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Problem Set 1: Consumer Theory

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1 Problem Set 1: Consumer Theory
Matthew Robson University of York Microeconomics 2

2 Question 1 Assuming that the weak preference relation 𝑅 is transitive, show that: π‘₯𝑃𝑦 & 𝑦𝐼𝑧⇒π‘₯𝑃𝑧 where x, y and z are consumption bundles.

3 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

4 Question 1 π‘₯𝑃𝑦 = π‘₯𝑅𝑦 & π‘›π‘œπ‘‘(𝑦𝑅π‘₯) 𝑦𝐼𝑧 = 𝑦𝑅𝑧 & 𝑧𝑅𝑦
Therefore, π‘₯𝑅𝑦 & 𝑦𝑅𝑧 β‡’ π‘₯𝑅𝑧 by transitivity of R. Need to show that the opposite is not true, π‘›π‘œπ‘‘(𝑧𝑅π‘₯). Suppose 𝑧𝑅π‘₯ was true. Then y𝑅𝑧 & 𝑧𝑅π‘₯ ⇒𝑦𝑅π‘₯ by transitivity of R, which contradicts π‘›π‘œπ‘‘(𝑦𝑅π‘₯). Thus, we have: π‘₯𝑅𝑧 & π‘›π‘œπ‘‘(𝑧𝑅π‘₯) = π‘₯𝑃𝑧

5 Question 2 Carefully illustrate in diagrams indifference curves for the following utility functions: π’Š π‘ˆ π‘₯ = 2 π‘₯ π‘₯ 2 π’Šπ’Š π‘ˆ π‘₯ =π‘šπ‘–π‘› π‘₯ 1 ,2 π‘₯ 2 π’Šπ’Šπ’Š π‘ˆ π‘₯ = 𝛼 π‘₯ 1 βˆ’π‘Ÿ π‘₯ 2 βˆ’π‘Ÿ βˆ’ 1 π‘Ÿ

6 Question 2i) π‘ˆ π‘₯ = 2 π‘₯ 1 + π‘₯ 2 2 π‘₯ 1 =π‘ˆβˆ’ π‘₯ 2 π‘₯ 1 = π‘ˆβˆ’ π‘₯ 2 2
Perfect Substitutes π‘ˆ π‘₯ = 2 π‘₯ π‘₯ 2 2 π‘₯ 1 =π‘ˆβˆ’ π‘₯ 2 π‘₯ 1 = π‘ˆβˆ’ π‘₯ 2 2

7 Question 2ii) π‘ˆ π‘₯ =π‘šπ‘–π‘› π‘₯ 1 ,2 π‘₯ 2 If π‘₯ 1 =π‘ˆ π‘₯ β‡’ π‘₯ 2 β‰₯ π‘ˆ π‘₯ 2
Perfect Complements π‘ˆ π‘₯ =π‘šπ‘–π‘› π‘₯ 1 ,2 π‘₯ 2 If π‘₯ 1 =π‘ˆ π‘₯ β‡’ π‘₯ 2 β‰₯ π‘ˆ π‘₯ 2 If π‘₯ 2 = π‘ˆ π‘₯ 2 β‡’ π‘₯ 1 β‰₯π‘ˆ(π‘₯)

8 Question 2iii) π‘ˆ π‘₯ = 0.5 𝛼 π‘₯ 1 βˆ’π‘Ÿ + 0.5 π‘₯ 2 βˆ’π‘Ÿ βˆ’ 1 π‘Ÿ
π‘ˆ π‘₯ = 𝛼 π‘₯ 1 βˆ’π‘Ÿ π‘₯ 2 βˆ’π‘Ÿ βˆ’ 1 π‘Ÿ π‘ˆ βˆ’π‘Ÿ = 0.5 𝛼 π‘₯ 1 βˆ’π‘Ÿ π‘₯ 2 βˆ’π‘Ÿ π‘ˆ βˆ’π‘Ÿ βˆ’ 0.5 π‘₯ 2 βˆ’π‘Ÿ = 0.5 𝛼 π‘₯ 1 βˆ’π‘Ÿ 2 π‘ˆ βˆ’π‘Ÿ βˆ’ π‘₯ 2 βˆ’π‘Ÿ = 𝛼 π‘₯ 1 βˆ’π‘Ÿ 2 π‘ˆ βˆ’π‘Ÿ βˆ’ π‘₯ 2 βˆ’π‘Ÿ βˆ’ 1 π‘Ÿ 𝛼 = π‘₯ 1 2 π‘ˆ βˆ’π‘Ÿ βˆ’ 𝛼π‘₯ 1 βˆ’π‘Ÿ βˆ’ 1 π‘Ÿ = π‘₯ 2

9 Question 3 Suppose the consumer’s utility function is π‘ˆ π‘₯ = 1 𝛼 π‘₯ 1 𝛼 + π‘₯ 2 , where 0<𝛼<1, she has income π‘š and faces prices 𝑝 1 and 𝑝 2 for commodities 1 and 2, respectively. Derive the consumer’s Marshallian demand functions.

10 Question 3 Lagrangian: β„’= 1 𝛼 π‘₯ 1 𝛼 + π‘₯ 2 +πœ† π‘šβˆ’ π‘₯ 1 𝑝 1 βˆ’ π‘₯ 2 𝑝 2
β„’= 1 𝛼 π‘₯ 1 𝛼 + π‘₯ 2 +πœ† π‘šβˆ’ π‘₯ 1 𝑝 1 βˆ’ π‘₯ 2 𝑝 2 FOC’s: 𝑑ℒ 𝑑 π‘₯ 1 = π‘₯ 1 π›Όβˆ’1 βˆ’πœ† 𝑝 1 =0 𝑑ℒ 𝑑 π‘₯ 2 =1βˆ’πœ† 𝑝 2 =0 𝑑ℒ π‘‘πœ† =π‘šβˆ’ π‘₯ 1 𝑝 1 βˆ’ π‘₯ 2 𝑝 2 =0 (1) (2) (3)

11 Question 3 Using (1) & (2): π‘₯ 1 π›Όβˆ’1 𝑝 1 =πœ†= 1 𝑝 2 π‘₯ 1 π›Όβˆ’1 = 𝑝 1 𝑝 β‡’ π‘₯ 1 = 𝑝 1 𝑝 π›Όβˆ’1 Plug into (3): π‘šβˆ’ 𝑝 1 𝑝 π›Όβˆ’1 𝑝 1 βˆ’ π‘₯ 2 𝑝 2 =0 π‘₯ 2 = π‘š 𝑝 2 βˆ’ 𝑝 1 𝑝 𝑝 1 𝑝 π›Όβˆ’1 π‘₯ 2 = π‘š 𝑝 2 βˆ’ 𝑝 1 𝑝 𝛼 π›Όβˆ’1

12 Question 4 For the utility function π‘ˆ π‘₯ = 1 𝛼 π‘₯ 1 𝛼 + π‘₯ 2 , derive the corresponding indirect utility function and the expenditure function.

13 Question 4 Indirect Utility Function: π‘ˆ π‘₯ = 1 𝛼 π‘₯ 1 𝛼 + π‘₯ 2
π‘ˆ π‘₯ = 1 𝛼 π‘₯ 1 𝛼 + π‘₯ 2 Indirect Utility Function: 𝑉 𝑝 1 , 𝑝 2 ,π‘š = 1 𝛼 𝑝 1 𝑝 π›Όβˆ’1 𝛼 + π‘š 𝑝 2 βˆ’ 𝑝 1 𝑝 𝛼 π›Όβˆ’1 𝑉 𝑝 1 , 𝑝 2 ,π‘š = 1 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 + π‘š 𝑝 2 βˆ’ 𝑝 1 𝑝 𝛼 π›Όβˆ’1 𝑉 𝑝 1 , 𝑝 2 ,π‘š = 1 𝛼 βˆ’1 𝑝 1 𝑝 𝛼 π›Όβˆ’1 + π‘š 𝑝 2 𝑉 𝑝 1 , 𝑝 2 ,π‘š = 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 + π‘š 𝑝 2 (1)

14 Question 4 β„’= π‘₯ 1 𝑝 1 + π‘₯ 2 𝑝 2 +πœ† 𝑒 βˆ’ 1 𝛼 π‘₯ 1 𝛼 βˆ’ π‘₯ 2 FOC’s:
Dual Problem – choose π‘₯ 1 and π‘₯ 2 to minimise expenditure, subject to a utility constraint. β„’= π‘₯ 1 𝑝 1 + π‘₯ 2 𝑝 2 +πœ† 𝑒 βˆ’ 1 𝛼 π‘₯ 1 𝛼 βˆ’ π‘₯ 2 FOC’s: 𝑑ℒ 𝑑 π‘₯ 1 = 𝑝 1 βˆ’πœ†π‘₯ 1 π›Όβˆ’1 =0 𝑑ℒ 𝑑 π‘₯ 2 = 𝑝 2 βˆ’πœ†=0 𝑑ℒ π‘‘πœ† = 𝑒 βˆ’ 1 𝛼 π‘₯ 1 𝛼 βˆ’ π‘₯ 2 =0 (1) (2) (3)

15 Question 4 Using (1) & (2): 𝑝 1 π‘₯ 1 π›Όβˆ’1 =πœ†= 𝑝 2 π‘₯ 1 π›Όβˆ’1 = 𝑝 1 𝑝 2
𝑝 1 π‘₯ 1 π›Όβˆ’1 =πœ†= 𝑝 2 π‘₯ 1 π›Όβˆ’1 = 𝑝 1 𝑝 2 π‘₯ 1 = 𝑝 1 𝑝 π›Όβˆ’1 Plug into (3): 𝑒 βˆ’ 1 𝛼 𝑝 1 𝑝 π›Όβˆ’1 𝛼 βˆ’ π‘₯ 2 =0 π‘₯ 2 = 𝑒 βˆ’ 1 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 (4) (5)

16 Question 4 𝐸 π‘₯, 𝑝 = π‘₯ 1 𝑝 1 + π‘₯ 2 𝑝 2 Expenditure Function:
𝐸 π‘₯, 𝑝 = π‘₯ 1 𝑝 1 + π‘₯ 2 𝑝 2 𝐸 𝑝 1 , 𝑝 2 , 𝑒 = 𝑝 1 𝑝 π›Όβˆ’1 𝑝 1 + 𝑒 βˆ’ 1 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 𝑝 2 𝐸 𝑝 1 , 𝑝 2 , 𝑒 = 𝑒 𝑝 𝑝 1 𝑝 π›Όβˆ’1 𝑝 1 βˆ’ 1 𝛼 𝑝 1 𝑝 2 𝑝 2 𝐸 𝑝 1 , 𝑝 2 , 𝑒 = 𝑒 𝑝 𝑝 1 𝑝 π›Όβˆ’1 (𝑝 1 𝛼) 𝛼 βˆ’ 𝑝 1 𝛼 𝐸 𝑝 1 , 𝑝 2 , 𝑒 = 𝑒 + 𝑝 1 π›Όβˆ’1 𝛼 𝑝 1 𝑝 π›Όβˆ’1 (6)

17 Question 4 Due to duality, the Indirect Utility Function and Expenditure Function are inverses of one another: 𝑉 𝑝 1 , 𝑝 2 ,π‘š = 𝑒 Therefore: 𝑒 = 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 + π‘š 𝑝 2 Rearrange to get: π‘š= 𝑒 𝑝 2 βˆ’ 𝑝 2 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 Or: 𝐸 𝑝 1 , 𝑝 2 , 𝑒 = 𝑒 𝑝 2 βˆ’ 𝑝 2 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 (7)

18 Question 4 Each method leads to slightly different equations… But they should be equivalent. 6 = 7 𝑒 𝑝 2 + 𝑝 1 π›Όβˆ’1 𝛼 𝑝 1 𝑝 π›Όβˆ’1 = 𝑒 𝑝 2 βˆ’ 𝑝 2 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 𝛼 π›Όβˆ’1 = 𝑒 𝑝 2 βˆ’ 𝑝 2 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 𝑝 1 𝑝 π›Όβˆ’1 = 𝑒 𝑝 2 βˆ’ 𝑝 1 1βˆ’π›Ό 𝛼 𝑝 1 𝑝 π›Όβˆ’1 = 𝑒 𝑝 2 + 𝑝 1 π›Όβˆ’1 𝛼 𝑝 1 𝑝 π›Όβˆ’1


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