## Assignment on Mixed Strategy

Handout 6

ECO 444

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Mixed Strategy

Take a game in the normal form, G = (N; S1, …,SN ; u1, …,uN ). We say that G is a finite game

if every set of strategies Si is a finite set. In this handout, we only consider finite games.

Take si ∈ Si. We say that si is a pure strategy of player i and Si is the set of pure strategies

of player i. A mixed strategy of player i is a probability distribution over Si. Let ∆(Si) denote

the set of mixed strategies of player i. A generic mixed strategy of player i is denoted by αi and

αi(si) is a probability assigned to pure strategy si by the mixed strategy αi. Note that a pure

strategy si is the same as a mixed strategy αi with αi(si) = 1 and αi(s ′ i) = 0 for s

′ i 6= si.

Consider the following example.

Bob

L R

Ann T 2,0 0,1

B 0,1 4,0

Let αA = (p, 1 − p) be Ann’s mixed strategy where p is the weight assigned to strategy T. Let

αB = (q, 1 − q) be Bob’s mixed strategy where q is the weight assigned to strategy L.1

A mixed extension of G is Γ = (N; ∆(S1), …, ∆(SN ); EU1, …,EUN ) where EUi is the expected

utility function of player i. A profile of mixed strategies is denoted by α = (α1, …,αN ) ∈ ∆(S1)×

…×∆(SN ). If we assume the independence of the randomizations, then a profile of mixed strategies

α = (α1, …,αN ) assigns probability λ(s) to the profile of pure strategies s = (s1, …,sN ),

λ(s) = Πj∈Nαj(sj) = α1(s1) · … ·αN (sN ). (1)

1For example, strategy αA = ( 1 3 , 2 3

) is Ann’s mixed strategy that assigns probability 1

3 to pure strategy T and

probability 2 3

to pure strategy B. As another example, consider αB = (0, 1) which is Bob’s mixed strategy that assigns probability 1 to pure strategy R. In other words, this specific mixed strategy is a pure strategy R.

1

The expected utility EUi is defined as

EUi(α) = Σs∈Sui(s) ·λ(s). (2)

A profile of mixed strategies α∗ is a mixed strategy Nash equilibrium (MSNE) in game Γ if

the following holds for each player i:

EUi(α ∗ i ,α

∗ −i) ≥ EUi(αi,α

∗ −i) ∀αi ∈ ∆(Si) (3)

That is, α∗ is an MSNE if and only if each player’s mixed strategy α∗i is a best response to α ∗ −i

where player i’s best-response correspondence is defined in the following way.

BRi(α−i) = {αi ∈ ∆(Si) such that EUi(αi,α−i) ≥ EUi(α′i,α−i) ∀α ′ i ∈ ∆(Si)} (4)

Nash Theorem

Every finite game has at least one MSNE.

In the 2 × 2 matrix on page 1 there is no Nash equilibrium in pure strategies. However, because

of the Nash theorem, there exists at least one MSNE. Let (α∗A,α ∗ B) denote an MSNE where

α∗A = (p ∗, 1 −p∗) and α∗B = (q

∗, 1 − q∗) is Ann’s and Bob’s equilibrium strategy, respectively.

In order to find (α∗A,α ∗ B), we need to (i) identify Ann’s and Bob’s best-response correspondences,

and (ii) find where they cross. Let BRA(αB) be Ann’s best-response correspondence: for each

mixed strategy of Bob αB, BRA(αB) is the collection of Ann’s mixed strategies that maximize her

expected utility. Similarly, we have BRB(αA).

We identify Ann’s best-response correspondence in the following way. Let EUA(sA,αB) be Ann’s

expected utility from playing a pure strategy sA when Bob’s mixed strategy is αB.

EUA(T,αB) = 2 · q + 0 · (1 − q) = 2q (5)

EUA(B,αB) = 0 · q + 4 · (1 − q) = 4 − 4q (6)

2

Hence, the following is true.

EUA(T,αB) > EUA(B,αB) ⇔ q > 2

3 (7)

EUA(T,αB) = EUA(B,αB) ⇔ q = 2

3 (8)

EUA(T,αB) < EUA(B,αB) ⇔ q < 2

3 (9)

The first row tells us that if q > 2 3 , then choosing strategy T with probability one generates strictly

higher expected utility for Ann than playing strategy B with probability one. In other words, in

this case playing T is Ann’s best response; more formally, (1, 0) is Ann’s unique best response.

The second row tells us that if q = 2 3 , then it does not really matter what Ann chooses: playing

T with probability one, or playing B with probability one, or mixing between T and B will yield

the same expected gain. The third row tells us that if q < 2 3 , the choosing B with probability one

is the best response. We formally describe Ann’s best-response correspondence.

BRA(αB) =

  {(1, 0)} if αB is such that q > 23

{(p, 1 −p) : p ∈ [0, 1]} if αB is such that q = 23

{(0, 1)} if αB is such that q < 23

(10)

We identify Bob’s best-response correspondence using the same approach. Let EUB(αA,sB) be

Bob’s expected utility from playing a pure strategy sB when Ann’s mixed strategy is αA.

EUB(αA,L) = 0 ·p + 1 · (1 −p) = 1 −p (11)

EUB(αA,R) = 1 · q + 0 · (1 −p) = p (12)

Hence, the following is true.

EUB(αA,L) > EUB(αA,R) ⇔ p < 1

2 (13)

EUB(αA,L) = EUB(αA,L) ⇔ p = 1

2 (14)

EUB(αA,L) < EUB(αA,L) ⇔ p > 1

2 (15)

3

We formally describe Bob’s best-response correspondence.

BRB(αA) =

  {(1, 0)} if αA is such that p < 12

{(q, 1 − q) : q ∈ [0, 1]} if αA is such that p = 12

{(0, 1)} if αA is such that p > 12

(16)

Ann’s best-response correspondence as described in (10) and Bob’s best-response correspondence

as described in (16) look rather complicated. However, in the case of 2 × 2 matrices, we can

simplify the notation and, in fact, graphically visualize best-response correspondences. Instead

of writing αA = (p, 1 − p) we can simply say that p is Ann’s mixed strategy. (Why?) Similarly,

instead of αB = (q, 1 − q) we can use q as Bob’s mixed strategy.

Hence, for each q ∈ [0, 1] (Bob’s mixed strategy, Ann’s best-response correspondence BRA(q) gives

a set of singleton. Similarly for Bob.

We re-write BRA(q) and BRB(p).

BRA(q) =

  {1} if q > 2

3

[0, 1] if q = 2 3

{0} if q < 2 3

(17)

BRB(p) =

  {1} if p < 1

2

[0, 1] if p = 1 2

{0} if p > 1 2

(18)

4

We graphically represent Ann’s and Bob’s best-response correspondences based on (17) and (18),

respectively.

p

q

1

1 ⅔

Figure 1: Ann’s best-response correspondence BRA as the solid line.

q

p

1

1 ½

Figure 2: Bob’s best-response correspondence BRB as the dashed line.

5

In order to find MSNE, we need to depict BRA and BRB in the same graph.

p

q

1

1 ⅔

½

Figure 3: BRA and BRB in the same graph.

A point at which the best-response correspondences intersect is an MSNE. In our case, p∗ = 1 2

and q∗ = 2 3 . Formally,

(( 1 2 , 1 2

) , ( 2 3 , 1 3

)) is the MSNE in our example.

6