ECON201. Intermediate Microeconomics
Oleg Rubanov
Problem Set 1
Problem 1
Come up with income, prices and a tax or subsidy that are described by the following
graph. You need to provide exact numbers. Assume that the per unit price of good 2 is
the same for any quantity that you buy.
good 2, x2
36
18
0
12
36
good 1, x1
Solution
Notice that there are two parts of the budget line. The slope of the left part of the budget
0−18
line is 18−36
= − 32 . The slope of the right part of the budget line is 36−12
= − 43 . So, the
12
p1
price ratio p2 is 32 to the left from the kink and is 34 to the right from the kink.
Then the prices of good 1 are p1 for x1 ≤ 12 and p′1 = p1 − t for x1 > 12 (the price
of extra units after x1 = 12, not all units). Let p2 = 1, then p1 = 32 and p′1 = 34 . The
(quantity) overconsumption subsidy is s = p1 − p′1 = 34 (the equivalent ad valorem subsidy
p′
is 1 − p11 = 50%).
How much money does the agent have? She should be able to buy, for example, 12 units
of good 1 and 18 units of good 2, so her income is m = 12 · p1 + 18 · p2 = 12 · 23 + 18 · 1 = 36.
You can check yourself that this is indeed the correct solution. What is the maximum
amount of good 1 the consumer can buy? She will buy 12 units at the price of 23 and for
1
the remaining 36 − 12 · 32 = 18 dollars she can buy 18
3 = 24 more units of good 1, bringing
4
the total number of units of good 1 to 12 + 24 = 36. The maximum number of units of
good 2 is naturally pm2 = 36
= 36.
1
Any other numbers proportional to (m, p1 , p2 , s) = (36, 23 , 1, 34 ) are also a solution. For
example, (m, p1 , p2 , t) = (1440, 60, 40, 30) is also a solution of the problem.
Problem 2
Consider the preferences on the set X = {k, l, m} given by the following table (each cell
indicates whether a ⪰ b (TRUE, denoted by T ) or not (FALSE, denoted by F ). However,
you do not know the entire table (some cells are empty).
a\b k
k
T
a)
l
F
m
T
l
T
T
T
m
F
F
T
b)
a\b k
k
F
l
F
m
F
l
F
T
T
m
F
T
F
a\b k
k
T
c)
l
T
m
F
l
T
T
F
m
F
F
T
For EACH of the three preferences given above, check if it satisfies EACH of the
following properties (so, you need to answer nine questions):
1. Complete?
2. Reflexive?
3. Transitive?
In each case, you either need to prove your statement, or provide an example where this
property is not satisfied.
Solution
a) 1. Completeness. Yes, we have T on the diagonal and for the remaining 3 pairs
(k, l), (k, m) and (l, m), we prefer at least one of the elements to the other in the
pair: k ⪰ l, m ⪰ k and m ⪰ l.
2. Reflexivity. Yes, we have T on the diagonal. Alternatively, we could use the
previous question, because all complete preferences are reflexive.
3. Transitivity. Yes, because if X ⪰ Y and Y ⪰ Z, then there are the following
possibilities:
ˆ If X = m, then automatically X ⪰ Z, because we prefer m to all other elements.
ˆ If X = l, then since we only prefer l to itself and X ⪰ Y , it must be that Y = l.
The same way, since Y ⪰ Z, it must be that Z = l. But then X ⪰ Z, because
X = Z = l and l ⪰ l.
ˆ If X = k, then since X ⪰ Y , then Y = k or Y = l. If Y ∈ {k, l} and Y ⪰ Z,
then Z ̸= m. But for all Z ̸= m, k ⪰ Z. So, X ⪰ Z.
2
In all cases, we proved that for all X, Y, Z, if X ⪰ Y and Y ⪰ Z, then X ⪰ Z.
Thus, transitivity holds.
b) 1. Completeness. No, because if X = Y = k, then neither X ⪰ Y , nor Y ⪰ X.
2. Reflexivity. No, because if X = Y = k, then neither X ⪰ Y , nor Y ⪰ X.
3. Transitivity. No, for example, if X = Z = m and Y = l, then X ⪰ Y and Y ⪰ Z,
but X ̸⪰ Z, because m ̸⪰ m.
c) 1. Completeness. No, because if X = k and Y = m, then neither X ⪰ Y , nor
Y ⪰ X.
2. Reflexivity. Yes, because we have only T on the diagonal.
3. Transitivity. Yes, because if X ⪰ Y and Y ⪰ Z, then there are the following
possibilities:
ˆ If X = k or X = l, then automatically Y ∈ {k, l} and therefore Z ∈ {k, l}, so
X ⪰ Z, because both k and l are weakly preferred to both k and l.
ˆ If X = m, then since X ⪰ Y , it must be that Y = m. The same way, Z = m,
so X ⪰ Z, because m ⪰ m.
In all cases, we proved that for all X, Y, Z, if X ⪰ Y and Y ⪰ Z, then X ⪰ Z.
Thus, transitivity holds.
Problem 3
There are other properties of binary relations that we did not cover in the lecture.
1. Irreflexivity. For any x ∈ X : NOT(x ⪰ x).
2. Symmetric. For any x, y ∈ X : if x ⪰ y then y ⪰ x.
3. Antisymmetric. For any x, y ∈ X , if (y ⪰ x)AND(x ⪰ y), THEN x = y (meaning
that x and y are the same element).
Also, answer the following questions:
a) Can a preference relation be both irreflexive and transitive?
b) Can a preference relation be both symmetric and antisymmetric?
c) Can a preference relation be both complete and not antisymmetric?
Solution
a) Yes, for example:
a\b
k
l
k
F
F
l
F
F
3
b) Yes, for example:
a\b
k
l
k
F
F
l
F
F
c) Yes, for example:
a\b
k
l
k
T
T
l
T
T
4