- January 22, 2021

DO NOT REMOVE THIS PAPER FROM THE EXAMINATION ROOM UNIVERSITY OF BRISTOL EXAMINATION FOR DEGREES IN THE SCHOOL OF ECONOMICS. January 2021 EFIMM0023 MRes Mathematics for Economics Start Time – 11:00 UK time, Wednesday 20th January 2021 End Time – 11:00 UK time, Wednesday 27th January 2021 Time allowed : SEVEN (7) DAYS This examination consists of TWO (2) Sections with TWO (2) questions in each section. Answer ALL questions in Section A [50 marks] Answer ALL questions in Section B [50 marks] IMPORTANT INSTRUCTIONS FOR THE ONLINE EXAM • Answers must be typed and submitted in a format which is accepted by Blackboard, such as MS Word, Google documents or PDF. There is a file size limit of 100 MB. Only one document can be submitted. • Please clearly indicate what questions you are attempting on your submitted answer. • This is an open-book exam. You can access websites, Blackboard, journals, everything and anything. You should not call anyone or work in groups. • If necessary, mathematical formulae and calculations can be handwritten and scanned into the document; written-response questions must be answered by typing • Word tables can be used for calculations if full workings are shown. • All answers will be subjected to originality checks. TURN OVER PAGE INTENTIONALLY LEFT BLANK SECTION A Answer ALL questions in this section. Please write answers clearly. Question 1 Part A: Consider a function f : R→ R defined as f(x) = x3 emax{x,−2x} . a) Find the first derivative, f ′(x), and the second derivative, f ′′(x), of the function f and verify that they are well-defined and continuous at x = 0. b) Find all critical points of f and classify them into local max, local min, and inflection points. c) Find all global max points of f if exist(s). Find all global min points of f if exist(s). Part B: Consider the following problem max (x,y)∈R2 f(xy) = (xy)3 emax{xy,−2xy} subject to h(x, y) = x2 + 2y2 − √ 72 ≤ 0. [B] d) Find all solutions to the problem [B]. Part C: Consider the following problem max (x,y)∈R2 f(x)f(y) = x3 emax{x,−2x} × y 3 emax{y,−2y} . [C] e) Find the solution to the problem [C]. f) Find the solution (x∗, y∗) to the problem [C] subject to y − ln(x + 1) = 0 in the following manner: (i) provide an equation in terms of x only, which x∗ solves and such that (x∗, y∗ = ln(x∗ + 1)) is the solution to [C], and (ii) provide a pair of intervals of lengths 0.1 such that x∗ belongs to the first one and y∗ belongs to the second one [ for example x∗ ∈ (0.1, 0.2) and y∗ ∈ (2.4, 2.5)]. (Total 25 marks) Question 2 Consider a consumer who has an initial wealth of w > 0 to spend across an infinite number of periods t = 0, 1, 2 · · · , to maximize his discounted sum of period utilities: max {c0,c1,··· } ∞∑ t=0 βtu(ct) subject to ∞∑ t=0 ct ≤ w and ct ≥ 0 ∀t = 0, 1, 2 · · · [2] where β ∈ (0, 1) and u : R+ → R is C2 with u′(x) > 0 and u′′(x) < 0. An infinite sequence ~c = {c0, c1, · · · } is a “feasible (consumption) plan” if it satisfies the constraints of the problem [2]. a) Show that if ~c = {c0, c1, · · · } solves [2], then ∑∞ t=0 ct = w. Lemma A: If ~c = {c0, c1, · · · } solves [2], then βtu′(ct) = βt+1u′(ct+1) for all t = 0, 1, 2 · · · . b) Prove the Lemma A above. [Hint: Try to prove that if a feasible plan doesn’t satisfy the equality of the Lemma for some t, then another feasible plan can be found that is better for the consumer.] Assume u(x) = √ x from now on. c) In the light of Lemma A, find the relationship between ct and ct+1 in any solution ~c to [2]. d) Find the feasible plan that satisfies the condition obtained in (c) for all t = 0, 1 · · · . e) Find the value of [2] from the feasible plan found in (d), denoted by v∗(w), as a function of w. f) The consumer starts each period t ≥ 1 with a wealth level wt = wt−1−ct−1. Determine whether the value function v∗ above satisfies the so-called Bellman equation, namely, v∗(wt) = u(ct) + βv∗(wt+1) where ct = wt − wt+1. (Total 25 marks) SECTION B Answer ALL questions in this section. Please write answers clearly. Question 3 Consider the following continuous time epidemiological model for a population of size 1. There are four type of citizens: • The ones that are susceptible to contagion at time t, s(t) • The ones that are infected at time t, i(t) • The ones that have died up to time t, d(t) • The ones that have recovered up to time t, r(t) By assumption, for all t ≥ 0, s(t) + i(t) + r(t) + d(t) = 1 and at time 0 we have the shock with s0 + i0 = 1, where i0 is the original proportion of the population infected. The disease has been well studied and it is well know that with instantaneous probability α > 0 some people recover, so · r(t) = αi(t) (1) and with instantaneous probability β > 0 some people die, so · d(t) = βi(t) (2) Those that are recovered do not get the disease again. It is also well known that if a susceptible citizen is in contact with an infected individual then the probability of contagion is pi > 0, and the probability that an infected individual is in contact with a susceptible one is c > 0. At the same time, this virus is airborne so a susceptible individual can get infected with probability m > 0 without contacting an infected person. The dynamics of infections are then given by · s(t) = −pici(t)−ms(t) (3) where we assume that α + β < 1 (they are proportions), and pic < 1. Furthermore, we assume that α+ β +m > pic so the disease does not spread fast. a) Express the system in terms of s(t) and i(t) b) Solve the model for arbitrary values of i0 ∈ (0, 1), assuming that r0 = d0 = 0. Hint: you can solve it as a system of differential equations or reduce the system to a single differential equation of second order. c) What happens with the system in the long run? How does the proportion of recovered people and dead people depend on parameters and the initial conditions? d) We have studied multiple SIR(D) models during the course. In some of them deaths and recoveries do not depend on parameters that determine how the disease spreads. In this model, those parameters are pi, c and m. Explain why that happens and potential avenues to improve the model. (Total 25 marks) Question 4 The Central bank of Narnia determines the policy according to the following loss function L = α(pt − pi∗) + (1− α)(µt − µ∗) where α ∈ [0, 1] is the policy weight assigned to each objective and L is a tolerance level. The policy objectives are an inflation rate pi∗ and unemployment rate µ∗ The economy functions relatively well, so in the short run the real sector responds to expansionary policies both on the monetary side and real side. The response of the real sector of the economy has been calculated to be µt+1 − µt = −δ(pt − pt−1) + γ(rt − rt−1) for γ > 0 and δ > 0. The financial sector reacts relatively fast to expansionary monetary policies: when the Central bank expands the supply of money, agents understand that this is to avoid issuing more debt, which will be paid in the future by higher taxes or more debt, and demand accordingly a higher return for lending to the Government. This implies that, whenever the inflation rate exceeds some value pˆi determined by the private sector’s expectation, the interest rate increases in lime with the parameter ν > 0: rt − rt−1 = ν(pt − pˆi) a) Describe the dynamics of the economy in terms of the inflation rate and/or unemployment for all t ≥ 0. b) Provide the general solution to the model. c) Provide conditions for convergence of the system, explain where it converges and how it converges. d) Economists usually complain about the IMF putting too much emphasis on price stability instead of unemployment. That, is, for any value of L, the IMF wants α1−α to be big. This is received with a lot of complaints from the different political forces in the country asking for help from the IMF. Using the previous model, can you explain why? e) For a given value of L, the Government faces a tradeoff between µ∗ and pi∗ in the long run when the system converges. Describe the tradeoff as function of the parameters of the system. In particular, what is the effect of pˆi in this tradeoff? What happens with this tradeoff when we change L? (Total 25 marks) END OF PAPER 欢迎咨询51作业君