compact.
Possible mix strategies for y1 , y2 , y3 is shown in the figure.
8-4
8.1.7
Lecture 8: August 26
Hogan’s Paper : Theorem 7
Theorem 8.3 Let g : X × Y ⇒ R. If g is a continous function (on both X and Y jointly) and Y is compact
then the function
h(x) = maxy∈Y g(x, y) is continous.
Z is compact.
(same argument as Y )
P
y T Az = ij aij yi zj is a polynomial in y1 , y2 , y3 , ..., ym and z1 , z2 , z3 , ..., zn . Hence it is continuous in theses
variables.
Therefore f (y) = max y T Az is a continuous function of y.
z∈Z
Therefore since Y is compact min max y T Az exists. (Weierstrass’ Theorem)
y∈Y z∈Z
⇒ ∃ a security strategy for row players and similarly ∃ a security strategy for column players.
8.2
Mixed Strategies for Z-S games
Theorem 8.4 Let A be a Z-S matrix game. Then
1. ∃ a security strategy for each player (y ∗ , z ∗ )
2. ∃ a unique security level for each player V m (A) = miny∈Y maxz∈Z y T Az
3. V (A) ≤ V m (A) ≤ V m (A) ≤ V (A)
Proof
The first two claims follow from the above proof.
We have to prove that V (A) ≤ V m (A) ≤ V m (A) ≤ V (A)
To prove this we first prove that V m (A) ≤ V m (A)
Lecture 8: August 26
8-5
min y T Az ≤ y T Az ≤ maxz∈Z y T Az
y∈Y
Put y = y ∗ and z = z ∗ and we get V m (A) ≤ V m (A)
Now we prove that V m (A) ≤ V (A)
V (A) = min max y T Az
y∈Y z∈Z
As z ≥ 0 and 1T z = 1, hence for a fixed y, max y T Az will be max(y T A)j .
z∈Z
Therefore, V m (A) = min max(y T A)j ≤ min max aij = V (A)
y∈Y
Similarly, V (A) ≤ V m (A).
Hence, proved.
j
i
j
j