Topic 5: The Difference Equation

Topic 5: The Difference Equation Yulei Luo Economics, HKU October 30, 2017 Luo, Y. (Economics, HKU) ME October 30, 2017 1 / 42

Discrete-time, Differences, and Difference Equations When time is taken to be a discrete variable, so that the variable t is allowed to take integer values only, the concept of the derivative will no longer appropriate (it involves infinitesimal changes, dt) and the change in variables must be described by so called differences ( t). Accordingly, the techniques of difference equations need to be developed. We may describe the pattern of change of y by the following difference equations: where y t+1 = y t+1 y t. y t+1 = 2 (1) or y t+1 = 0.1y t (2) Luo, Y. (Economics, HKU) ME October 30, 2017 2 / 42

Solving a FO Difference Equation Iterative Method. For the FO case, the difference equation describes the pattern of y between two consecutive periods only. Given an initial value y 0, a time path can be obtained by iteration. Consider y t+1 y t = 2 with y 0 = 15, y 1 = y 0 + 2 y 2 = y 1 + 2 = y 0 + 2 (2) and in general, for any period t, y t = y 0 + t (2) = 15 + 2t. (3) Luo, Y. (Economics, HKU) ME October 30, 2017 3 / 42

. Consider y t+1 = 0.9y t with y 0. By iteration, y 1 = 0.9y 0, y 2 = (0.9) 2 y 0 y t = (0.9) t y 0 (4) Consider the following homogeneous difference equation ( n ) t my t+1 ny t = 0 = y t+1 = y0, (5) m which can be written as a more general form y t = Ab t. (6) Luo, Y. (Economics, HKU) ME October 30, 2017 4 / 42

General Method to Solve FO Difference Equation Suppose that we are solving the FO DE: y t+1 + ay t = c (7) The general solution is the sum of the two components: a particular solution y p (which is any solution of the above DE) and a complementary function y c. Let s first consider the CF. We try a solution of the form y t = Ab t, Ab t+1 + aab t = 0 = b + a = 0 or b = a, which means that the CF should be y c = A ( a) t (8) Consider now the particular solution. We try the simplest form y t = k, k + ak = c = k = c 1 + a = the PI is y p = c 1 + a (a = 1). (9) Luo, Y. (Economics, HKU) ME October 30, 2017 5 / 42

(Conti.) If it happens that a = 1, try another solution form y t = kt, k (t + 1) + akt = c c = k = t + 1 + at = c. y p = ct (a = 1). (10) The general solution is then y t = A ( a) t + c 1 + a or y t = A ( a) t + ct (a = 1) (11) Using the initial condition y t = y 0 when t = 0, we can easily determine the definite solution: y 0 = A + c 1 + a = A = y 0 c (a = 1) or 1 + a y 0 = A + c 0 = A = y 0 (a = 1) Luo, Y. (Economics, HKU) ME October 30, 2017 6 / 42

The Dynamic Stability of Equilibrium In the discrete-time case, the dynamic stability depends on the Ab t term. The dynamic stability of equilibrium depends on whether or not the CF (Ab t ) will tend to zero as t. We can divide the range of b into seven distinct regions: see Figure 17.1. { Nonoscillatory if b > 0 Oscillatory if b < 0 ; { Divergent if b > 1 Convergent if b < 1 The role of A. First, it can produce a scale effect without changing the basic configuration of the time path. Second, the sign of A can affect the shape of the path: a negative A can produce a mirror effect as well as a scale effect. Luo, Y. (Economics, HKU) ME October 30, 2017 7 / 42

The Cobweb Model A variant of the market model: it treats Q s as a function not of the current price but of the price of the preceding time period, that is, the supply function is lagged or delayed. Q s,t = S (P t 1 ) (12) When this function interacts with a demand function of the form Q d,t = D (P t ),interesting price dynamics will appear. Assuming linear supply and demand functions, and the market equilibrium implies Q s,t = Q d,t (13) Q d,t = α βp t (α, β > 0) (14) Q s,t = γ + δp t 1 (γ, δ > 0). (15) Luo, Y. (Economics, HKU) ME October 30, 2017 8 / 42

(Conti.) In equilibrium, the model can be reduced to the following FO DE βp t + δp t 1 = α + δ = P t+1 + δ β P t = α + δ β (16) Consequently, we have P t = ( P 0 α + γ ) ( δ ) t + α + γ β + δ β β + δ. (17) The particular integral P = α+γ β+δ is the intertemporal equilibrium price of the model. We can rewrite the price dynamics as follows P t = ( P 0 P ) ( δ β ) t + P. (18) Luo, Y. (Economics, HKU) ME October 30, 2017 9 / 42

(Conti.) P 0 P can have both the scale effect and the mirror effect on the price dynamics. Given our model specification (δ, β > 0), we can deduce an oscillatory time path because δ β < 0. That s why we call the model the Cobweb model. The model has three possibilities of oscillation patterns: Explosive if δ > β Uniform if δ = β Damped if δ < β See Figure 17.2. Luo, Y. (Economics, HKU) ME October 30, 2017 10 / 42

Nonlinear Difference Equations-The Qualitative-Graphic Approach When nonlinearity occurs in the case of FO DE models, we can use the graphic approach (Phase diagram) to analyze the properties of the DE. Consider the following nonlinear DEs y t+1 + y 3 t = 5 or y t+1 + sin y t ln y t = 3 = y t+1 = f (y t ) when y t+1 and y t are plotted against each other, the resulting diagram is a phase diagram and the curve corresponding to f is a phase line. See Figure 17.4. The first two phase lines, f 1 and f 2, are characterized by positive slopes f 1 (0, 1) and f 2 > 1 and the remaining two, f 3 and f 4, are negatively sloped f 3 ( 1, 0) and f 4 < 1 Luo, Y. (Economics, HKU) ME October 30, 2017 11 / 42

(Conti.) For the phase line f 1, the iterative process leads from y 0 to y in a steady path, without oscillation. For the phase line f 2 (whose slope is greater than 1), a divergent path appears. For phase lines, f 3 and f 4, the slopes are negative. The oscillatory time paths appear. Summary: The algebraic sign of the slope of the phase line determines whether there will be oscillation, and the absolution value of its slope governs the question of convergence. Luo, Y. (Economics, HKU) ME October 30, 2017 12 / 42

The Solow Model Solow (1956): What do simple neoclassical assumptions imply about growth? The model is the starting point for almost all analyses of growth. Key assumptions include: The production function F has three factors: capital K, labor L, and technology A: Y = F (K, AL) (K, L, A > 0), (19) where F K, F L > 0, and F KK, F LL < 0 (diminishing returns to capital and labor). AL means effective labor. F is assumed to be constant return to scale in K and L: ( ) K Y = ALF AL, 1 = ALf (k) or y = f (k), (20) where y = Y / (AL) and f (k) = F (k, 1). F also satisfy: lim K (K, AL) K 0 =, lim F L (K, AL) =, L 0 lim K (K, AL) K = 0, lim L (K, AL) = 0, L F (0, AL) = 0 for all A and L. Luo, Y. (Economics, HKU) ME October 30, 2017 13 / 42

Solving the Solow Model Assume that there is only a representative agent in the economy. The Solow model can be formulated as follows: K t+1 = (1 δ) K t + sf (K t, A t L t ), (21) A t+1 = (1 + g) A t, (22) L t+1 = (1 + n) L t, (23) given K 0, A 0, and L 0. Note that in the continuous-time version, we get a system of differential equations. Define k = K AL and y = Y AL, we have the following first-order nonlinear difference equation about capital per effective labor unit (k): k t+1 = (1 δ) k t + sf (k t ). (1 + g) (1 + n) Luo, Y. (Economics, HKU) ME October 30, 2017 14 / 42

Steady States There exist two steady states (SS): k = 0 and k = k > 0 that satisfies: (1 δ) k + sf (k) k = (1 + g) (1 + n). When f (k) = k α, [ ] s 1/(1 α) k =. (1 + g) (1 + n) (1 δ) Graphic Analysis [see the phase diagram.] Policy implications: The effects of s, α, and δ on the steady state stock of capital per effective labor unit. Alternatively, we can use the linearization method to approximate the original nonlinear difference equation around the intertemporal equilibrium (the steady state) and then solve the resulting linear difference equation. Luo, Y. (Economics, HKU) ME October 30, 2017 15 / 42

Linear Approximation How to linearize a difference equation (DE): x t+1 = f (x t ), given an initial condition, x 0? Typically, linearize the equation around a SS x satisfying x = f (x): x t+1 x + f (x) (x t x). (24) If we treat this approximation as exact, we have the following first-order DE: x t+1 = ax t + b, (25) where a = f (x) and b = (1 a) x. The solution to this DE is x t = ( 1 a t ) x + a t x 0. If a < 1, then a t 0 as t so that x t x. Hence, the original nonlinear DE is locally stable as it is approximated around the SS. For the Solow model, we have k t = ( 1 a t ) k + a t (1 α) (1 δ) k 0, where a = α + (0, 1). (1 + g) (1 + n) Luo, Y. (Economics, HKU) ME October 30, 2017 16 / 42

Speed of Convergence How does the initial level of capital per capita affect growth rates? Convergence: Poor countries grow faster than rich countries. Divergence: Rich countries grow faster than poor countries. The Solow model predicts that poor countries with low k will grow fast because of decreasing returns to capital: g t,t+1 = k [ ] t+1 k t s f (k t ) (1 δ) = + k t (1 + g) (1 + n) k t (1 + g) (1 + n) 1, where ( ) f (kt ) d /dk t = f (k t ) k t f (k t ) k t kt 2 < 0 because f (k t ) = kt α is concave (i.e., f (k t ) is decreasing). This convergence is only observed among U.S. states, Canadian provinces, European regions, etc, but not observed among the countries of the world. Luo, Y. (Economics, HKU) ME October 30, 2017 17 / 42

Summary k is called globally asymptotically stable as k t converges to it from any initial positive capital stock, k 0. What does this imply about the aggregate variables: K t A t L t Since is constant, K t and A t L t grow at the same rate (1 + g) (1 + n) 1 g + n. Since F (, ) is homogeneous of degree 1, Y t also grows at the same rate g + n. Per capita output Y t L t grows at rate g. The steady state for k t corresponds to a balanced growth path (a sequence in which all of the variables grow at a constant rate) for the original variables. Luo, Y. (Economics, HKU) ME October 30, 2017 18 / 42

Second Order Difference Equation A second-order difference equation involves the second difference of y : 2 y t+2 = ( y t+2 ) = (y t+2 y t+1 ) where is the first difference. = (y t+2 y t+1 ) (y t+1 y t ) = y t+2 2y t+1 + y t, (26) Luo, Y. (Economics, HKU) ME October 30, 2017 19 / 42

SO Linear DEs with Constant Coeffi cients and Constant Term A simple variety of SO equation takes the form y t+2 + a 1 y t+1 + a 2 y t = c (27) We first discuss particular solution. As usual, try the simplest solution form y t = k, which means that y p = k = c 1 + a 1 + a 2 (1 + a 1 + a 2 = 0) (28) In case a 1 + a 2 = 1, try another solution form y t = kt, which means that y p = kt = c a 1 + 2 t (29) Luo, Y. (Economics, HKU) ME October 30, 2017 20 / 42

(Conti.) We next discuss the complementary function which is the solution of the reduced homogenous equation (c = 0). As in the FO DE case, try the following solution form y t = Ab t = (30) Ab t+2 + a 1 Ab t+1 + a 2 Ab t = 0 = (31) b 2 + a 1 b + a 2 = 0 (32) This quadratic characteristic equation have two roots: b 1, b 2 = a 1 ± a1 2 4a 2 2 and both should appear in the general solution of the reduced DE. There are three possibilities. (33) Luo, Y. (Economics, HKU) ME October 30, 2017 21 / 42

Case 1 (distinct real roots) When a 2 1 4a 2 > 0, the CF can be written as y c = A 1 b t 1 + A 2 b t 2. (34) Example: Consider which means that b 1 = 1, b 2 = 2, y t+2 + y t+1 2y t = 12, (35) y t = A 1 + A 2 ( 2) t + 4t (36) where A 1 and A 2 can be determined by two initial conditions y 0 = 4 and y 1 = 5 : 4 = A 1 + A 2 and 5 = A 1 + A 2 ( 2) + 4 = A 1 = 3 and A 2 = 1. Luo, Y. (Economics, HKU) ME October 30, 2017 22 / 42

Case 2 (repeated real roots) When a 2 1 4a 2 = 0, the CF can be written as y c = A 3 b t + A 4 tb t. (37) Example: Consider which means that b 1 = b 2 = 3, y t+2 + 6y t+1 + 9y t = 4, (38) y t = A 3 ( 3) t + A 4 t ( 3) t + 1 4 (39) where A 1 and A 2 can be determined by two initial conditions y 0 and y 1. Luo, Y. (Economics, HKU) ME October 30, 2017 23 / 42

Case 3 (complex roots) When a 2 1 4a 2 < 0, b 1, b 2 = h ± vi where h = a 1 a 2 2 and v = 1 +4a 2 2. The CF is y c = A 1 b t 1 + A 2 b t 2 = A 1 (h + vi) t + A 2 (h vi) t. (40) De Moivre theorem implies that (h ± vi) t = R t (cos θt ± i sin θt) where R = h 2 + v 2 = a 2, cos θ = h R, sin θ = v R (41) Luo, Y. (Economics, HKU) ME October 30, 2017 24 / 42

(Conti.) The CF can be rewritten as y c = A 1 R t (cos θt + i sin θt) + A 2 R t (cos θt i sin θt) (42) = R t [(A 1 + A 2 ) cos θt + (A 1 A 2 ) i sin θt] = R t (A 5 cos θt + A 6 i sin θt) (43) where R and θ can be determined once h and v become known. Example: Consider y t+2 + 1 4 y t = 5,which means that h = 0, v = 1 2, R = y c = ( ) 1 2 0 + = 1 2 2, cos θ = 0, sin θ = 1, θ = π = (44) 2 ( ) 1 t ( A 5 cos π 2 2 t + A 6i sin π ) 2 t. (45) Luo, Y. (Economics, HKU) ME October 30, 2017 25 / 42

The Convergence of Time Path The convergence of time path y is determined by the two characteristic roots of the SO DE. In Case 1 if b 1 > 1 and b 2 > 1, then both components in the CF will be explosive and y c must be divergent. if b 1 < 1 and b 2 < 1, then both components in the CF will converge to 0 as t goes to infinity, as will y c also. if b 1 > 1 and b 2 < 1, then A 2 b t 2 tend to converge to 0, while A 1b t 1 tends to deviate further from 0 and will eventually render the path divergent. Call the root with higher absolute value the dominant root since this root sets the tone of the time path. A time path will converge iff the dominant root is less than 1 in absolute value. The nondominant root also affects the time path, at least in the beginning periods. Luo, Y. (Economics, HKU) ME October 30, 2017 26 / 42

In Case 2 (repeated roots), for the term A 4 tb t, if b > 1, the b t term will be explosive. and the multiplicative t term also serves to intensify the explosiveness as t increases. if b < 1, the b t term will be converge. and the multiplicative t will offset the convergence as t increases. It turns out the damping force b t of will eventually dominant the exploding force t. Hence, the basic requirement for convergence is still that the root be less than 1 in absolution value. Luo, Y. (Economics, HKU) ME October 30, 2017 27 / 42

In Case 3 (complex roots), The term A 5 cos θt + A 6 i sin θt produces a fluctuation pattern of a periodic nature. Since time is discrete, the resulting path displays a sort of stepped fluctuation. The term R t determines the convergence of y : determines whether the stepped fluctuation is to be intensified or mitigated as t increases. Hence, the basic requirement for convergence is still that the root be less than 1 in absolution value. The fluctuation can be gradually narrowed down iff R < 1 (Note that R is just the absolute value of the complex roots h ± vi). Luo, Y. (Economics, HKU) ME October 30, 2017 28 / 42

Difference Equations System So far our dynamic analysis has focused on a single difference equation. However, some economic models may include a system of simultaneous dynamic equations in which several variables need to be handled. Hence, the solution method to solve such dynamic system need to be introduced. The dynamic system with several dynamic equations and several variables can be equivalent with a single higher order equation with a single variable. Hence, the solution of a dynamic system would still include a set of PI and CF, and the dynamic stability of the system would still depend on the absolution values (for difference equation system) of the characteristic roots in the CF. Luo, Y. (Economics, HKU) ME October 30, 2017 29 / 42

The Transformation of a Higher-order Dynamic Equation In particular, a SO difference equation can be rewritten as two simultaneous FOC equations in two variables. Consider the following example: y t+2 + a 1 y t+1 + a 2 y t = c (46) If we introduce an artificial new variable x, defined as x t = y t+1, we can then express the original SO equation by the following two FO DE x t+1 + a 1 x t + a 2 y t = c (47) y t+1 x t = 0 (48) Luo, Y. (Economics, HKU) ME October 30, 2017 30 / 42

Solving Simultaneous Dynamic Equations Suppose that we are given x t+1 + 6x t + 9y t = 4 (49) y t+1 x t = 0 (50) To solve this two-de system, we still need to seek the PI and the CF, and sum them to obtain the desired time paths of the two variables x and y. We first solve for the PI. As usual, try the constant solution: y t+1 = y t = y and x t+1 = x t = x = x = y = 1 4. Luo, Y. (Economics, HKU) ME October 30, 2017 31 / 42

For the CF, try the following function forms x t = mb t and y t = nb t where m and n are arbitrary constants and b represents the characteristic root. Next, we need to find the values of m, n, and b that satisfy the reduced version. Substituting these guessed solutions into the above dynamic system and cancelling out the common term b t gives (b + 6) m + 9n = 0 (51) m + bn = 0 (52) which is a linear homogeneous-equation system in m and n. We can rule out the uninteresting trivial solution (m = n = 0) by requiring that b + 6 9 1 b = b2 + 6b + 9 = 0 (53) This characteristic equation have two roots b (= b 1 = b 2 ) = 3. Luo, Y. (Economics, HKU) ME October 30, 2017 32 / 42

(Conti.) Given each b i (i = 1, 2), the above homogeneous equation implies that there will have an infinite number of solutions for (m, n) For this repeated-root case, we have m i = k i n i (54) x t = m 1 ( 3) t + m 2 t ( 3) t, y t = n 1 ( 3) t + n 2 t ( 3) t which must satisfy y t+1 = x t : n 1 ( 3) t+1 + n 2 (t + 1) ( 3) t+1 = m 1 ( 3) t + m 2 t ( 3) t = Setting n 1 = A 3 and n 2 = A 4 gives m 1 = 3 (n 1 + n 2 ), m 2 = 3n 2 x c = 3A 3 ( 3) t 3A 4 (t + 1) ( 3) t (55) y c = A 3 ( 3) t + A 4 t ( 3) t (56) Note that both time paths have the same ( 3) t term, so they both explosive oscillation. Luo, Y. (Economics, HKU) ME October 30, 2017 33 / 42

Matrix Notation We can analyze the above dynamic system by using matrix. The above two-equation system can be written as [ ][ ] [ ][ ] [ ] 1 0 xt+1 6 9 xt 4 + = 0 1 y t+1 1 0 y t 0 }{{}}{{}}{{}}{{}}{{} I u K v d (57) Try a constant PI first, [ ] x (I + K ) = d = y [ x y ] = (I + K ) 1 d = [ 1/4 1/4 ]. Luo, Y. (Economics, HKU) ME October 30, 2017 34 / 42

(Conti.) Next, try the CF [ ] [ ] [ mb t+1 m u = nb t+1 = b t+1 m and v = n n [ ] m (bi + K ) = 0 n To avoid the trivial solution, we must have where A i are arbitrary constants. bi + K = 0 = b = 3 = m i = k i n i, where n i = A i, m i = k i A i ] b t = Luo, Y. (Economics, HKU) ME October 30, 2017 35 / 42

(Conti.) With distinct real roots, [ ] [ xc m1 b = 1 t + m 2b t ] [ 2 k1 A y c n 1 b1 t + n 2b2 t = 1 b1 t + k 2A 2 b2 t A 1 b1 t + A 2b2 t ]. (58) With repeated roots, [ xc y c ] [ m1 b = 1 t + m 2tb2 t n 1 b1 t + n 2tb2 t ] (59) The general solution can be written as [ ] [ xt xc = y t y c ] + [ x y ]. (60) Luo, Y. (Economics, HKU) ME October 30, 2017 36 / 42

Two-Variable Phase Diagram: Discrete-time Case Now we shall discuss the qualitative-graphic (phase-diagram) analysis of a nonlinear difference equation system. Specifically, we focus on the following two-equation system x t+1 x t = f (x t, y t ) (61) y t+1 y t = g (x t, y t ) (62) which is called autonomous system (t is not an explicit argument in f and g). The two-variable phase diagram (PD) can answer the qualitative questions: the location and the dynamic stability of the intertemporal equilibrium. The most crucial task of the PD is to determine the direction of movement of the two variables over time. In the two-variable case, we can also draw the PD in the space of (x, y). Luo, Y. (Economics, HKU) ME October 30, 2017 37 / 42

(Conti.) In this case, we have two demarcation lines: x t+1 = x t+1 x t = f (x t, y t ) = 0 (63) y t+1 = y t+1 y t = g (x t, y t ) = 0 (64) which interact at point E representing the intertemporal equilibrium ( x t+1 = 0 and y t+1 = 0) and divide the space into 4 regions. (will be specified later.) If the demarcation line can be solved for y in terms of x, we can plot the line in the (x, y) space. Otherwise, we can use the implicit-function theorem to derive: slope of x t+1 = dy dx f / x x t+1 =0 = f / y = f x ; (65) f y slope of y t+1 = dy dx y t+1 =0 = g/ x g/ y = g x g y. (66) Specifically, we assume that f x < 0, f y > 0, g x > 0, g y < 0,which means that both slopes are positive. Further assume that f x fy > g x g y. Luo, Y. (Economics, HKU) ME October 30, 2017 38 / 42

(Conti.) The two curves, at any other point, either x or y changes over time according to the signs of x t+1 and y t+1 at that point: d ( x t+1 ) dx = f x < 0, (67) which means that as we move from west to east in the space (as x increases), x t+1 decrease so that the sign of x t+1 must pass through three stages, in the order: +, 0,. Similarly, d ( y t+1 ) dy = g y < 0, (68) which means that as we move from south to north in the space (as y increases), y t+1 decreases so that the sign of y t+1 must pass through three stages, in the order: +, 0,. Luo, Y. (Economics, HKU) ME October 30, 2017 39 / 42

Linearization of a Nonlinearization Difference-Equation System Another qualitative technique of analyzing a nonlinear difference equation system is to examine its linear approximation which is derived by using the Taylor expansion of the system around its intertemporal equilibrium. At the point of expansion (i.e., the IE), the linear approximation has the same equilibrium as the original nonlinear system. In a suffi ciently small neighborhood of E, the linear approximation should have the same general streamline configuration as the original system. As long as we confine our stability analysis to the immediate neighborhood of the IE, the approximated system can include enough information from the original nonlinear system. This analysis is called local stability analysis. Luo, Y. (Economics, HKU) ME October 30, 2017 40 / 42

(Conti.) For the two difference equation system, we have x t+1 = f (x 0, y 0 ) + f x (x 0, y 0 ) (x x 0 ) + f y (x 0, y 0 ) (y y 0 ) y t+1 = g (x 0, y 0 ) + g x (x 0, y 0 ) (x x 0 ) + g y (x 0, y 0 ) (y y 0 ) For purpose of local stability analysis, the above linearization can be put a simpler form. First, the expansion point is the IE, (x, y) and f (x, y) = g (x, y) = 0. We then have another form of linearization x t+1 (1 + f x (x, y)) x f y (x, y) y = f x (x, y) x f y (x, y) y y t+1 g x (x, y) x (1 + g y (x, y)) y = g x (x, y) x g y (x, y) y which means that [ xt+1 x y t+1 y ] [ ] 1 + fx f y g x 1 + g y (x,y ) }{{} J E [ xt x y t y ] = [ 0 0 ]. Luo, Y. (Economics, HKU) ME October 30, 2017 41 / 42

(Conti.) The Jacobian matrix JE in the above reduced system can determine the local stability of the equilibrium. Denote [ ] [ ] 1 + fx f J E = y a b = (69) g x 1 + g y c d (x,y ) The characteristic roots of the reduced linearization is r a b c r d = r 2 (a + d) r + (ad bc) = 0 = tr (J E ) = r 1 + r 2 = a + d = 2 + f x + g y (70) det (J E ) = r 1 r 2 = ad bc = (1 + f x ) (1 + g y ) f y g x = (71) r 1, r 2 = tr (J E ) ± (tr (J E )) 2 4 det (J E ) 2 There are also four cases for the local stability of the above system, but here we only focus on the most popular economic case: The saddle-point case in which r 1 > 1 and r 2 < 1. We will discuss the application in the next lecture on dynamic optimization. Luo, Y. (Economics, HKU) ME October 30, 2017 42 / 42