ECOS2903 Final Exam Solution 11 June 2020
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ECOS2903 Final Exam Solution
11 June 2020
Q0 Please type this sentence at the beginning of your submitted answer:
“I declare that I am undertaking the exam in the prescribed exam conditions with no assistance from a third party or the use of prohibited resources. ”
Q1 (10 points) Find a matrix A that solves the following equation:
┌ 1 2 3 ┐ x ! 1(1)
2
3(3) ; x A = 1
3 |
Note: The answer is not unique. It’s sufficient to give only one example of A that satisfies the equation. Please clearly specify the entries of A and use calculation to justify why the equation holds.
Answer:
┌ 1 2 3 ┐ x ! 1(1) 2(2) 3(3) ; =┌ 6 12 18 ┐
A is a 3x1 vector
! a2(a1) ;
ì a3 |
that satisfies
6a1 + 12a2 + 18a3 = 1
Q2 (15 points) Sets
(a) Write down a subset of R that is closed but not bounded. (3 points) (b) Write down a subset of R2 that is bounded but not closed. (3 points) (c) Write down a subset of R3 that is closed and bounded. (3 points)
(d) Is it possible to write down a subset of R that is not bounded, not closed, and not
open? If so, give an example. If not, please prove why. (6 points)
You don’t need to justify your answer to (a), (b), or (c).
Answer (d) Yes, it is possible: (-o , 0) u [2, 3) is an example of such set.
However, it is not possible to write down a single interval that is unbounded, neither open nor closed. An unbounded interval looks like (-o , o ), (-o , a), (-o , a], [a, o ), or (a, o ). The firstone is both open and closed. The second and the last ones are open. The third and the fourth are closed.
In addition, the empty set is not a correct answer for (d). The empty set is bounded, closed, and open. Therefore, it is incorrect to say that the empty set is unbounded, not closed, or not open.
Q3 (20 points) Let D = {(α1 , α2 , α3 ) I α1 > 1, α2 > 2, α3 > 3}. Consider the utility function U : D → R s.t.
U (α1 , α2 , α3 ) = ln (α1 - 1) + 2 ln (α2 - 2) + 3 ln (α3 - 3) .
(a) Write down the gradient vector and the Hessian matrix of U. (10 points) (b) Is U concave? Prove your answer. (10 points)
Answer:
(a) The gradient vector of U is
α1 -1 2 |
α2 -2 3 |
α3 -3
The Hessian matrix of U is
;
|
! - H = (
ì
1 (α1 -1)2
0
0
0
2
-
(α2 -2)2
0
0
0
3
- (α3 -3)2
;
)
|
(b) The leading principal minors of H are
-6
IHI = (α1 - 1)2 (α2 - 2)2 (α3 - 3)2 < 0
│
1
-
(α1 -1)2
0
0 2 - (α2 -2)2 1 |
(α1 - 1)2 |
= 2
< 0
> 0
Therefore, H is negative definite. This proves that U is strictly concave.
Q4 (30 points) Let D = {(α1 , α2 ) I α1 > 1, α2 > 2}. Suppose that Gwen’s preference over (α1 , α2 ) - quantities of goods 1 and 2 - is described by the utility function U : D → R s.t.
U (α1 , α2 ) = (α1 - 1) (α2 - 2)2 .
Gwen has a total budget of M dollars to spend on goods 1 and 2. The unit price for good 1 is 夕1 and the unit price for good 2 is 夕2. Gwen’s goal is to maximise her utility subject to her budget constraint.
(a) Write down Gwen’s constrained maximisation problem and set up the Lagrangian function associated with it. (2 points)
(b) Find the solution (α1(*), α2(*)) to Gwen’s maximisation problem. Please first express α1(*)
and α2(*) as functions of 夕1 , 夕2 , M , and then calculate their values when 夕1 = 1, 夕2 = 2,
and M = 10 (rounded to 3 decimal places). (10 points)
(c) Check the sufficient conditions to verify that your solution in (b) is indeed a maximum point (and not a minimum or saddle point). (8 points)
(d) How does the maximised utility U* = U (α1(*), α2(*)) change with income M? How does
it change with 夕1 ? Please first express calculate their values when 夕1 = 1, 夕2 = 2, and M = 10 (rounded to 3 decimal places). (10 points)
Answer:
(a) Because U strictly increases with both α1 and α2 , Gwen’s budget constraint always binds at the optimum. This means that we can set up a maximisation problem with an equality constraint:
x1(m);x(a) (α1 - 1) (α2 - 2)2
5.≠ . 夕1 α 1 + 夕2 α2 = M
Set up the Lagrangian
x1;(m)x2(a) c (α1 , α2 , A)
5.≠. c (α1 , α2 , A) = (α1 - 1) (α2 - 2)2 - A (夕1 α 1 + 夕2 α2 - M)
(b) FOC:
[α1] (α2 - 2)2 = A夕1
[α2] 2 (α1 - 1) (α2 - 2) = A夕2
[A] 夕1 α 1 + 夕2 α2 = M
Take the ratio of the first two equations:
= 夕(夕)2(1)
α2 = 2 (α1 - 1) 夕(夕)2(1) + 2
Substitute this into the budget constraint:
夕1 α 1 + 夕2 α2 = M
夕1 α 1 + 夕2 ┌2 (α1 - 1) 夕(夕)2(1) + 2! = M
夕1 α 1 + 2 (α1 - 1) 夕1 + 2夕2 = M
3夕1 α 1 = M - 2夕2 + 2夕1
α* =
α2(*) = 2 (α1(*) - 1)夕(夕)2(1) + 2
= 2 ╱ M - 3夕1(2夕2)+ 2夕1 - 1、夕(夕)2(1) + 2 = 2 ╱ M - 3(2)夕(夕)1(2) - 夕1 、夕(夕)2(1) + 2
= + 2
3夕2
= 2M +3夕(2夕)2(2) - 2夕1
A* = (α2(*)夕1(-) 2)2
(2M - 4夕2 - 2夕1 )2
9夕1夕2(2)
When 夕1 = 1, 夕2 = 2, and M = 10,
α1(*) = s 2.667
α2(*) = s 3.667
A* = s 2.778
(c) Note that
ln U (α1 , α2 ) = ln ┌(α1 - 1) (α2 - 2)2 ┐= ln (α1 - 1) + 2 ln (α2 - 2)
It’s easy to verify that this is a strictly concave function with an approach similar to Q3. Therefore, U is an increasing transformation of a strictly concave function. Because we’re maximising an increasing transformation of a concave function subject to a linear (and
hence, quasi-convex) constraint, the solution to the FOC must solve the maximisation
problem.
(d) By the Envelope Theorem:
?U* ?M ?U*
?p1
When p1 = 1, p2 = 2, and M = 10,
从1(*) =
λ* =
Therefore,
= λ*
= -λ* 从1(*)
s 2.667
s 2.778
= s 2.778
?U* |
?p1 |
200 |
27 |
s -7.407
Q5 (25 points) Please maximise
f : ,(从, g) e R2 I 从 持 0, g 持 0、→ R s.t. f (从, g) = 85从 + 25g subject to the following constraints:
2从 + g < 3
从 + 2g < 3
(a) Find the solution (从* , g* ) to the maximisation problem. (12 points) (b) Which of the constraints are binding? (3 points)
(c) Check the sufficient conditions to verify that your solution in (a) is indeed a maximum point (and not a minimum or saddle point). (10 points)
Answer:
(a)
max
α←g←λ1 ←λ2
S.t. c (从,g, λ1 , λ2 ) =
c (从,g, λ1 , λ2 )
85从 + 25g - λ 1 (2从 + g - 3) - λ2 (从 + 2g - 3)
FOCs for 从 and
g:
[从] [g]
4从-
g-
= 2λ1 + λ2
= λ1 + 2λ2
Complimentary slackness conditions:
[λ1] λ 1 2 0, 2从 + g < 3, λ1 (2从 + g - 3) = 0
[λ2] λ2 2 0, 从 + 2g < 3, λ1 (从 + 2g - 3) = 0
Case 1: λ1 = λ2 = 0.
Then, the FOCs imply that 4① - = 0 and g- = 0, both of which are infeasible. Therefore, this case does not produce any solution candidate. 。
Case 2: λ1 持 0, λ2 = 0.
[λ1] 2① + g = 3
[①]&[g] 4①- 2 = 2g 2 ÷ ① = 4g
The solution to these two equations is ①
λ 1 = 4① - - g-
and
① + 2g < 3
Both inequalities hold when ① . √
Case 3: λ1 = 0, λ2 持 0.
[λ2] ① + 2g = 3
[①]&[g] 8①- = g- ÷ ① = 64g
The solution to these two equations is ① = 1(3)1(2) and g = 22(1). We still need to verify that
λ2 = g- - 4①- 持 0
and
2① + g < 3
Unfortunately, the second inequality does not hold because
2① + g = 1(6)1(4) + 22(1) = 22(129) s 5.86 持 3.
Therefore, ① = 11(32) and g =
Case 4: λ1 持 0, λ2 持 0
is not a solution candidate. 。
[λ1] ÷ 2① + g = 3
[λ2] ÷ ① + 2g = 3
The solution to these equations is
① = g = 1.
Plug this solution back into the FOCs:
[①] 4 = 2λ1 + λ2
[g] 1 = λ1 + 2λ2
This yield λ1 it must be positive. Therefore, ① = g = 1 is not a solution candidate. 。
Conclusion: The solution is x = |
4 3 |
and |
y = |
1 3 |
(Case 2). |
(b) From part (a) we know that 1 > 0 and 2 = 0. Therefore, this means that the first constraint (2x + y < 3) is binding. The second constraint (x + 2y < 3) is not binding. (c) The Hessian matrix off is
H =
-2x-
0
!
When x and y are non-negative, IHI 2 0, -2x- < 0, - is concave. Because we’re maximising a concave function over linear (and hence, quasi- convex) constraints, the solution to FOCs indeed solves the maximisation problem.
2023-06-16