Lecture 6: Discrete-Time Dynamic Optimization

Transcription

1 Lecture 6: Discrete-Time Dynamic Optimization Yulei Luo Economics, HKU November 13, 2017 Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

2 The Nature of Optimal Control In static optimization, the task is to find a single value for each control variable, such that the objective function will be maximized or minimized. In contrast, in a dynamic setting, time enters explicitly and we encounter a dynamic optimization problem. In such a problem, we need to find the optimal time path of control and state during an entire planning period. Specifically, at any time t, we have to choose the value of some control variable, c(t), which will then affect the value of the state variable, x(t + 1), via a state transition equation given the current state x(t). Two leading methods to solve DO problems are: optimal control (OC) and dynamic programming (DP). They can be applied in deterministic or stochastic and discrete-time or continuous-time settings. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

3 The Setting of Optimal Control Since our objective is to maximize the profit over the entire period, the objective should take the form of summation from t = 0 to t = T in the discrete-time setting. The problem also specifies the initial value of the state variable x, x(0), and the terminal condition of y, x(t + 1): subject to for any t 0. T max {c(t),x (t+1)} r (c (t), x (t), t), (1) x (t + 1) x (t) = f (c (t), x (t), t), (2) x 0 = x (0), x (T + 1) 0, (3) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

4 (Conti.) As in the static optimization case, the Lagrangian for this intertemporal (dynamic) problem is T L = {r (c t, x t, t) + λ t [f (c t, x t, t) + x t x t+1 ]} + λ T +1 x T +1, (4) where λ t is the shadow price (i.e., the Lagrange multiplier). The FOCs for c t are: L c t = r c (c t, x t, t) + λ t f c (c t, x t, t) = 0 for t = 0, 1,, T. (5) The FOCs for x t+1 are complicated because each x t appears in two terms in the Lagrangian function. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

5 (Conti.) Consider some relevant terms in the above Lagrangian function: T T 1 λ t (x t x t+1 ) = (λ 0 x 0 λ T x T +1 ) + x t+1 (λ t+1 λ t ). The FOCs for x t+1 : L x t+1 = r x (c t+1, x t+1, t + 1) + λ t+1 f x (c t+1, x t+1, t + 1) for t = 0,, T 1. +λ t+1 λ t = 0, Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

6 Theorem (The Maximum Principle) The necessary FOCs for maximizing (1) subject to (2) are: (i) L t c t = 0 for all t = 0,, T. (ii) x t+1 x t = f (c t, x t, t) for all t (The state transition equation) (iii) r x (c t+1, x t+1, t + 1) + λ t+1 f x (c t+1, x t+1, t + 1) + λ t+1 λ t = 0 for all t (The costate (λ t ) equation) (iv) x T +1 0, λ T +1 0, x T +1 λ T +1 = 0 (The transversality condition, or the complementary-slackness condition) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

7 Example: Optimal Consumption-Saving Model Suppose that a consumer has the utility function u (c t ), where c t is consumption at time t. The consumer s utility is concave: u > 0, u < 0, (6) and u (c t ) satisfies the Inada conditions (They ensure that consumption will always be interior): lim c 0 u (c) = and lim u (c) = 0. (7) c The consumer is also endowed with a 0 = a (0), and has income stream derived from holding the asset: y t = ra t, where r is the interest rate. The consumer uses the income to purchase c. Any income not consumed is added to the asset holdings as savings: a t+1 = (1 + r) a t c t. (8) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

8 The Model Setup The consumer s lifetime utility maximization problem is to subject to T max {c t,a t+1 } β t u (c t ) (9) a t+1 = Ra t c t, t = 0,, T, (10) a 0 = a (0), a T +1 0, (11) where β is the consumer s rate of time preference (β 1) and R = 1 + r is the gross interest rate. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

9 (Conti.) Since lim c 0 u (c) =, c (t) > 0 for all t T. Further, to ensure that a solution exists, assets are constrained to be nonnegative after the terminal period T (i.e., when the consumer dies), a T Note that this nonnegativity constraint must bind, i.e., a T +1 = 0. In fact, this constraint serves two important purposes: First, without this constraint a T +1 0, the consumer would want to set a T +1 = and die with outstanding debts. This is clearly not feasible. Second, the fact that the constraint is binding in the optimal solution guarantees that resources are not being thrown away after T. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

10 Using Optimal Control to Solve the Model Setting up the Lagrangian: L = T β t {u(c t ) + λ t [Ra t c t a t+1 ]} + λ T +1 a T +1 (12) The FOC for c t is L c t = 0: The FOC for a t+1 is u (c t ) λ t = 0 for t = 0, 1,, T. (13) L a t+1 = 0: λ t + βrλ t+1 = 0 for t = 0,, T 1. (14) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

11 (Conti.) The terminal condition a T +1 0 implies that the transversality condition (the complementary slackness condition) is λ T +1 0, a T +1 0, a T +1 λ T +1 = 0, (15) which means that either the asset holdings (a) must be exhausted on the terminal date, or the shadow price of capital (λ t ) must be 0 on the terminal date. Since u > 0, the marginal value of capital (λ) cannot be 0 and thus the capital stock should optimally be exhausted by the terminal date T + 1, i.e., a T +1 = 0. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

12 (Conti.) Combining (13) with (14) gives u (c t ) = βru (c t+1 ), (16) which is called the Euler equation characterizing optimal consumption dynamics. Suppose that the utility function has an isoelastic form u(c t ) = c1 γ t 1 1 γ, (17) where γ 0 (Note that when γ = 1, the function reduces to ln c t ). We have c γ t = βrc γ t+1 = c t+1 = (βr) 1/γ. (18) c t Therefore, if βr > 1, the optimal consumption will rise over time; if βr < 1, the optimal consumption will decline over time. When βr = 1, c t = c t+1. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

13 (Conti.) (18) implies that the elasticity of intertemporal substitution (EIS) is ( ) ( ) ( ) d ct+1 c t / ct+1 c t = d ln ct+1 c t = 1 γ. (19) Hence, increasing γ dr/r ( i.e., reducing 1 γ d ln R ) make the agent more unwilling to postpone consumption (i.e., more unwilling to save). That s why we call this type of utility functions the isoelastic utility function. The optimal problem is pinned down by a given initial condition (a 0 ) and by a terminal condition (a T +1 = 0). The sets of (T 1) Euler equations (18) and constraints (10) then determine the time path of c as well as a. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

14 Obtaining the Intertemporal Budget Constraint The original budget constraint (10) implies that a 1 = Ra 0 c 0, a 2 = Ra 1 c 1, a T +1 = Ra T c T. a T +1 R T ( + ct R T + + c ) 1 R + c 0 T = Ra 0 = c t R t = Ra 0, (20) where we use the fact that a T +1 = 0. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

15 (Conti.) For simplicity, assume that γ = 1. Combining (18) with (20) gives c R (βrc 0) ( ) R T β T R T c 0 = Ra 0 = R c 0 = 1 + β + + β T a 0. (21) Hence, the time path of optimal consumption is c t = R t+1 β t 1 + β + + β T a 0 (22) where t = 0,, T. Given c t, it is straightforward to determine the time path of a t : a t+1 = Ra t c t = Ra t where t = 0,, T and a T +1 = 0. R t+1 β t 1 + β + + β T a 0, (23) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

16 (Conti.) Consider an important special case, βr = 1. In this case, (18) implies that c 0 = c 1 = = c T, ( c ) R T = Ra 0 = where t = 0,, T. c t = R 1 1 R (T +1) a 0. (24) Next, the optimal path of a t can be derived by solving the following difference equation: r a t+1 = Ra t c t = Ra t 1 R (T +1) a 0 = ( ) a t = 1 a 0 R t R (T +1) 1 1 R (T +1) a 0, (25) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

17 Alternative Way to Solve Optimal Consumption Instead of defining the Lagrange multiplier for each flow budget constraint, we may define a Lagrange multiplier for the intertemporal budget constraint (that is, the lifetime budget constraint): T c t R t = Ra 0, where we have used the fact that a T +1 = 0. That is, we may write the Lagrangian as follows ( ) T L = β t T c t u(c t ) + λ Ra 0 R t, where λ is the constant Lagrange multiplier for the lifetime budget constraint. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

18 (Conti.) The FOCs for an optimum are then β t u (c t ) = λ 1, where t = 0,, T. (26) Rt Since λ is a constant, the above FOCs implies that the Euler equations are u (c t ) = βru (c t+1 ), where t = 0,, T 1, which is identical to those Euler equations we derived above. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

19 The Infinite Horizon Case There are some reasons to consider the infinite horizon (T = ) case: People don t live forever, but they may care about their offspring (a bequest motive). Assuming an infinite horizon eliminates the terminal date complications (stop saving, consume their entire value, etc. There is no real world counterpart for these actions because the real economy continue forever). Many economic models with a long time horizon tend to show very similar results to IH models if the horizon is long enough. IH models are stationary in nature: the remaining time horizon doesn t change as we move forward in time. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

20 The Model Setup The consumer s lifetime utility maximization problem max c(t) β t u (c t ) (27) subject to a t+1 = Ra t c t, t 0, given the initial asset holdings a 0 = a (0). In addition, we need to impose the following institutional assumption: no Ponzi game condition (npg): ( at+1 ) lim t R t 0, (28) means that in present value terms, the agent cannot engage in borrowing and lending so that his terminal asset holdings are negative. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

21 (Conti.) (36) can help obtain an intertemporal budget constraint: Applying (36), we have = lim T ( T T c t R t + a T +1 R T c t R t + a T +1 R T = Ra 0. ) T T = lim c ( t R t + lim at ) +1 T R T = c t R t Ra 0. c ( t R t + lim at ) +1 T R T (29) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

22 (Conti.) In an optimum, (37) must be binding: c t R t = Ra 0 (30a) ( because (36) is binding, i.e., lim at+1 ) T R = 0 (Otherwise, the t agent can consume the left amount and reach a higher utility level, which contradicts optimum). Set up the Lagrangian: ( ) L = β t c t u (c t ) + λ Ra 0 R t The FOCs w.r.t. c t for any t are β t u (c t ) λ 1 = 0, t 0, (31) Rt where λ is the Lagrangian multiplier associated with the intertemporal budget constraint. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

23 (Conti.) Note that the FOCs imply the following Euler equation linking consumption in two consecutive periods u (c t ) = βru (c t+1 ), t 0. (32) Suppose that the utility function is log, u (c t ) = log c t. (39) implies that c t = (βr) t c 0, t 1. Substituting them into the (38a) gives (βr) R t c 0 = ( ) Ra 0 = β t c 0 = Ra 0 = c 0 = R (1 β) a 0, (33) and consumption in periods t 1 can be recovered: c t = (βr) t c 0 = (βr) t R (1 β) a 0, t 1. (34) Similarly, we can determine the optimal time path of asset holdings: a t+1 = Ra t (βr) t R (1 β) a 0. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

24 The Infinite Horizon Case There are some reasons to consider the infinite horizon (T = ) case: People don t live forever, but they may care about their offspring (a bequest motive). Assuming an infinite horizon eliminates the terminal date complications (stop saving, consume their entire value, etc. There is no real world counterpart for these actions because the real economy continue forever). Many economic models with a long time horizon tend to show very similar results to IH models if the horizon is long enough. IH models are stationary in nature: the remaining time horizon doesn t change as we move forward in time. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

25 The Model Setup The consumer s lifetime utility maximization problem max c(t) β t u (c t ) (35) subject to a t+1 = Ra t c t, t 0, given the initial asset holdings a 0 = a (0). In addition, we need to impose the following institutional assumption: no Ponzi game condition (npg): ( at+1 ) lim t R t 0, (36) means that in present value terms, the agent cannot engage in borrowing and lending so that his terminal asset holdings are negative. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

26 (Conti.) (36) can help obtain an intertemporal budget constraint: Applying (36), we have = lim T ( T T c t R t + a T +1 R T c t R t + a T +1 R T = Ra 0. ) T T = lim c ( t R t + lim at ) +1 T R T = c t R t Ra 0. c ( t R t + lim at ) +1 T R T (37) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

27 (Conti.) In an optimum, (37) must be binding: c t R t = Ra 0 (38a) ( because (36) is binding, i.e., lim at+1 ) T R = 0 (Otherwise, the t agent can consume the left amount and reach a higher utility level, which contradicts optimum). Set up the Lagrangian: ( ) L = β t c t u (c t ) + λ Ra 0 R t The FOCs w.r.t. c t for any t are β t u (c t ) λ 1 = 0, t 0, (39) Rt where λ is the Lagrangian multiplier associated with the intertemporal budget constraint. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

28 (Conti.) Note that the FOCs imply the following Euler equation linking consumption in two consecutive periods u (c t ) = βru (c t+1 ), t 0. (40) Suppose that the utility function is log, u (c t ) = log c t. (39) implies that c t = (βr) t c 0, t 1. Substituting them into the (38a) gives (βr) R t c 0 = ( ) Ra 0 = β t c 0 = Ra 0 = c 0 = R (1 β) a 0, (41) and consumption in periods t 1 can be recovered: c t = (βr) t c 0 = (βr) t R (1 β) a 0, t 1. (42) Similarly, we can determine the optimal time path of asset holdings: a t+1 = Ra t (βr) t R (1 β) a 0. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

29 The Ramsey-Cass-Koopmans Model (Optimal Growth Model) The key difference bw the RCK model and the Solow model is that the saving rate is endogenized in the RCK model. We posit a single representative agent in the economy and has the following preference max {c t,k t+1 } β t u (c t ), (43) where u (c t ) also satisfies the same conditions as before. Abstracting from population growth or technological progress, the resource constraint facing the representative consumer at any t are k t+1 = (1 δ) k t + f (k t ) c t, (44) where k 0 is given. δ (0, 1) is the depreciation rate, the production function f (k t ) satisfies the usual conditions: f (k t ) > 0 and f (k t ) < 0. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

30 Solving the Model As usual, we first set up the Lagrangian: L = β t {u(c t ) + λ t [(1 δ) k t + f (k t ) c t k t+1 ]} (45) where λ t 0 denote the multiplier on the resource constraint at time t. The FOCs w.r.t. c t and k t+1 are u (c t ) = λ t (46) β [ 1 δ + f (k t+1 ) ] λ t+1 = λ t (47) Combining both FOCs can generate the Euler equation β [ 1 δ + f (k t+1 ) ] u (c t+1 ) = u (c t ) (48) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

31 (conti.) We also need the transversality condition (TVC) to guarantee that asymptotically the shadow value of more capital is zero: lim t βt λ t k t+1 = lim β t u (c t )k t+1 = 0, (49) t where β t λ t is the present-value utility evaluation of an additional unit of resources in period t. Hence, the TVC says that the value (discounted into PV utilities) of each additional unit of capital at infinity times the actual amount of capital has to be zero. Otherwise, the consumer can modify such a capital path and increase consumption for an overall increase in utility without violating feasibility. In the finite horizon case, the corresponding condition is that k T +1 = 0. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

32 (conti.) The npg condition discussed in the last lecture and the TVC here play a very similar roles in dynamic optimization in a purely mechanical sense. In fact, they are the same condition when the FOCs are satisfied: [( lim t βt λ t k t+1 = 0 lim R 1 t 1 ) ( R 1 2 where R t+1 = 1 δ + f (k t+1 ) and λ 0 = u (c 0 ) is finite. ) ( )] R 1 t λ0 k t+1 = λ 0 lim t However, the two conditions are conceptually very different: The npg is a restriction on the choices of the agent. In contrast, the TVC is a prescription how to behave optimally, given a choice set. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

33 Intertemporal Equilibrium (Steady State) We have formed a nonlinear difference equation system in terms of c t and k t that can characterize the economy completely: u (c t ) = β [ 1 δ + f (k t+1 ) ] u (c t+1 ) (50) k t+1 = (1 δ) k t + f (k t ) c t with two boundary conditions: the initial condition k 0 = k (0) and the transversality condition (TVC) (49). Now we can determine the intertemporal equilibrium (the steady state) of the above dynamic system, i.e., k t+1 = k t = k and c t+1 = c t = c: 1 = β [ 1 δ + f ( k )] (51) 0 = δk + f ( k ) c (52) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

34 (conti.) Note that the first equation is independent of c and can easily determine k : f ( k ) = 1 β 1 + δ = ρ + δ, where ρ = 1 β 1 is called the time discount rate. Note that if f (k t ) = Akt α, ( ) Aα 1/(1 α) k =. ρ + δ When k is determined, c = f ( k ) δk. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

35 Qualitative Analysis (The Phase Diagram) For this nonlinear system, we can use the phase diagram to analyze its qualitative dynamics. More specifically, assume that the utility function is ln c t, and rewrite the dynamic system as: which means ( ) d ct+1 c t c t+1 c t = β ( 1 δ + f (k t+1 ) ), (53) k t+1 k t = f (k t ) δk t c t, (54) dk t+1 = βf (k t+1 ) < 0 and d (k t+1 k t ) dc t = 1 < 0 and from which we can have two demarcation lines separating the space. It turns out the intertemporal equilibrium is saddle-point equilibrium. [Insert Figure here] Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

36 Linear Approximation Alternatively, we can use the linearization method to approximate the original nonlinear system around the intertemporal equilibrium and then solve the resulting linear difference equation system. Linearizing the above dynamic system around ( k, c ) gives c t+1 = ( c + β 1 δ + Aαk α 1) (c t c) + (α 1) βcaαk α 2 ( kt+1 k t+1 = ( k (c t c) + 1 δ + Aαk α 1) ( kt k ), where 1 β = 1 δ + Aαk α 1 and which can be written (denote x t+1 = x t+1 x, x = k, c) [ ] 1 (α 1) βcaαk α 2 [ ] [ ][ ] [ ] ct ct k t = β k t 0 }{{}}{{}}{{}}{{}}{{} J u M v d (57) Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

37 (conti.) Multiplying J 1 on both sides gives [ ][ ] [ ][ ] [ ] 1 0 ct J 0 1 k 1 ct 0 t = β k t 0 }{{}}{{}}{{}}{{}}{{} I u M v d Iu + Kv = 0 (58) where we have substituted k = K = J 1 M = = [ [ ( Aα ρ+δ ) 1/(1 α) into the matrix M and 1 (α 1) βcaαk α (α 1) βcaαk α 2 ] [ β (α 1) caαk α β ] ]. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

38 (conti.) The characteristic roots b can be obtained by solving the equation: b 1 + (α 1) βcaαk α 2 (α 1) caαk α 2 bi + K = 0 = 1 b 1 β = [ ] b (1 α) βcaαk α 2 + (1 + ρ) b [ (1 α) βcaαk α 2] (1 + ρ) + (α 1) caαk α 2 = 0 = trace = b 1 + b 2 = 1 + (1 α) βcaαk α β > 2, det = b 1 b 2 = = 1 β > 1 [ 1 + (1 α) βcaαk α 2] 1 α 2 + (α 1) caαk β Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

39 (conti.) Hence, the discriminant should be positive because [ = trace (J E ) 2 4 det (J E ) = 1 + (1 α) βcaαk α ] β β [ = 1 + (1 α) βcaαk α 2 1 ] 2 > 0 β which means that both roots are real. Also, because det = 1 β > 1 and trace > 2, the two roots must individually be positive. We can also judge the magnitudes of the two roots as follows: bi + K = 0 p (b) = (b b 1 ) (b b 2 ) = 0 = p (1) = (1 b 1 ) (1 b 2 ) = 1 trace + det = (1 α) βcaαk α 2 < 0 This can only be true if one root (say b 1 ) is less than 1 and the other root is greater than 1. We can then conclude (and confirm the predictions of the PD) that the equilibrium is saddle-point. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

40 Transitional Dynamics After obtaining b 1 < 1 and b 2 > 1, we can have the general solution for this system: [ ] [ ct K1 A = 1 b1 t + K 2A 2 b t ] 2 k t A 1 b1 t + A 2b2 t. (59) where K 1 and K 2 are two constants determined by the roots (here we ignore their detailed values). Given k 0, k 0 = k 0 k = A 1 + A 2 c 0 = c 0 c = K 1 A 1 + K 2 A 2 Note that b t 2 as t because b 2 > 1. To guarantee a convergent time path to the I.E. (i.e., to kill the explosive path), the endogenous c 0 need to be set in the right way and make A 2 be 0: c 0 c = K 1 ( k0 k ). (60) Hence, given any k 0, we can find c 0 s.t. the economy jump to the pair of stable branches and then move to the saddle point equilibrium. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

41 Determining the Undetermined Coeffi cients (Optional) The following ) are about the details about the transitional dynamics of ( c t, k t we obtained above. Given that the economy is characterized [ ] [ ] ct+1 ct by = K, where k t+1 k [ t ] 1 + (α 1) βcaαk α 2 (α 1) caαk α 2 K = 1 1, and β 0 < b 1 < 1 and b 2 > 1 are two corresponding eigenvalues. Given the model parameters, we can decompose the 2-by-2 matrix K: K = QΛQ 1, where Q has the eigenvectors of K as its columns, [ and ] Λ has b2 0 eigenvectors of K down its diagonal: Λ =. 0 b 1 Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

42 (Conti.) The solution to the above system can bee written [ ] [ ] [ ] ct+1 b t = Q 2 0 k t+1 0 b1 t Q 1 c0, (61) k 0 where k 0 is given. Since b 2 > 1, we have one explosive eigenvalue and the system will blow-up as t unless the endogenous variable c 0 jumps in[ the right way ] so as to kill the explosive dynamics. q11 q Denote Q = 12. = = = [ ] ct+1 k t+1 1 det (Q) 1 det (Q) 1 det (Q) q 21 q 22 [ ] [ ] [ ] [ ] q11 q 12 b t 2 0 q22 q 12 c0 q 21 q 22 0 b1 t, q 21 q 11 k 0 [ q11 b2 t q 12 b1 t ] [ ] q22 c 0 q 12 k 0 q 21 b2 t q 22 b1 t, q 21 c 0 + q 11 k 0 ) ) q 11b2 (q t 22 c 0 q 12 k 0 + q 12 b1 ( q t 21 c 0 + q 11 k 0 ) ). q 21 b2 (q t 22 c 0 q 12 k 0 + q 22 b1 ( q t 21 c 0 + q 11 k 0 Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43

43 Since b2 t as t, we have to shut these explosive paths down by setting c 0 just right. In other words, we choose consumption so that we are not on the explosive path that violates the transversality condition (TVC). Apparently, we should set to rule out the unstable dynamics. Hence our complete solution is [ ] ct+1 1 = k t+1 det (Q) = 1 det (Q) c 0 = q 12 q 22 k 0 (62) ( ) q 12b1 t q q q 22 k 0 + q 11 k 0 ( ) q 22 b1 t q q q 22 k 0 + q 11 k 0 ( ) q q q 22 + q 11 q 12 q 22 ( q 21 q 12 q 22 + q 11 ) b t 1 k 0. In practice we can first compute the matrix K, then compute its eigenvalues and eigenvectors to get the matrix Q and pick the stable eigenvalue so that we have a bounded solution. Luo, Y. (Economics, HKU) ECON0703: ME November 13, / 43