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:
𝑥𝑃𝑦 & 𝑦𝐼𝑧⇒𝑥𝑃𝑧
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 𝑥𝑃𝑦 = 𝑥𝑅𝑦 & 𝑛𝑜𝑡(𝑦𝑅𝑥) 𝑦𝐼𝑧 = 𝑦𝑅𝑧 & 𝑧𝑅𝑦
Therefore, 𝑥𝑅𝑦 & 𝑦𝑅𝑧 ⇒ 𝑥𝑅𝑧 by transitivity of R.
Need to show that the opposite is not true, 𝑛𝑜𝑡(𝑧𝑅𝑥).
Suppose 𝑧𝑅𝑥 was true. Then y𝑅𝑧 & 𝑧𝑅𝑥 ⇒𝑦𝑅𝑥 by transitivity of R, which contradicts 𝑛𝑜𝑡(𝑦𝑅𝑥).
Thus, we have: 𝑥𝑅𝑧 & 𝑛𝑜𝑡(𝑧𝑅𝑥) = 𝑥𝑃𝑧

5. Question 2
Carefully illustrate in diagrams indifference curves for the following utility functions:
𝒊 𝑈 𝑥 = 2 𝑥 𝑥 2
𝒊𝒊 𝑈 𝑥 =𝑚𝑖𝑛 𝑥 1 ,2 𝑥 2
𝒊𝒊𝒊 𝑈 𝑥 = 𝛼 𝑥 1 −𝑟 𝑥 2 −𝑟 − 1 𝑟

6. Question 2i) 𝑈 𝑥 = 2 𝑥 1 + 𝑥 2 2 𝑥 1 =𝑈− 𝑥 2 𝑥 1 = 𝑈− 𝑥 2 2
Perfect Substitutes
𝑈 𝑥 = 2 𝑥 𝑥 2
2 𝑥 1 =𝑈− 𝑥 2
𝑥 1 = 𝑈− 𝑥 2 2

7. Question 2ii) 𝑈 𝑥 =𝑚𝑖𝑛 𝑥 1 ,2 𝑥 2 If 𝑥 1 =𝑈 𝑥 ⇒ 𝑥 2 ≥ 𝑈 𝑥 2
Perfect Complements
𝑈 𝑥 =𝑚𝑖𝑛 𝑥 1 ,2 𝑥 2
If 𝑥 1 =𝑈 𝑥 ⇒ 𝑥 2 ≥ 𝑈 𝑥 2
If 𝑥 2 = 𝑈 𝑥 2 ⇒ 𝑥 1 ≥𝑈(𝑥)

8. Question 2iii) 𝑈 𝑥 = 0.5 𝛼 𝑥 1 −𝑟 + 0.5 𝑥 2 −𝑟 − 1 𝑟
𝑈 𝑥 = 𝛼 𝑥 1 −𝑟 𝑥 2 −𝑟 − 1 𝑟
𝑈 −𝑟 = 0.5 𝛼 𝑥 1 −𝑟 𝑥 2 −𝑟
𝑈 −𝑟 − 0.5 𝑥 2 −𝑟 = 0.5 𝛼 𝑥 1 −𝑟
2 𝑈 −𝑟 − 𝑥 2 −𝑟 = 𝛼 𝑥 1 −𝑟
2 𝑈 −𝑟 − 𝑥 2 −𝑟 − 1 𝑟 𝛼 = 𝑥 1
2 𝑈 −𝑟 − 𝛼𝑥 1 −𝑟 − 1 𝑟 = 𝑥 2

9. Question 3
Suppose the consumer’s utility function is 𝑈 𝑥 = 1 𝛼 𝑥 1 𝛼 + 𝑥 2 , where 0<𝛼<1, she has income 𝑚 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: ℒ= 1 𝛼 𝑥 1 𝛼 + 𝑥 2 +𝜆 𝑚− 𝑥 1 𝑝 1 − 𝑥 2 𝑝 2
ℒ= 1 𝛼 𝑥 1 𝛼 + 𝑥 2 +𝜆 𝑚− 𝑥 1 𝑝 1 − 𝑥 2 𝑝 2
FOC’s:
𝑑ℒ 𝑑 𝑥 1 = 𝑥 1 𝛼−1 −𝜆 𝑝 1 =0
𝑑ℒ 𝑑 𝑥 2 =1−𝜆 𝑝 2 =0
𝑑ℒ 𝑑𝜆 =𝑚− 𝑥 1 𝑝 1 − 𝑥 2 𝑝 2 =0
(1)
(2)
(3)

11. Question 3
Using (1) & (2):
𝑥 1 𝛼−1 𝑝 1 =𝜆= 1 𝑝 2
𝑥 1 𝛼−1 = 𝑝 1 𝑝 ⇒ 𝑥 1 = 𝑝 1 𝑝 𝛼−1
Plug into (3):
𝑚− 𝑝 1 𝑝 𝛼−1 𝑝 1 − 𝑥 2 𝑝 2 =0
𝑥 2 = 𝑚 𝑝 2 − 𝑝 1 𝑝 𝑝 1 𝑝 𝛼−1
𝑥 2 = 𝑚 𝑝 2 − 𝑝 1 𝑝 𝛼 𝛼−1

12. Question 4
For the utility function 𝑈 𝑥 = 1 𝛼 𝑥 1 𝛼 + 𝑥 2 , derive the corresponding indirect utility function and the expenditure function.

13. Question 4 Indirect Utility Function: 𝑈 𝑥 = 1 𝛼 𝑥 1 𝛼 + 𝑥 2
𝑈 𝑥 = 1 𝛼 𝑥 1 𝛼 + 𝑥 2
Indirect Utility Function:
𝑉 𝑝 1 , 𝑝 2 ,𝑚 = 1 𝛼 𝑝 1 𝑝 𝛼−1 𝛼 + 𝑚 𝑝 2 − 𝑝 1 𝑝 𝛼 𝛼−1
𝑉 𝑝 1 , 𝑝 2 ,𝑚 = 1 𝛼 𝑝 1 𝑝 𝛼 𝛼−1 + 𝑚 𝑝 2 − 𝑝 1 𝑝 𝛼 𝛼−1
𝑉 𝑝 1 , 𝑝 2 ,𝑚 = 1 𝛼 −1 𝑝 1 𝑝 𝛼 𝛼−1 + 𝑚 𝑝 2
𝑉 𝑝 1 , 𝑝 2 ,𝑚 = 1−𝛼 𝛼 𝑝 1 𝑝 𝛼 𝛼−1 + 𝑚 𝑝 2
(1)

14. Question 4 ℒ= 𝑥 1 𝑝 1 + 𝑥 2 𝑝 2 +𝜆 𝑢 − 1 𝛼 𝑥 1 𝛼 − 𝑥 2 FOC’s:
Dual Problem – choose 𝑥 1 and 𝑥 2 to minimise expenditure, subject to a utility constraint.
ℒ= 𝑥 1 𝑝 1 + 𝑥 2 𝑝 2 +𝜆 𝑢 − 1 𝛼 𝑥 1 𝛼 − 𝑥 2
FOC’s:
𝑑ℒ 𝑑 𝑥 1 = 𝑝 1 −𝜆𝑥 1 𝛼−1 =0
𝑑ℒ 𝑑 𝑥 2 = 𝑝 2 −𝜆=0
𝑑ℒ 𝑑𝜆 = 𝑢 − 1 𝛼 𝑥 1 𝛼 − 𝑥 2 =0
(1)
(2)
(3)

15. Question 4 Using (1) & (2): 𝑝 1 𝑥 1 𝛼−1 =𝜆= 𝑝 2 𝑥 1 𝛼−1 = 𝑝 1 𝑝 2
𝑝 1 𝑥 1 𝛼−1 =𝜆= 𝑝 2
𝑥 1 𝛼−1 = 𝑝 1 𝑝 2
𝑥 1 = 𝑝 1 𝑝 𝛼−1
Plug into (3):
𝑢 − 1 𝛼 𝑝 1 𝑝 𝛼−1 𝛼 − 𝑥 2 =0
𝑥 2 = 𝑢 − 1 𝛼 𝑝 1 𝑝 𝛼 𝛼−1
(4)
(5)

16. Question 4 𝐸 𝑥, 𝑝 = 𝑥 1 𝑝 1 + 𝑥 2 𝑝 2 Expenditure Function:
𝐸 𝑥, 𝑝 = 𝑥 1 𝑝 1 + 𝑥 2 𝑝 2
𝐸 𝑝 1 , 𝑝 2 , 𝑢 = 𝑝 1 𝑝 𝛼−1 𝑝 1 + 𝑢 − 1 𝛼 𝑝 1 𝑝 𝛼 𝛼−1 𝑝 2
𝐸 𝑝 1 , 𝑝 2 , 𝑢 = 𝑢 𝑝 𝑝 1 𝑝 𝛼−1 𝑝 1 − 1 𝛼 𝑝 1 𝑝 2 𝑝 2
𝐸 𝑝 1 , 𝑝 2 , 𝑢 = 𝑢 𝑝 𝑝 1 𝑝 𝛼−1 (𝑝 1 𝛼) 𝛼 − 𝑝 1 𝛼
𝐸 𝑝 1 , 𝑝 2 , 𝑢 = 𝑢 + 𝑝 1 𝛼−1 𝛼 𝑝 1 𝑝 𝛼−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 ,𝑚 = 𝑢
Therefore:
𝑢 = 1−𝛼 𝛼 𝑝 1 𝑝 𝛼 𝛼−1 + 𝑚 𝑝 2
Rearrange to get:
𝑚= 𝑢 𝑝 2 − 𝑝 2 1−𝛼 𝛼 𝑝 1 𝑝 𝛼 𝛼−1
Or:
𝐸 𝑝 1 , 𝑝 2 , 𝑢 = 𝑢 𝑝 2 − 𝑝 2 1−𝛼 𝛼 𝑝 1 𝑝 𝛼 𝛼−1
(7)

18. Question 4
Each method leads to slightly different equations… But they should be equivalent.
6 = 7
𝑢 𝑝 2 + 𝑝 1 𝛼−1 𝛼 𝑝 1 𝑝 𝛼−1 = 𝑢 𝑝 2 − 𝑝 2 1−𝛼 𝛼 𝑝 1 𝑝 𝛼 𝛼−1
= 𝑢 𝑝 2 − 𝑝 2 1−𝛼 𝛼 𝑝 1 𝑝 𝑝 1 𝑝 𝛼−1
= 𝑢 𝑝 2 − 𝑝 1 1−𝛼 𝛼 𝑝 1 𝑝 𝛼−1
= 𝑢 𝑝 2 + 𝑝 1 𝛼−1 𝛼 𝑝 1 𝑝 𝛼−1