Chapters 2 and 3: Common Belief in Rationality Andrs Perea - - PowerPoint PPT Presentation

Chapters 2 and 3: Common Belief in Rationality

Andrés Perea

Maastricht University

Period 5, 2012/2013

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 1 / 78

What is game theory about?

In game theory, we study situations where you must make a choice, but where the …nal outcome also depends on the choices of others. Examples are everywhere: Negotiating about the price of a car, choosing a marketing strategy for your …rm, bidding in an auction, discussing with your partner about what TV program to watch this evening. Key question: What choice would you make, and why? This depends crucially on how you reason about the opponent!

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Example: Where to locate my pub?

v v v v v v v

a b c d e f g 100 100 100 100 100 100 Story In a village called Longstreet, you and your opponent must both choose a location for your pub. Possible locations: a, b, c, d, e, f , g. Between every two locations, there are 100 thirsty men living. All at equal distance from each other. Every man will visit the pub that is nearest to his house. You wish to attract as many customers as possible. What location would you choose, and why?

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 3 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 Which location is optimal for you, depends on your belief about the

• pponent’s location:

If you believe your opponent chooses a, then location b is optimal for you. If you believe your opponent chooses b, then location c is optimal for you. If you believe your opponent chooses c, d or e, then location d is

• ptimal for you.

If you believe your opponent chooses f , then location e is optimal for you. If you believe your opponent chooses g, then location f is optimal for you.

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Beliefs diagram

a b c d e f g a b c d e f g

• 1
• 1
• PPPPPPPP

P q PPPPPPPP P q

Your choices Opponent’s choices

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 5 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 We call b, c, d, e and f rational choices for you, since they are all

• ptimal for some belief about the opponent’s choice.

Location a can never be optimal for you: Whatever location your

• pponent chooses, choosing b is always better for you!

We say that your choice a is strictly dominated by your choice b. Similarly, your choice g is strictly dominated by your choice f .

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 6 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 Consider your choice b. Choosing b is optimal if you believe that your opponent chooses a. But choosing a is irrational for your opponent. So, this belief is unreasonable! If you take your opponent seriously, then you must believe that your

• pponent chooses rationally too.

So, you must believe that your opponent does not choose a or g. But then, choosing b can no longer be optimal for you, as choosing c is always better!

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 7 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 So, if you believe that the opponent chooses rationally, then you believe that he will not choose a or g, and then b is no longer

• ptimal for you.

Similarly, if you believe that the opponent chooses rationally, then choosing f is no longer optimal for you, as e would always be better. Hence, the choices b and f are rational for you, but unreasonable.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 8 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 On the other hand, c, d and e can still be optimal if you believe that the

• pponent chooses rationally.

Choosing c is optimal for you if you believe that the opponent rationally chooses b. Choosing d is optimal for you if you believe that the opponent rationally chooses d. Choosing e is optimal for you if you believe that the opponent rationally chooses f .

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Beliefs diagram

You Opponent You a b c d e f g a b c d e f g a b c d e f g

• 1
• 1
• PPPPPPPP

q PPPPPPPP q

• 1
• 1
• PPPPPPPP

q PPPPPPPP q

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 10 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 But are all of the choices c, d and e reasonable? No! If you believe that the opponent chooses rationally, it seems reasonable to believe that the opponent believes that you choose rationally! If the opponent believes that you choose rationally, then he could rationally choose c, d and e, but not b and f . So, to believe that the opponent chooses b or f is unreasonable. But if you believe that the opponent will only choose c, d or e, you must choose d!

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You Opponent You a b c d e f g a b c d e f g a b c d e f g

• 1
• 1
• PPPPP

P q PPPPP P q

• 1
• 1
• PPPPP

P q PPPPP P q

So, location d is the only candidate for a reasonable choice. But is d really reasonable? Yes! Consider, namely, the belief hierarchy that starts at your choice d.

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In this belief hierarchy: You believe that the opponent chooses d. You believe that the opponent believes that you choose d. You believe that the opponent believes that you believe that the

• pponent chooses d.

And so on. It consists of a …rst-order belief, a second-order belief, a third-order belief, and so on. If you hold this belief hierarchy, then you believe that the opponent chooses rationally, you believe that the opponent believes that you choose rationally, you believe that the opponent believes that you believe that the

• pponent chooses rationally,

and so on. We say that you express common belief in rationality.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 13 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 Summarizing Your choices a and g are irrational. Your choices b and f are rational, but can no longer be optimal if you believe that the opponent chooses rationally. Your choices c and e can be optimal if you believe that the opponent chooses rationally, but can no longer be optimal if you believe, in addition, that the opponent believes that you choose rationally. You can rationally choose d if you express common belief in rationality. In particular, under common belief in rationality there is only one choice you can rationally make, namely d!

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Example: Going to a party

blue green red yellow same color as Barbara 4 3 2 1 Story This evening, you are going to a party together with your friend Barbara. You must both decide which color to wear: blue, green, red or yellow. Your preferences for wearing these colors are as in the table. These numbers are called utilities. You hate wearing the same color as Barbara: If you both would wear the same color, your utility would be 0. What color should you choose, and why?

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blue green red yellow same color as Barbara 4 3 2 1 Again, what color is optimal for you depends on your belief about Barbara’s choice: If you believe that Barbara wears blue, then green is optimal for you. If you believe that Barbara wears green, then blue is optimal for you. If you believe that Barbara wears red, then blue is optimal for you. If you believe that Barbara wears yellow, then blue is optimal for you. Does this mean that red and yellow are irrational for you?

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blue green red yellow same color as Barbara 4 3 2 1 Suppose you believe that, with probability 0.6, Barbara chooses blue, and that, with probability 0.4, she chooses green. If you would choose blue, your expected utility would be (0.6) 0 + (0.4) 4 = 1.6. If you would choose green, your expected utility would be (0.6) 3 + (0.4) 0 = 1.8. If you would choose red, your utility would be 2. If you would choose yellow, your utility would be 1. So, choosing red is optimal for you if you hold this probabilistic belief about Barbara’s choice. In particular, red is a rational choice for you.

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blue green red yellow same color as Barbara 4 3 2 1 Choosing yellow can never be optimal for you, even if you hold a probabilistic belief about Barbara’s choice. If you assign probability less than 0.5 to Barbara’s choice blue, then by choosing blue yourself, your expected utility will be at least (0.5) 4 = 2. If you assign probability at least 0.5 to Barbara’s choice blue, then by choosing green yourself your expected utility will be at least (0.5) 3 = 1.8. Hence, whatever your belief about Barbara, you can always guarantee an expected utility of at least 1.8. So, yellow can never be optimal for you, and is therefore an irrational choice for you.

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Beliefs diagram

blue green red yellow same color as Barbara 4 3 2 1 Your choices blue green red yellow Barbara’s choices blue green red yellow

HHHHHHHH H j

• *
• 7
• 1

0.6 0.4

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Your choices blue green red yellow Barbara’s choices blue green red yellow

HHHHHHHH j

• *
• 7
• 1

0.6 0.4 The choices blue, green and red are rational for you. But are all of these choices also reasonable? This depends on Barbara’s preferences!

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blue green red yellow same color as friend you 4 3 2 1 Barbara 2 1 4 3 For Barbara, the choices red, yellow and blue are rational, whereas green is irrational. Choosing red is optimal for her if she believes that you choose yellow. Choosing yellow is optimal for her if she believes that you choose red. Choosing blue is optimal for her is she believes that, with probability 0.6, you choose red, and with probability 0.4 you choose yellow.

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blue green red yellow same color as friend you 4 3 2 1 Barbara 2 1 4 3 So, if you believe that Barbara chooses rationally, you believe that Barbara will choose red, yellow or blue. But then, choosing red will no longer be optimal for you, as choosing green will always be better in this case. Choosing blue is optimal for you if you believe that Barbara rationally chooses red. Choosing green is optimal for you if you believe that Barbara rationally chooses blue.

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Beliefs diagram

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *
• >
• 0.6

0.4

HHHHHHH H j

• *

L L L L L L L LXXXX X z @ @ @ @ @ R

0.6 0.4

@ @ @ @ @ @ @ @ @ R

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 23 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 2 1 4 3 The color yellow is irrational for you. The color red is rational for you, but you can no longer rationally choose it if you believe that Barbara chooses rationally. If you believe that Barbara chooses rationally, you can still rationally choose the colors blue and green. But are blue and green still optimal if you express common belief in rationality?

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You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *
• >
• 0.6

0.4

HHHHHH j

• *

L L L L LLXXX X z @ @ @ @ R

0.6 0.4

@ @ @ @ @ @ @ R

Consider the belief hierarchy that starts at your choice blue: You believe that Barbara chooses red. You believe that Barbara believes that you choose yellow. You believe that Barbara believes that you choose irrationally (yellow), so this belief hierarchy does not express common belief in rationality.

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This does not mean that you cannot rationally choose blue under common belief in rationality! You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *
• >
• 0.6

0.4

• *

L L L L LLXXX X z @ @ @ @ R

0.6 0.4

@ @ @ @ @ @ @ R

• The belief hierarchy that starts at your choice blue expresses

common belief in rationality, and supports your choice blue. So, you can rationally choose blue under common belief in rationality!

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 26 / 78

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *
• >
• 0.6

0.4

HHHHHH j

• *

L L L L LLXXX X z @ @ @ @ R

0.6 0.4

@ @ @ @ @ @ @ R

What about your choice green? Consider the belief hierarchy that starts at your choice green. You believe that Barbara chooses blue. You believe that Barbara believes that, with probability 0.6, you choose red, and with probability 0.4 you irrationally choose yellow. It does not express common belief in rationality.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 27 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 2 1 4 3 In fact, you cannot rationally choose green under common belief in rationality: If Barbara believes that you choose rationally, then she believes that you will not choose yellow. But then, she cannot rationally choose blue, as yellow would always be better for her. So, if you believe that Barbara chooses rationally, and that Barbara believes that you choose rationally, you must believe that she will only choose red or yellow. But then, you should choose blue, and not green.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 28 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 2 1 4 3 Summarizing Your choice yellow is irrational. Yoir choice red is rational, but can no longer be optimal if you believe that Barbara chooses rationally. You can rationally choose green if you believe that Barbara chooses rationally, but not if you believe, in addition, that Barbara believes that you choose rationally. You can rationally choose blue under common belief in rationality. In fact, blue is the only color you can rationally choose under common belief in rationality.

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New Scenario

Barbara has same preferences over colors as you. Barbara likes to wear the same color as you, whereas you hate this. blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 Which color(s) can you rationally choose under common belief in rationality?

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Beliefs diagram

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHHH H j

• >
• 0.6

0.4

• blue

green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5

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You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHHH H j

• >
• 0.6

0.4

• The belief hierarchy that starts at your choice blue expresses

common belief in rationality. Similarly, the belief hierarchies that start at your choices green and red also express common belief in rationality. So, you can rationally choose blue, green and red under common belief in rationality.

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Choosing rationally

We will now de…ne formally what we mean by a rational choice. I = f1, 2, ..., ng : set of players. Ci: set of choices for player i. A choice-combination for i’s opponents is a combination (c1, ..., ci1, ci+1, ..., cn). By Ci := C1 ... Ci1 Ci+1 ... Cn we denote the set of all choice-combinations for i’s opponents. A belief for player i about his opponents’ choices is a probability distribution bi over the set Ci of opponents’ choice-combinations. For every choice-combination (c1, ..., ci1, ci+1, ..., cn), the number bi(c1, ..., ci1, ci+1, ..., cn) speci…es the probability that player i assigns to the event that his opponents make precisely this combination of choices.

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A utility function for player i is a function ui that assigns to every combination of choices (c1, ..., cn) some number ui(c1, ..., cn). The number ui(c1, ..., cn) indicates how desirable player i …nds the

• utcome induced by (c1, ..., cn).

In the example “Going to a party”: u1(green, red) = 3, u1(green, blue) = 3, u1(green, green) = 0, u1(blue, red) = 4.

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Suppose that player i holds a belief bi about the opponents’ choices. The expected utility of making choice ci, while having the belief bi, is ui(ci, bi) =

∑

ci 2Ci

bi(ci) ui(ci, ci). The choice ci is optimal for player i given his belief bi, if ui(ci, bi) ui(c0

i , bi)

for all other choices c0

i 2 Ci.

The choice ci is rational for player i if it is optimal for some belief bi about the opponents’ choices.

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Belief hierarchies

In order to judge whether a given choice is reasonable for you, you must hold a belief about the opponents’ choices, a belief about what the opponents believe about their opponents’ choices, a belief about what the opponents believe that their opponents believe about the other players’ choices, and so on. This is a belief hierarchy. It consists of a …rst-order belief, a second-order belief, a third-order belief, and so on. Belief hierarchies can be constructed from an extended beliefs diagram.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 36 / 78

Extended beliefs diagram

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHHHH H j

• >
• 0.6

0.4

• Andrés Perea (Maastricht University)

Chapters 2 and 3 Period 5, 2012/2013 37 / 78

Epistemic model

Writing down a belief hierarchy for you is a lot of work. You must write down your belief about the opponents’ choices your belief about what your opponents believe about their opponents’ choices, a belief about what the opponents believe that their opponents believe about the other players’ choices, and so on, ad in…nitum. Is there a shorter way to represent a belief hierarchy?

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You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHHH H j

• >
• 0.6

0.4

• Writing down the complete belief hierarchy that starts at your choice red

is di¢cult!

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 39 / 78

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHH H j

• >
• 0.6

0.4

• Idea: Denote by tred

1

your belief hierarchy that starts at your choice red. Denote by tblue

2

and tgreen

2

the belief hierarchies for Barbara that start at her choices blue and green. Then, tred

1

believes that, with prob. 0.6, Barbara chooses blue and has belief hierarchy tblue

2

, and believes that, with prob. 0.4, Barbara chooses green and has belief hierarchy tgreen

2

.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 40 / 78

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHHH H j

• >
• 0.6

0.4

• Formally: We call the belief hierarchies tred

1

, tblue

2

and tgreen

2

types. Type tred

1

has belief b1(tred

1

) = (0.6) (blue, tblue

2

) + (0.4) (green, tgreen

2

).

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 41 / 78

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHHH H j

• >
• 0.6

0.4

• Also, b1(tblue

1

) = (green, tgreen

2

) and b1(tgreen

1

) = (blue, tblue

2

) and …nally b1(tyellow

1

) = (yellow, tyellow

2

). We can do the same for Barbara’s belief hierarchies. This leads to an epistemic model.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 42 / 78

Epistemic model for “Going to a party”

Types T1 = ftblue

1

, tgreen

1

, tred

1

, tyellow

1

g T2 = ftblue

2

, tgreen

2

, tred

2

, tyellow

2

g Beliefs for player 1 b1(tblue

1

) = (green, tgreen

2

) b1(tgreen

1

) = (blue, tblue

2

) b1(tred

1

) = (0.6) (blue, tblue

2

) + (0.4) (green, tgreen

2

) b1(tyellow

1

) = (yellow, tyellow

2

) Beliefs for player 2 b2(tblue

2

) = (blue, tblue

1

) b2(tgreen

2

) = (green, tgreen

1

) b2(tred

2

) = (red, tred

1

) b2(tyellow

2

) = (yellow, tyellow

1

)

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 43 / 78

In an epistemic model, we can derive for every type the …rst-order belief, second-order belief, and so on. So, we can derive for every type the complete belief hierarchy it has. Types T1 = ftblue

1

, tgreen

1

, tred

1

, tyellow

1

g T2 = ftblue

2

, tgreen

2

, tred

2

, tyellow

2

g Beliefs for player 1 b1(tblue

1

) = (green, tgreen

2

) b1(tgreen

1

) = (blue, tblue

2

) b1(tred

1

) = (0.6) (blue, tblue

2

) + (0.4) (green, tgreen

2

) b1(tyellow

1

) = (yellow, tyellow

2

) Beliefs for player 2 b2(tblue

2

) = (blue, tblue

1

) b2(tgreen

2

) = (green, tgreen

1

) b2(tred

2

) = (red, tred

1

) b2(tyellow

2

) = (yellow, tyellow

1

)

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 44 / 78

De…nition (Epistemic model)

An epistemic model speci…es for every player i a set Ti of possible types. Moreover, for every type ti it speci…es a probability distribution bi(ti) over the set Ci Ti of opponents’ choice-type combinations. Here, Ci Ti is the set of combinations ((c1, t1), ..., (ci1, ti1), (ci+1, ti+1), ..., (cn, tn))

• f opponents’ choices and opponents’ types.

For every such combination, bi(ti)((c1, t1), ..., (ci1, ti1), (ci+1, ti+1), ..., (cn, tn)) represents the probability that type ti assigns to the event that the

• pponents’ choices and belief hierarchies are exactly given by

((c1, t1), ..., (ci1, ti1), (ci+1, ti+1), ..., (cn, tn)).

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 45 / 78

Common belief in rationality

Intuitively, common belief in rationality means that you believe that your opponents choose rationally, you believe that your opponents believe that their opponents choose rationally, and so on, ad in…nitum. How can we state common belief in rationality formally, within an epistemic model?

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 46 / 78

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• *

HHHHH H j

• >
• 0.6

0.4

• Your type tred

1

has belief b1(tred

1

) = (0.6) (blue, tblue

2

) + (0.4) (green, tgreen

2

). For Barbara, blue is optimal for type tblue

2

, and green is optimal for type tgreen

2

. So, type tred

1

• nly assigns positive probability to choice-type pairs

for Barbara where the choice is optimal for the type. Hence, tred

1

believes in Barbara’s rationality.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 47 / 78

De…nition (Belief in the opponents’ rationality)

Type ti believes in the opponents’ rationality if his belief bi(ti) only assigns positive probability to choice-type combinations ((c1, t1), ..., (ci1, ti1), (ci+1, ti+1), ..., (cn, tn)) where choice c1 is optimal for type t1, ..., choice cn is optimal for type tn.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 48 / 78

De…nition (Common belief in rationality)

Type ti expresses 1-fold belief in rationality if ti believes in the opponents’ rationality. Type ti expresses 2-fold belief in rationality if ti only assigns positive probability to opponents’ types that express 1-fold belief in rationality. Type ti expresses 3-fold belief in rationality if ti only assigns positive probability to opponents’ types that express 2-fold belief in rationality. And so on. Type ti expresses common belief in rationality if ti expresses k-fold belief in rationality for all k. In the literature, this concept is also known as rationalizability.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 49 / 78

Type ti expresses up to k-fold belief in rationality if ti expresses 1-fold, 2-fold, ..., k-fold belief in rationality. So, if ti expresses up to k-fold belief in rationality, then ti only assigns positive probability to opponents’ choice-type pairs (cj, tj) where cj is

• ptimal for tj, and tj expresses up to (k 1)-fold belief in rationality.

De…nition

Player i can rationally make choice ci under common belief in rationality if there is some epistemic model, and some type ti within this epistemic model, such that type ti expresses common belief in rationality, and choice ci is optimal for type ti.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 50 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 Beliefs for player 1 b1(tblue

1

) = (green, tgreen

2

) b1(tgreen

1

) = (blue, tblue

2

) b1(tred

1

) = (0.6) (blue, tblue

2

) + (0.4) (green, tgreen

2

) b1(tyellow

1

) = (yellow, tyellow

2

) Beliefs for player 2 b2(tblue

2

) = (blue, tblue

1

) b2(tgreen

2

) = (green, tgreen

1

) b2(tred

2

) = (red, tred

1

) b2(tyellow

2

) = (yellow, tyellow

1

) Each of your types tblue

1

, tgreen

1

and tred

1

expresses common belief in rationality.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 51 / 78

Existence

So far, we have discussed the idea of common belief in rationality, which states that you believe that your opponents choose rationally, you believe that your opponents believe that their opponents choose rationally, and so on, ad in…nitum. Question: Is common belief in rationality always possible? So, can we …nd, for every game and every player, at least one type for this player that expresses common belief in rationality? Answer: Yes, provided that every player only has …nitely many choices.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 52 / 78

Example: Going to a party

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 You blue green red yellow Barbara blue green red yellow You blue green red yellow

• Andrés Perea (Maastricht University)

Chapters 2 and 3 Period 5, 2012/2013 53 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 You blue green red yellow Barbara blue green red yellow You blue green red yellow

• J

J J J J J J J J J ^

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 54 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 You blue green red yellow Barbara blue green red yellow You blue green red yellow

• J

J J J J J J J J J ^

• Andrés Perea (Maastricht University)

Chapters 2 and 3 Period 5, 2012/2013 55 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 You blue green red yellow Barbara blue green red yellow You blue green red yellow

• J

J J J J J J J J J ^

• 3

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 56 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 You blue green red yellow Barbara blue green red yellow You blue green red yellow

• J

J J J J J J J J J ^

• 3
• Andrés Perea (Maastricht University)

Chapters 2 and 3 Period 5, 2012/2013 57 / 78

blue green red yellow same color as friend you 4 3 2 1 Barbara 4 3 2 1 5 You blue green red yellow Barbara blue green red yellow You blue green red yellow

• J

J J J J J J J J J ^

• 3
• QQQQ

Q s

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 58 / 78

You blue green red yellow Barbara blue green red yellow You blue green red yellow

• J

J J J J J J J J J ^

• 3
• QQQQ

Q s

By starting at your choice blue, we enter a cycle of arrows that only reaches rational choices. So, the belief hierarchy starting at your choice blue expresses common belief in rationality.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 59 / 78

For any game, with two or more players, we can construct such a chain of beliefs as follows: Start with an arbitrary choice combination (c1

1 , ..., c1 n).

For every player i, let c2

i be a choice for player i that is optimal if he

believes that his opponents choose according to (c1

1 , ..., c1 n).

This leads to a new choice-combination (c2

1 , ..., c2 n).

For every player i, let c3

i be a choice for player i that is optimal if he

believes that his opponents choose according to (c2

1 , ..., c2 n).

This leads to a new choice-combination (c3

1 , ..., c3 n).

And so on. If there are only …nitely many choices, this chain of beliefs must run into a cycle (c1

1 , ..., c1 n)

! (c2

1 , ..., c2 n)

! ... ! (cK

1 , ..., cK n )

| {z }

=(c1

1 ,...,c1 n ) Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 60 / 78

Consider the cycle of beliefs (c1

1 , ..., c1 n)

! (c2

1 , ..., c2 n)

! ... ! (cK

1 , ..., cK n )

| {z }

=(c1

1 ,...,c1 n )

Construct for every player i a set of types Ti = ftc1

i

i , ..., tcK 1

i

i

g. Every type tck

i

i

believes that, with probability 1, every opponent j chooses ck1

j

and is of type t

ck1

j

j

. By construction, choice ck

i is optimal for type tck

i

i .

So, every type in the epistemic model believes in the opponents’ rationality. Then, every type in this epistemic model will express common belief in rationality. Hence, every choice in this cycle can rationally be chosen under common belief in rationality!

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 61 / 78

Theorem (Common belief in rationality is always possible)

Consider a static game with …nitely many choices for every player. Then, we can construct an epistemic model in which (1) every type expresses common belief in rationality, and (2) every type assigns, for every opponent, probability 1 to one speci…c choice and one speci…c type for that opponent.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 62 / 78

Algorithm

We look for an algorithm that helps us …nd those choices you can rationally make under common belief in rationality. Start with more basic question: Can we characterize those choices that are rational – that is, optimal for some belief?

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 63 / 78

Consider the example “Going to a party”. blue green red yellow same color as Barbara 4 3 2 1 Only your choice yellow is irrational. Your choice yellow is strictly dominated by the randomized choice in which you choose blue and green with probability 0.5. blue green red yellow yellow 1 1 1 randomized choice 1.5 2 3.5 3.5

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 64 / 78

v v v v v v v

a b c d e f g 100 100 100 100 100 100 In the example “Where to locate my pub”, only your choices a and g are irrational. Your choice a is strictly dominated by b, and your choice g is strictly dominated by f .

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 65 / 78

So, in each of the examples considered so far we see the following: A choice is irrational precisely when it is strictly dominated by another choice, or strictly dominated by a randomized choice. In fact, this is always true!

Theorem

A choice is irrational, if and only if, it is strictly dominated by another choice, or strictly dominated by a randomized choice. Or, equivalently:

Theorem

A choice is rational, if and only if, it is not strictly dominated by another choice, nor strictly dominated by a randomized choice.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 66 / 78

Formally: A choice ci is strictly dominated by a choice c0

i if

ui(ci, ci) < ui(c0

i , ci)

for every opponents’ choice-combination ci. A randomized choice for player i is a probability distribution ri over his set of choices Ci. A choice ci is strictly dominated by a randomized choice ri if ui(ci, ci) < ui(ri, ci) for every opponents’ choice-combination ci.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 67 / 78

Step 1: 1-fold belief in rationality Which choices are rational for a type that expresses 1-fold belief in rationality? If you believe in the opponents’ rationality, then you assign positive probability only to opponents’ choices that are rational. Remember: A choice is rational precisely when it is not strictly dominated. So, if you believe in the opponents’ rationality, then you assign positive probability only to opponents’ choices that are not strictly dominated.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 68 / 78

Step 1: 1-fold belief in rationality So, if you believe in the opponents’ rationality, then you assign positive probability only to opponents’ choices that are not strictly dominated. In a sense, you eliminate the opponents’ strictly dominated choices from the game, and concentrate on the reduced game that remains. The choices that you can rationally make if you believe in your

• pponents’ rationality, are exactly the choices that are optimal for you

for some belief within this reduced game. But these are exactly the choices that are not strictly dominated for you within this reduced game! Hence, these are the choices that survive 2-fold elimination of strictly dominated choices.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 69 / 78

Step 2: Up to 2-fold belief in rationality Which choices are rational for a type that expresses up to 2-fold belief in rationality? Consider a type ti that expresses up to 2-fold belief in rationality. Then, ti only assigns positive probability to opponents’ choice-type pairs (cj, tj) where cj is optimal for tj, and tj expresses 1-fold belief in rationality. So, type ti only assigns positive probability to opponents’ choices cj which are optimal for a type that expresses 1-fold belief in rationality. Hence, type ti only assigns positive probability to opponents’ choices cj which survive 2-fold elimination of strictly dominated choices.

Andrés Perea (Maastricht University) Chapters 2 and 3 Period 5, 2012/2013 70 / 78

Step 2: Up to 2-fold belief in rationality Hence, type ti only assigns positive probability to opponents’ choices cj which survive 2-fold elimination of strictly dominated choices. Then, every choice ci which is optimal for ti must be optimal for some belief within the reduced game obtained after 2-fold elimination

• f strictly dominated choices.

So, every choice ci which is optimal for ti must not be strictly dominated within the reduced game obtained after 2-fold elimination

• f strictly dominated choices.

Conclusion: Every choice that is optimal for a type that expresses up to 2-fold belief in rationality, must survive 3-fold elimination of strictly dominated choices.

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Algorithm (Iterated elimination of strictly dominated choices)

Step 1. Within the original game, eliminate all choices that are strictly dominated. Step 2. Within the reduced game obtained after step 1, eliminate all choices that are strictly dominated. Step 3. Within the reduced game obtained after step 2, eliminate all choices that are strictly dominated. . . . Continue in this fashion until no further choices can be eliminated.

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Theorem (Algorithm “works”)

(1) For every k 1, the choices that are optimal for a type that expresses up to k-fold belief in rationality are exactly those choices that survive (k + 1)-fold elimination of strictly dominated choices. (2) The choices that can rationally be made under common belief in rationality are exactly those choices that survive iterated elimination of strictly dominated choices.

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Properties of the algorithm

Algorithm (Iterated elimination of strictly dominated choices)

Step 1. Within the original game, eliminate all choices that are strictly dominated. Step 2. Within the reduced game obtained after step 1, eliminate all choices that are strictly dominated. Step 3. Within the reduced game obtained after step 2, eliminate all choices that are strictly dominated. . . . Continue in this fashion until no further choices can be eliminated. This algorithm always stops after …nitely many steps. It always yields a nonempty output for every player. The order and speed in which you eliminate choices is not relevant for the eventual output.

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Example: The traveler’s dilemma

Story You and Barbara have been on a trip to China, where you both have bought the same Chinese vase. When arriving at the airport, both your vases are broken. You negotiate with the airport manager about a reimbursement. He proposes the following procedure: You both must write a price from 100, 200, 300, 400 on a piece of paper. The person with the lowest price receives the average of the prices + 100. The person with the highest price receives the average of the prices

• 225.

If you write down the same price, you both receive that price.

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Barbara’s choice Your choice 100 200 300 400 100 100 250 300 350 200 75 200 350 400 300 25 25 300 450 400 25 75 125 400 What price(s) can you write down under common belief in rationality? Use the algorithm. Step 1: For you and Barbara, the choice 400 is strictly dominated by the randomized choice (0.45) 100 + (0.55) 300. Eliminate 400 for you and Barbara.

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Barbara You 100 200 300 100 100, 100 250, 75 300, 25 200 75, 250 200, 200 350, 25 300 25, 300 25, 350 300, 300 Step 2: Within reduced game, the choice 300 for you and Barbara is strictly dominated by the randomized choice (0.5) 100 + (0.5) 200. Eliminate 300 for you and Barbara.

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Barbara You 100 200 100 100, 100 250, 75 200 75, 250 200, 200 Step 3: Within reduced game, the choice 200 for you and Barbara is strictly dominated by the choice 100. Eliminate the choice 200 for you and Barbara. So, only the choice 100 survives for you. Hence, you can only rationally write down a price of 100 under common belief in rationality!

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