ECON30020 Mathematical Economics Tutorial 3. Review of Fundamentals
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ECON30020 Mathematical Economics
Tutorial 3. Review of Fundamentals
Pages for both the second and third editions of HLMRS are given for questions that are adapted from that book.
Question 1. Convex combination (HLMRS 2.4 q.3, p.58 or 56)
Find the convex combinations of the following pairs of points and, when possible, show them graphically:
(a) -2 and 4
(b) (-1, 1) and (3, 4)
(c) (-2, 0, 1) and (1, -2, 2)
Answer: Throughout, assume that λ e [0, 1].
(a) the convex combinations of -2 and 4:
λ = λ(-2) + (1 - λ)4 = 4 - 6λ
(b) the convex combinations of (-1, 1) and (3, 4):
λ = λ(-1, 1) + (1 - λ)(3, 4) = (-λ + (1 - λ)3, λ + (1 - λ)4) = (3 - 4λ, 4 - 3λ)
(c) We calculate this in a similar fashion to before:
λ = λ(-2, 0, 1) + (1 - λ)(1, -2, 2) = (-2λ, 0, λ) + (1 - λ, -2 + 2λ, 2 - 2λ) = (1 - 3λ, 2λ - 2, 2 - λ)
Question 2. Convex sets
Are the following sets convex? If yes, provide an illustration (formal proof is not required). If no, prove it!
(a) A budget set BS = {(x1 , x2 ) e R : p1x1 + p2x2 < Y }, where p1 , p2 > 0 are prices and
Y > 0 is income.
(b) A budget line BL = {(x1 , x2 ) e R : p1x1 + p2x2 = Y }, where p1 , p2 > 0 are prices and
Y > 0 is income.
Answer:
(a) Yes, the set is convex. If you fix p1 , p2 and Y , you will find that the set BS is a triangle (simply draw a graph with x1 on the horizontal axis and x2 on the vertical). You can convince yourself that the line connecting any two points in the triangle also lies within the triangle.
x2
Y/p2
the budget line: p1x1 + p2x2 = Y
Y/p1
Aside: Proving convexity of the budget set
The question does not ask for this, but it may be informative if you wish to know how to formally show that the budget set is convex. Suppose that (x, x) and (x, x) e BS. This means that
p1x + p2x < Y and p1x + p2x < Y
or, with some rearrangement,
λp1x + λp2x < λY and (1 - λ)p1x + (1 - λ)p2x < (1 - λ)Y
When added together, these inequalities imply
λp1x + λp2x + (1 - λ)p1x + (1 - λ)p2x < λY + (1 - λ)Y =÷ p1 (λx + (1 - λ)x) + p2 (λx + (1 - λ)x) < Y
Why is this important? Consider the convex combination of the two points for some λ e [0, 1]:
λ = λ(x, x) + (1 - λ)(x, x)
= (λx + (1 - λ)x, λx + (1 - λ)x)
Note that this matches what we have above for the condition in BS. The only thing remaining now is to show that the convex combination λ is an element of R . This follows from the fact that (x, x) and (x, x) e R, from which we have that
(λx, λx), ((1 - λ)x , (1 - λ)x) e R since λ, (1 - λ) e R+
and adding them together yields λ e R since the sum of two non-negative numbers is also non-negative and hence is in R+ . Therefore, the set is a convex set. The steps to show that the budget line below is a convex set are identical, except that we just replace the weak inequality with an equality sign instead.
(b) Yes, the set is convex. The budget line, as the name suggests, is a straight line. The line
connecting any two points on the budget line is also part of the budget line itself.
Question 3. Concave functions (HLMRS 2.4 q.10 and 11, p.60 or 58)
Consider the function y = x1/2, x > 0.
(a) Using the points x\ = 1, x\\ = 9 and λ = 5/8, illustrate definition 2.28/2.25 of a concave
function for the above function. Use a graph in your answer.
(b) Show that the function is strictly concave according to the definition 2.28/2.25.
Answer:
(a) The convex combination is
λ = λx\ + (1 - λ)x\\ = x 1 + (1 - ) x 9 = x 1 + x 9 = 4
The value of the function at this point is
f(λ ) = ^4 = 2
The convex combination of values is given by
λf(北\ ) + (1 - λ)f(北\\ ) = ^1 + ^9 = = 1.75
Thus, we have
f(λ ) = 2 > λf(北\ ) + (1 - λ)f(北\\ ) = 1.75,
as must be the case for a concave function.
(b) In order to show that the function is strictly concave, we need to verify the following for
any 北\ 北\\ and λ e (0, 1) :
f(λ ) > λf (北\ ) + (1 - λ)f(北\\ ).
Thus, we have
(λ北\ + (1 - λ)北\\ )0.5 > λ ^北\ + (1 - λ)^北\\ .
Squaring both sides (as both sides are positive) and applying (a + b)2 = a2 + 2ab + b2 , we have:
λ北\ + (1 - λ)北\\ > λ2北\ + 2λ(1 - λ)^北\ ^北\\ + (1 - λ)2北\\ .
Rearranging and simplifying, we get
λ(1 - λ)北\ + λ(1 - λ)北\\ > 2λ(1 - λ)^北\ ^北\\ .
As both λ and (1 - λ) are positive, this simplifies to
北\ + 北\\ > 2^北\ ^北\\ , or
北\ - 2^北\ ^北\\ + 北\\ = (^北\ - ^北\\ )2 > 0
which is always true for distinct 北\ and 北\\ . Therefore, f(北) = ^北 is strictly concave.
Question 4. Properties of Concave and Convex Functions
Let f : R → R be a concave function and g : R → R be a convex function. For any x, y, z satisfying x < y < z, show that
and |
f (y) - f (x) y - x g(y) - g(x) y - x |
>
< |
f (z) - f (x) z - x g(z) - g(x) z - x |
>
< |
f (z) - f (y) z - y .
g(z) - g(y) z - y . |
(1)
(2) |
Answer: Note that y is a convex combination of x and z, so we can write, for some λ e [0, 1],
y = λx + (1 - λ)z =÷ y - x = λx + (1 - λ)z - x = (1 - λ)(z - x).
By concavity of f , we have:
f (y) > λf (x)+(1-λ)f (z) =÷ f (y)-f (x) > λf (x)+(1-λ)f (z)-f (x) = (1-λ)(f (z)-f (x)).
Thus,
f (y) - f (x) (1 - λ)(f (z) - f (x)) f (z) - f (x)
> =
which is the first part of (1).
To show the second part of (1), note that
z - y = z - λx - (1 - λ)z = λ(z - x).
By concavity of f , we have
f (y) > λf (x) + (1 - λ)f (z), or f (z) - f (y) < λ(f (z) - f (x)).
Therefore,
f (z) - f (y) λ(f (z) - f (x)) f (z) - f (x)
< =
as required.
Geometrically, this is a relationship between the slopes of secant lines. You can think of it as ”the average slope of a concave function is decreasing when you are moving to the right” .
f(z) f(y)
f(x)
slope =
x y z
For the second inequality: y is a convex combination of x and z, so
y - x = (1 - λ)(z - x) and z - y = λ(z - x). (3)
By convexity of g, we have
g(y) < λg(x) + (1 - λ)g(z).
Rearranging this inequality in two different ways, we obtain
g(y) - g(x) < (1 - λ)(g(z) - g(x)) and g(z) - g(y) > λ(g(z) - g(x)).
Dividing the first by y - x and the second by z - y and making use of (3) yields
g(y) - g(x) g(z) - g(x) g(z) - g(y)
< <
which proves the result for g .
Question 5. Power Set
For any set Y with cardinality n, i.e. n = |Y |, show that the power set 伫(Y)—the set of all possible subsets of Y—has 2n elements if the empty set 0 is included. (In the lecture, the cardinality of 伫(Y) was given as 2n - 1, which excludes the empty set.)
Answer: Label the elements of Y by i with i e {1, . . . , n}.
For each i, let ai e {0, 1} be the binary variable that takes value 1 if i is included in a subset, and 0 if it is excluded. Thus, any subset of Y can be described as a list of ai’s of length n, {a1 , . . . , an}. For example, subset {1, 1, 1, 0, . . . , 0} consists of the first three elements of Y .
Since each ai can take one of the two values, there are 2n possible subsets. Note that the set {0, . . . , 0} is the empty set and the set {1, . . . , 1} is the set Y itself.
2022-06-18