Problem Set #2: Overlapping Generations Models Suggested Solutions - Q2 revised

Transcription

1 University of Warwick EC9A Advanced Macroeconomic Analysis Problem Set #: Overlapping Generations Models Suggested Solutions - Q revised Jorge F. Chavez December 6, 0 Question Consider the following production function 6K t Kt /L t if K t /L t Y t = F (K t, L t ) = K t + 4L t if K t /L t > (a) Show that this function is characterized by constant returns to scale (CRS) and that output per worker is given by: 6k t kt if k t y t = f(k t ) = k t + 4 if k t > Solution. This implies showing that F : R R is homogeneous of degree. That s it, we need to show that for λ > 0, F (λk t, λl t ) = λf (K t, L t ). Then: For k t : F (λk t, λl t ) = 6 (λk t ) (λk t) λl t = λ ( 6K t Kt ) /L t = λf (Kt, L t ) for k t > : F (λk t, λl t ) = λk t + 4λL t = λ (K t + 4L) = λf (K t, L t )T hen j.chavez-cotrado@warwick.ac.uk

2 EC9A (Fall 0) Problem Set # Finally, to show the formula for f(k t ): F (K t, L t ) L t = { 6 Kt L t K t L t t K t /L t Kt L t + 4 K t /L t > = { 6kt k t k t k t + 4 k t > For the following parts assume that population and technology are constant and that capital fully depreciates at the end of each period. (b) Calculate the share of labor income as a function of k t. Show your results in a figure. Solution. Recall that w t = f (k t ) k t f (k t ), then: w t = f (k t ) k t f (k t ) = = { 6kt kt 6 t k t + kt if k t k t + 4 k t if k t > { k t if k t 4 if k t > Finally, the share of labor income is: s L = w { t f (k t ) = kt / (6 k t ) if k t / ( + k) if k t > Figure plots the share of labor income as a function of k t. (c) Find the dynamical system for the Solow model under the assumption that the saving rate s = /. Show that there exists a unique, globally stable, non-trivial steady state. Solution. Because δ = 0: { ( ) k t+ = sf (k t ) = 6kt kt k t (k t + 4) k t > ϕ (k t ) The steady-states for this dynamical system are: (a) For k : 6k k k = k = k k(k ) = 0 k = 0, k = Because this function is defined for k, we rule out k =. Then the only steady-state here is k = 0 Jorge F. Chávez

3 EC9A (Fall 0) Problem Set # 0.5 s L = k 6 k s L = k+ 0 4 Figure : Share of labor income (b) For k > : k = k + 4 k + 4 = k = 4 Stability of the steady-state k ss requires ϕ (k ss ) <. The derivative of the policy rule φ(k t ) is: { ϕ (k t ) = (6 k t) if k t if k t > Then ϕ (0) = > which implies that this steady-state is unstable. Finally, ϕ (4) = < which implies that this steady state is globally stable. Therefore, there exists a unique non-trivial steady-state. Remark. If you would like to be more formal, you can always try to show uniqueness by contradiction. Start assuming that there is another feasible non-trivial steady-state, and then look for a clear contradiction. (d) Find the dynamical system of the OLG model (where individuals work in their first life period and retire in the second). Assume utility is a function of only second period consumption (that is, individuals save their entire first period income). Is the non-trivial steady state unique and globally stable? Solution. Because individuals only consume in their second year of life (when old): s t = w t = k t+ Jorge F. Chávez

4 EC9A (Fall 0) Problem Set # 45 o (k t +4) 8 (6k t k t) Figure : Law of motion for k t+ in the Solow model Because δ = : Then: s t = s y t + s o t = k t+ ( δ)k t = k t+ k t+ = w t { k t if k t 4 if k t > In steady-state: k = { k k t 4 k t > Then, k = 4 if k >, k = 0 and k = if k (e) Why do the two models generate results that are qualitatively different? Solution. In the Solow model savings per capita (s t ) are a fraction of y t = f(k t ), which is concave. In contrast, in the OLG setting, savings are a constant fraction of wages. Jorge F. Chávez 4

5 EC9A (Fall 0) Problem Set # 45 o k t (k t +4) (6k t k t) Figure : Law of motion for k t+ : Solow model vs. OLG model Question Consider the following overlapping generations model (OLG). Technology and population are constant. Output per worker is y t = Ak α t k t = K t /L t Utility of individuals who are born in t is U(c t, c t+ ) = c t + ρ c t+ Capital fully depreciates at the end of each period: δ =. (a) Find the dynamical system. (hint: the system is characterized by two qualitatively different regimes, separated by a threshold level of k t+, where, for low levels of k t+ there is no consumption in the first period). Solution. A household born in time t solves: { } c t t + ρ ct t+ max (c t t,ct t+) R + s.t. c t t + s t t w t c t t+ R t+ s t t The budget constraints can give way to a life-time budget constraint: c t t + ct t+ R t+ w t Jorge F. Chávez 5

6 EC9A (Fall 0) Problem Set # Note that this is a particular case as the utility is linear, implying that goods a perfect substitutes. Appendix shows how to analyze this kind of settings using the Karush-Kuhn-Tucker generalization of the so-called method of Lagrangean multipliers for a problem with non-negativity constraints, like the one stated above. However this problem can be tacked directly. In principle, there will be three cases regarding the relationships between R t+ and ρ. First, if the slope of the budget constraint is higher than the slope of the indifference curve (R t+ < ρ) then the problem will have a corner solution (c t t > 0, c t t+ = 0) as can be seen in figure 4a. Likewise, if R t+ > ρ then we will have a second potential corner solution (c t t = 0, c t t+ > 0) as is shown in figure 4b. The third case is the case in which the slope of the indifference curve is exactly equal to the slope of the budget constraint R t+ = ρ, in which case we will have a continuum of interior solutions. U c t t+ c t t+ U ( 0,c t t+ ) ) U ( 0,c t t+ ) ) U 0 (c t t,0)) c t t 0 (c t t,0)) U U c t t { ( c t t,c t t+) R + s.th. c t t + ct t+ Rt+ = w t} (a) If R t+ < ρ { ( c t t,c t t+) R + s.th. c t t + ct t+ Rt+ = w t} (b) If R t+ > ρ Figure 4: Linear utilities We can summarize the results in terms of savings s t : w t if R t+ > ρ (because c t t = 0 & c t t+ > 0) s t = [0, w t ] if R t+ = ρ (because c t t > 0 & c t t+ > 0) 0 if R t+ < ρ (because c t t > 0 & c t t+ = 0) Recall that R t+ = αak α t+. Then, if R t+ = ρ: R t+ = αak α t+ = ρ }{{} constant which implies a particular and constant level of k t+ : k = ( ) /( α) αa > 0 ρ Jorge F. Chávez 6

7 EC9A (Fall 0) Problem Set # Then we can define the following if and only if statements: R t+ > ρ αak α t+ > ρ k t+ < k () R t+ = ρ αak α t+ = ρ k t+ = k () R t+ < ρ αak α t+ < ρ k t+ > k () Clearly k is an important landmark to understand the law of motion that governs the evolution of k t. Two important things to note are: i. The individual born in t s saving/investment decision (s t t = s t = k t+ ) affects the return to this investment. Since the return to capital tomorrow will be R t+ = αakt+ α we can see that this return is a function of k t+. ii. Because the individuals budget constraint in the first period is c t t + s t t w t, then it is clear that the wage at time t (which by the way depends on the stock of capital in the economy when the individual from the t generation was young, w t = ( α)akt α will be an upper bound for the stock of capital: s t t = s t = k t+ w t : the individual cannot save more than your income in the first period of your life. 45 o C w t = ( α)ak α t k B A ϕ(k t ) k k 0 ˆk k k k t Figure 5: Dynamic system At first sight, the iff statements (), () and () stated above seem to sketch the dynamical system for k t+. However, a closer examination of the last condition reveals that it implies a huge contradiction: Jorge F. Chávez 7

8 EC9A (Fall 0) Problem Set # if s t = k t+ > k then R t+ < ρ and the economy will consume everything in the first period and therefore savings in the first period will be zero s t = 0. So, the second corner c t t = w t > 0, c t t+ = 0 is not feasible in an equilibrium. Then as figure 5 shows, there is a critical value for k t (in the first period) defined by: w t = ( α) Aˆk α = k t+ = k which yields: [ k ˆk = ( α) A ] /α If the individual is born in an economy with starting capital k 0 < ˆk, then she will save all her wage w t = ( α)ak α 0 = k < ˆk. Clearly k > k 0. If k < ˆk then we repeat the whole decision process. However, if k 0 is high enough so that k > ˆk, as depicted in figure??, now the upper bound w t will be above k: w = ( α)ak α > k. Then the individual will have three alternatives for her saving decision, and here is where the fact that the individual s saving decision determines the return to her savings (capital) will play a crucial role: (a) If s = w > k (save all, consumer nothing at t = ), then s = k > k and therefore the return to the savings R will be less than ρ. This implies that s = 0 was the optimal strategy. (b) If s < k (save less than k), then the return will be R > ρ which implies that saving the whole wage w was the optimal strategy. (c) Finally the individual could choose to save s = k. In this case, R = ρ which implies that k = k. One can think of this decision-making process as a trial and error process. First the individual considers saving everything. Then, realizing that it is not optimal, it could consider saving less than the whole w and will do this until it reaches k, in which there is not contradictions. Likewise, if the individual starts by considering saving less than k, later will realize that it reaches an equilibrium if and only if k = k. Finally, we can defined the dynamical system k t+ φ(k t ) as: k t+ = { Figure 5 plots φ(k t ). ( α) Akt α if k t < ˆk k if k t ˆk (b) Find sufficient conditions on the parameters that assure that the steady state is characterized by consumption in the first period. Solution. In order to ensure that the steady state has positive consumption for agents when young (c t t > 0) we need ˆk < k. That is, that point B in 5 is to the left of point C. Otherwise the steady-state will be characterizing as having c t t = 0 all the time. Indeed the case in which s t is suspicious as the fact that the economy accumulates no capital (given the assumption full depreciation) implies that the economy will be stuck in the trivial steady-state: no capital, no production, no consumption. That is, the economy will virtually disappear. Jorge F. Chávez 8

9 EC9A (Fall 0) Problem Set # Then, ˆk < k requires: ( [( α) A] /( α) > k αa = ρ ( ) αa ( α) A > ρ Which yields the condition: ρ > α α ) /( α) Jorge F. Chávez 9

10 EC9A (Fall 0) Problem Set # Appendix A KKT conditions for the case of linear utlity A household born in time t solves the following problem with non-negativity constraints: max c t (c t t,ct t+) R t + ρ ct t+ + s.t. c t t + ct t+ R t+ w t c t t 0, c t t+ 0 The lagrangean is: L = c t t + ( ) ρ ct t+ + λ w t c t t ct t+ R t+ The Karush-Kuhn-Tucker (KKT) FONCs are: L c t t = λ 0 (4) L c t t+ = ρ λ R t+ 0 (5) L λ = w t c t t ct t+ R t+ 0 (6) c t t 0 (7) c t t+ 0 (8) λ 0 (9) Complementary slackness conditions are c t tl c t t = 0, c t t+ L c t t+ = 0, λl λ = 0 and must be satisfied simultaneously with the KKT conditions. i. For interior solutions: c t t > 0, c t t+ > 0, λ > 0 Now (4), (5) and (6) bind so that λ = and then (5) implies that R t+ = ρ. That is, if and only if when this last condition is satisfied we can achieve an interior solution. ii. Corner solution : c t t > 0, c t t+ = 0, λ > 0 Now (4) binds which implies that λ = > 0 while (5) does not bind. Therefore ρ λ < 0 R t+ < ρ R t+ That is, for corner solution to be a solution we need R t+ < ρ iii. Corner solution : c t t = 0, c t t+ > 0, λ > 0 Note that whenever c t t > 0 (4) binds and so it must be the case that λ > 0 Jorge F. Chávez 0

11 EC9A (Fall 0) Problem Set # Now (4) does not bind, which implies that λ >. 4 Then: ρ λ = 0 R t+ = ρλ R t+ > ρ R t+ That is, for corner solution to be a solution we need R t+ > ρ A: An alternative method to solve the UMP There is a third alternative to analytically solve the optimization problem which entails reducing the budget constraint and the non-negativity constraints into a single restriction on the domain of the objective function. This is by far the most efficient method in the cases in which the objective is a non-linear function, as you may have already seen in other problems. Here, the standard non-linear programming problem, with non-negativity constraints, in A reduces to a single-variable problem in which the objective has a domain bounded from above and from below. This means that we need to check for both type corners. The agent solves: {( ) max w t ct t+ + } 0 c t t+ R t+w t R t+ ρ ct t+ FONCs now reduce to a condition on parameters: i. Corner : R t+ + ρ < 0 for ct+ = 0 ii. Interior: R t+ + ρ = 0 for ct+ (0, R t+ w t ) iii. Corner : R t+ + ρ < 0 for ct+ = R t+ w t 4 Therefore we can rule out the case for λ = 0. Jorge F. Chávez