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Solutions Manual

Foundations of Mathematical Economics

Michael Carter

November 15, 2002

Solutions for Foundations of Mathematical Economics cβ 2001 Michael Carter

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Chapter 1: Sets and Spaces

1.1

{ 1, 3, 5, 7 . . .} or {π β π : π is odd } 1.2 Every π₯ β π΄ also belongs to π΅. Every π₯ β π΅ also belongs to π΄. Hence π΄,π΅ have precisely the same elements.

1.3 Examples of finite sets are

β the letters of the alphabet {A, B, C, . . . , Z } β the set of consumers in an economy β the set of goods in an economy β the set of players in a game.

Examples of infinite sets are

β the real numbers β β the natural numbers π β the set of all possible colors β the set of possible prices of copper on the world market β the set of possible temperatures of liquid water.

1.4 π = { 1, 2, 3, 4, 5, 6 }, πΈ = { 2, 4, 6 }. 1.5 The player set is π = { Jenny,Chris }. Their action spaces are

π΄π = {Rock, Scissors,Paper } π = Jenny,Chris 1.6 The set of players is π = {1, 2, . . . , π }. The strategy space of each player is the set of feasible outputs

π΄π = { ππ β β+ : ππ β€ ππ } where ππ is the output of dam π.

1.7 The player set is π = {1, 2, 3}. There are 23 = 8 coalitions, namely π«(π) = {β , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

There are 210 coalitions in a ten player game.

1.8 Assume that π₯ β (π βͺ π )π. That is π₯ /β π βͺ π . This implies π₯ /β π and π₯ /β π , or π₯ β ππ and π₯ β π π. Consequently, π₯ β ππ β© π π. Conversely, assume π₯ β ππ β© π π. This implies that π₯ β ππ and π₯ β π π. Consequently π₯ /β π and π₯ /β π and therefore π₯ /β π βͺ π . This implies that π₯ β (π βͺ π )π. The other identity is proved similarly. 1.9 βͺ

πβπ π = π

β© πβπ π = β

1

Solutions for Foundations of Mathematical Economics cβ 2001 Michael Carter

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0-1 1 π₯1

-1

1 π₯2

Figure 1.1: The relation { (π₯, π¦) : π₯2 + π¦2 = 1 }

1.10 The sample space of a single coin toss is {π»,π }. The set of possible outcomes in three tosses is the product

{π»,π } Γ {π»,π } Γ {π»,π } ={(π»,π»,π»), (π»,π», π ), (π»,π,π»), (π»,π, π ), (π,π»,π»), (π,π», π ), (π, π,π»), (π, π, π )

}

A typical outcome is the sequence (π»,π», π ) of two heads followed by a tail.

1.11

π β© βπ+ = {0}

where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs. To see this, first note that 0 is a feasible production plan. Therefore, 0 β π . Also, 0 β βπ+ and therefore 0 β π β© βπ+. To show that there is no other feasible production plan in βπ+, we assume the contrary. That is, we assume there is some feasible production plan y β βπ+ β {0}. This implies the existence of a plan producing a positive output with no inputs. This technological infeasible, so that π¦ /β π . 1.12 1. Let x β π (π¦). This implies that (π¦,βx) β π . Let xβ² β₯ x. Then (π¦,βxβ²) β€

(π¦,βx) and free disposability implies that (π¦,βxβ²) β π . Therefore xβ² β π (π¦). 2. Again assume x β π (π¦). This implies that (π¦,βx) β π . By free disposal,

(π¦β²,βx) β π for every π¦β² β€ π¦, which implies that x β π (π¦β²). π (π¦β²) β π (π¦). 1.13 The domain of β

Solutions for Foundations of Mathematical Economics cβ 2001 Michael Carter

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1.16 The following table lists their respective properties.

< β€ = reflexive Γ β β transitive

β β β symmetric Γ β β asymmetric

β Γ Γ anti-symmetric

β β β complete

β β Γ Note that the properties of symmetry and anti-symmetry are not mutually exclusive.

1.17 Let βΌ be an equivalence relation of a set π β= β . That is, the relation βΌ is reflexive, symmetric and transitive. We first show that every π₯ β π belongs to some equivalence class. Let π be any element in π and let βΌ (π) be the class of elements equivalent to π, that is

βΌ(π) β‘ { π₯ β π : π₯ βΌ π } Since βΌ is reflexive, π βΌ π and so π β βΌ(π). Every π β π belongs to some equivalence class and therefore

π = βͺ πβπ βΌ(π)

Next, we show that the equivalence classes are either disjoint or identical, that is βΌ(π) β= βΌ(π) if and only if fβΌ(π) β© βΌ(π) = β . First, assume βΌ(π) β© βΌ(π) = β . Then π β βΌ(π) but π /β βΌ(π). Therefore βΌ(π) β= βΌ(π). Conversely, assume βΌ(π) β© βΌ(π) β= β and let π₯ β βΌ(π) β© βΌ(π). Then π₯ βΌ π and by symmetry π βΌ π₯. Also π₯ βΌ π and so by transitivity π βΌ π. Let π¦ be any element in βΌ(π) so that π¦ βΌ π. Again by transitivity π¦ βΌ π and therefore π¦ β βΌ(π). Hence βΌ(π) β βΌ(π). Similar reasoning implies that βΌ(π) β βΌ(π). Therefore βΌ(π) = βΌ(π). We conclude that the equivalence classes partition π .

1.18 The set of proper coalitions is not a partition of the set of players, since any player can belong to more than one coalition. For example, player 1 belongs to the coalitions {1}, {1, 2} and so on. 1.19

π₯ β» π¦ =β π₯ βΏ π¦ and π¦ ββΏ π₯ π¦ βΌ π§ =β π¦ βΏ π§ and π§ βΏ π¦

Transitivity of βΏ implies π₯ βΏ π§. We need to show that π§ ββΏ π₯. Assume otherwise, that is assume π§ βΏ π₯ This implies π§ βΌ π₯ and by transitivity π¦ βΌ π₯. But this implies that π¦ βΏ π₯ which contradicts the assumption that π₯ β» π¦. Therefore we conclude that π§ ββΏ π₯ and therefore π₯ β» π§. The other result is proved in similar fashion. 1.20 asymmetric Assume π₯ β» π¦.

π₯ β» π¦ =β π¦ ββΏ π₯ while

π¦ β» π₯ =β π¦ βΏ π₯ Therefore

π₯ β» π¦ =β π¦ ββ» π₯

3

Solutions for Foundations of Mathematical Economics cβ 2001 Michael Carter

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transitive Assume π₯ β» π¦ and π¦ β» π§. π₯ β» π¦ =β π₯ βΏ π¦ and π¦ ββΏ π₯ π¦ β» π§ =β π¦ βΏ π§ and π§ ββΏ π¦

Since βΏ is transitive, we conclude that π₯ βΏ π§. It remains to show that π§ ββΏ π₯. Assume otherwise, that is assume π§ βΏ π₯. We know that π₯ βΏ π¦ and transitivity implies that π§ βΏ π¦, contrary to the assumption that π¦ β» π§. We conclude that π§ ββΏ π₯ and

π₯ βΏ π§ and π§ ββΏ π₯ =β π₯ β» π§ This shows that β» is transitive.

1.21 reflexive Since βΏ is reflexive, π₯ βΏ π₯ which implies π₯ βΌ π₯. transitive Assume π₯ βΌ π¦ and π¦ βΌ π§. Now

π₯ βΌ π¦ ββ π₯ βΏ π¦ and π¦ βΏ π₯

π¦ βΌ π§ ββ π¦ βΏ π§ and π§ βΏ π¦ Transitivity of βΏ implies

π₯ βΏ π¦ and π¦ βΏ π§ =β π₯ βΏ π§

π§ βΏ π¦ and π¦ βΏ π₯ =β π§ βΏ π₯ Combining

π₯ βΏ π§ and π§ βΏ π₯ =β π₯ βΌ π§

symmetric

π₯ βΌ π¦ ββ π₯ βΏ π¦ and π¦ βΏ π₯ ββ π¦ βΏ π₯ and π₯ βΏ π¦ ββ π¦ βΌ π₯

1.22 reflexive Every integer is a multiple of itself, that is π = 1π.

transitive Assume π = ππ and π = ππ where π, π β π . Then π = πππ so that π is a multiple of π.

not symmetric If π = ππ, π β π , then π = 1ππ and π /β π . For example, 4 is a multiple of 2 but 2 is not a multiple of 4.

1.23

[π, π] = { π, π¦, π, π§ } (π, π) = { π¦ }

1.24

βΏ (π¦) = {π, π¦, π§ } β» (π¦) = {π, π§ } βΎ (π¦) = {π, π₯, π¦ } βΊ (π¦) = {π, π₯ }

4

Solutions for Foundations of Mathematical Economics cβ 2001 Michael Carter

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1.25 Let π be ordered by βΏ. π₯ β π is a minimal element there is no element which strictly precedes it, that is there is no element π¦ β π such that π¦ βΊ π₯. π₯ β π is the first element if it precedes every other element, that is π₯ βΎ π¦ for all π¦ β π . 1.26 The maximal elements of π are π and π§. The minimal element of π is π₯. These are also best and worst elements respectively.

1.27 Assume that π₯ is a best element in π ordered by βΏ. That is, π₯ βΏ π¦ for all π¦ β π . This implies that there is no π¦ β π which strictly dominates π₯. Therefore, π₯ is maximal in π . In Example 1.23, the numbers 5, 6, 7, 8, 9 are all maximal elements, but none of them is a best element.

1.28 Assume that the elements are denoted π₯1, π₯2, . . . , π₯π. We can identify the maximal element by constructing another list using the following recursive algorithm

π1 = π₯1

ππ =

{ π₯π if π₯π β» ππβ1 ππβ1 otherwise

By construction, there is no π₯π which strictly succedes ππ. ππ is a maximal element.