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Problem Set 1: Consumer Theory
Matthew Robson University of York Microeconomics 2
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Question 1 Assuming that the weak preference relation π
is transitive, show that: π₯ππ¦ & π¦πΌπ§βπ₯ππ§ where x, y and z are consumption bundles.
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Question 1 Weak Preference Relation: xRy - x is at least as good as y
Strict Preference Relation: xPy - x is better than y. Indifference Relation: xIy - x and y are equally good. Assumptions: R is reflexive, i.e. xRx R is transitive, i.e. if xRy and yRz, then xRz R is connected (or complete), i.e. xRy or yRx
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Question 1 π₯ππ¦ = π₯π
π¦ & πππ‘(π¦π
π₯) π¦πΌπ§ = π¦π
π§ & π§π
π¦
Therefore, π₯π
π¦ & π¦π
π§ β π₯π
π§ by transitivity of R. Need to show that the opposite is not true, πππ‘(π§π
π₯). Suppose π§π
π₯ was true. Then yπ
π§ & π§π
π₯ βπ¦π
π₯ by transitivity of R, which contradicts πππ‘(π¦π
π₯). Thus, we have: π₯π
π§ & πππ‘(π§π
π₯) = π₯ππ§
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Question 2 Carefully illustrate in diagrams indifference curves for the following utility functions: π π π₯ = 2 π₯ π₯ 2 ππ π π₯ =πππ π₯ 1 ,2 π₯ 2 πππ π π₯ = πΌ π₯ 1 βπ π₯ 2 βπ β 1 π
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Question 2i) π π₯ = 2 π₯ 1 + π₯ 2 2 π₯ 1 =πβ π₯ 2 π₯ 1 = πβ π₯ 2 2
Perfect Substitutes π π₯ = 2 π₯ π₯ 2 2 π₯ 1 =πβ π₯ 2 π₯ 1 = πβ π₯ 2 2
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Question 2ii) π π₯ =πππ π₯ 1 ,2 π₯ 2 If π₯ 1 =π π₯ β π₯ 2 β₯ π π₯ 2
Perfect Complements π π₯ =πππ π₯ 1 ,2 π₯ 2 If π₯ 1 =π π₯ β π₯ 2 β₯ π π₯ 2 If π₯ 2 = π π₯ 2 β π₯ 1 β₯π(π₯)
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Question 2iii) π π₯ = 0.5 πΌ π₯ 1 βπ + 0.5 π₯ 2 βπ β 1 π
π π₯ = πΌ π₯ 1 βπ π₯ 2 βπ β 1 π π βπ = 0.5 πΌ π₯ 1 βπ π₯ 2 βπ π βπ β 0.5 π₯ 2 βπ = 0.5 πΌ π₯ 1 βπ 2 π βπ β π₯ 2 βπ = πΌ π₯ 1 βπ 2 π βπ β π₯ 2 βπ β 1 π πΌ = π₯ 1 2 π βπ β πΌπ₯ 1 βπ β 1 π = π₯ 2
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Question 3 Suppose the consumerβs utility function is π π₯ = 1 πΌ π₯ 1 πΌ + π₯ 2 , where 0<πΌ<1, she has income π and faces prices π 1 and π 2 for commodities 1 and 2, respectively. Derive the consumerβs Marshallian demand functions.
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Question 3 Lagrangian: β= 1 πΌ π₯ 1 πΌ + π₯ 2 +π πβ π₯ 1 π 1 β π₯ 2 π 2
β= 1 πΌ π₯ 1 πΌ + π₯ 2 +π πβ π₯ 1 π 1 β π₯ 2 π 2 FOCβs: πβ π π₯ 1 = π₯ 1 πΌβ1 βπ π 1 =0 πβ π π₯ 2 =1βπ π 2 =0 πβ ππ =πβ π₯ 1 π 1 β π₯ 2 π 2 =0 (1) (2) (3)
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Question 3 Using (1) & (2): π₯ 1 πΌβ1 π 1 =π= 1 π 2 π₯ 1 πΌβ1 = π 1 π β π₯ 1 = π 1 π πΌβ1 Plug into (3): πβ π 1 π πΌβ1 π 1 β π₯ 2 π 2 =0 π₯ 2 = π π 2 β π 1 π π 1 π πΌβ1 π₯ 2 = π π 2 β π 1 π πΌ πΌβ1
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Question 4 For the utility function π π₯ = 1 πΌ π₯ 1 πΌ + π₯ 2 , derive the corresponding indirect utility function and the expenditure function.
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Question 4 Indirect Utility Function: π π₯ = 1 πΌ π₯ 1 πΌ + π₯ 2
π π₯ = 1 πΌ π₯ 1 πΌ + π₯ 2 Indirect Utility Function: π π 1 , π 2 ,π = 1 πΌ π 1 π πΌβ1 πΌ + π π 2 β π 1 π πΌ πΌβ1 π π 1 , π 2 ,π = 1 πΌ π 1 π πΌ πΌβ1 + π π 2 β π 1 π πΌ πΌβ1 π π 1 , π 2 ,π = 1 πΌ β1 π 1 π πΌ πΌβ1 + π π 2 π π 1 , π 2 ,π = 1βπΌ πΌ π 1 π πΌ πΌβ1 + π π 2 (1)
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Question 4 β= π₯ 1 π 1 + π₯ 2 π 2 +π π’ β 1 πΌ π₯ 1 πΌ β π₯ 2 FOCβs:
Dual Problem β choose π₯ 1 and π₯ 2 to minimise expenditure, subject to a utility constraint. β= π₯ 1 π 1 + π₯ 2 π 2 +π π’ β 1 πΌ π₯ 1 πΌ β π₯ 2 FOCβs: πβ π π₯ 1 = π 1 βππ₯ 1 πΌβ1 =0 πβ π π₯ 2 = π 2 βπ=0 πβ ππ = π’ β 1 πΌ π₯ 1 πΌ β π₯ 2 =0 (1) (2) (3)
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Question 4 Using (1) & (2): π 1 π₯ 1 πΌβ1 =π= π 2 π₯ 1 πΌβ1 = π 1 π 2
π 1 π₯ 1 πΌβ1 =π= π 2 π₯ 1 πΌβ1 = π 1 π 2 π₯ 1 = π 1 π πΌβ1 Plug into (3): π’ β 1 πΌ π 1 π πΌβ1 πΌ β π₯ 2 =0 π₯ 2 = π’ β 1 πΌ π 1 π πΌ πΌβ1 (4) (5)
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Question 4 πΈ π₯, π = π₯ 1 π 1 + π₯ 2 π 2 Expenditure Function:
πΈ π₯, π = π₯ 1 π 1 + π₯ 2 π 2 πΈ π 1 , π 2 , π’ = π 1 π πΌβ1 π 1 + π’ β 1 πΌ π 1 π πΌ πΌβ1 π 2 πΈ π 1 , π 2 , π’ = π’ π π 1 π πΌβ1 π 1 β 1 πΌ π 1 π 2 π 2 πΈ π 1 , π 2 , π’ = π’ π π 1 π πΌβ1 (π 1 πΌ) πΌ β π 1 πΌ πΈ π 1 , π 2 , π’ = π’ + π 1 πΌβ1 πΌ π 1 π πΌβ1 (6)
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Question 4 Due to duality, the Indirect Utility Function and Expenditure Function are inverses of one another: π π 1 , π 2 ,π = π’ Therefore: π’ = 1βπΌ πΌ π 1 π πΌ πΌβ1 + π π 2 Rearrange to get: π= π’ π 2 β π 2 1βπΌ πΌ π 1 π πΌ πΌβ1 Or: πΈ π 1 , π 2 , π’ = π’ π 2 β π 2 1βπΌ πΌ π 1 π πΌ πΌβ1 (7)
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Question 4 Each method leads to slightly different equationsβ¦ But they should be equivalent. 6 = 7 π’ π 2 + π 1 πΌβ1 πΌ π 1 π πΌβ1 = π’ π 2 β π 2 1βπΌ πΌ π 1 π πΌ πΌβ1 = π’ π 2 β π 2 1βπΌ πΌ π 1 π π 1 π πΌβ1 = π’ π 2 β π 1 1βπΌ πΌ π 1 π πΌβ1 = π’ π 2 + π 1 πΌβ1 πΌ π 1 π πΌβ1
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