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Examiners’ commentaries 2015

Examiners’ commentaries 2015 EC3120 Mathematical economics Important note This commentary reflects the examination and assessment arrangements for this course in the academic year 2014–15. The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE).

Information about the subject guide and the Essential reading references Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2014). You should always attempt to use the most recent edition of any Essential reading textbook, even if the commentary and/or online reading list and/or subject guide refer to an earlier edition. If different editions of Essential reading are listed, please check the VLE for reading supplements – if none are available, please use the contents list and index of the new edition to find the relevant section.

General remarks Learning outcomes At the end of this course and having completed the essential reading and activities you should be able to: •

use and explain the underlying principles, terminology, methods, techniques and conventions used in the subject

•

solve economic problems using the mathematical methods described in the subject.

Planning your time in the examination The examination is in two sections: A and B. Section A contains questions that require minimum or no manipulation with respect to what you can find in the essential readings. A section A question, for example, may ask you to provide a formal definition, a short proof, or to derive a simple implication of established results. Section B contains problems which require the application of mathematical techniques which you will have acquired by having completed the essential readings and activities. Section A contains 5 questions and you have to answer them all. In Section B you have to answer 3 out of 5 questions. Section A carries 40% of the final mark (8 marks per question); Section B carries 60% of the final mark (20 marks per question). Ideally you should plan to spend not much more than one hour on Section A and almost two hours on Section B. This not only takes into account that Section B carries more weight, but also allows

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EC3120 Mathematical economics

for the fact that, while you have no choice among section A questions, you will have to read through and choose among the problems in section B.

What are the examiners looking for? Examiners need to make sure that you can solve the economic problems posed by using the mathematical methods that you have acquired. Also, they need to make sure that you can explain the underlying principles and that you can correctly use the terminology and the conventions used in mathematical economics. Getting the right numerical answer is neither sufficient nor necessary in order to obtain a very good mark. It is not sufficient because sometimes an answer hits the right number but is not sufficiently clear on how it got there. It is not strictly necessary because the examiners may look favourably at those scripts where although a clear and correct procedure has been followed to address a question, a small computation mistake has resulted in the wrong numerical answer. Computational mistakes are usually penalised, but typically not very heavily. For a full (100%) mark, on the other hand, the examiners require both a correct numerical answer and clarity of exposition as far as the methodologies used are concerned.

Common mistakes made and how to correct for these A common mistake is a less than careful time planning which results in candidates answering fewer questions than strictly required. A missed answer is marked at 0, with a strong impact on the final mark. Equally, you should not waste time in answering more than the strictly needed number of questions because you will not receive extra marks for attempting more questions and you will instead reduce the time that you have available to complete and add detail to the rest of your answers. Note that calculators are now allowed for this examination.

Examination revision strategy Many candidates are disappointed to find that their examination performance is poorer than they expected. This may be due to a number of reasons. The Examiners’ commentaries suggest ways of addressing common problems and improving your performance. One particular failing is ‘question spotting’, that is, confining your examination preparation to a few questions and/or topics which have come up in past papers for the course. This can have serious consequences. We recognise that candidates may not cover all topics in the syllabus in the same depth, but you need to be aware that examiners are free to set questions on any aspect of the syllabus. This means that you need to study enough of the syllabus to enable you to answer the required number of examination questions. The syllabus can be found in the Course information sheet in the section of the VLE dedicated to each course. You should read the syllabus carefully and ensure that you cover sufficient material in preparation for the examination. Examiners will vary the topics and questions from year to year and may well set questions that have not appeared in past papers. Examination papers may legitimately include questions on any topic in the syllabus. So, although past papers can be helpful during your revision, you cannot assume that topics or specific questions that have come up in past examinations will occur again.

2

Examiners’ commentaries 2015

If you rely on a question-spotting strategy, it is likely you will find yourself in difficulties when you sit the examination. We strongly advise you not to adopt this strategy.

3

EC3120 Mathematical economics

Examiners’ commentaries 2015 EC3120 Mathematical economics Important note This commentary reflects the examination and assessment arrangements for this course in the academic year 2014–15. The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE).

Information about the subject guide and the Essential reading references Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2014). You should always attempt to use the most recent edition of any Essential reading textbook, even if the commentary and/or online reading list and/or subject guide refer to an earlier edition. If different editions of Essential reading are listed, please check the VLE for reading supplements – if none are available, please use the contents list and index of the new edition to find the relevant section.

Comments on specific questions – Zone A Candidates should answer EIGHT of the following TEN questions: all FIVE from Section A (8 marks each) and any THREE from Section B (20 marks each). Candidates are strongly advised to divide their time accordingly. Workings should be submitted for all questions requiring calculations. Any necessary assumptions introduced in answering a question are to be stated. Section A Answer all five questions from this section. Question 1 State the envelope theorem and prove it for the simple case of an unconstrained maximisation problem. Approaching the question The envelope theorem can be stated as: ∂ d f (x(b), y(b); b) = L(x(b), y (b), λ(b); b). ∂b db We can prove this for the simple case of an unconstrainted problem. Let f (x; a) be a continuous function of x ∈ Rn and the scalar a. For any a consider the problem of finding max f (x; a). Let x∗ (a) be the maximiser which we assume is differentiable with respect to a. We can show that: d ∂ f (x∗ (a); a). f (x∗ (a); a) = ∂a da

4

Examiners’ commentaries 2015

Apply the chain rule: X ∂f ∂x∗ ∂f ∗ d (x∗ (a); a) i (a) + f (x∗ (a); a) = (x (a); a). da ∂xi ∂a ∂a i Given that (∂f /∂xi )(x∗ (a); a) = 0 for all i by the first-order conditions, this gives: ∂ d f (x∗ (a); a) = f (x∗ (a); a). da ∂a Intuitively, when we are already at a maximum, changing slightly the parameters of the problem or the constraints does not affect the optimal solution.

Question 2 Consider the firm’s cost minimisation problem. (a) Define the cost function. (b) State Shephard’s Lemma for the cost function. (c) Compute conditional factor demands and cost function for a Cobb–Douglas production function f (K, L) = K α Lβ with α + β = 1, and verify that the cost function is homogeneous of degree one in input prices.

Approaching the question (a) The cost function is the value function for the firm’s cost minimisation problem. The cost function c(w, r, q) is the minimum cost of producing output q: c(w, r, q ) = rK(r, w, q ) + wL(r, w, q) where K (r, w, q ) and L(r, w, q ) are conditional factor demands. (b) Shepherd’s Lemma for the cost function states that: ∂c(r, w, q) ∂r

=

K (r, w, q)

∂c(r, w, q) ∂w

=

L(r, w, q ).

(c) The cost minimisation problem for the firm is as follows: min rK + wL s.t. K α Lβ = q. L,K

This can be restated as: max −rK − wL s.t. K α Lβ = q. L,K

The Lagrangian is: L = −rK − wL + λ(K α Lβ − q). The first-order conditions give: r λαK α−1 Lβ αL = = w λβK α Lβ−1 βK

⇒L=

βr K. αw

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EC3120 Mathematical economics

Substituting into the constraint: β βr K αw β  βr K α+β K αw β  βr K K αw Kα



=

q

=

q

=

q

⇒K =

βr K= αw

L = Hence: c(w, r, q) = r



αw βr

β





βr αw

αw K βr 1−β

β



α

q+w

βr αw

q

q=



βr αw

α

q.

q.

Check that this is homogeneous of degree one in input prices: c(sw, sr, q) =

= =

   βsr α αsw β q + sw q αsw βsr β α   αw βr sr q + sw q βr αw

sr



sc(w, r, q ).

Question 3 Answer all parts of this question. (a) Define a quadratic form. (b) When do we call a symmetric matrix positive definite?   1 2 (c) Check the definiteness of A = . 3 4 Approaching the question (a) A quadratic form is a real-valued function: Q(x1 , x2 , . . . , xn ) =

X

aij xi xj

i leqj

or, equivalently: Q(x) = xT Ax where A is a symmetric n × n matrix.

(b) We call A positive definite when xT Ax > 0 for all x 6= 0 in Rn . Equivalently, we can say that a matrix A is positive definite when all its n leading principal minors are strictly positive. (c) Since |A1 | = 1 and |A2 | = 4 − 6 = −1, the matrix is indefinite.

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Examiners’ commentaries 2015

Question 4 Solve the following ordinary differential equations: (a) y ′′ + 4y ′ − 5y = 0, where y(0) = 1 and y ′ (0) = 0. (b) y ′′ + 4y ′ + 4y = 0, where y(0) = 1 and y ′ (0) = 0. Approaching the question (a) The auxiliary equation is: 0 = r2 + 4r − 5 = (r + 5)(r − 1) so the general solution is:

y(t) = ae−5t + bet .

The initial conditions give the equations a + b = 1 and −5a + b = 0. Subtracting these equations gives 6a = 1, i.e. a = 1/6, and the first equation then yields b = 5/6. (b) The auxiliary equation is: 0 = r2 + 4r + 4 = (r + 2)2 . Thus the general solution, given coincident roots, is y(t) = (a + bt)e−2t . Setting t = 0 gives a = 1, i.e. y(t) = (1 + bt)e−2t . Hence: y′ (t) = (−2(1 + bt) + b)e−2t . Then y′ (0) = 0 implies b = 2.

Question 5 Consider the control problem max

Z

∞

0

  1 e−rt x(t) − u2 (t) dt 2

subject to x˙ = u − x and x(0) = x0 , where x˙ denotes the time derivative. (a) Set up the Hamiltonian and state the necessary conditions for a maximum. (b) Eliminate u(t) to obtain a system of first order linear differential equations in x and the co-state variable. Compute the family of solutions to this system explicitly. (c) Find the subfamily of solutions consistent with the optimality principle. Approaching the question Standard question: this is a slight variation on Example 9.1 in the subject guide.

Section B Answer three questions from this section. Question 6 Consider the utility function 1

1 ln x1 + ln x2 2 2 where x1 and x2 are the quantities consumed of goods 1 and 2 respectively. Denote by m the consumer’s income and by u the consumer’s desired utility level; finally, denote by p1 and p2 the prices of good 1 and 2 respectively. u(x1 , x2 ) =

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EC3120 Mathematical economics

(a) By solving the utility maximisation problem, find the uncompensated demand functions and the indirect utility. (b) By using duality, find the expenditure function and compensated demands. (c) State and verify Roy’s identity. Approaching the question (a) Set up the utility maximisation problem: max

x1 ,x2

1 1 ln x1 + ln x2 2 2

such that: p1 x1 + p2 x2 = m. The Lagrangian is:

1 1 ln x1 + ln x2 − λ(p1 x1 + p2 x2 − m). 2 2 The first-order conditions are: L=

∂L ∂x1

=

1 − λp1 = 0 2x1

∂L ∂x2

=

1 − λp2 = 0 2x2

∂L ∂λ

=

−(p1 x1 + p2 x2 − m) = 0.

Hence: 1 2x1

=

λp1

1 2x2

=

λp2

p1 x1 + p2 x2

=

m.

Therefore: 1 2p1 x1

=

1 2p2 x2

x2

=

p1 x1 . p2

p1 x1 p2

=

m

2p1 x1

=

m.

Replacing in the budget constraint: p1 x1 + p2

Hence the uncompensated demand functions are: x1 (p, m) =

m 2p1

x2 (p, m) =

m 2p2

and the indirect utility function is: v(p, m) =

8

1 1 m m + ln . ln 2p1 2 2p2 2

Examiners’ commentaries 2015

(b) By duality: v(p, e(p, u)) = u. Here: 1 e(p, u) 1 e(p, u) + ln ln 2p2 2p1 2 2   1 e(p, u) e(p, u) = + ln ln 2p1 2 2p2

v(p, e(p, u)) =

=

1 e(p, u)2 ln 4p1 p2 2

=

e(p, u) ln √ 2 p1 p2

=

u.

Hence the expenditure function is: e(p, u) = eu √ 2 p1 p2

√ e(p, u) = 2eu p1 p2 .

⇒

We can now find the compensated demand functions by Shepherd’s Lemma: ∂e(p, u) ∂p1

=

h1 (p, u)

∂e(p, u) ∂p2

=

h2 (p, u).

Here: h1 (p, u) = h2 (p, u) =

p2 p1 r p1 eu . p2 eu

r

(c) Roy’s identity can be stated as follows: xi (p, m) = −

∂v(p, m)/∂pi . ∂v (p, m)/∂m

We can verify the identity for this case, where: v(p, m) =

1 1 m m + ln ln 2 2p2 2 2p1

=

m2 1 ln 2 4p1 p2

=

1 1 1 − ln p1 + ln m + ln . 2 2 4p2

We have: ∂v(p, m) ∂p1

=

−

∂v(p, m) ∂m

=

1 . m

x1 (p, m) =

m . 2p1

By Roy’s identity:

1 2p1

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EC3120 Mathematical economics

Similarly, one can verify that: x2 (p, m) =

m . 2p2

The uncompensated demand functions correspond to the ones we computed in (a), hence the identity is verified.

Question 7 Consider the utility maximisation problem with a quasi-linear utility function 1/2

+ ax2

u(x1 , x2 ) = x1

with a > 0. Denote by m the consumer’s income and by p1 and p2 the prices of good 1 and 2, respectively. (a) Write the utility maximisation problem as a constrained optimisation problem. (b) Check if the Kuhn–Tucker theorem applies. (c) Solve the problem by using the Kuhn–Tucker conditions. Approaching the question (a) We can write the utility maximisation problem as follows: 1/2

max x1 x1 ,x2

+ ax2

such that: p1 x1 + p2 x2

≤

m

−x1

≤

0

−x2

≤

0.

(b) First note that the Kuhn–Tucker theorem applies when u(x) is concave and differentiable and the constraint is convex and differentiable. Here the constraint is linear, hence convex; the objective function is concave when the Hessian matrix is negative semi-definite. Here the Hessian:   −1/2−1 −x1 /4 0 H= 0 0 with |H1 | < 0 and |H2 | = 0, hence the Hessian is negative semi-definite and the objective function is concave. The Kuhn–Tucker theorem applies. (c) The Lagrangian is: 1/2

L = x1

+ ax2 − λ0 (p1 x1 + p2 x2 − m) − λ1 (−x1 ) − λ2 (−x2 ).

The Kuhn–Tucker conditions are:

10

∂L ∂x1

=

1 −1/2 x − λ0 p1 + λ1 = 0 2 1

∂L ∂x2

=

a − λ0 p2 + λ2 = 0

λ0 , λ1 , λ2

≥

0

λ0 (p1 x1 + p2 x2 − m) =

0

λ1 (−x1 ) =

0

λ2 (−x2 ) =

0.

Examiners’ commentaries 2015

Because the objective function is strictly increasing in x1 and x2 the budget constraint is binding, so λ0 > 0 and p1 x1 + p2 x2 = m. Also, the constraint −x1 ≤ 0 cannot bind, for then the marginal utility with respect to x1 would be infinitely large. Hence x1 > 0 and λ1 = 0. The other non-negativity constraint may or may not bind. Hence the Kuhn–Tucker conditions become: ∂L ∂x1

=

1 −1/2 − λ0 p1 = 0 x 2 1

∂L ∂x2

=

a − λ0 p2 + λ2 = 0

λ2

≥

0

p1 x1 + p2 x2

=

m

λ2 (−x2 ) =

0.

Consider now the case x2 = 0. The problem becomes: ∂L ∂x1 ∂L ∂x2 p1 x1

=

1 −1/2 x − λ0 p1 = 0 2 1

=

a − λ0 p2 + λ2 = 0

=

m.

From the budget constraint:

m p1 and replacing in the first first-order condition we get:   1 m −1/2 1 ∂L − λ p = = x−1/2 − λ0 p1 = 0 0 1 ∂x1 2 1 2 p1 x1 =

from which we obtain: λ0 =

1 1  p1 1/2 . = √ 2p1 m 2 p1 m

Replacing this value in the second first-order condition we obtain: 1 ∂L p2 + λ2 = 0 = a − λ0 p2 + λ2 = a − √ 2 p1 m ∂x2 which gives:

p2 − a. λ2 = √ 2 p1 m

The solution: x1

=

m p1

x2

=

0

is a valid solution when λ2 > 0, hence when: p2 . a< √ 2 p1 m For larger values of a, we have the interior solution with x1 > 0 and x2 > 0. The Kuhn–Tucker conditions become: ∂L ∂x1

=

1 −1/2 − λ0 p1 = 0 x 2 1

∂L ∂x2

=

a − λ0 p2 = 0

p1 x1 + p2 x2

=

m.

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EC3120 Mathematical economics

From the second first-order condition we obtain: λ0 =

a p2

and replacing in the first first-order condition: 1 −1/2 p1 x −a =0 p2 2 1 which gives:

p22 . 4a2 p12 Finally, we solve for x2 by substituting in the budget constraint: x1 =

p1

p22 + p2 x2 = m 4a2 p21

which gives: x2 = √ Notice that x2 > 0 for a > p2 /2 p1 m.

p2 m − 2 . p2 4a p1

Hence the solution to the constrained optimisation problem is: p2 m for a ≤ √ x2 = 0 , x1 = , p1 2 p1 m for a ≥

p2 , √ 2 p1 ...